Masa-masa kesetressan.. (x_x)

bener-bener deh minggu-minggu ini lagi sress berat....!!!!!
mulai dari menumpuknya tugas-tugas (ya walaupun itu dah biasa... haghag), tapi kenapa harus barengan sama ujian-ujian juga!!!!
Nah, ternyata bukan hanya jadwal ujian yang bentur sama jadwal ngumpulin tugas, mlah nambah lagi sama remidial-remidial... mana aku hampir semua remidial pula.... GILAAAAA!!!!!!!!!!!!!!


Hope: "cepet liburan lah dan aku bisa nyelesein semuanya dengan tepat waktu.... (Say??!!! AAAMMMMIIIIINNNN.....)"

Shinju

Akhir-akhir ini aku lagi pengen banget baca..... Nah buku yang baru aku geluti yaitu adalaaaahhhh... jeng jeng jeng (hehehehe)
Shinju!!!!
Awalnya aku beli buku ini sudah beberapa bulan lalu dan aku juga sudah memulai membacanya saat aku mulai membelinya, tapiiiiii cerita awalnya itu lho.... ngeboseniiiinnn bangeeeet, tapi kemaren aku kesepian di kost and mau nggak mau aku harus ngelakuin something yang nggak bikin boring, nah kutemukan juga tuh buku lama ku Shinju yang belum juga habis halamannya aku baca.

Kucoba baca satu per satu katanya, ya sama seperti awalnya ceritanya ngebosenin tapi aku baca lagi dan baca lagi, lama kelamaan ceritanya jadi bagus, sampe'-sampe' dah larut malam pun aku usahain buat bacanya..... tapi sampe' hari ini pun aku belum menyelesein buku tersebut, tapi semakin jauh aku baca, aku jadi semakin penasaran akhir ceritanya....

cerita novel ini berkisah tentang Shinju: Bunuh diri akibat cintanya ditentang, tapi seorang yoriki yang bernama Sano Ichiro mencium gelagat janggal bahwa kematian Noriyoshi yang seorang seniman pinggiran dan seorang wanita yang berasal dari klan Nui...
Wawancara ke berbagai tempat dan orang pun dia tempuh untuk mencari kebenaran. Dari tempat yang sangat agung sampai ke rumah-rumah bordil... dari mewawancarai seorang wanita penghibur, Kikunojo seorang aktor yang berperan apik sebagai wanita, pesumo pojok (sumo ilegal) dan akhirnya membawanya sehingga dapat mengetahui bahwa ada hubungan kematian mereka berdua dengan keluarga terhiormat yaitu klan Nui sendiri....

ceritanya seru kan....

Dalam buku ini penggambaran settingnya sangat meyakinkan dan sangat mendetail.... And try this....

Chm 1045 (sections 15-20)
Fall 2000
Lecture: MWF 10:10 - 11:00 a.m.
275 Fisher Lecture Hall
Office Hours: MW 4:00-5:00 p.m.
201 Molecular Biophysics Building
Faculty Mailbox: First Floor Chemistry Building (Ditmer)
Course URL:
http://wine1.sb.fsu.edu/chm1045/

Syllabus
• Description/Objectives
• Requirements
• Grading/Evaluation
• Contact with Instructor
• Office Hours
• Honor Code
• ADA Requirements
• Course Calendar
________________________________________
Syllabus
Description/Objectives
General Chemistry I. Lecture 3 hours per week, and recitation 1 hour per week.
Chemical symbols, formulas, and equations; the states of matter; electronic structure and bonding.
Text: General Chemistry - An Integrated Approach. Hill and Petrucci. Second Edition. Prentice Hall Publishers.
ISBN: 0-13-010318-7
Requirements

Prerequisite: MAC 1105 with a grade of "C-" or higher, or placement beyond MAC 1105 on the University's Math department exam.
Co-requisite: Chm 1045 L (General Chemistry I Laboratory)
Note: Students taking Chm 1045 after taking Chm 1020 and/or Chm 1030 should contact a Chemistry advisor to register for reduced credit.
Grading/Evaluation
There will be three midterm exams and a final. Each exam will be worth 100 points. In addition, there will be a series of 14 homework assignments that will be available in recitation, and are to be handed in at the recitation. Late homework will not be accepted - get a time stamp from the dept. of chemistry main office if you are turning in homework outside of recitation, put this homework in the instructors mailbox in the chemistry department. Each homework will be worth 10 points. The following grade adjustments will be made:
• The lowest homework score will be dropped.
• There will be three extra-credit assignments available during the semester - each worth 10 points. These assignments will be handed out after each of the three midterm exams.
• Thus, the student's grade will be determined from a possible point total of 530.
• The grading will follow a standard percentile scale:
93-100% A
90-92% A-
87-89% B+
83-86% B
80-82% B-
77-79% C+
73-76% C
70-72% C-
60-69% D
<60% F The four exams will cover the following material: Exam Material Date Exam 1 Chapters 1-4 Chemistry: Matter and Measurement Atoms, Molecules and Ions Stoichiometry: Chemical Calculations Chemical Reactions in Aqueous Solutions Fri Sep 29 Exam 2 Chapters 5-6 Gases Thermochemistry Fri Oct 20 Exam 3 Chapters 7-8 Atomic Structure Electron Configurations, Atomic Properties, and the Periodic Table Mon Nov 13 Final Exam Chapters 9-10 Chemical Bonds Bonding Theory and Molecular Structure Tues Dec 12 7:30 - 9:30 a.m. Location: TBA Contact with Instructor Dr. Michael Blaber (850) 644-1863 blaber@sb.fsu.edu Basic Concepts Hypothesis versus Theory versus Fact ________________________________________ Hypothesis: • A tentative explanation or idea about how things work • A hypothesis guides you in further work to get a better answer Example of a hypothesis: "The moon is made of cheese" (note: this is the kind of hypothesis my wife would come up with). How could we test this hypothesis? • Construct a rocket to go to the moon and return with samples • Make a cheese pizza substituting the moon samples for the cheese • Ask people (e.g. graduate students. No, wait, they aren't real people…) to eat the pizza and see if they can tell any difference from pizza made with real cheeseTM (from Wisconsin) • Most likely conclusion: Hey, this pizza tastes like dirt New hypothesis (altered to include additional information from above experiment): The moon is not made of cheese, but is made of dirt, sort of like the Earth Theory: • A theory is an explanation of the general principles of certain phenomena with considerable facts to support it • A theory remains valid only if every new piece of information supports it • If a single piece of available information does not support a theory, then the theory (as proposed) is disproved Fact: • An indisputable truth ________________________________________ Example: • It is a fact that on June 30, 1908 in Tunguska, Siberia, an explosion equivalent to about 15 million tons of TNT occurred. • It is a theory that this explosion was due to a natural, extra-terrestrial phenomenon and not to an activity associated with man. • One hypothesis is that a comet collided with the Earth (a competing hypothesis is that a small black hole collided with the Earth) ________________________________________ 1998 Dr. Michael Blaber Dimensional Analysis ________________________________________ What is dimensional analysis and how can we use it to help us solve problems in chemistry? Dimensional Analysis for Party Planning… If you have every planned a party, you have used dimensional analysis. The amount of beer and munchies you will need depends on the number of people you expect. For example, if you are planning a Friday night party and expect 30 people you might estimate you need to go out and buy 120 bottles of beer and 10 large pizza's. How did you arrive at these numbers? The following indicates the type of dimensional analysis solution to party problem: Finally, in going to buy the beer, you perform another dimensional analysis: should you buy the beer in six-packs or in cases? Realizing that carrying around 20 six packs is a real headache, you get 5 cases instead. In this party problem, we have used dimensional analysis in two different ways: • In the first application, we have used dimensional analysis to calculate how much beer we need based on knowing 1) how much beer we need for one person, and 2) how many people we expect. Likewise for the Pizza. • In the second application we used dimensional analysis to convert units (i.e. from individual beers to the equivalent amount of six packs or cases) First, let's take a look at dimensional analysis as it relates to converting units: • If we ignore the numbers for a moment, and just look at the units (i.e. dimensions), we have: beers * (six pack/beer) • We can treat the dimensions in a similar fashion as other numerical analyses, i.e. any number divided by itself is 1. Therefore: beers * (six pack/beer) = beers * (six pack/beer) = six pack • So, the dimensions of the numerical answer will be "six packs". How can we use dimensional analysis to be sure we have set up our equation correctly? • Consider the following alternative way to set up the above unit conversion analysis: • While it is correct that there are 6 beers in one six pack, the above equation yields a value of 720 with units of beers2/six pack. • These rather bizarre units indicate that the equation has been setup incorrectly (and as a consequence you will have a ton of extra beer at the party). Next, let’s take a look at dimensional analysis as it relates to calculations: • In the above case it was relatively straightforward keeping track of units during the calculation. • What if the calculation involves powers, etc? For example, the equation relating kinetic energy to mass and velocity is: • An example of units of mass is kilograms (kg) and velocity might be in meters/second (m/s) • What are the dimensions of Ekinetic? • Since velocity is squared, the dimensions associated with the numerical value of the velocity are also squared • We can double check this by knowing the the Joule (J) is a measure of energy, and has units of kg m2 s-2 (1J = 1 kg m2 s-2) Another example • Pressure (P) is a measure of the Force (F) per unit area (A): • Force, in turn, is a measure of the acceleration (a) on a mass (m): • Thus, pressure (P) can be written as: • What are the units of pressure from this relationship? (Note: acceleration is the change in velocity per unit time) • We can simplify this description of the units of Pressure by dividing numerator and denominator by m: In fact, these are the units of a Pascal (Pa), a measure of pressure In summary… • Dimensional analysis is used in numerical calculations, and in converting units • Dimensional analysis can help us identify whether an equation is set up correctly (i.e. the resulting units should be as expected) • Units are treated similarly to the associated numerical values, i.e. if a variable in an equation is supposed to be squared, then the associated dimensions are squared, etc. ________________________________________ 1998 Dr. Michael Blaber Basic Concepts Introduction to matter ________________________________________ Chemistry "The study of the properties of materials and the changes that materials undergo" Introduction to Matter Matter "Matter is the physical material of the universe; it is anything that occupies space and has mass" Matter can exist in three physical states: 1. gas or vapor 2. liquid 3. solid Gas No fixed volume or shape - it conforms to the volume and shape of its container. Gases can be compressed or expanded to occupy different volumes. Liquid A liquid has a distinct volume, independent of its container, but it has no specific shape. It assumes the shape of the container it is in. Liquids cannot be appreciably compressed. Solid A solid has a definite shape and volume; it is rigid. Solids cannot be appreciably compressed. Substances A pure substance has a fixed composition and distinct properties. Most matter we come in contact with in our daily lives is not a pure substance, but a mixture of substances. Physical and Chemical Properties Every pure substance has a unique set of properties - characteristics which allow us to distinguish it from other substances. These properties fall into two general categories: physical and chemical. Physical properties - properties we can measure without changing the basic identity of the substance. Chemical properties - describe the way a substance may change or "react" to form other substances. Physical and Chemical Changes Substances can undergo various changes in properties, these changes may be classified as either physical or chemical. Physical changes - a substance changes its physical appearance but not its basic identity. All changes of state (e.g. solid to liquid to gas) are physical changes. Chemical changes - also known as chemical reactions, a substance is transformed into a chemically different substance. Mixtures Mixtures refer to combinations of two or more substances in which each substance retains its own chemical identity and hence its own properties. Heterogenous mixtures are not uniform throughout the sample, and have regions of different appearance and properties Homogenous mixtures are uniform throughout the sample, however, the individual substances retain their individual chemical and physical nature. Homogenous mixtures are also called solutions, however, the most common type of solution is described by a solid (the solute) dissolved in a liquid (the solvent). An important characteristic of mixtures is that the individual components retain their physical and chemical properties. Thus, it is possible to separate the components based on their different properties. For example, we can separate ethanol from water by making use of their different boiling temperatures, in a process known as distillation. ________________________________________ 1996 Michael Blaber Elements and Compounds Pure substances have an invariable composition and are composed of either elements or compounds. Elements "Substances which cannot be decomposed into simpler substances by chemical means". Compounds Can be decomposed into two or more elements. Elements Elements are the basic substances out of which all matter is composed. • Everything in the world is made up from only 109 different elements. • 90% of the human body is composed of only three elements: Oxygen, Carbon and Hydrogen Elements are known by common names as well as by their abbreviations. These consisting of one or two letters, with the first one capitalized. These abbreviations are derived from English or foreign words (e.g. Latin, German). Element Abbreviation Carbon C Fluorine F Hydrogen H Iodine I Nitrogen N Oxygen O Phosphorus P Sulfur S Aluminum Al Barium Ba Calcium Ca Chlorine Cl Helium He Magnesium Mg Platinum Pt Silicon Si Copper Cu (from cuprum) Iron Fe (from ferrum) Lead Pb (from plumbum) Mercury Hg (from hydrargyrum) Potassium K (from kalium) Silver Ag (from argentum) Sodium Na (from natrium) Tin Sn (from stannum) Compounds Compounds are substances of two or more elements united chemically in definite proportions by mass. For example, pure water is composed of the elements hydrogen (H) and oxygen (O) at the defined ratio of 11 % hydrogen and 89 % oxygen by mass. The observation that the elemental composition of a pure compound is always the same is known as the law of constant composition (or the law of definite proportions). It is credited to the French chemist Joseph Louis Proust (1754-1826). ________________________________________ 1996 Michael Blaber Atoms, Molecules and Ions The Atomic Theory of Matter ________________________________________ Chemists make their observations in the macroscopic world and seek to understand the fundamental properties of matter at the level of the microscopic world (i.e. molecules and atoms). The reason why certain chemicals react the way they do is a direct consequence of their atomic structure. The Atomic Theory of Matter The word "atom" is derived from the Greek word "atomos", meaning indivisible. The philosopher Democritus (460-370 B.C.) believed that matter was composed of fundamentally indivisible particles, called "atomos". Dalton's atomic theory of 1803: 1. Each element is composed of extremely small particles called atoms 2. All atoms of a given element are identical; the atoms of different elements are different and have different properties (including different masses) 3. Atoms of an element are not changed into different types of atoms by chemical reactions; atoms are neither created nor destroyed in chemical reactions 4. Compounds are formed when atoms of more than one element combine; a given compound always has the same relative number and kind of atoms. Atoms are the basic building blocks of matter; they are the smallest units of an element: • An element is composed of only one kind of atom • In compounds the atoms of two or more elements combine in definite arrangements • Mixtures do not involve the specific interactions between elements found in compounds, and the elements which comprise the mixture can be of varying ratios Atoms are the smallest particle of an element which retains the chemical properties of that element Simple "laws" (i.e. theories) of chemical combination which were known at the time of Dalton: 1. The law of constant composition (in a given compound the relative number and kind of atoms are constant) 2. The law of conservation of mass (the total mass of materials present after a chemical reaction is the same as the total mass before the reaction) Dalton used these "laws" to derive another "law" - the law of multiple proportions (if two elements, A and B, can combine to form more than one compound, then the ratios of the relative masses of each element which can combine can be represented by characteristically small whole numbers). ________________________________________ Where's the water? Hydrogen and oxygen gas can react to form water (H2O). This is a violent reaction which liberates considerable amounts of heat. A large child's balloon will hold about four liters of gas. This would represent about 5 grams of oxygen and 0.3 grams of hydrogen if we mixed them in the appropriate (molar) ratios. Thus, we would expect, at most, to form about 5.3 grams of water (about 1 tablespoon full). Since some gasses escape before reacting, and since we may not add the gasses in exactly the correct ratio, the amount of water formed may be far less. The water that is formed is going to be in the vapor phase. This demonstrates one of the attractive aspects of hydrogen as a fuel: the only pollution (i.e. product) is water! ________________________________________ 1996 Michael Blaber Atoms, Molecules and Ions The Discovery of Atomic Structure ________________________________________ The Discovery of Atomic Structure • 1803 Dalton - the atom is a indivisible, indestructible, tiny ball • 1850 Evidence is accumulating that the atom is itself composed of smaller particles • The current model... The behavior of electrically charged particles Like charges repel each other, unlike charges attract Behavior of moving charge in magnetic field • A charged particle moving though a magnetic field will feel a force perpendicular to the plane described by the velocity vector and magnetic field vector • This deflects the moving charged particle according to the "right hand rule" (based on a positive charge) • A negative charge will be deflected in the opposite direction Cathode rays and electrons Electrical discharge through partially evacuated tubes produced radiation. This radiation originated from the negative electrode, known as the cathode (thus, these rays were termed cathode rays). • The "rays" traveled towards, or were attracted to the positive electrode (anode) • Not directly visible but could be detected by their ability to cause other materials to glow, or fluoresce • Traveled in a straight line • Their path could be "bent" by the influence of magnetic or electrical fields • A metal plate in the path of the "cathode rays" aquired a negative charge • The "cathode rays" produced by cathodes of different materials appeared to have the same properties These observations indicated that the cathode ray radiation was composed of negatively charged particles (now known as electrons). J.J. Thompson (1897) measured the charge to mass ratio for a stream of electrons (using a cathode ray tube apparatus) at 1.76 x 108 coulombs/gram. • Charged particle stream can be deflected by both an electric charge and by a magnetic field • An electric field can be used to compensate for the magnetic deflection - the resulting beam thus behaves as if it were neutral • The required current needed to "neutralize" the magnetic field indicates the charge of the beam • The loss of mass of the cathode indicated the "mass" of the stream of electrons Thompson determined the charge to mass ratio for the electron, but was not able to determine the mass of the electron. However, from his data, if the charge of a single electron could be determined, then the mass of a single electron could determined. Robert Millikan (1909) was able to successfully measure the charge on a single electron (the "Milliken oil drop experiment"). This value was determined to be 1.60 x 10-19 coulombs. Thus, the mass of a single electron was determined to be: (1 gram/1.76 x 108 coulombs)*(1.60 x 10-19 coulombs) = 9.10 x 10-28 grams Note: the currently accepted value for the mass of the electron is 9.10939 x 10-28 grams. Radioactivity Wilhelm Roentgen (1895) discovered that when cathode rays struck certain materials (copper for example) a different type of ray was emitted. This new type of ray, called the "x" ray had the following properties: • The could pass unimpeded through many objects • They were unaffected by magnetic or electric fields • They produced an image on photographic plates (i.e. they interacted with silver emulsions like visible light) Henri Becquerel (1896) was studying materials which would emit light after being exposed to sunlight (i.e. phosphorescent materials). The discovery by Roentgen made Becquerel wonder if the phosphorescent materials might also emit x- rays. He discovered that uranium containing minerals produced x-ray radiation (i.e. high energy photons). Marie and Pierre Curie set about to isolate the radioactive components in the uranium mineral. Ernest Rutherford studied alpha rays, beta rays and gamma rays, emitted by certain radioactive substances. He noticed that each behaved differently in response to an electric field: • The -rays were attracted to the anode • The -rays were attracted to the cathode • The -rays were not affected by the electric field The  and  "rays" were composed of (charged) particles and the -"ray" was high energy radiation (photons) similar to x-rays • -particles are high speed electrons (charge = -1) • -particles are the positively charged core of the helium atom (charge = +2) The nuclear atom J.J. Thompson model of the atom (1900) • The atom consists of a sphere of positive charge within which was buried negatively charged electrons • Also known as the "plum pudding" model of the atom Rutherford model of the atom (1910) • Most of the mass of the atom, and all its positive charge, reside in a very small dense centrally located region called the "nucleus" • Most of the total volume of the atom is empty space within which the negatively charged electrons move around the nucleus Rutherford (1919) discovers protons - positively charged particles in the nucleus Chadwick (1932) discovers neutron - neutral charge particles in the nucleus ________________________________________ 1996 Michael Blaber Atoms, Molecules and Ions Modern view of atomic structure ________________________________________ The Modern View of Atomic Structure Physicists have identified a long list of particles which make up the atomic nucleus. Chemists, however, are primarily concerned with the following sub-atomic particles: • electron • proton • neutron Electron The electron is negatively charged, with a charge of -1.602 x 10-19 Coulombs (C). For convenience, the charge of atomic and sub-atomic particles are usually described as a multiple of this value (also known as the electronic charge). Thus, the charge of the electron is usually simply refered to as -1. Proton The proton has a charge of +1 electron charge (or, +1.602 x 10-19 C) Neutrons Neutrons have no charge, they are electrically neutral. Note: Because atoms have an equal number of electrons and protons, they have no net electrical charge Protons and neutrons are located in the nucleus (center) of the atom. The nucleus is small compared to the overall size of the atom. The majority of the space of an atom is the space in which the electrons move around. Electrons are attracted to the protons in the nucleus by the force of attraction between particles of opposite charge. Note: The strength of attraction between electrons and protons in the nuclei for different atoms is the basis of many of the unique properties of different atoms. The electrons play a major role in chemical reactions. In atomic models, the electrons are represented as a diffuse electron cloud The mass of an atom is extremely small. The units of mass used to describe atomic particles is the atomic mass unit (or amu). An atomic mass unit is equal to 1.66054 x 10-24 grams. How do the different sub-atomic particles compare as far as their mass? Proton = 1.0073 amu Neutron = 1.0087 amu Electron = 5.486 x 10-4 amu From this comparison we can see that: • The mass of the proton and neutron are nearly identical • The nucleus (protons plus neutrons) contains virtually all of the mass of the atom • The electrons, while equal and opposite in charge to the protons, have only 0.05% the mass The size of an atom is quite small also, the typical range for atomic diameters is between 1 x 10-10 and 5 x 10-10 meters. Note: a convenient unit of measurement for atomic distances is the angstrom (Å). The angstrom is equal to 1 x 10-10 meters. Thus, most atoms are between 1 and 5 angstroms in diameter. ________________________________________ Pinheads and Bullion Cubes Pinheads have a diameter of about 1 x 10-3 meters (a millimeter across). If an atom had a diameter of 2.5 x 10-10 meters, then (1 atom/2.5 x 10-10 meters) * (1 x 10-3 meters) = 4 x 106 atoms i.e. four million of them could line up across the head of a pin. The diameter of atomic nuclei are about 10-4Å. Thus, the nuclei is about 0.01% the diameter of the atom as a whole. If the nucleus had a diameter equal to that of a pinhead, then the atom itself would have a diameter of some 10 meters (about 39 and a half feet). The nucleus of an atom is therefore quite dense. Consider a simple case of a nucleus containing 1 neutron and 1 proton: mass of nucleus = ~2.0 amu = 2 * (1.66 x 10-24 grams) = 3.32 x 10-24 grams diameter of nucleus = (approximately) 1 x 10-4 Å = 1 x 10-14 meters radius of nucleus = 1 x 10-14 meters/2 = 0.5 x 10-14 meters volume of nucleus = (4/3)(radius of nucleus)3 volume of nucleus = 5.24 x 10-43 meters3 mass/volume = 3.32 x 10-24 grams/5.24 x 10-43 meters3 mass/volume = 6.34 x 1018 grams/meter3 a bouillon cube (stuff you can make soup out of if you are really broke) is about one cubic centimeter, or 1 x 10-6 meters3 and if it were made up of atomic nuclei it would weigh: (1 x 1018 grams/meter3) * (1 x 10-6 meters3) = 6.34 x 1012 grams or about six billion kilograms, or about 2.8 billion tons. ________________________________________ Isotopes, Atomic Numbers and Mass Numbers What characteristic feature of sub-atomic particles distinguishes one element from another? • All atoms of an element have the same number of protons in the nucleus • Since the net charge on an atom is 0, the atom must have an equal number of electrons. • What about the neutrons? Although usually equal to the number of protons, the number of neutrons can vary somewhat. Atoms which differ only in the number of neutrons are called isotopes. Since the neutron is about 1.0087 amu (the proton is 1.0073), different isotopes have different masses. Your friend, Carbon All atoms of the element Carbon (C) have 6 protons and 6 electrons. The number of protons in the carbon atom are denoted by a subscript on the left of the atomic symbol: This is called the atomic number, and since it is always 6 for carbon, it is somewhat redundant and usually omitted. Another number, the "Mass Number" is a superscript on the left of the atomic symbol. It denotes the sum of the number of protons and neutrons in the particular isotope being described. For example: refers to an isotope of carbon which has (as expected for the element carbon) six protons, and six neutrons. The following isotope of carbon: has 6 protons (atomic number) and 8 neutrons (8=14-6). This isotope is also known simply as "carbon 14". Carbon 12 is the most common form of carbon (~99% of all carbon). An atom of a specific isotope is called a nuclide. Since all atoms are composed of protons, electrons and neutrons, all chemical and physical differences between elements are due to the differences in the number of these sub-atomic particles. Therefore, an atom is the smallest sample of an element, because dividing an atom further (into sub-atomic particles) destroys the element's unique identitity. ________________________________________ 1996 Michael Blaber Atoms, Molecules and Ions The Periodic Table ________________________________________ The Periodic Table As more and more elements were discovered and characterized, efforts were made to see whether they could be grouped, or classified, according to their chemical behavior. This effort resulted, in 1869, in the development of the Periodic Table. Certain elements show similar characteristics: • Lithium (Li), Sodium (Na) and Potassium (K) are all soft, very reactive metals • Helium (He), Neon (Ne) and Argon (Ar) are very non-reactive gasses If the elements are arranged in order of increasing atomic number, their chemical and physical properties are found to show a repeating, or periodic pattern. Note: This table lists the atomic number (number of protons) in the upper left corner of each box. The atomic number is formally placed as a subscript preceding the atom name. click on picture for larger image As an example of the periodic nature of the atoms (when arranged by atomic number), each of the soft reactive metals comes immediately after one of the nonreactive gasses. The elements in a column of the periodic table are known as a family or group. The labeling of the families are somewhat arbitrary, but are usually divided into the general groups of: • Metals (everything on the left and middle region) • Non-metals (upper diagonal on the right hand side - green, salmon and red) • Metaloids (atoms in the boundary between the metals and metaloids: Boron(B), Silicon(Si), Germainium(Ge), Arsenic(As), Antimony(Sb), Tellurium(Te), Astatine(At)). These are some of the more useful materials for semi-conductors. or, another convention is the 'A' and 'B' designators with column number labels (either in Roman or Arabic numerals). These columns have different types of classifications: Group Name Elements 1A Alkali metals Li, Na, K, Rb, Cs, Fr 2A Alkaline earth metals Be, Mg, Ca, Sr, Ba, Ra 6A Chalcogens ("chalk formers") O, S, Se, Te, Po 7A Halogens ("salt formers") F, Cl, Br, I, At 8A Noble gases (or inert, or rare gases) He, Ne, Ar, Kr, Xe, Rn The elements in a family of the periodic table have similar properties because they have the same type of arrangement of electrons at the periphery of their atoms. The majority of elements are metals: • high luster • high electrical conductivity • high heat conductivity • solid at room temperature (except Mercury [Hg]) Note: hydrogen is a non-metal (at left hand side of the periodic table) Non-metals • solid, liquid or gas at room temp ________________________________________ 1996 Michael Blaber Atoms, Molecules and Ions Molecules and ions ________________________________________ Molecules and Ions Although atoms are the smallest unique unit of a particular element, in nature only the noble gases can be found as isolated atoms. Most matter is in the form of ions, or compounds. Molecules and chemical formulas A molecule is comprised of two or more chemically bonded atoms. The atoms may be of the same type of element, or they may be different. Many elements are found in nature in molecular form - two or more atoms (of the same type of element) are bonded together. Oxygen, for example, is most commonly found in its molecular form "O2" (two oxygen atoms chemically bonded together). Oxygen can also exist in another molecular form where three atoms are chemically bonded. O3 is also known as ozone. Although O2 and O3 are both compounds of oxygen, they are quite different in their chemical and physical properties. There are seven elements which commonly occur as diatomic molecules. These include H, N, O, F, Cl, Br, I. An example of a commonly occurring compound that is composed of two different types of atoms is pure water, or "H2O". The chemical formula for water illustrates the method of describing such compounds in atomic terms: there are two atoms of hydrogen and one atom of oxygen (the "1" subscript is omitted) in the compound known as "water". There is another compound of Hydrogen and Oxygen with the chemical formula H2O2 , also known as hydrogen peroxide. Again, although both compounds are composed of the same types of atoms, they are chemically quite different: hydrogen peroxide is quite reactive and has been used as a rocket fuel (it powered Evil Kenievel part way over the Snake River canyon). Most molecular compounds (i.e. involving chemical bonds) contain only non-metallic elements. Molecular, Empirical, and Structural Formulas Empirical vs. Molecular formulas • Molecular formulas refer to the actual number of the different atoms which comprise a single molecule of a compound. • Empirical formulas refer to the smallest whole number ratios of atoms in a particular compound. Compound Molecular Formula Empirical Formula Water H2O H2O Hydrogen Peroxide H2O2 HO Ethylene C2H4 CH2 Ethane C2H6 CH3 Molecular formulas provide more information, however, sometimes a substance is actually a collection of molecules with different sizes but the same empirical formula. For example, carbon is commonly found as a collection of three dimensional structures (carbon chemically bonded to carbon). In this form, it is most easily represented simply by the empirical formula "C" (the elemental name). Structural formulas Sometimes the molecular formulas are drawn out as structural formulas to give some idea of the actual chemical bonds which unite the atoms. Structural formulas give an idea about the connections between atoms, but they don't necessarily give information about the actual geometry of such bonds. Ions The nucleus of an atom (containing protons and neutrons) remains unchanged after ordinary chemical reactions, but atoms can readily gain or lose electrons. If electrons are lost or gained by a neutral atom, then the result is that a charged particle is formed - called an ion. For example, Sodium (Na) has 11 protons and 11 electrons. However, it can easily lose 1 electron. The resulting cation has 11 protons and 10 electrons, for an overall net charge of 1+ (the units are electron charge). The ionic state of an atom or compound is represented by a superscript to the right of the chemical formula: Na+, Mg2+ (note the in the case of 1+, or 1-, the '1'is omitted). In contrast to the Na atom, the Chlorine atom (Cl) easily gains 1 electron to yield the chloride ion Cl- (i.e. 17 protons and 18 electrons). In general, metal atoms tend to lose electrons, and nonmetal atoms tend to gain electrons. Na+ and Cl- are simple ions, in contrast to polyatomic ions such as NO3- (nitrate ion) and SO42- (sulfate ion). These are compounds made up of chemically bonded atoms, but have a net positive or negative charge. The chemical properties of an ion are greatly different from those of the atom from which it was derived. Predicting ionic charges Many atoms gain or lose electrons such that they end up with the same number of electrons as the noble gas closest to them in the periodic table. The noble gasses are generally chemically non-reactive, they would appear to have a stable arrangement of electrons. Other elements must gain or lose electrons, to end up with the same arrangement of electrons as the noble gases, in order to achieve the same kind of electron stability. ________________________________________ Example: Nitrogen Nitrogen has an atomic number of 7; the neutral Nitrogen atom has 7 protons and 7 electrons. If Nitrogen gained three electrons it would have 10 electrons, like the Noble gas Neon (10 protons, 10 electrons). However, unlike Neon, the resulting Nitrogen ion would have a net charge of N3- (7 protons, 10 electrons). The location of the elements on the Periodic table can help in predicting the expected charge of ionic forms of the elements. This is mainly true for the elements on either side of the chart. ________________________________________ Ionic compounds Ions form when one or more electrons transfer from one neutral atom to another. For example, when elemental sodium is allowed to react with elemental chlorine an electron transfers from a neutral sodium to a neutral chlorine. The result is a sodium ion (Na+) and a chlorine ion, chloride (Cl-): The oppositely charged ions attract one another and bind together to form NaCl (sodium chloride) an ionic compound. An ionic compound contains positively and negatively charged ions It should be pointed out that the Na+ and Cl- ions are not chemically bonded together. Whereas atoms in molecular compounds, such as H2O, are chemically bonded. Ionic compounds are generally combinations of metals and non-metals. Molecular compounds are general combinations of non-metals only. Pure ionic compounds typically have their atoms in an organized three dimensional arrangement (a crystal). Therefore, we cannot describe them using molecular formulas. We can describe them using empirical formulas. If we know the charges of the ions comprising an ionic compound, then we can determine the empirical formula. The key is knowing that ionic compounds are always electrically neutral overall. Therefore, the concentration of ions in an ionic compound are such that the overall charge is neutral. In the NaCl example, there will be one positively charged Na+ ion for each negatively charged Cl- ion. ________________________________________ What about the ionic compound involving Barium ion (Ba2+) and the Chlorine ion (Cl-)? 1 (Ba2+) + 2 (Cl-) = neutral charge Resulting empirical formula: BaCl2 ________________________________________ 1996 Michael Blaber Atoms, Molecules and Ions Naming inorganic compounds ________________________________________ Naming Inorganic Compounds With over 10 million known chemicals, and potentially dangerous results if chemicals are combined in an incorrect manner, imagine the problem if you are in the lab and say "mix 10 grams of that stuff in with this stuff". We need to be very clear on identification of chemicals. Two early classifications of chemical compounds: 1. Organic compounds. These contain the element Carbon (C). "Life on earth is carbon based" 2. Inorganic compounds. All other compounds Organic compounds were associated with living organisms, however, a large number of organic compounds have been synthesized which do not occur in nature, so this distinction is no longer valid. Ionic compounds: (an association of a cation and an anion) The positive ion (cation) is always named first and listed first in writing the formula for the compound. The vast majority of monatomic (composed of a single atom) cations are formed from metallic elements: • Na+ Sodium ion • Zn2+ Zinc ion • Al3+ Aluminum ion If an element can form more than one positive ion, the positive charge of the ion is indicated by a Roman numeral in parentheses following the name of the metal: • Fe2+ iron(II) ion • Fe3+ iron(III) ion • Cu+ copper(I) ion • Cu2+ copper(II) ion Iron and copper are examples of transition metals. They occur in the block of elements from IIIB to IIB of the periodic table. The transition metals often form two or more different monoatomic cations. click on picture for larger image An older nomenclature for distinguishing between the different ions of a metal is to use the suffixes -ous and -ic. The suffix -ic will indicate the ion of higher ionic charge: • Fe2+ ferrous ion • Fe3+ ferric ion • Cu+ cuprous ion • Cu2+ cupric ion Note that the different ions of the same element often have quite different chemical properties (again, pointing to the importance of electrons in determining chemical reactivity). Ionic compounds: Anions Monatomic anions are usually formed from non-metallic elements. They are named by dropping the ending of the element name and adding -ide: • Cl- chloride ion • F- flouride ion • S2- sulfide ion • O2- oxide ion Some common polyatomic anions include: • OH- hydroxide ion • CN- cyanide ion Many polyatomic anions contain oxygen, and are referred to as oxyanions. When an element can form two different oxyanions the name of the one that contains more oxygen ends in -ate, the one with less ends in -ite: • NO2- nitrite ion • NO3- nitrate ion • SO32- sulfite ion • SO42- sulfate ion Note that unlike the -ous and -ic suffix nomenclature to distinguish the different cations of a metal, the -ite and -ate suffix is used to distinguish the relative amounts of the oxygen atoms in a (polyatomic) oxyanion (in the above examples the ionic charge is the same for the -ite and -ate ions of a specific oxyanion). Now to get really perverse.... Some compounds can have multiple oxyanion forms (the oxyanions involving the halogens, for example): • ClO- • ClO2- • ClO3- • ClO4- Note again, that the number of Oxygens relative to the Chlorine is changing, but that the ionic charge is not. How do we name these? The -ite and -ate suffixes are still used, but we have to add an additional modification to allow us to distinguish between the four forms: • ClO- hypochlorite ion • ClO2- chlorite ion • ClO3- chlorate ion • ClO4- perchlorate ion It should be pointed out that some of the naming of ions is historical and is not necessarily systematic. It may be frustrating and confusing, but its all part of chemistry's rich history. Many polyatomic anions that have high (negative) charges can add one or more hydrogen cations (H+) to form anions of lower effective charge. The naming of these anions reflects whether the H+ addition involves one or more hydrogen ions: • HSO4- hydrogen sulfate ion • H2PO4- dihydrogen phosphate ion Acids • An acid is a substance whose molecules yield hydrogen (H+) ions when dissolved in water. • The formula of an acid consists of an anionic group whose charged is balanced by one or more H+ ions. • The name of the acid is related to the name of the anion • Anions whose names end in -ide have associated acids that have the hydro- prefix and an -ic suffix: Cl- chloride anion HCl hydrochloric acid S2- sulfide anion H2S hydrosulfuric acid Using the -ic suffix here may seem a bit inconsistent since it was used in naming metal cations to indicate the form which had the higher positive charge. However, when you think about it, the acid compound has a higher net positive charge than the anion from which it is derived (the anion is negatively charge and the associated acid is neutral). Again, things get complicated when we consider the acids of oxyanions: • If the anion has an -ate ending, the corresponding acid is given an -ic ending • If the anion has an -ite ending, the corresponding acid has an -ous ending. • Prefixes in the name of the anion are kept in naming the acid ClO- hypochlorite ion HClO hypochlorous acid ClO2- chlorite ion HClO2 chlorous acid ClO3- chlorate ion HClO3 chloric acid ClO4- perchlorate ion HClO4 perchloric acid This is confusing: we previously had used the -ous and -ic suffixes to indicate the ionic charge differences in metal cations (-ic had a higher positive charge). Although in comparison to the ionic form, the -ic and -ous acid forms have a higher net positive charge, the -ic suffix would indicate forms with a higher oxygen content, and not an apparent charge difference. Molecular compounds Although they may not be ionic compounds, chemically bonded compounds of two different elements can be thought of as being made up of an element with a more positive chemical nature, and one that has a more negative nature in comparison. Elements on the left hand side of the periodic table prefer to donate electrons (thus taking on a more positive chemical nature), and elements on the right hand side prefer to accept electrons (thus taking on a more negative chemical nature). The element with the more positive nature in a compound is named first. The second element is named with an -ide ending. Often a pair of elements can form several different molecular compounds. For example, Carbon and Oxygen can form CO and CO2. Prefixes are used to identify the relative number of atoms in such compounds: • CO carbon monoxide (carbon mono oxide) • CO2 carbon dioxide Such prefixes can extend for quite a way for some organic and polymeric compounds (a common detergent in shampoos is sodium dodecylsulfate, or "SDS", also known as Sodium Laurel Sulfate because it sounds more benign and holistic). The list of such prefixes includes: Prefix Meaning Mono- 1 Di- 2 Tri- 3 Tetra- 4 Penta- 5 Hexa- 6 Hepta- 7 Octa- 8 Nona- 9 Deca- 10 Undeca- 11 Dodeca- 12 ________________________________________ 1996 Michael Blaber Stoichiometry: Chemical Formulas and Equations Chemical equations ________________________________________ What happens to matter when it undergoes chemical changes? The law of conservation of mass: Atoms are neither created, nor distroyed, during any chemical reaction Thus, the same collection of atoms is present after a reaction as before the reaction. The changes that occur during a reaction just involve the rearrangement of atoms. In this section we will discuss stoichiometry (the "measurement of elements"). Chemical equations Chemical reactions are represented on paper by chemical equations. For example, hydrogen gas (H2) can react (burn) with oxygen gas (O2) to form water (H20). The chemical equation for this reaction is written as: The '+' is read as 'reacts with' and the arrow '' means 'produces'. The chemical formulas on the left represent the starting substances, called reactants. The substances produced by the reaction are shown on the right, and are called products. The numbers in front of the formulas are called coefficients (the number '1' is usually omitted). Because atoms are neither created nor destroyed in a reaction, a chemical equation must have an equal number of atoms of each element on each side of the arrow (i.e. the equation is said to be 'balanced'). Steps involved in writing a 'balanced' equation for a chemical reaction: 1. Experimentally determine reactants and products 2. Write 'un-balanced' equation using formulas of reactants and products 3. Write 'balanced' equation by determining coefficients that provide equal numbers of each type of atom on each side of the equation (generally, whole number values) Note! Subscripts should never be changed when trying to balance a chemical equation. Changing a subscript changes the actual identity of a product or reactant. Balancing a chemical equation only involves changing the relative amounts of each product or reactant. ________________________________________ Consider the reaction of burning the gas methane (CH4) in air. We know experimentally that this reaction consumes oxygen (O2) and produces water (H2O) and carbon dioxide (CO2). Thus, we have accomplished step #1 above. We now write the unbalanced chemical equation (step #2): Now lets count up the atoms in the reactants and products: We seem to be o.k. with our number of carbon atoms in both the reactants and products, but we have only half the hydrogens in our products as in our reactants. We can fix this by doubling the relative number of water molecules in the list of products: Note that while this has balanced our carbon and hydrogen atoms, we now have 4 oxygen atoms in our products, and only have 2 in our reactants. We can balance our oxygen atoms by doubling the number of oxygen atoms in our reactants: We now have fulfilled step #3, we have a balance chemical equation for the reaction of methane with oxygen. Thus, one molecule of methane reacts with two molecules of oxygen to produce one molecule of carbon dioxide and two molecules of water. ________________________________________ The physical state of each chemical can be indicated by using the symbols (g), (l), and (s) (for gas, liquid and solid, respectively): ________________________________________ 1996 Michael Blaber Stoichiometry: Chemical Formulas and Equations Patterns of chemical reactivity ________________________________________ Patterns of Chemical Reactivity Using the periodic table We can often predict a reaction if we have seen a similar reaction before. For example, sodium (Na) reacts with water (H20) to form sodium hydroxide (NaOH) and H2 gas: note: (aq) indicates aqueous liquid Potassium (K) is in the same family (column) of elements in the periodic table. Therefore, one might predict that the reaction of K with H2O would be similar to that of Na: In fact, all alkali metals react with water to form their hydroxide compounds and hydrogen. Combustion in air Combustion reactions are rapid reactions that produce a flame. Most common combustion reactions involve oxygen (O2) from the air as a reactant. A common class of compounds which can participate in combustion reactions are hydrocarbons (compounds that contain only carbon and hydrogen). Examples of common hydrocarbons: Name Molecular formula methane CH4 propane C3H8 butane C4H10 octane C8H18 When hydrocarbons are combusted they react with oxygen (O2) to form carbon dioxide (CO2) and water (H2O). For example, when propane is burned the reaction is: Other compounds which contain carbon, hydrogen and oxygen (e.g. the alcohol methanol CH3OH, and the sugar glucose C6H12O6) also combust in the presence of oxygen (O2) to produce CO2 and H2O. Combination and decomposition reactions In combination reactions two or more compounds react to form one product: In decomposition reactions one substance undergoes a reaction to form two or more products. For example, many metal carbonates undergo a heat dependent decomposition to the corresponding oxide plus CO2: ________________________________________ 1996 Michael Blaber Stoichiometry: Chemical Formulas and Equations Atomic and Molecular Weights ________________________________________ Atomic and Molecular Weights The subscripts in chemical formulas, and the coefficients in chemical equations represent exact quantities. H2O, for example, indicates that a water molecule comprises exactly two atoms of hydrogen and one atom of oxygen. The following equation: not only tells us that propane reacts with oxygen to produce carbon dioxide and water, but that 1 molecule of propane reacts with 5 molecules of oxygen to produce 3 molecules of carbon dioxide and 4 molecules of water. Since counting individual atoms or molecules is a little difficult, quantitative aspects of chemistry rely on knowing the masses of the compounds involved. The atomic mass scale Atoms of different elements have different masses. Early work on the separation of water into its constituent elements (hydrogen and oxygen) indicated that 100 grams of water contained 11.1 grams of hydrogen and 88.9 grams of oxygen: 100 grams Water -> 11.1 grams Hydrogen + 88.9 grams Oxygen

Later, scientists discovered that water was composed of two atoms of hydrogen for each atom of oxygen. Therefore, in the above analysis, in the 11.1 grams of hydrogen there were twice as many atoms as in the 88.9 grams of oxygen. Therefore, an oxygen atom must weigh about 16 times as much as a hydrogen atom:



Hydrogen, the lightest element, was assigned a relative mass of '1', and the other elements were assigned 'atomic masses' relative to this value for hydrogen. Thus, oxygen was assigned an atomic mass of 16.
We now know that a hydrogen atom has a mass of 1.6735 x 10-24 grams, and that the oxygen atom has a mass of 2.6561 X 10-23 grams. As we saw earlier, it is convenient to use a reference unit when dealing with such small numbers: the atomic mass unit. The atomic mass unit (amu) was not standardized against hydrogen, but rather, against the 12C isotope of carbon (amu = 12).
Thus, the mass of the hydrogen atom (1H) is 1.0080 amu, and the mass of an oxygen atom (16O) is 15.995 amu. Once the masses of atoms were determined, the amu could be assigned an actual value:
1 amu = 1.66054 x 10-24 grams
conversely:
1 gram = 6.02214 x 1023 amu
Average atomic mass
Most elements occur in nature as a mixture of isotopes (i.e. populations of atoms with different numbers of neutrons, and therefore, mass). We can calculate the average atomic mass of an element by knowing the relative abundance of each isotope, as well as the mass of each isotope.
Example: Naturally occurring carbon is 98.892% 12C and 1.108% 13C. The mass of 12C is 12 amu, and that of 13C is 13.00335 amu. Therefore, the average atomic mass of carbon is:
(0.98892)*(12 amu) + (0.01108)*(13.00335 amu) = 12.011 amu

The average atomic mass of each element (in amu) is also referred to as its atomic weight. Values for the atomic weights of each of the elements are commonly listed in periodic tables.
Formula and Molecular Weights
The formula weight of a substance is the sum of the atomic weights of each atom in its chemical formula.
For example, water (H2O) has a formula weight of:
2*(1.0079 amu) + 1*(15.9994 amu) = 18.01528 amu

If a substance exists as discrete molecules (as with atoms that are chemically bonded together) then the chemical formula is the molecular formula, and the formula weight is the molecular weight. For example, carbon, hydrogen and oxygen can chemically bond to form a molecule of the sugar glucose with the chemical and molecular formula of C6H12O6. The formula weight and the molecular weight of glucose is thus:

6*(12 amu) + 12*(1.00794 amu) + 6*(15.9994 amu)
= 180.0 amu

Ionic substances are not chemically bonded and do not exist as discrete molecules. However, they do associate in discrete ratios of ions. Thus, we can describe their formula weights, but not their molecular weights. Table salt (NaCl), for example, has a formula weight of:
23.0 amu + 35.5 amu
= 58.5 amu

Percentage composition from formulas
In some types of analyses of it is important to know the percentage by mass of each type of element in a compound. Take for example methane:
CH4
Formula and molecular weight: 1*(12.011 amu) + 4*(1.008) = 16.043 amu
%C = 1*(12.011 amu)/16.043 amu = 0.749 = 74.9%
%H = 4*(1.008 amu)/16.043 amu = 0.251 = 25.1%
________________________________________
1996 Michael Blaber
Stoichiometry: Chemical Formulas and Equations
The Mole
________________________________________
The Mole
Even tiny samples of chemicals contain huge numbers of atoms, ions or molecules. For convenience sake, some kind of reference for a collection of a large number of these objects would be very useful (e.g. a "dozen" is a reference to a collection of 12 objects). In chemistry we use a unit called a mole (abbreviated mol).
A mole is defined as the amount of matter that contains as many objects as the number of atoms in exactly 12 grams of 12C.
Various experiments have determined that this number is...
6.0221367 x 1023
This is usually abbreviated to simply 6.02 x 1023, and is known as Avogadro's number.
One mole of atoms, volkswagens, people, etc. contains 6.02 x 1023 of these objects. Just how big is this number? One mole of marbles spread over the earth would result in a layer three miles thick.
Molar Mass
A single 12C atom has a mass of 12 amu. A single 24Mg atom has a mass of 24 amu, or twice the mass of a 12C atom. Thus, one mole of 24Mg atoms should have twice the mass as one mole of 12C atoms. Since one mole of 12C atoms weighs 12 grams (by definition), one mole of 24Mg atoms must weigh 24 grams.
Note that the mass of one atom in atomic mass units (amu) is numerically equal to the mass of one mole of the same atoms in grams (g).
The mass in grams of 1 mole (mol) of a substance is called its molar mass.
The molar mass (in grams) of any substance is always numerically equal to its formula weight (in amu).
One H2O molecule weighs 18.0 amu; 1 mol of H2O weighs 18.0 grams
One NaCl ion pair weighs 58.5 amu; 1 mol of NaCl weighs 58.5 grams
Interconverting masses, moles, and numbers of particles
Keeping track of units in calculations is necessary when interconverting masses and moles. This is formally known as dimensional analysis.
________________________________________
"Igor! bring me 1.5 moles of calcium chloride"
Chemical formula of calcium chloride = CaCl2
Molecular mass of Ca = 40.078 amu
Molecular mass of Cl = 35.453 amu
Therefore, the formula weight of CaCl2 = (40.078) + 2(35.453) = 110.984 amu (remember, this compound is ionic, so there is no "molecular" weight).
Therefore, one mole of CaCl2 would have a mass of 110.984 grams.
so, 1.5 moles of CaCl2 would be:
(1.5 mole)(110.984 grams/mole) = 166.476 grams
________________________________________
"Igor! I have 2.8 grams of gold, how many atoms do I have?"
Molecular formula of gold is: Au
Molecular weight of Au = 196.9665 amu
Therefore, 1 mole of gold weighs 196.9665 grams. So, in 2.8 grams of gold we would have:
(2.8 gram)(1 mole/196.9665 gram) = 0.0142 mole
From Avogadro's number, we know that there are approximately 6.02 x 1023 atoms/mole. Therefore, in 0.0142 moles we would have:
(0.0142 mole)(6.02 x 1023 atoms/mole) = 8.56 x 1021 atoms
________________________________________
1996 Michael Blaber
Stoichiometry: Chemical Formulas and Equations
Empirical formulas from analyses
________________________________________
Empirical Formulas from Analyses
An empirical formula tells us the relative ratios of different atoms in a compound. The ratios hold true on the molar level as well. Thus, H2O is composed of two atoms of hydrogen and 1 atom of oxygen. Likewise, 1.0 mole of H2O is composed of 2.0 moles of hydrogen and 1.0 mole of oxygen.
We can also work backwards from molar ratios:
if we know the molar amounts of each element in a compound we can determine the empirical formula.
________________________________________
Mercury forms a compound with chlorine that is 73.9% mercury and 26.1% chlorine by mass. What is the empirical formula?
Let's say we had a 100 gram sample of this compound. The sample would therefore contain 73.9 grams of mercury and 26.1 grams of chlorine. How many moles of each atom do the individual masses represent?
For Mercury:
(73.9 g)*(1 mol/200.59 g) = 0.368 moles
For Chlorine:
(26.1 g)*(1 mol/35.45 g) = 0.736 mol
What is the molar ratio between the two elements?
( 0.736 mol Cl/0.368 mol Hg) = 2.0
Thus, we have twice as many moles (i.e. atoms) of Cl as Hg. The empirical formula would thus be (remember to list cation first, anion last):
HgCl2
________________________________________
Molecular formula from empirical formula
The chemical formula for a compound obtained by composition analysis is always the empirical formula. We can obtain the chemical formula from the empirical formula if we know the molecular weight of the compound.
The chemical formula will always be some integer multiple of the empirical formula (i.e. integer multiples of the subscripts of the empirical formula).
________________________________________
Vitamin C (ascorbic acid) contains 40.92 % C, 4.58 % H, and 54.50 % O, by mass. The experimentally determined molecular mass is 176 amu. What is the empirical and chemical formula for ascorbic acid?
In 100 grams of ascorbic acid we would have:
40.92 grams C
4.58 grams H
54.50 grams O
This would give us how many moles of each element?



Determine the simplest whole number ratio by dividing by the smallest molar amount (3.406 moles in this case - see Oxygen):



The relative molar amounts of carbon and oxygen appear to be equal, but the relative molar amount of hydrogen is higher. Since we cannot have "fractional" atoms in a compound, we need to normalize the relative amount of hydrogen to be equal to an integer. 1.333 would appear to be 1 and 1/3, so if we multiply the relative amounts of each atom by '3', we should be able to get integer values for each atom.
C = (1.0)*3 = 3
H = (1.333)*3 = 4
O = (1.0)*3 = 3
or, C3H4O3
This is our empirical formula for ascorbic acid. What about the chemical formula? We are told that the experimentally determined molecular mass is 176 amu. What is the molecular mass of our empirical formula?
(3*12.011) + (4*1.008) + (3*15.999) = 88.062 amu
The molecular mass from our empirical formula is signficantly lower than the experimentally determined value. What is the ratio between the two values?
(176 amu/88.062 amu) = 2.0
Thus, it would appear that our empirical formula is essentially one half the mass of the actual molecular mass. If we multiplied our empirical formula by '2', then the molecular mass would be correct. Thus, the actual molecular formula is:
2* C3H4O3 = C6H8O6
________________________________________
The general flow chart for solving empirical formulas from known mass percentages is:

Combustion analysis
When a compound containing carbon and hydrogen is subject to combustion with oxygen in a special combustion apparatus all the carbon is converted to CO2 and the hydrogen to H2O.

The amount of carbon produced can be determined by measuring the amount of CO2 produced. This is trapped by the sodium hydroxide, and thus we can monitor the mass of CO2 produced by determining the increase in mass of the CO2 trap. Likewise, we can determine the amount of H produced by the amount of H2O trapped by the magnesium perchlorate.
________________________________________
Consider the combustion of isopropyl alcohol. The sample is known to contain only C, H and O. Combustion of 0.255 grams of isopropyl alcohol produces 0.561 grams of CO2 and 0.306 grams of H2O. From this information we can quantitate the amount of C and H in the sample:

Since one mole of CO2 is made up of one mole of C and two moles of O, if we have 0.0128 moles of CO2 in our sample, then we know we have 0.0128 moles of C in the sample. How many grams of C is this?

How about the hydrogen?

Since one mole of H2O is made up of one mole of oxygen and two moles of hydrogen, if we have 0.017 moles of H2O, then we have 2*(0.017) = 0.034 moles of hydrogen. Since hydrogen is about 1 gram/mole, we must have 0.034 grams of hydrogen in our original sample.
When we add our carbon and hydrogen together we get:
0.154 grams (C) + 0.034 grams (H) = 0.188 grams
But we know we combusted 0.255 grams of isopropyl alcohol. The 'missing' mass must be from the oxygen atoms in the isopropyl alcohol:
0.255 grams - 0.188 grams = 0.067 grams oxygen
This much oxygen is how many moles?

Overall therfore, we have:
0.0128 moles Carbon
0.0340 moles Hydrogen
0.0042 moles Oxygen
Divide by the smallest molar amount to normalize:
C = 3.05 atoms
H = 8.1 atoms
O = 1 atom
Within experimental error, the most likely empirical formula for propanol would be:
C3H8O
________________________________________
1996 Michael Blaber


Stoichiometry: Chemical Formulas and Equations
Quantitative information from balanced equations
________________________________________
Quantitative Information from Balanced Equations
The coefficients in a balanced chemical equation can be interpreted both as the relative numbers of molecules involved in the reaction and as the relative number of moles.
For example, in the balanced equation:
2H2(g) + O2(g)-> 2H2O(l)
the production of two moles of water would require the consumption of 2 moles of H2 and one mole of O2. Therefore, when considering this particular reaction
2 moles of H2
1 mole of O2
and
2 moles of H2O
would be considered to be stoichiometrically equivalent quantitites. Represented as:
2 mol H2 1 mol O2 2 mol H2O
Where ' ' means "stoichiometrically equivalent to".
These stoichiometric relationships, derived from balanced equations, can be used to determine expected amounts of products given amounts of reactants. For example, how many moles of H2O would be produced from 1.57 moles of O2 (assuming the hydrogen gas is not a limiting reactant)?

The ratio is the stoichiometric relationship between H2O and O2 from the balanced equation for this reaction.
________________________________________
For the combustion of butane (C4H10) the balanced equation is:

Calculate the mass of CO2 that is produced in burning 1.00 gram of C4H10.
First of all we need to calculate how many moles of butane we have in a 100 gram sample:

now, the stoichiometric relationship between C4H10 and CO2 is: , therefore:

The question called for the determination of the mass of CO2 produced, thus we have to convert moles of CO2 into grams (by using the molecular weight of CO2):

________________________________________
Thus, the overall sequence of steps to solve this problem were:

In a similar way we could determine the mass of water produced, or oxygen consumed, etc.
________________________________________
1996 Michael Blaber


Stoichiometry: Chemical Formulas and Equations
Limiting reactants
________________________________________
Limiting Reactants
Suppose you are a chef preparing a breakfast for a group of people, and are planning to cook French toast. You make French toast the way you have always made it: one egg for every three slices of toast. You never waiver from this recipe, because the French toast will turn out to be either too soggy or too dry (arguably, you are anal retentive). There are 8 eggs and 30 slices of bread in the pantry. Thus, you conclude that you will be able to make 24 slices of French toast and not one slice more.
This is a similar situation with chemical reactions in which one of the reactants is used up before the others - the reaction stops as soon as one of the reactants is consumed. For example, in the production of water from hydrogen and oxygen gas suppose we have 10 moles of H2 and 7 moles of O2.

Because the stoichiometry of the reaction is such that 1 mol of O2 2 moles of H2, the number of moles of O2 needed to react with all of the H2 is:

Thus, after all the hydrogen reactant has been consumed, there will be 2 moles of O2 reactant left.
The reactant that is completely consumed in a chemical reaction is called the limiting reactant (or limiting reagent) because it determines (or limits) the amount of product formed. In the example above, the H2 is the limiting reactant, and because the stoichiometry is 2H2 2H2O (i.e. H2 H2O), it limits the amount of product formed (H2O) to 10 moles. We actually have enough oxygen (O2) to form 14 moles of H2O (1O2 2H2O).
One approach to solving the question of which reactant is the limiting reactant (given an initial amount for each reactant) is to calculate the amount of product that could be formed from each amount of reactant, assuming all other reactants are available in unlimited quantities. In this case, the limiting reactant will be the one that produces the least amount of potential product.
________________________________________
Consider the following reaction:

Suppose that a solution containing 3.50 grams of Na3PO4 is mixed with a solution containing 6.40 grams of Ba(NO3)2. How many grams of Ba3(PO4)2 can be formed?
1. First we need to convert the grams of reactants into moles:


2. Now we need to define the stoichiometric ratios between the reactants and the product of interest (Ba3(PO4)2):
2 Na3PO4 Ba3(PO4)2
3 Ba (NO3)2 Ba3(PO4)2
3. We can now determine the moles of product that would be formed if reactant were to be consumed in its entirety during the course of the reaction:


4. The limiting reactant is the Ba (NO3)2 and we could thus make at most 0.0082 moles of the Ba3(PO4)2 product.
5. 0.0082 moles of the Ba3(PO4)2 product would be equal to:

________________________________________
Theoretical yields
The quantity of product that is calculated to form when all of the limiting reactant is consumed in a reaction is called the theoretical yield.
The amount of product actually obtained is called the actual yield.
Actual yield < Theoretical yield for the following reasons: • for some reason not all the reactants may react • there maybe some significant side reactions • physical recovery of 100% of the sample may be impossible (like getting all the peanut butter out of the jar) The percent yield of a reaction relates the actual yield to the theoretical yield: Percent yield = x 100 For example, in the previous exercise we calculated that 4.94 grams of Ba3(PO4)2 product should be formed. This is the theoretical yield. If the actual yield were 4.02 grams the percent yield would be: ________________________________________ 1996 Michael Blaber Aqueous Reactions and Solution Stoichiometry Solution Composition ________________________________________ Water possesses many unusual properties. One of the most important properties of water is its ability to dissolve a wide variety of substances. It may sound strange, but absolutely pure water can be considered corrosive due to its capacity to absorb other compounds and ions. Solutions in which water is the dissolving medium are called aqueous solutions. Limestone caves, for example, are formed by the dissolving action of water, and dissolved CO2, on solid Calcium Carbonate. The dissolved mineral is then deposited as stalagtites and stalagmites as the water evaporates: CaCO3(s) + H2O(l) + CO2(aq) - Ca(HCO3)2(aq) Many physiological chemical reactions occur in aqueous solutions. How do we express solution composition? What are the chemical forms in which substances occur in aqueous solutions? Solution Composition A solution is a homogenous mixture of two or more substances, consisting of 1. The solvent - usually the substance in greater concentration 2. The other component(s) is (are) called the solute(s) - they are said to be dissolved in the solvent When a small amount of NaCl is dissolved in a large quantity of water, we refer to the water as the solvent and the NaCl as the solute. Molarity The term concentration is used to indicate the amount of solute dissolved in a given quantity of solvent or solution. The most widely used way of quantifying concentration in chemistry is molarity. The molarity (symbol M) of a solution is defined as the number of moles of solute in a liter volume of solution: For example, a 1.0 molar solution (1.0 M) contains 1.00 mol of solute in every liter of solution. What is the molarity of a solution made by dissolving 20 grams of NaCl in 100 mls of water? M solution If we know the molarity of a solution we can calculate the number of moles of solute in a given volume. Thus, molarity is a conversion factor between volume of solution and moles of solute: Calculate the number of moles of CaCl2 in 0.78 liters of a 3.5 M solution: CaCl2 How many liters of a 2.0 M solution of HNO3 do we need to have 5 moles of HNO3? Note: we had to invert the stock solution (i.e. convert to liters per mole) to be able to calculate the needed volume (i.e. to keep the dimensional analysis correct) Dilution For convenience, solutions are either purchased or prepared in concentrated stock solutions which must be diluted prior to use. When we take a sample of a stock solution we have a certain number of moles of molecules in that sample. Dilution alters the molarity (i.e. concentration) of the solution but not the total number of moles of molecules in the solution (in other words, dilution does not create or destroy molecules). One of the standard equations for determining the effects of dilution upon a sample is to set up an equation comparing (concentration)*(volume) before and after dilution. Since (concentration)*(volume) gives us the total number of moles in the sample, and since this does not change, this value before and after dilution are equal: (concentration)*(volume) = (concentration)*(volume) (moles/liter)*(liter) = (moles/liter)*(liter) moles = moles How much of a 5 M stock solution of NaCl will you need to make up 250 mls of a 1.5 M solution? X liters = 0.075 liters (or 75 mls) Thus, we would need 0.075 liters of our 5M NaCl stock solution. The rest of the 0.25 liter volume is made up by the addition of water: 0.25 liters - 0.075 liters = 0.175 liters So we would take 0.075 liters of stock 5M NaCl solution and add to that 0.175 liters of water for a final volume of 0.25 liters with a final concentration of 1.5 moles/liter (i.e. 1.5 M) ________________________________________ What is the concentration of water? Molecular weight of H2O = 18.0g/mole Density of H2O = 1g/ml or 1000g/L Pure water is 55.6M H2O Aqueous Reactions and Solution Stoichiometry Solution Stoichiometry ________________________________________ Solution Stoichiometry For balanced chemical equations involving solutions we calculate the number of moles by knowing the concentration (moles/liter, or Molarity) and volume (in liters). How many moles of water form when 25.0 mls of 0.100 M HNO3 (nitric acid) solution is completely neutralized by NaOH (a base)? 1. Let's begin by writing the balanced equation for the reaction: 2. The stoichiometric relationship between HNO3 and H2O is HNO3 H2O, therefore, for one mole of HNO3 that is completely consumed (i.e. neutralized) in the reaction, one mole of H2O is produced. 3. How many moles of HNO3 are we starting with? HNO3 4. Therefore, we should have 0.0025 moles of H2O produced Titrations How can we know the concentration of some solution of interest? One answer to this problem lies in the method of titration. In titration we will make use of a second solution known as a standard solution which has the following characteristics: 1. The second solution contains a chemical which reacts in a defined way, with known stoichiometry, with the solute of the first solution 2. The concentration of the solute in this second solution is known. Classic titrations include so-called acid-base titrations. In these experiments a solution of an acid with an unknown concentration is titrated with a solution of known concentration of base (or vice versa). For example, we may have a solution of hydrochloric acid (HCl) of unknown concentration and a standard solution of NaOH. To a fixed amount of the HCl solution is added incremental amounts of the NaOH solution until the acid is completely neutralized - i.e. a stoichiometrically equivalent quantity of HCl and NaOH have been combined. This is known as the equivalence point in the titration. By knowing the concentration of the standard solution, and the amount added to achieve stoichiometric equivalency, we can determine the amount of moles of HCl in the original sample volume. How do we know when we have reached the equivalence point in such a titration experiment? In this type of acid-base titration, so called indicator-dyes are used. For example phenolphthalein is colorless in acidic solutions and turns red in basic solutions. Thus, in the above experiment we will add a small amount of this indicator-dye and add base until we barely begin to see a color change to red. Mike's Handy Science Tip Phenolphthalein is not only useful in acid-base titrations, but its also a darn good laxative used to be the active ingredient in ExLax. Just thought you'd like to know... 25 mls of a solution of HCl with an unknown concentration is titrated with a standard solution of 0.5 M NaOH. The phenolphthalein indicator dye begins to turn color after the addition of 2.8 mls of standard solution. What is the concentration of the HCl? Balanced equation for the reaction: HCl + NaOH -> NaCL + H2O
of NaOH was added
Since the stoichiometry of the NaOH and HCl is 1:1, the sample of HCl must have contained 0.0014 moles of HCl. The concentration of the HCl solution is therefore:
, or 0.056 M
________________________________________
1996 Michael Blaber

Energy Relations in Chemistry: Thermochemistry
The Nature of Energy
________________________________________
Sugar you eat is "combusted" by your body to produce CO2 and H2O. During this process energy is also released.
This energy is used (among other things) to:
• Operate your muscles
• Maintain your body temperature
Chemical reactions involve changes in energy:
• Some reactions produce energy
• Some reactions require energy
Our society as an "organism" requires energy: 90% of our energy comes from chemical reactions involving the combustion of petroleum products.
The study of energy and its transformations is known as thermodynamics
This area of study began when steam engines were developed during the industrial revolution and the relationships between heat, work and energy for different fuels was being studied.
The relationship between chemical reactions and energy changes is known as thermochemistry
The Nature of Energy
A Force is any kind of push or pull exerted on an object.
• Gravity is a force which keeps us stuck to the earth.
• The Electrostatic force attracts electrons to protons in an atom.
If you move an object against some force, work is being done.
The amount of work (w) being done is relative to the distance (d) the object is moved and the strength of the force (F) against the object:
w = F * d
Energy, in the form of work, must be used to move an object against a force.
When we do work, our body temperature increases (and we sweat to cool us down). Our bodies are generating Heat energy.
Heat is an energy which is transferred from one object to another depending on the relative temperature:
• Heat energy flows from an object towards other objects of lower temperature

Energy is the capacity to do work or to transfer heat
Objects can possess energy due to their motions and positions, as kinetic energy and potential energy.
Kinetic and Potential Energy
Kinetic energy is the energy of motion. The magnitude of the kinetic energy (Ek) of an object depends upon its mass (m) and velocity (v):

In other words, both the mass and the speed of an object determines how much energy it has, and thus, how much work it can accomplish.
An object can also possess energy based upon its position relative to other objects - a type of stored up energy, or "potential energy"
Potential energy is the result of the attractions and repulsion between objects. An electron has potential energy when located near a proton due to the attractive electrostatic force between them.
Chemical and thermal energy are terms which relate to potential and kinetic energy at the atomic level
• Chemical energy is the potential energy stored in the arrangement of electrons and protons
• Thermal energy reflects the kinetic energy of the molecules of a substance.
Energy Units
The SI unit for energy is the joule ("J"). In honor of James Prescot Joule (1818-1889) a British Scientist who investigated work and heat. (Note: SI is short for the French term Systeme International d'Unites. Which defines metric standards).
Kinetic energy for example is defined as:

Thus, the joule must have units of:
kg*(meters/second)2
and, in fact, 1 joule is defined as:

Traditionally, energy changes accompanying chemical reactions have been expressed in calories, which is a non-SI unit (though still widely used).
1 calorie = 4.184 J
Systems and surrounding
When we focus on a study of energy changes we look at a small, well defined and isolated part of the universe - the flask or container the reactants are in. This is called the system.
Everything else is called the surroundings.
Usually the system is isolated from its surroundings such that there will be an exchange of energy between system and surroundings, but not matter.
Thus, the system will contain the same mass after an experiment, but the system can lose or gain energy (in the form of heat, work, or both).
Lowering the energy of the system
Systems tend to attain as low an energy as possible
Systems with a high potential energy are less stable and more likely to undergo change than systems with a low potential energy.
Like a shopping cart at the top of a hill, chemical reactants move spontaneously toward a lower potential energy when possible.
________________________________________
1996 Michael Blaber

Energy Relations in Chemistry: Thermochemistry
Enthalpies of Reaction
________________________________________
Heat and Enthalpy Changes
When a chemical reaction occurs in an open container most of the energy gained or lost is in the form of heat. Almost no work is done (i.e. nothing is being moved).
Heat flows between the system and surroundings until the two are at the same temperature.
• When a chemical reaction occurs in which the system absorbs heat, the process is endothermic (it feels cold)
• When a chemical reaction occurs in which the system produces heat it is exothermic (it feels hot)
Enthalpy
Under conditions of constant pressure (e.g. most biological processes under constant atmospheric pressure) the heat absorbed or released is termed enthalpy (or "heat content").
We do not measure enthalpy directly, rather we are concerned about the heat added or lost by the system, which is the change in enthalpy (or H).
In formal terms: The change in enthalpy, H, equals the heat, qp, added to or lost by the system when the process occurs under constant pressure:
H=qp
H represents the difference between the enthalpy of the system at the beginning of the reaction compared to what it is at the end of the reaction:
H = Hfinal - Hinitial
We are considering the enthalpic state of the system. Thus:
• if the system has higher enthalpy at the end of the reaction, then it absorbed heat from the surroundings (endothermic reaction)
• if the system has a lower enthalpy at the end of the reaction, then it gave off heat during the reaction (exothermic reaction)
Therefore:
• For endothermic reactions Hfinal > Hinitial and H is positive (+H)
• For exothermic reactions Hfinal < Hinitial and H is negative (-H) ________________________________________ Enthalpies of Reaction Because the enthalpy change for a reaction is described by the final and initial enthalpies: H = Hfinal - Hinitial we can also describe H for a reaction by comparing the enthalpies of the products and the reactants: H = H(products) - H(reactants) The enthalpy change that accompanies a reaction is called the enthalpy of reaction (Hrxn). It is sometimes convenient to provide the value for Hrxn along with the balanced chemical equation for a reaction (also known as a thermochemical equation): 2H2(g) + O2(g) -> 2H2O(g) H = -483.6 kJ
Note the following:
• H is negative, indicating that this reaction results in the release of heat (exothermic)
• The reaction gives of 483.6 kilo Joules of energy when 2 moles of H2 combine with 1 mole of O2 to produce 2 moles of H2O.
The relative enthalpies of the reactants and products can also be shown on an energy diagram:

Properties of enthalpy:
1. Enthalpy is an extensive property. The magnitude of H is dependent upon the amounts of reactants consumed. Doubling the reactants, doubles the amount of enthalpy.
2. Reversing a chemical reaction results in the same magnitude of enthalpy but of the opposite sign. For example, splitting two moles of water to produce 2 moles of H2 and 1 mole of O2 gas requires the input of +483.6 kJ of energy.
3. The enthalpy change for a reaction depends upon the state of the reactants and products. The states (i.e. g, l, s or aq) must be specified.
CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g) H = -802 kJ
Given the above thermochemical equation for the combustion of methane, how much heat energy is released when 4.5 grams of methane is burned (in a constant pressure system)?


The negative sign (exothermic) indicates that 225.5 kJ of energy are given off by the system into the surroundings.
________________________________________
1996 Michael Blaber

Energy Relations in Chemistry: Thermochemistry
Calorimetry
________________________________________
Calorimetry
Experimentally, we can determine the heat flow (Hrxn) associated with a chemical reaction by measuring the temperature change it produces.
• The measurement of heat flow is called calorimetry
• An apparatus that measures heat flow is called a calorimeter
Heat capacity and specific heat
The temperature change experienced by an object when it absorbs a certain amount of energy is determined by its heat capacity.
• The heat capacity of an object is defined as the amount of heat energy required to raise its temperature by 1 K (or °C)
• The greater the heat capacity of an object, the more heat energy is required to raise the temperature of the object
For pure substances the heat capacity is usually given for a specified amount of the substance
• The heat capacity of 1 mol of a substance is called its molar heat capacity
• The heat capacity of 1 gram of a substance is called its specific heat
The specific heat of a substance can be determined experimentally by measuring the temperature change (T) that a known mass (m) of the substance undergoes when it gains or loses a specific quantity of heat (q):


209 J of energy are required to increase the temperature of 50.0 g of water by 1.00 K. What is the specific heat of water?


Specific heat values of some substances (J g-1K-1)
Al (s) 0.90 CaCO3 (s) 0.85
C (s) 0.71 CCl4 (l) 0.86
Fe (s) 0.45 H2O (l) 4.18
Hg (l) 0.14
We can calculate the quantity of heat that a substance has gained or lost by using its specific heat together with its measured mass and temperature change:

rearrange:

________________________________________
How much heat is required to raise the temperature of 250g of water from 22 °C to 98 °C? (specific heat of water is 4.18 J g-1 K-1).
q = (4.18 J g-1 K-1)*(250g)*(371-295 K)
q = 79,420 J (79.420 kJ, or 7.942 x 104 J)
What is the molar heat capacity of water?
Molar heat capacity = 4.18 J g-1 K-1 * (18 grams/1.0 mole)
= 75.2 J mole-1 K-1
________________________________________
Constant-Pressure Calorimetry
Recall that H is defined as the quantity of heat transferred under constant pressure (H = qp).
A calorimeter for such measurements would have the following general construction:

Note that the pressure regulator could be just a vent to allow the pressure to be maintained at atmospheric pressure
• For reactions which involve dilute aqueous solutions, the specific heat of the solution will be approximately that of water (4.18 J g-1 K-1)
• The heat absorbed by an aqueous solvent is equal to the heat given off by the reaction of the solutes:
qaq solvent = -qrxn
Remember: If a reaction gives off heat, it is exothermic and H is negative. The enthalpy of the products is less than the enthalpy of the reactants
H = Hproducts-Hreactants
In our calorimeter with an aqueous solution, if the reaction of the solutes is exothermic, the solution will absorb this heat and increase in temperature.
Thus, for an exothermic reaction:
• The solutes have a lower final enthalpy after the reaction (H negative)
• The solution has a higher final enthalpy after the reaction (H positive)
So, to determine the actual Hrxn we would invert the sign of the Hsoln (the actual value we will measure)
________________________________________
50 ml of 1.0 M HCl and 50 ml of NaOH are combined in a constant pressure calorimeter. The temperature of the solution is observed to rise from 21.0 °C to 27.5 °C. Calculate the enthalpy change for the reaction (assume density is 1.0 gram/ml, and that the specific heat of the solution is that of water).
Tsolution = 27.5 - 21.0 °C = 6.5 °C (K)
specific heat = 4.18 J g-1 K-1
mass = (100 ml)*(1 gram/ml) = 100 grams
Hsolution = qp = (4.18 J g-1 K-1)*(100 g)*(6.5 K) = 2,717 J
We know that Hrxn = -Hsolution therefore,
Hrxn = - 2,717 J
Note: exothermic reactions have negative values for Hrxn thus, if the reaction gave off heat (i.e. raised the temperature of the solution) you know that the sign for Hrxn must be negative.
What is the enthalpy change on a molar basis?
HCl + NaOH -> NaCl + H2O
This is the balanced equation, and we combined:
(0.05 liters HCl)*(1.0 mol/liter) = 0.05 moles HCl
and
(0.050 liters NaOH)*(1.0 mol/liter) = 0.05 moles NaOH
The stoichiometry for HCl and NaOH in this reaction is 1:1, so they are combining in stoichiometrically equivalent amounts and will produce 0.05 moles of NaCl (and H2O).
So, the enthalpy change for the production of 0.05 moles of NaCl in the above reaction would be:
-2717 J for each 0.05 moles NaCl, or
-2717 J/.05 moles = 54,340 J/mole = 54.34 kJ/mole
________________________________________
Bomb Calorimetry (constant-volume calorimetry)
Since combustion reactions involve dramatic increases in pressure, they are typically studied under conditions of constant volume in a device known as a bomb calorimeter.
The bomb calorimeter is essentially a sealed insulated instrument with no pressure regulation.
The sealed reaction chamber is surrounded by water, and the energy released, or absorbed, by the sample is measured indirectly by monitoring the temperature change of the water.

Analyses under constant volume allow determination of E (energy change at constant volume), not H (energy change at constant pressure).
________________________________________
1996 Michael Blaber

Energy Relations in Chemistry: Thermochemistry
Hess's Law
________________________________________
Hess's Law
It is often possible to calculate H for a reaction from listed H values of other reactions (i.e. you can avoid having to do an experiment)
Enthalpy is a state function
• It depends only upon the initial and final state of the reactants/products and not on the specific pathway taken to get from the reactants to the products
• Whether one can arrive at the products via either a single step or multi-step mechanism is unimportant as far as the enthalpy of reaction is concerned - they should be equal
Consider the combustion reaction of methane to form CO2 and liquid H2O
CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)
This reaction can be thought of as occurring in two steps:
• In the first step methane is combusted to produce water vapor:
CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g)
• In the second step water vapor condenses from the gas phase to the liquid phase:
2H2O(g) -> 2H2O(l)
Each of these reactions is associated with a specific enthalpy change:
CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g) H = -802 kJ
2H2O(g) -> 2H2O(l) H = -88 Kj
(Note: under conditions of standard temperature and pressure the liquid state of water is the normal state. Thus, the gas would be expected to condense. This is an exothermic process under these conditions. In a related process, it should get warmer when it rains)
Combining these equations yields the following:
CH4(g)+2O2(g)+2H2O(g) -> CO2(g)+2H2O(g)+2H2O(l)
H = (-802) kJ + (-88) kJ= -890 kJ
Hess's Law
if a reaction is carried out in a series of steps, H for the reaction will be equal to the sum of the enthalpy changes for the individual steps
the overall enthalpy change for the process is independent of the number of steps or the particular nature of the path by which the reaction is carried out.
Thus we can use information tabulated for a relatively small number of reactions to calculate H for a large number of different reactions
________________________________________
Determining H for the reaction

is difficult because some CO2 is also typically produced.
However, complete oxidation of either C or CO to yield CO2 is experimentally pretty easy to do:


We can invert reaction number 2 (making it endothermic) and have CO(g) as a product. (This describes the decomposition of CO2 to produce CO and O2)

Thus, we now have two equations with known enthalpies of reaction: the first describes the combustion of carbon and oxygen to produce CO2 and the second describes how CO2 can be decomposed to produce carbon monoxide (and oxygen). We can combine these together to describe the production of carbon monoxide from the combustion of carbon and oxygen:

The overall reaction, going from left-hand side reactant(s) to the right-hand side product(s) would be:

We can algebraically subtract the one-half O2 from both sides to yield the following equation with the associated overall enthalpy:

Another way to look at the method of combining reactions would be as follows:

plus

gives:

H = (-393.5 kJ) + (283.0 kJ) = -110.5 kJ
canceling out identical compounds from the left and right hand sides of this reaction gives

________________________________________
Carbon occurs in two forms: graphite and diamond. The enthalpy of combustion of graphite is -393.5 kJ, and that of diamond is -395.4 kJ
C(graphite) + O2(g) -> CO2(g) H = -393.5 kJ
C(diamond) + O2(g) -> CO2(g) H = -395.4 kJ
Calculate H for the conversion of graphite to diamond
What we want is H for the reaction:
C(graphite) -> C(diamond)
C(graphite) + O2(g) -> CO2(g) H = -393.5 kJ
CO2(g) -> C(diamond) + O2(g) H = +395.4 kJ
C(graphite) -> C(diamond) H = +1.9 kJ
________________________________________
We can never expect to obtain more or less energy from a chemical reaction by changing the method of carrying out the reaction ("conservation of energy").
Another way of saying this is that the particular pathway chosen to arrive at the same reactants yields the same H for the overall reaction.
Consider the previous example of the combustion of methane to produce gaseous H2O and then the condensation of the gaseous H2O to the liquid state. How is this represented in an Enthalpy Diagram?

The key features are the following:
1. Each line represents a set of reactants or products for a balanced chemical reaction. When going from one line to another, the atoms must balance. For example, if we were to ask what is the enthalpy associated with the condensation of water we would have (from the above data):

The CO2(g) on both sides will cancel to yield:

2. The relative distance of each line must reflect the relative enthalpy difference (H) between the reactants/products. If the enthalpy change in going from reactants to products is negative, then the line for the products must be below the reactants. Furthermore, the length of the distance must be proportional. For example, the distance reflecting the enthalpy associated with the condensation of water (H = -88 kJ)is only about 10% as long as the distance between the reactants and products for the combustion of methane to CO2 and liquid water (H = -890 kJ)
________________________________________
1996 Michael Blaber

Energy Relations in Chemistry: Thermochemistry
Enthalpies of Formation
________________________________________
Enthalpies of Formation
Using Hess's Law we can calculate reaction enthalpies for a variety of reactions using tables of known enthalpies
Many experimentally determined enthalpies are listed by the type of process
• H for converting various liquids to the gas phase are listed in tables of enthalpies of vaporization
• H for melting solids to liquids are listed in tables of enthalpies of fusion
• H for for combusting a substance in oxygen are listed in tables of enthalpies of combustion
The enthalpy change associated with the formation of a compound from its constituent elements is called the enthalpy of formation (Hf )
Conditions which influence enthalpy changes include:
• temperature
• pressure
• state of reactants and products (s, g, l, aq)
The standard state of a substance is the form most stable at 298 °K (25 °C, or standard "room temperature") and 1 atmosphere (1 atm) of pressure
When a reaction occurs with all reactants and products in their standard states, the enthalpy change is the standard enthalpy of reaction (H°)
Thus, the standard enthalpy of formation (H°f)of a compound is the change in enthalpy that accompanies the formation of 1 mole of that substance from its elements, with all substances in their standard states
The standard enthalpy of formation for ethanol (C2H5OH) is the enthalpy change for the following reaction

Notes:
• Elemental source of oxygen is O2 and not O because O2 is the stable form of oxygen at 25 °C and 1 atm, likewise with H2
• Elemental source of carbon is specified as graphite (and not, for example, diamond) because graphite is the lowest energy form of carbon at room temp and 1 atm
• Why is the O2 stoichiometry left at "1/2"? The stoichiometry of formation reactions always indicates the formation of 1 mol of product. Thus, H°f values are reported as kJ / mole of the substance produced
If C(graphite) is the lowest energy form of carbon under standard conditions, then what is the H°f for C(graphite)?
• By definition, the standard enthalpy of formation of the most stable form of any element is zero because there is no formation reaction needed when the element is already in its standard state
• H°f for C(graphite), H2(g) and O2(g) = 0
Using enthalpies of formation (H°f) to calculate enthalpies of reactionunder standard conditions H°rxn)
We can determine the standard enthalpy change for any reaction (H°rxn) by using standard enthalpies of formation (H°f) and Hess's Law
Consider the following combustion reaction of propane:
C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l)
The reactants:
• The standard heat of formation (H°f) of propane gas from its elemental constituents in the standard state is ¬103.85 kJ/mole
3C(graphite) + 4H2(g) -> C3H8(g) H°f = -103.85 kJ
• The standard heat of formation (H°f) for O2(g) is zero
O2(g) -> O2(g)H°f = 0 kJ
and so...
5O2(g) -> 5O2(g)H°f = 0 kJ
Overall, therefore, the standard heat of formation (H°f) for the reactants is:
3C(graphite)+4H2(g)+5O2(g) -> C3H8(g)+5O2(g)
H°f = -103.85 kJ
The products:
• The standard heat of formation (H°f) of CO2(g) from its elemental constituents in the standard state is ¬393.5 kJ/mole
C(graphite) + O2(g) -> CO2(g) H°f = ¬393.5 kJ
so, for 3 moles of CO2 molecules the standard heat of formation would be:
3C(graphite) + 3O2(g) -> 3CO2(g) H°f = ¬1180.5 kJ
• The standard heat of formation (H°f) of H2O(l) from its elemental constituents in the standard state is ¬285.8 kJ/mole

and so the H°f for 4 waters would be:

combining the H°f for both products yields:
3C(graphite)+4H2(g)+5O2(g) -> 3CO2(g)+4H2O(l)
with H°f = (-1180.5) + (-1143.2) = -2323.7 kJ
Let's summarize what we have determined so far:
Overall the standard heat of formation (H°f) for the reactants is:
3C(graphite)+4H2(g)+5O2(g) -> C3H8(g)+5O2(g)
H°f = -103.9 kJ
Overall the standard heat of formation (H°f) for the products is:
3C(graphite)+4H2(g)+5O2(g) -> 3CO2(g)+4H2O(l)
H°f = - 2323.7 kJ
Lets plot these on a relative scale of enthalpy:

This information can be used to determine the relative enthalpy difference under standard conditions (H°) between the reactants and products:

This enthalpy difference (-2219.8 kJ) is the enthalpy of the reaction for the combustion of propane under standard conditions (H°rxn)
Calculate the enthalpy change (H°rxn) for the combustion of 1 mol of ethanol
C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O(l)
heat of formation for reactants
2C(graphite)+3H2(g)+(1/2)O2(g) -> C2H5OH(l)
H°f = -277.7 kJ
plus
3O2(g) -> 3O2(g) H°f = 0 kJ
gives:
2C(graphite)+3H2(g)+(7/2)O2(g) -> C2H5OH(l)+ 3O2(g)
H°f = -277.7 kJ
heat of formation for products
C(graphite)+O2(g) -> CO2(g) H°f = -393.5 kJ
therefore
2C(graphite)+2O2(g) -> 2CO2(g) H°f = -787 kJ
H2(g) + (1/2)O2(g) -> H2O(l) H°f = -285.8 kJ
therefore
3H2(g) + (3/2)O2(g) -> 3H2O(l) H°f = -857.4 kJ
combining gives:
2C(graphite)+3H2(g)+(7/2)O2(g) -> 2CO2(g)+ 3H2O(l)
H°f = (-857.4)+(-787) = -1644.4kJ
H°rxn
H°rxn = H°f (products) - H°f (reactants)
(-1644.4) - (-277.7) = -1366.7 kJ
________________________________________
1996 Michael Blaber

Energy Relations in Chemistry: Thermochemistry
Foods and Fuels
________________________________________
Foods and Fuels
The energy released when 1 gram of material is combusted is called its fuel value
Note: since all heats of combustion are exothermic it is customary to leave off the negative sign when reporting fuel values.
Foods
Most of the energy our body needs comes from fats and carbohydrates.
Carbohydrates are broken down in the intestines to glucose. Glucose is transported in the blood to cells where it is oxidized to produce CO2, H2O and energy:
C6H12O6(s)+6O2(g) -> 6CO2(g)+6H2O(l) H°rxn=-2816 kJ
The breakdown of fats also produces CO2 and H2O
Any excess energy in the body is stored as fats
• Insoluble in water so they can be separated and stored in the body
• They produce more energy per gram than carbohydrates or proteins
Compound Fuel Value (kJ/gram)
Fats 38
Carbohydrates 17
Proteins 17
About 100 kJ per kilogram of body weight per day is required to keep the body functioning at a minimum level
Exercise
• Light exercise burns up about 800 kJ/hour of energy
• Heavy exercise burns up about 2000 kJ/hour
Fuels
Different types of fuels contain varying amounts of carbon, hydrogen and oxygen.
The greater the percentage of carbon and hydrogen in the fuel the higher the fuel value
Fuel C
(%) H
(%) O
(%) Fuel Value
(kJ/g)
Wood 50 6 44 18
Bituminous Coal 77 5 7 32
Gasoline 85 15 0 48
Hydrogen 0 100 0 142
The average daily consumption of energy per person in the U.S. is about 8.8 x 105 kJ, or about 100 times greater than our food energy requirements
Our energy comes primarily from the combustion of fossil fuels (hydrocarbons derived from ancient plants and animals)
Coal represents 90% of the fossil fuels on earth. However, it typically contains sulfur, which when combusted can lead to environmental pollution (acid rain)
Solar energy: on a clear day the sun's energy which strikes the earth equals 1kJ per square meter per second.
If we could utilize the solar energy which strikes 0.1% of the U.S. land mass we would have enough power to run the country
Hydrogen: clean burning (produces only water) and high fuel value. Hydrogen can be made from coal as well as methane
C(coal) + H2O(g) -> CO(g)+H2(g)
CH4(g) + H2O(g) -> CO(g) + 3H2(g)
________________________________________
1996 Michael Blaber


























Electronic Structure of Atoms
The Wave Nature of Light
________________________________________
Electrons hold the key to understanding why substances behave as they do. When atoms react it is their outer pars, their electrons, that interact.
We refer to the arrangements of electrons in atoms as their electronic structure.
• Number of electrons
• Where they can be found
• The energies they possess
Be warned!: electrons to not behave like anything we are familiar with in the macroscopic world
The Wave Nature of Light
Much of our present understanding of the electronic structure of atoms has come from analysis of the light emitted or absorbed by substances
Electromagnetic radiation
• Carries energy through space (also known as radiant energy)
• Includes visible light, dental x-rays, radio waves, heat radiation from a fireplace
• Share certain fundamental characteristics
• All move through a vacuum at 3.00 x 108 m/s ("speed of light")
• Have "wave-like" characteristics

• The number of complete wavelengths, or cycles, that pass a given point in 1 second is the frequency of the wave
(frequency=cycles/second)
Electromagnetic radiation has both electric and magnetic properties. The wave-like property of electromagnetic radiation is due to the periodic oscillations of these components.
We can assign a frequency and a wavelength to electromagnetic radiation
Because all electromagnetic radiation moves at the same speed (speed of light) wavelength and frequency are related

• If the wavelength is long, there will be fewer cycles passing a given point per second, thus the frequency will be low
• If the wavelength is short, there will be more cycles passing a given point per second, and the frequency will be high
• Thus, there is an inverse relationship between wavelength and frequency

• (frequency [nu] * wavelength[lambda]) is a constant (c)
________________________________________
What is the speed of a wave?
Imagine you are on the beach watching the ocean waves go by, and you want to know the speed of the waves. There is an island offshore with a palm tree that will serve as a convenient frame of reference. You count the number of waves that pass by the tree in one minute:

In this case, two peaks (two wavelengths) pass by the tree in one minute. Thus, the frequency is 2 wavelengths/minute. If we measure the distance between the peaks (i.e. the wavelength) we can determine the speed of the wave:
Speed of the wave = (distance between peaks) * (frequency)
= (wavelength) * (frequency)

________________________________________

The unit of length chosen to describe a particular wavelength is typically dependent on the type of electromagnetic radiation
Unit Symbol Length (m) Type of Radiation
Angstrom Å 10-10 X-ray
Nanometer nm 10-9 UV, visible
Micrometer m 10-6 Infrared
Millimeter mm 10-3 Infrared
Centimeter cm 10-2 Microwave
Meter m 1 TV, radio
The range of EM wavelengths is dramatic
• The wavelengths of gamma-rays (<0.1 Å) are similar to the diameter of atomic nuclei • The wavelengths of some radio waves can be larger than a football field Frequency • Frequency is expressed in cycles per second, also known as hertz (Hz) • Usually the dimension 'cycles' is omitted and frequencies thus have the dimension of s-1 Sodium vapor lamps are sometimes used for public lighting. They give off a yellowish light with a wavelength of 589 nm. What is the frequency of this radiation? frequency*wavelength = speed of light frequency = speed of light/wavelength = (3.00x108 m/s)/(589x10-9m) = 5.09 x 1014 s-1 = 5.09 x 1014 cycles per second or 5.09 x 1014 hertz ________________________________________ Electronic Structure of Atoms Quantum Effects and Photons ________________________________________ Quantum Effects and Photons What's the difference between a "red hot" poker and a "white hot" poker? • The pokers are different temperatures ("white hot" poker has a higher temperature) • The pokers emit different intensities and wavelengths of electromagnetic radiation (especially in the visible spectrum Max Planck (1900) Energy can be released (or absorbed) by atoms only in "packets" of some minimum size • This minimum energy packet is called a quantum • The energy (E) of a quantum is related to its frequency () by some constant (h): E = h • h is known as "Planck's constant", and has a value of 6.63 x 10-34 Joule seconds (Js) • Electromagnetic energy is always emitted or absorbed in whole number multiples of (h*) Calculate the smallest amount of energy (i.e. one quantum) that an object can absorb from yellow light with a wavelength of 589 nm Energy quantum = h so we need to know the frequency  = c = c/ = (3.00 x 108 m/s)/(589 x 10-9 m) = 5.09 x 1014 s-1 plugging into Planck's equation: E = (6.63 x 10-34 Js)*( 5.09 x 1014 s-1) E (1 quanta) = 3.37 x 10-19 J Note that a quanta is quite small. When we receive infrared radiation from a fireplace we absorb it in quanta according to Planck's Law. However, we can't detect that the energy absorbtion is incremental. On the atomic scale, however, the quantum effects have a profound influence The Photoelectric Effect Light shining on a metallic surface can cause the surface to emit electrons • For each metal there is a minimum frequency of light below which no electrons are emitted, regardless of the intensity of the light • The higher the light's frequency above this minimum value, the greater the kinetic energy of the released electron(s) Using Planck's results Einstein (1905) was able to deduce the basis of the photoelectric effect • Einstein assumed that the light was a stream of tiny energy packets called Photons • Each photon has an energy proportional to its frequency (E=h) • When a photon strikes the metal its energy is transferred to an electron • A certain amount of energy is needed to overcome the attractive force between the electron and the protons in the atom Thus, if the quanta of light energy absorbed by the electron is insufficient for the electron to overcome the attractive forces in the atom, the electron will not be ejected - regardless of the intensity of the light. If the quanta of light energy absorbed is greater than the energy needed for the electron to overcome the attractive forces of the atom, then the excess energy becomes kinetic energy of the released electron. Since different metals have different atomic structure (number of protons, different electronic structure) the quanta of light needed to overcome the attractive forces within the atom differs for each element ________________________________________ The Photoelectric effect as a carnival game: A popular carnival game is where you are given a giant mallet and have to hit a pad on the ground. This sends a small metal slug shooting up a vertical track and, if you hit it hard enough, it will hit a bell at the top. This is like the photoelectric effect - if the electron will be released from the atom if it absorbs a photon with enough energy. Imagine that you have a 10 year old child and Arnold Schwartenegger. The child is too weak to hit the pad hard enough to ring the bell. It doesn't matter if you have an army of 10 year olds lined up to take their turn - none of them will ever hit it hard enough to ring the bell. However, Arnold (being Arnold) will have no problem ringing the bell. Thus, if the light shining on a metal does not have photons with the necessary energy to cause an electron to be ejected, then it does not matter how bright the light is. The key thing is to increase the energy of the individual photons, and this is achieved by increasing the frequency (i.e. decreasing the wavelength) ________________________________________ High energy photons, from x-rays for example, can cause electrons from many atoms to be ejected and with high kinetic energy as well. The release of such high energy electrons can cause tissue damage (cancer). Radio waves have such a low quanta of energy that even though we are bombarded by them, they do not cause the release of electrons. Einstein's interpretation of the photoelectric effect suggests that light has characteristics of particles. Is light a wave or does it consist of particles? ________________________________________ 1996 Michael Blaber Electronic Structure of Atoms Bohr's model of the hydrogen atom ________________________________________ In 1913 Niels Bohr developed a theoretical explanation for a phenomenon known as line spectra. Bohr's Model of the Hydrogen Atom Line Spectra Lasers emit radiation which is composed of a single wavelength. However, most common sources of emitted radiation (i.e. the sun, a lightbulb) produce radiation containing many different wavelengths. When the different wavelengths of radiation are separated from such a source a spectrum is produced. A rainbow represents the spectrum of wavelengths of light contained in the light emitted by the sun • Sun light passing through a prism (or raindrops) is separated into its component wavelengths • Sunlight is made up of a continuous spectrum of wavelengths (from red to violet) - there are no gaps Not all radiation sources emit a continuous spectrum of wavelengths of light • When high voltage is applied to a glass tube containing various gasses under low pressure different colored light is emitted o Neon gas produces a red-orange glow o Sodium gas produces a yellow glow • When such light is passed through a prism only a few wavelengths are present in the resulting spectra o These appear as lines separated by dark areas, and thus are called line spectra When the spectrum emitted by hydrogen gas was passed through a prism and separated into its constituent wavelengths four lines appeared at characteristic wavelengths In 1885 a Swiss school teacher figured out that the frequencies of the light corresponding to these wavelengths fit a relatively simple mathematical formula: where C = 3.29 x 1015 s-1 (not the 'c' used for the speed of light) However, the physical basis for this relationship was unknown. Bohr's Model • Bohr began with the assumption that electrons were orbiting the nucleus, much like the earth orbits the sun. • From classical physics, a charge traveling in a circular path should lose energy by emitting electromagnetic radiation • If the "orbiting" electron loses energy, it should end up spiraling into the nucleus (which it does not). Therefore, classical physical laws either don't apply or are inadequate to explain the inner workings of the atom • Bohr borrowed the idea of quantized energy from Planck o He proposed that only orbits of certain radii, corresponding to defined energies, are "permitted" o An electron orbiting in one of these "allowed" orbits:  Has a defined energy state  Will not radiate energy  Will not spiral into the nucleus If the orbits of the electron are restricted, the energies that the electron can possess are likewise restricted and are defined by the equation: Where RH is a constant called the Rydberg constant and has the value 2.18 x 10-18 J 'n' is an integer, called the principle quantum number and corresponds to the different allowed orbits for the electron. Thus, an electron in the first allowed orbit (closest to the nucleus) has n=1, an electron in the next allowed orbit further from the nuclei has n=2, and so on. Thus, the relative energies of these allowed orbits for the electrons can be diagrammed as follows: All the relative energies are negative • The lower the energy, the more stable the atom • The lowest energy state (n=1) is called the ground state of the atom • When an electron is in a higher (less negative) energy orbit (i.e. n=2 or higher) the atom is said to be in an excited state • As n becomes larger, we reach a point at which the electron is completely separated from the nucleus o E = (-2.18 x 10-18 J)(1/infinity) = 0 o Thus, the state in which the electron is separated from the nucleus is the reference or zero energy state (actually higher in energy than other states) Bohr also assumed that the electron can change from one allowed orbit to another • Energy must be absorbed for an electron to move to a higher state (one with a higher n value) • Energy is emitted when the electron moves to an orbit of lower energy (one with a lower n value) • The overall change in energy associated with "orbit jumping" is the difference in energy levels between the ending (final) and initial orbits: E = Ef - Ei Substituting in for the previously defined energy equation: When an electron "falls" from a higher orbit to a lower one the energy difference is a defined amount and results in emitted electromagnetic radiation of a defined energy (E) • Planck had deduced that the energy of the photons comprising EM radiation is a function of its frequency (E = h) • Therefore, if the emitted radiation from a falling electron had a defined energy, then it must have a correspondingly defined frequency Note: • E is positive when nf is greater than ni, this occurs when energy is absorbed and an electron moves up to a higher energy level (i.e. orbit). • When E is negative, radiant energy is emitted and an electron has fallen down to a lower energy state Revisiting Balmer's equation: In 1885 a Swiss school teacher figured out that the frequencies of the light corresponding to these wavelengths fit a relatively simple mathematical formula: where C = 3.29 x 1015 s-1 (not the 'c' used for the speed of light) Since energy lost by the electrons is energy "gained" by the emitted EM energy, the EM energy from Bohr's equation would be: Thus, Balmer's constant 'C' = (RH/h) (Rydberg constant divided by Planck's constant), and nf = 2. Thus, the only emitted energies which fall in the visible spectrum are from those electrons which fell down to the second quantum orbital. Those which fell down to the first orbital have a higher energy (frequency) than can be seen in the visible spectrum. Calculate the wavelength of light that corresponds to the transition of the electron from the n=4 to the n=2 state of the hydrogen atom. Is the light absorbed or emitted by the atom? Since the electron is "falling" from level 4 down to level 2, energy will be given up and manifested as emitted electromagnetic radiation: E = (2.18 x 10-18 J)((1/16)-(1/4)) = -4.09 x 10-19 J (light is emitted) 4.09 x 10-19 J = (6.63 x 10-34 Js) * () 6.17 x 1014 s-1 =   = (3.00 x 108 m s-1)/ (6.17 x 1014 s-1) = 4.87 x 10-7m = 487 nm Bohr's model of the atom was important because it introduced quantized energy states for the electrons. However, as a model it was only useful for predicting the behavior of atoms with a single electron (H, He+, and Li2+ ions). Thus, a different model of the atom eventually replaced Bohr's model. However, we will retain the concept of quantized energy states ________________________________________ 1996 Michael Blaber Electronic Structure of Atoms The dual nature of the electron ________________________________________ The Dual Nature of the Electron Depending on the experimental circumstances, EM radiation appears to have either a wavelike or a particlelike (photon) character. Louis de Broglie (1892-1987) who was working on his Ph.D. degree at the time, made a daring hypothesis: if radiant energy could, under appropriate circumstances behave as though it were a stream of particles, then could matter, under appropriate circumstances, exhibit wave-like properties? For example, the electron in orbit around a nucleus. DeBroglie suggested that the electron could be thought of as a wave with a characteristic wavelength. He proposed that the wavelength of the electron was a function of its mass (m) and its velocity (): i.e. the wavelength for "matter waves", where h is Planck's constant and is  velocity (not, the frequency). The quantity m for any object is its momentum (mass * velocity). What is the characteristic wavelength of an electron with a velocity of 5.97 x 106 m/s? (the mass of the electron is 9.11 x 10-28 g) Planck's constant (h) is 6.63 x 10-34 J s (also, recall that 1J = 1 kg m2/s2) converting g to kg: Converting from kg m2/s2 to Joules: ________________________________________ • The relationship between energy (E) and frequency ( ) for electromagnetic radiation (Planck's quantum of energy) • The relationship between wavelength ( ) and frequency ( ) for electromagnetic radiation From these relationships, we can determine the relationship between energy and wavelength: or, rearranging: • The relationship between wavelength ( ) and momentum (m*v) for DeBroglie's "particle wave" From the above relationships, we can calculate the relationship between energy (E) and momentum (m*v) Simplify, and solve for E: The highest velocity (v) attainable by matter is the speed of light (c), therefore, the maximum energy would seem to be: or ________________________________________ Why do nuclear bombs make such a loud "pop"? • The Fuel value of hydrogen is 142 x 103 J g-1 • If all matter were converted into energy E = 1x10-3kg * (3x108m/s)2 E = 9x1013 kg m2 s-2 E = 9x1013 J In other words, we can get out about 9 orders of magnitude greater energy if the hydrogen is converted directly into energy, rather than combusting it. Nuclear fission and fusion reactions convert a fraction of their matter into energy. The bomb that was dropped on Hiroshima contained about 15 kg of the 235Uranium isotope, a fissionable material. The actual amount of mass that was converted into energy is estimated at about 1 kg (releasing around 1 x 1017 J of energy in a split second). The estimated temperature at the moment of detonation is estimated to have been around 5 million degrees. In addition to the sun-like heat, much of the damage was due to the pressure wave that was produced. ________________________________________ The Uncertainty Principle For a relatively large solid object, like a bowling ball, we can determine its position and velocity at any given moment with a high degree of accuracy. However, if an object (like an electron) has wave-like properties then how can we accurately define its' position? Werner Heisenberg (1901-1976) concluded that due to the dual nature of matter (both particle and wavelike properties) it is impossible to simultaneously know both the position and momentum of an object as small as an electron. Thus, it is not appropriate to imagine the electrons as moving in well-defined circular orbits about the nucleus. ________________________________________ 1996 Michael Blaber Electronic Structure of Atoms Quantum Mechanics and Atomic Orbitals ________________________________________ Quantum Mechanics and Atomic Orbitals 1926 Erwin Schrödinger Schrödinger's wave equation incorporates both wave- and particle-like behaviors for the electron. Opened a new way of thinking about sub-atomic particles, leading the area of study known as wave mechanics, or quantum mechanics. Schrödinger's equation results in a series of so called wave functions, represented by the letter  (psi). Although has no actual physical meaning, the value of 2 describes the probability distribution of an electron. From Heisenberg's uncertainty principle, we cannot know both the location and velocity of an electron. Thus, Schrödinger's equation does not tell us the exact location of the electron, rather it describes the probability that an electron will be at a certain location in the atom. Departure from the Bohr model of the atom In the Bohr model, the electron is in a defined orbit, in the Schrödinger model we can speak only of probability distributions for a given energy level of the electron. For example, an electron in the ground state in a Hydrogen atom would have a probability distribution which looks something like this (a more intense color indicates a greater value for 2, a higher probability of finding the electron in this region, and consequently, greater electron density): Orbitals and quantum numbers Solving Schrödinger's equation for the hydrogen atom results in a series of wave functions (electron probability distributions) and associated energy levels. These wave functions are called orbitals and have a characteristic energy and shape (distribution). The lowest energy orbital of the hydrogen atom has an energy of -2.18 x 10¬18 J and the shape in the above figure. Note that in the Bohr model we had the same energy for the electron in the ground state, but that it was described as being in a defined orbit. The Bohr model used a single quantum number (n) to describe an orbit, the Schrödinger model uses three quantum numbers: n, l and ml to describe an orbital. The principle quantum number 'n' • Has integral values of 1, 2, 3, etc. • As n increases the electron density is further away from the nucleus • As n increases the electron has a higher energy and is less tightly bound to the nucleus The azimuthal (second) quantum number 'l' • Has integral values from 0 to (n-1) for each value of n • Instead of being listed as a numerical value, typically 'l' is referred to by a letter ('s'=0, 'p'=1, 'd'=2, 'f'=3) • Defines the shape of the orbital The magnetic (third) quantum number 'ml' • Has integral values between 'l' and -'l', including 0 • Describes the orientation of the orbital in space For example, the electron orbitals with a principle quantum number of 3 (i.e. n=3) would have the following available values of 'l' and 'ml': n (principle quantum number) l (azimuthal) (defines shape) Subshell Designation ml (magnetic) (defines orientation) Number of Orbitals in Subshell 3 0 3s 0 1 1 3p -1,0,1 3 2 3d -2,-1,0,1,2 5 • A collection of orbitals with the same value of 'n' is called an electron shell • A collection of orbitals with the same value of 'n' and 'l' belong to the same subshell Thus: • the third electron shell (i.e. 'n'=3) consists of the 3s, 3p and 3d subshells (each with a different shape) • The 3s subshell contains 1 orbital, the 3p subshell contains 3 orbitals and the 3d subshell contains 5 orbitals. (within each subshell, the different orbitals have different orientations in space) • Thus, the third electron shell is comprised of nine distinctly different orbitals, although each orbital has the same energy (that associated with the third electron shell) Note: remember, this is for hydrogen only. Restrictions on the possible values for the different quantum numbers (n, l and ml) gives rise to the following patterns for the different shells: • Each shell is divided into a number of subshells equal to the principle quantum number (e.g. the fourth shell is divided into four subshells: s, p, d,and f; whereas the first shell has a single subshell: s) • Each subshell is divided into orbitals (increasing by odd numbers): Subshell Number of orbitals s 1 p 3 d 5 f 7 The number and relative energies of all hydrogen electron orbitals through n=3 are shown below: • At ordinary temperatures essentially all hydrogen atoms are in their ground states • The electron may be promoted to an excited state by the absorbtion of a photon with appropriate quantum of energy ________________________________________ 1996 Michael Blaber Electronic Structure of Atoms Representations of Orbitals ________________________________________ Representations of Orbitals The s Orbitals The 1s orbital is spherically symmetrical. A plot of 2 versus distance (r) from the nucleus shows a dramatic reduction in probability of finding the electron very far from the nucleus: This indicates that in the ground state the electrostatic attraction of the electron for the proton in the nucleus is such that the electron is unlikely to be found far from the nucleus. The higher energy s orbitals are also spherically symmetrical, however, they exhibit distinct nodes in the distribution probability: • In the higher s orbitals there exists node regions where the electron density approaches zero (2s has 1 node, 3s has 2 nodes, etc) • The higher s orbitals (excited states) have electron density distributions which indicate that there is a higher probability of finding the electron further away from the nucleus The size of the orbital increases as n increases The most widely used representation of the Schrödinger orbits is to draw a boundary which represents 90% of the total electron density distribution. For the s orbitals this would be a sphere representation. p Orbitals • The p orbitals are 'dumbbell' shaped orbitals of electron density, with a node at the nucleus. • There are three distinct p orbitals, they differ in their orientations • There is no fixed correlation between the three orientations and the three magnetic quantum numbers (ml) The d and f orbitals In the third shell and beyond there are five d orbitals, each has a different orientation in space: Although the 3dz2 orbital looks different, it has the same energy as the other d orbitals. There are 7 equivalent f orbitals (for each value of n 4 or greater). They are pretty difficult to represent on a 3-d contour diagram. Understanding orbital shapes is key to understanding the molecules formed by combining atoms ________________________________________ 1996 Michael Blaber Electronic Structure of Atoms Orbitals in many-electron atoms ________________________________________ Orbitals in many-electron atoms The hydrogen atom is a simple system having only one electron. The quantum mechanical description of the hydrogen atoms places all subshells (i.e. l quantum number, or the s, p, d and f subshells) with the same principle quantum number (n) on the same energetic level. An atom with more than 1 electron is called a many-electron atom. Although the shape of electronic orbitals for many-electron atoms are the same as those for the hydrogen atom, the presence of more than 1 electron influences the energy levels of the orbitals (due to electron-electron replusion). For example, the 2s orbital is a lower energy state than the 2p orbital in a many-electron atom: (note: this is a qualitative representation for an "average" many-electron atom) Effective Nuclear Charge In a many-electron atom, each electron is simultaneously: • attracted to the protons in the nucleus • repelled by other electrons (like-charge repulsion) What is the average environment, created by the nucleus and all the other electrons in the atom, which is "felt" by a particular electron in the atom? The net positive charge attracting the electron is called the effective nuclear charge • Any electron density between the nucleus and the electron of interest will reduce the nuclear charge acting on that electron • The effective nuclear charge (Zeff) equals the number of protons in the nucleus (Z), minus the average number of electrons (S) that are between the electron in question and the nucleus • The positive charge "felt" by the outer electrons is always less than the full nuclear charge (inner electrons "screen" the nuclear charge). Energies of orbitals The extent to which an electron will be screened by the other electrons depends on the shape of the electron distribution as we move out from the nucleus • Probability of being closer to the nucleus (based on orbital shapes) is as follows: • For the 3rd principle quantum number, for example, the 3s electrons experience the least shielding and the 3d electrons the most • Conversely, the 3s electrons experience a greater Zeff and the 3d electrons the least In a many-electron atom, for a given principle quantum number ('n'), Zeff decreases with increasing 'l' The energy of an electron depends on Zeff • Because Zeff is larger for 3s electrons (in the above n=3 example) they have a lower energy than 3p electrons (which in turn have lower energy than 3d electrons) In a many-electron atom, for a given principle quantum number ('n'), the energy level of an orbital increases with increasing 'l' Note: all the orbitals of a give sub-shell still have the same energy level (e.g. all the 3d orbitals (with different ml quantum values) The sodium atom has 11 electrons, two in a 1s orbital, two in a 2s orbital, six in 2p orbitals and one in a 3s orbital. As far as the electrons in s type subshells, which experiences the smallest effective nuclear charge (Zeff)? Answer: the outermost electron, or the one in the 3s orbital Electron spin and the Pauli exclusion principle • What determines the orbitals in which the electrons reside? • How do the electrons populate the available orbitals? Line spectra revisited... • Lines which were thought to be single lines actually were composed of two very closely spaced lines • Thus, there were twice as many energy levels as there were "supposed" to be It was proposed (Uhlenbeck and Goudsmit, 1925) that electrons have yet another quantum property called electron spin: • A new quantum number for the electron called the electron spin quantum number, or ms • ms has a value of +1/2 or -1/2 • The electron spin quantum number characterized the "direction of spin" of the electron o A spinning charge produces a magnetic field o The opposite spins produce opposite magnetic fields which results in the splitting of the line spectrum into two closely spaced lines Electron spin is crucial for understanding the electron structures of atoms: • The Pauli exclusion principle (Wolfgang Pauli, 1925) states that no two electrons in an atom can have the same set of four quantum numbers (n, l, ml and ms) • For a given orbital (e.g. 2pz) the values of n, l and ml are fixed. Thus, if we want to put more than one electron into an orbital we must assign unique values to the magnetic spin (ms quantum number) • the ms quantum number can only have two values (+1/2, and -1/2) therefore, only two electrons at most can occupy the same orbital, and they have opposite values for magnetic spin What are the consequences of magnetic spin quantum number and the Pauli exclusion principle? • If we know the number of electrons in an atom we can assign probable quantum numbers and know something about their orbital shapes • Provides an understanding for the periodic nature of the elements ________________________________________ 1996 Michael Blaber Electronic Structure of Atoms Electron configurations ________________________________________ Electron Configurations The way in which electrons are distributed among the various orbitals is called the electron configuration Orbitals are filled in order of increasing energy, with no more than two electrons per orbital Lithium This element has 3 electrons. We would thus begin by placing two electrons in the 1s ground state, or lowest energy, orbital. These two electrons would have opposite magnetic spin quantum numbers. We would then place the third electron in the next highest energy level orbital - the 2s orbital: • The arrows indicate the value of the magnetic spin (ms) quantum number (up for +1/2 and down for -1/2) • The description of the occupation of the orbitals would be described in the following way: 1s22s1 or, "1s two, 2s one". • Electrons having opposite spins are said to be "paired" electrons, as with the electrons occupying the Li 1s orbital • Likewise, the single electron in the 2s orbital (for Li) is said to be "unpaired" Writing electronic configurations • The two electrons in He represent the complete filling of the first electronic shell. Thus, the electrons in He are in a very stable configuration • For Boron (5 electrons) the 5th electron must be placed in a 2p orbital because the 2s orbital is filled. Because the 2p orbitals are equal energy, it doesn't matter which 2p orbital is filled What do we do now with the next element, Carbon (6 electrons)? Do we pair it with the single 2p electron (but with opposite spin)? Or, do we place it in another 2p orbital? The second 2p electron in Carbon is placed in another 2p orbital, but with the same spin as the first 2p electron: Hund's rule: for degenerate orbitals, the lowest energy is attained when the number of electrons with the same spin is maximized Electrons repel each other, by occupying different orbitals the electrons remain as far as possible from one another • A carbon atom in its lowest energy (ground state) has two unpaired electrons • Ne has filled up the n=2 shell, and has a stable electronic configuration Electronic configurations can also be written in a short hand which references the last completed orbital shell (i.e. all orbitals with the same principle quantum number 'n' have been filled) • The electronic configuration of Na can be written as [Ne]3s1 • The electronic configuration of Li can be written as [He]2s1 The electrons in the stable (Noble gas) configuration are termed the core electrons The electrons in the outer shell (beyond the stable core) are called the valence electrons Something curious The noble gas Argon (18 electrons) marks the end of the row started by Sodium Will the next element (K with 19 electrons) put the next electron one of the 3d orbitals? • Chemically, we know Potassium is a lot like Lithium and Sodium • What these elements (the alkali metals) have in common is an unpaired valence electron in an s orbital • If Potassium has an unpaired electron in an s orbital it would mean that it is in the 4s orbital • Thus, the 4s orbital would appear to be of lower energy than the 3d orbital(s) The 4s orbital would be filled when we have an element with 20 electrons (Calcium). Then we go back and fill up the 3d orbitals, which can hold a maximum of 10 electrons Thus, the 4th row of the periodic table is 10 elements wider than the previous row - we have available five 'd' orbitals we can fill (with 10 electrons). These 10 elements are the Transition Elements, or Transition Metals. With Cerium (element 58) the 'f' orbitals enter the picture. These orbitals can hold 14 electrons. • The first 'f' orbitals are the 4f orbitals (n=4; l=0(s), 1(p), 2(d), 3(f)) • These additional elements are represented by the 14 lanthanide (4f orbital filling) and actinide (5f orbitals) series of elements • The energy of the 5d and 4f orbitals are very close ________________________________________ 1996 Michael Blaber Electronic Structure of Atoms Electron configurations and the periodic table ________________________________________ Electron Configurations and the Periodic Table The periodic table is structured so that elements with the same type of valence electron configuration are arranged in columns. • The left-most columns include the alkali metals and the alkaline earth metals. In these elements the valence s orbitals are being filled • On the right hand side, the right-most block of six elements are those in which the valence p orbitals are being filled These two groups comprise the main-group elements • In the middle is a block of ten columns that contain transition metals. These are elements in which d orbitals are being filled • Below this group are two rows with 14 columns. These are commonly referred to the f-block metals. In these columns the f orbitals are being filled Important facts to remember: 1. 2, 6, 10 and 14 are the number of electrons that can fill the s, p, d and f subshells (the l=0,1,2,3 azimuthal quantum number) 2. The 1s subshell is the first s subshell, the 2p is the first p subshell (n=2, l=1, 3d is the first d subshell, and the 4f is the first f subshell What is the electron configuration for the element Niobium? (41) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d3 Niobium is actually: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s1 4d4 The reason is that the 5s and 4d energy levels are quite close and certain electronic arrangements can result in the levels being slightly different than expected. What is the electron configuration of the element Nickel? (28) 1s2 2s2 2p6 3s2 3p6 4s2 3d8 What is the electron configuration for Nickel in terms of the nearest noble gas? [Ar] 4s2 3d8 How would the last valence orbital be filled? ________________________________________ 1996 Michael Blaber Periodic Properties of the Elements Development of the Periodic Table ________________________________________ The most significant tool for organizing and remembering chemical facts is the periodic table • Based on the periodic nature of electron configurations • Elements in the same column have the same number of valence electrons • Similarities in chemical nature due to similarities in valence electron configuration Development of the Periodic Table • Certain elements, such as gold and silver, can be found naturally in their elemental form and were discovered thousands of years ago. • Some radioactive elements are quite unstable and their isolation is dependent upon modern technology • The majority of elements are stable, but commonly present in compound form with other elements In the 1800's methods were developed to isolate various elements from compound form • 1800: 31 elements identified • 1865: 63 elements identified 1869: Dmitri Mendeleev and Lothar Meyer published schemes for classifying elements The elements could be ordered according to their atomic weight (i.e. grams/mole for the naturally occuring mixture of isotopic forms) which resulted in periodic characteristics Mendeleev's insistence on ordering elements by atomic weight, and grouping them by characteristics resulted in several "gaps" in the table • Both Gallium (Ga) and Germanium (Ge) were unknown at the time • Thus there was a gap under Aluminum (Al) and a gap under Silicon (Si) • Mendeleev concluded therefore that there must be two elements, which he called "eka-Aluminum" and "eka-Silicon" which must fill these gaps Mendeleev predicted not only that Ga and Ge must exist, but also described some of their general physical properties • Their approximate atomic weight • The stoichiometric relationship for compounds involving oxygen and chlorine Ga and Ge were discovered decades later, but their physical and chemical characteristics as predicted by Mendeleev were correct The accuracy of Mendeleev's predictions for undiscovered elements, based on his periodic table, convinced scientists of its validity 1911 Rutherford model of the atom: • most of the mass of the atom was located in a dense nucleus • the nucleus had a net positive charge • beyond the nucleus was a mostly empty space which contained electrons with a net negative charge 1913 Henry Moseley (killed at Gallipoli at age 28) • investigated the characteristic frequencies of X-rays produced by bombarding each of the elements in turn by high energy cathode rays (electrons). • He discovered a mathematical relationship between the frequency and the atomic number (the "serial number" in the periodic table). • This must mean that the atomic number is more than a serial number; that it has some physical basis. • Moseley proposed that the atomic number was the number of electrons in the atom of the specific element. • This also means that the atomic number is the number of positive charges carried by the nucleus. ________________________________________ Moseley's experiments: Moseley was using various elements (metals) as targets in cathode ray tubes. He noticed that when struck by the cathode rays, different metals gave off x-rays with distinct wavelengths. Essentially what was happening was that the cathode rays (high energy electrons) were knocking out the inner-most electrons of the metal targets. X-rays were emitted when an outer electron "fell" down into this inner shell. Since the inner-most electrons are not shielded by other electrons, the energy required to knock them out is dependent upon the number of protons in the nucleus. Thus, the emitted x-ray frequency associated with an outer electron falling down into this shell is related to the number of protons in the nucleus. Moseley realized that the atomic numbers were not just a convenient numbering scheme for the elements, but had a real physical meaning - ultimately realized as being the number of protons (and electrons) in a (neutral) element. ________________________________________ 1996 Michael Blaber Periodic Properties of the Elements Electron Shells in Atoms ________________________________________ Electron Shells in Atoms When we move down the column of the periodic table, we change the principle quantum number 'n' of the valence electrons of the atom We have referred to all orbitals with the same principle quantum number 'n' as a shell What does the quantum mechanical description of probability distributions for all electrons in an atom look like? He 1s2 Ne 1s2 2s2 2p6 Ar 1s2 2s2 2p6 3s2 3p6 The overall electron distribution can be calculated using supercomputers and the result is a spherically symmetrical distribution termed radial electron density • Helium shows a single "shell" • Neon shows two "shells" • Argon shows three "shells" Each of these maxima corresponds to electrons that have the same principle quantum number 'n' • In He the 1s electrons have a maximum probability distribution at around 0.3 Å from the nucleus • In Ne the 1s electrons have a maximum at around 0.08 Å, and the 2s and 2p electrons combine to form another maximum at around 0.35 Å (the n=2 "shell") • In Ar the 1s electrons have a maximum at around 0.02 Å, the 2s and 2p electrons combine to form a maximum at around 0.18 Å and the 3s and 3p electrons combine to form a maximum at around 0.7 Å Why is the 1s shell in Argon so much closer to the nucleus than the 1s shell in Neon, and why is that closer than the 1s shell in helium? • The nuclear charge (Z) of He = 2+, Ne = 10+, Ar = 18+ • The inner most electrons (1s shell) are not shielded by other electrons, therefore the attraction to the nucleus is greater with higher number of protons • Likewise, the n=2 shell of Ar is closer to the nucleus than the n=2 shell of Ne. The Zeff for the 2s subshell of Ne would be (10-2) = 8+, and for Ar would be (18-2) = 16+. Thus, the 2s subshell electrons in Ar would be closer to the nucleus due to the greater effective nuclear charge. Three properties that provide important insights into chemical behavior include: 1. atomic size 2. ionization energy 3. electron affinity ________________________________________ 1996 Michael Blaber Periodic Properties of the Elements Sizes of Atoms ________________________________________ Sizes of atoms From the quantum mechanical model of atoms we can conclude that an atom does not have a sharply defined boundary. This leads to a conceptual problem - just what exactly is the "size" of an atom? It is possible to estimate the atomic radius of an atom by assuming that atoms are spherical objects that touch each other when they are bonded together in molecules • The Br-Br distance in Br2 is 2.28 Å, thus the radius of the Br atom is 1.14 Å • The C-C bond distance is 1.54 Å, thus the radius of Carbon is 0.77 Å What about a C-Br bond distance? • To have useful predictive values, the determined atomic radii should hold true (i.e. be additive) when considering other possible compounds • Predict C-Br bond distance to be equal to 1.14 + 0.77 = 1.91 Å • In many different compounds with C-Br bonds, the distance is approximately this length What are the general characteristics of atomic bond lengths as determined from bond-bond distances (small molecule crystallography, NMR, other methods) • Within the columns of the periodic table, the atomic radii increase as you go down the column • Within the rows of the periodic table, the atomic radii decrease as you move to the right What is the basis for these observations? Two general factors affect the size of the outermost orbital: • The principle quantum number • The effective nuclear charge Proceeding across a row: • The number of core (shielding) electrons remains constant (only the valence electrons are increasing) • The number of protons is increasing • If the number of protons is increasing, but the core shielding electrons remain constant then the effective nuclear charge (Zeff) on the valence electrons is increasing, and they will be pulled tighter towards the nucleus, thus the atomic radii will decrease For elements where we are filling the 3p subshell • 12 shielding electrons (1s22s22p63s2) in each case Element Atomic # Zeff Al Si P S Cl Ar 13 14 15 16 17 18 1+ 2+ 3+ 4+ 5+ 6+ Proceeding down a column: • The valence electrons are remaining constant • The principle quantum number is increasing • The shielding electrons are increasing, but so is the nuclear charge - the end result is that essentially the effective nuclear charge on the valence electrons is relatively constant • Since the major effect is that the principle quantum number increases as you go down a column, the atomic radius increases For example, the group 1A elements (valence e' are in blue, shielding [core electrons in this case] e' for the valence e' are in red) Atomic # Element Electron Configuration Zeff Valence n quantum # 3 Li 1s22s1 1+ 2 11 Na 1s22s22p63s1 1+ 3 19 K 1s22s22p63s23p64s1 1+ 4 37 Rb 1s22s22p63s23p64s23d104p65s1 1+ 5 ________________________________________ 1996 Michael Blaber Periodic Properties of the Elements Ionization Energy ________________________________________ Ionization Energy The ionization energy of an atom measures how strongly an atom holds its electrons The ionization energy is the minimum energy required to remove an electron from the ground state of the isolated gaseous atom Note that this does not mean the energy required to remove an electron from the n=1 shell (i.e the ground state orbital), the ground state here refers to the lowest energy electron configuration for the element in question The first ionization energy, I1, is the energy needed to remove the first electron from the atom: Na(g) -> Na+(g) + 1e-
The second ionization energy, I2, is the energy needed to remove the next (i.e. the second) electron from the atom
Na+(g) -> Na2+(g) + 1e-
The higher the value of the ionization energy, the more difficult it is to remove the electron
As electrons are removed, the positive charge from the nucleus remains unchanged, however, there is less repulsion between the remaining electrons
• Zeff increases with removal of electrons
• Greater energy is needed to remove remaining electrons (i.e. the ionization energy is higher for each subsequent electron)
Ionization energies (kJ/mol)
Element I1 I2 I3 I4
Na 496 4560
Mg 738 1450 7730
Al 577 1816 2744 11,600
• There is also a big increase in ionization energy for removal of an electron from an inner shell (lower n value).

This is due to the fact that when you move to an orbital with a lower principle quantum number, you are removing an electron which is much closer to the nucleus (and has a higher attraction for the nucleus)

The inner shell electrons are too tightly bound to be ionized or shared with another atom, and thus do not participate in chemical bonding
Periodic trends in ionization energies
First ionization energies as a function of atomic number
• Within each period (row) the ionization energy typically increases with atomic number
• Within each group (column) the ionization energy typically decreases with increasing atomic number

The basis for these observations:
• As the effective charge increases, or as the distance of the electron from the nucleus decreases, the greater the attraction between the nucleus and the electron. The effective charge increases across a period, in addition, the atomic radius decreases
• As we move down a group the distance from the nucleus increases and the attraction of the electrons for the nucleus decreases
Which of the following elements has the lowest ionization energy? B, Al, C and Si
Probably Al. Its valence electrons have a higher principle quantum number, and are therefore further away from the nucleus, than C or B. Furthermore, its nucleus would have a lower effective nuclear charge than Si.
________________________________________
1996 Michael Blaber
Periodic Properties of the Elements
Electronic Affinities
________________________________________
Electron Affinities
Atoms can also gain electrons to form negatively charged ions (anions)
The electron affinity is the energy change associated with an atom or ion in the gas state gaining an electron.
• For all positively charged ions, and most neutral atoms, energy is released when an electron is added
Cl(g) + e- -> Cl-(g) E = -328 kJ/mol
Thus, we say that chlorine has an electron affinity of -328 kJ/mol.
The greater the attraction for the electron, the more exothermic the process
For anions and some neutral atoms, added an electron is an endothermic process, i.e. work must be done to force an electron onto the atom. This results in the formation of an unstable anion.
• The halogens, which are one electron short of a completely filled p subshell have the greatest attraction for an electron (i.e. the electron affinity has the largest negative magnitude)

In adding an electron they achieve a stable electron configuration like the noble gases
• The 2A and 8A groups have filled subshells (s, and p, respectively) and therefore, an additional electron must reside in a higher energy orbital. Adding an electron to these groups is an endothermic process

The general trend is for the electron affinity to become increasingly negative (stronger binding of an electron) as we move across each period toward the halogens.
Electron affinities do not change much as we move down a group
• The distance from the nucleus is increasing with greater n (less attraction)
however,
• The electrons in the subshell are more diffuse, reducing electron-electron repulsions
Element Ion E (kJ/mol)
F F- -328
Cl Cl- -349
Br Br- -325
I I- -295


________________________________________
1996 Michael Blaber

Periodic Properties of the Elements
Metals, Nonmetals and Metalloids
________________________________________
Metals, Nonmetals and Metalloids
Characteristic properties of metallic and non-metallic elements:
Metallic Elements Nonmetallic elements
Distinguishing luster (shine) Non-lustrous, various colors
Malleable and ductile (flexible) as solids Brittle, hard or soft
Conduct heat and electricity Poor conductors
Metallic oxides are basic, ionic Nonmetallic oxides are acidic, compounds
Cations in aqueous solution Anions, oxyanions in aqueous solution
Metals
• Most metals are malleable (can be pounded into thin sheets; a sugar cube chunk of gold can be pounded into a thin sheet which will cover a football field), and ductile (can be drawn out into a thin wire).
• All are solids at room temp (except Mercury, which is a liquid)
• Metals tend to have low ionization energies, and typically lose electrons (i.e. are oxidized) when they undergo chemical reactions
o Alkali metals are always 1+ (lose the electron in s subshell)
o Alkaline earth metals are always 2+ (lose both electrons in s subshell)
o Transition metal ions do not follow an obvious pattern, 2+ is common, and 1+ and 3+ are also observed
• Compounds of metals with non-metals tend to be ionic in nature
• Most metal oxides are basic oxides; those that dissolve in water react to form metal hydroxides:
Metal oxide + water -> metal hydroxide
Na2O(s) + H2O(l) -> 2NaOH(aq)
CaO(s) + H2O(l) -> Ca(OH)2(aq)
• Metal oxides exhibit their basic chemical nature by reacting with acids to form salts and water:
Metal oxide + acid -> salt + water
MgO(s) + HCl(aq) -> MgCl2(aq) + H2O(l)
NiO(s) + H2SO4(aq) -> NiSO4(aq) + H2O(l)
________________________________________
• What is the chemical formula for aluminum oxide?
Al has 3+ charge, the oxide ion is O2-, thus Al2O3
• Would you expect it to be solid, liquid or gas at room temp?
Oxides of metals are characteristically solid at room temp
• Write the balanced chemical equation for the reaction of aluminum oxide with nitric acid:
Metal oxide + acid -> salt + water
Al2O3(s) + 6HNO3(aq) -> 2Al(NO3)3(aq) + 3H2O(l)
________________________________________
Nonmetals
• Vary greatly in appearance
• Non-lustrous
• Poor conductors of heat and electricity
• The melting points of non-metals are generally lower than metals
• Seven non-metals exist under standard conditions as diatomic molecules:
1. H2(g)
2. N2(g)
3. O2(g)
4. F2(g)
5. Cl2(g)
6. Br2(l)
7. I2(l) (volatile liquid - evaporates readily)
• Nonmetals, when reacting with metals, tend to gain electrons (typically attaining noble gas electron configuration) and become anions:
Nonmetal + Metal -> Salt
3Br2(l) + 2Al(s) -> 2AlBr3(s)
• Compounds composed entirely of nonmetals are molecular substances (not ionic)
• Most nonmetal oxides are acidic oxides. Those that dissolve in water react to form acids:
Nonmetal oxide + water -> acid
CO2(g) + H2O(l) -> H2CO3(aq) [carbonic acid]
(carbonated water is slightly acidic)
• Nonmetal oxides can combine with bases to form salts
Nonmetal oxide + base -> salt
CO2(g) + 2NaOH(aq) -> Na2CO3(aq) + H2O(l)
Metalloids
Properties intermediate between the metals and nonmetals.
Silicon for example appears lustrous, but is not malleable or ductile (it is brittle - a characteristic of some nonmetals). It is a much poorer conductor of heat and electricity than the metals. Metalloids are useful in the semiconductor industry.
Trends in Metallic and Nonmetallic Character
• Metallic character is strongest for the elements in the leftmost part of the periodic table, and tends to decrease as we move to the right in any period (nonmetallic character increases with increasing ionization values)
• Within any group of elements (columns), the metallic character increases from top to bottom (the ionization values generally decrease as we move down a group). This general trend is not necessarily observed with the transition metals.

________________________________________
1996 Michael Blaber

Periodic Properties of the Elements
Group Trends: The Active Metals
________________________________________
Group Trends: The Active Metals
Group 1A: The Alkali Metals
1A
3
Li
11
Na
19
K
37
Rb
55
Cs
87
Fr
The word "alkali" is derived from an Arabic word meaning "ashes". Many sodium and postassium compounds were isolated from wood ashes (Na2CO3 and K2CO3 are still occasionally referred to as "soda ash" and "potash").
As we move down the group (from Li to Fr) we find the following trends:
• All have a single electron in an 's' valence orbital
• The melting point decreases
• The density increases
• The atomic radius increases
• The ionization energy decreases (first ionization energy)
The alkali metals have the lowest I1 values of the elements
This represents the relative ease with which the lone electron in the outer 's' orbital can be removed.
The alkali metals are very reactive, readily losing 1 electron to form an ion with a 1+ charge:
M -> M+ + e-
Due to this reactivity, the alkali metals are found in nature only as compounds. The alkali metals combine directly with most nonmetals:
• React with hydrogen to form solid hydrides
2M(s) + H2(g) -> 2MH(s)
(Note: hydrogen is present in the metal hydride as the hydride H- ion)
• React with sulfur to form solid sulfides
2M(s) + S(s) -> M2S(s)
• React with chlorine to form solid chlorides
2M(s) + Cl2(g) -> 2MCl(s)
• Alkali metals react with water to produce hydrogen gas and alkali metal hydroxides (very exothermic)
2M(s) + 2H2O(l) -> 2MOH(aq) + H2(g)
The reaction between alkali metals and oxygen is more complex:
• A common reaction is to form metal oxides which contain the O2- ion
4Li(s) + O2 (g) -> 2Li2O(s) (lithium oxide)
• Other alkali metals can form metal peroxides (contains O22- ion)
2Na(s) + O2 (g) -> Na2O2(s) (sodium peroxide)
• K, Rb and Cs can also form superoxides (O2- ion)
K(s) + O2 (g) -> KO2(s) (potassium superoxide)
Note:
• The color of a chemical is produced when a valence electron in an atom is excited from one energy level to another by visible radiation. In this case, the particular frequency of light that excites the electron is absorbed. Thus, the remaining light that you see is white light devoid of one or more wavelengths (thus appearing colored). Alkali metals, having lost their outermost electrons, have no electrons that can be excited by visible radiation. Alkali metal salts and their aqueous solution are colorless unless they contain a colored anion.
• When alkali metals are placed in a flame the ions are reduced (gain an electron) in the lower part of the flame. The electron is excited (jumps to a higher orbital) by the high temperature of the flame. When the excited electron falls back down to a lower orbital a photon is released. The transition of the valence electron of sodium from the 3p down to the 3s subshell results in release of a photon with a wavelength of 589 nm (yellow)
Flame colors:
• Lithium: crimson red
• Sodium: yellow
• Potassium: lilac
Group 2A: The Alkaline Earth Metals
2A
4
Be
12
Mg
20
Ca
38
Sr
56
Ba
88
Ra
Compared with the alkali metals, the alkaline earth metals are typically:
• harder
• more dense
• melt at a higher temperature
The first ionization values (I1) of the alkaline earth metals are not as low as the alkali metals:
the alkaline earth metals are therefore less reactive than the alkali metals (Be and Mg are the least reactive of the alkaline earth metals)
Calcium, and elements below it, react readily with water at room temperature:
Ca(s) + 2H2O(l) -> Ca(OH)2(aq) + H2(g)
The tendency of the alkaline earths to lose their two valence electrons is demonstrated in the reactivity of Mg towards chlorine gas and oxygen:
Mg(s) + Cl2(g) -> MgCl2(s)
2Mg(s) + O2(g) -> 2MgO(s)
The 2+ ions of the alkaline earth metals have a noble gas like electron configuration and are thus form colorless or white compounds (unless the anion is itself colored).
Flame colors:
• Calcium: brick red
• Strontium: crimson red
• Barium: green
________________________________________
1996 Michael Blaber

Periodic Properties of the Elements
Group Trends: Selected Nonmetals
________________________________________
Group Trends: Selected Nonmetals
Hydrogen
Hydrogen has a 1s1 electron configuration and is placed above the alkali metal group.
Hydrogen is a non-metal, which occurs as a gas (H2) under normal conditions.
• Its ionization energy is considerably higher (due to lack of shielding, and thus higher Zeff) than the metals and is more like the nonmetals
• Hydrogen generally reacts with other nonmetals to form molecular compounds (typically highly exothermic)
• Hydrogen reacts with active metals to form metal hydrides which contain the H- hydride ion:
2Na(s) + H2(g) -> 2NaH(s)
• Hydrogen can also lose an electron to yield the aqueous H+(aq) ion
Group 6A: The Oxygen Family
6A
8
O
16
S
34
Se
52
Te
84
Po
As we proceed down group 6A the elements become more metallic in nature:
• Oxygen is a gas, the rest are solids
• Oxygen, sulfur and selenium are nonmetals
• Tellurium is a metalloid with some metal properties
• Polonium is a metal
Oxygen can be found in two molecular forms, O2 and O3 (ozone). These two forms of oxygen are called allotropes (different forms of the same element in the same state)
3O2(g) -> 2O3(g) H = 284.6 kJ
the reaction is endothermic, thus ozone is less stable that O2
Oxygen has a great tendency to attract electrons from other elements (i.e. to "oxidize" them)
• Oxygen in combination with metals is almost always present as the O2- ion (which has noble gas electronic configuration and is particularly stable)
• Two other oxygen anions are observed: peroxide (O22-) and superoxide (O2-)
Sulfur
Sulfur also exists in several allotropic forms, the most common stable allotrope is the yellow solid S8 (an 8 member ring of sulfur atoms)
Like oxygen, sulfur has a tendency to gain electrons from other elements, and to form sulfides (which contain the S2- ion). This is particular true for the active metals:
16Na(s) + S8(s) -> 8Na2S(s)
Note: most sulfur in nature is present as a metal-sulfur compound
Sulfur chemistry is more complex than that of oxygen
Group 7A: The Halogens
7A
9
F
17
Cl
35
Br
53
I
85
At
• "Halogen" is derived from the Greek meaning "salt formers"
• Astatine is radioactive and rare, and some of its properties are unknown
• All the halogens are nonmetals
• Each element consists of diatomic molecules under standard conditions
Colors of diatomic halogens: (not flame colors)
Fluorine: pale yellow
Chlorine: yellow green
Bromine: reddish brown
Iodine: violet vapor
The halogens have some of the most negative electron affinities (i.e. large exothermic process in gaining an electron from another element)
The chemistry of the halogens is dominated by their tendency to gain electrons from other elements (forming a halide ion)
X2 + 2e- -> 2X-
• Fluorine and chlorine are the most reactive halogens (highest electron affinities). Fluorine will remove electrons from almost any substance
In 1992 22.3 billion pounds of chlorine was produced. Both chlorine and sodium can be produced by electrolysis of molten sodium chloride (table salt). The electricity is used to strip electrons from chloride ions and tranfer them to sodium ions to produce chlorine gas and solid sodium metal
Chlorine reacts slowly with water to produce hydrochloric acid and hypochlorous acid:
Cl2(g) + H2O(l) -> HCl(aq) + HOCl(aq)
Hypochlorous acid is a disinfectant, thus chlorine is a useful addition to swimming pool water
The halogens react with most metals to form ionic halides:
Cl2(g) + 2Na(s) -> 2NaCl(s)
Group 8A: The Noble Gases
8A
2
He
10
Ne
18
Ar
36
Kr
54
Xe
86
Rn
• Nonmetals
• Gases at room temperature
• monoatomic
• completely filled 's' and 'p' subshells
• large first ionization energy, but this decreases somewhat as we move down the group
Rn is highly radioactive and some of its properties are unknown
They are exceptionally unreactive. It was reasoned that if any of these were reactive, they would most likely be Rn, Xe or Kr where the first ionization energies were lower.
In order to react, they would have to be combined with an element which had a high tendency to remove electrons from other atoms. Such as fluorine.
Compounds of noble gases to date:
XeF2
XeF4
XeF6
only one compound with Kr has been made
KrF2
No compounds observed with He, Ne, or Ar (truly inert gases)
________________________________________
1996 Michael Blaber






Basic Concepts of Chemical Bonding
Lewis Symbols and the Octet Rule
________________________________________
Why are some substances chemically bonded molecules and others are an association of ions?
• depends upon the electronic structures of the atoms
• nature of the chemical forces within the compounds
A broad classification of chemical forces:
1. Ionic bonds
2. Covalent bonds
3. Metallic bonds
Ionic bonds - electrostatic forces that exist between ions of opposite charge
• typically involves a metal with a nonmetal
Covalent bonds - results from the sharing of electrons between two atoms
• typically involves one nonmetallic element with another
Metallic bonds
• found in solid metals (copper, iron, aluminum)
• each metal bonded to several neighboring groups
• bonding electrons free to move throughout the 3-dimensional structure
Lets look at the preferred arrangements of electrons in atoms when they form chemical compounds
Lewis Symbols and the Octet Rule
Valence electrons reside in the outer shell and are the electrons which are going to be involved in chemical interactions and bonding (valence comes from the Latin valere, "to be strong").
Electron-dot symbols (Lewis symbols):
• convenient representation of valence electrons
• allows you to keep track of valence electrons during bond formation
• consists of the chemical symbol for the element plus a dot for each valence electron
Sulfur
Electron configuration is [Ne]3s23p4, thus there are six valence electrons. Its Lewis symbol would therefore be:

Note:
• The dots (representing electrons) are placed on the four sides of the atomic symbol (top, bottom, left, right)
• Each side can accommodate up to 2 electrons
• The number of valence electrons in the table below is the same as the column number of the element in the periodic table (for representative elements only)
Atoms often gain, lose, or share electrons to achieve the same number of electrons as the noble gas closest to them in the periodic table
Because all noble gasses (except He) have filled s and p valence orbitals (8 electrons), many atoms undergoing reactions also end up with 8 valence electrons. This observation has led to the Octet Rule:
Atoms tend to lose, gain, or share electrons until they are surrounded by 8 valence electrons
Note: there are many exceptions to the octet rule (He and H, for example), but it provides a useful model for understanding the basis of chemical bonding.

________________________________________
1996 Michael Blaber


Basic Concepts of Chemical Bonding
Ionic Bonding
________________________________________
Ionic Bonding
Sodium metal reacts with chlorine gas in a violently exothermic reaction to produce NaCl (composed of Na+ and Cl- ions):
2Na(s) + Cl2(g) -> 2NaCl(s)
These ions are arranged in solid NaCl in a regular three-dimensional arrangement (or lattice):

The chlorine has a high affinity for electrons, and the sodium has a low ionization potential. Thus the chlorine gains an electron from the sodium atom. This can be represented using electron-dot symbols (here we will consider one chlorine atom, rather than Cl2):

The arrow indicates the transfer of the electron from sodium to chlorine to form the Na+ metal ion and the Cl- chloride ion. Each ion now has an octet of electrons in its valence shell:
Na+ 2s22p6
Cl- 3s23p6
Energetics of Ionic Bond Formation
The formation of ionic compounds (like the addition of sodium metal and chlorine gas to form NaCl) are usually extremely exothermic.
The loss of an electron from an element:
• Always endothermic (takes energy to strip the e' from the atom)
• Na(g) -> Na+(g) + 1e- H = 496 kJ/mol
The gain of an electron by a nonmetal:
• Generally exothermic (energy released)
• Cl(g) + 1e- -> Cl-(g) H = -349 kJ/mol
The formation of NaCl from Na and Cl would thus require the input of 147 kJ/mol. However, it appears to be a highly exothermic reaction.
Ionic compounds are stable due to the attraction between unlike charges:
• The ions are drawn together
• Energy is released
• Ions form solid lattice
Lattice energy:
the energy required to separate completely a mole of a solid ionic compound into its gaseous ions
It is a measure of just how much stabilization results from the arranging of oppositely charged ions in an ionic solid.
To completely break up a salt crystal:
NaCl(s) -> Na+(g) + Cl-(g) Hlattice = +788 kJ/mol
Thus, -788 kJ/mol is given off as heat energy when 1 mol of NaCl is incorporated into the salt lattice.
So, forming the ions from Na(g) and Cl(g) requires the input of +147 kJ/mol, these ions incorporate into the salt lattice liberating -788 kJ/mol, for an overall highly exothermic release of -641 kJ/mol.
The magnitude of the lattice energy depends upon the charges of the ions, their size and the particular lattice arrangement.
The potential energy of two interacting charged particles is:

Q1 = charge on first particle
Q2 = charge on second particle
d = distance between centers of particles
k = 8.99 x 109 J m/C2
Thus, the interaction increases:
• As the charges increase
• As the two charges are brought closer together
The minimum distance between oppositely charged ions is the sum of the atomic (ionic) radii. Although atomic radii do vary, it is not over a considerable range, thus, the attraction between two ions is determined primarily by the charge of the ions.
Electron configuration of ions
How does the energy released in lattice formation compare to the energy required to strip away another electron from the Na+ ion?
Since the Na+ ion has a noble gas electron configuration, stripping away the next electron from this stable arrangement would take far more energy than what is released during lattice formation (Sodium I2 = 4,560 kJ/mol). Thus, sodium is present in ionic compounds as Na+ and not Na2+.
Likewise, adding an electron to fill a valence shell (and achieve noble gas electron configuration) is exothermic or only slightly endothermic. To add an additional electron into a new subshell requires tremendous energy - more than the lattice energy. Thus, we find Cl- in ionic compounds, but not Cl2-.
Lattice energies range from around 700 kJ/mol to 4000 kJ/mol:
Compound Lattice Energy
(kJ/mol)
LiF 1024
LiI 744
NaF 911
NaCl 788
NaI 693
KF 815
KBr 682
KI 641
MgF2 2910
SrCl2 2130
MgO 3938
This amount of energy can compensate for values as large as I3 for valence electrons (i.e. can strip away up to 3 electrons).
Because most transition metals would require the removal of more than 3 electrons to attain a noble gas core, they are not found in ionic compounds with a noble gas core (thus they may have color). Some examples which can form a noble gas core (and be colorless):
Ag: [Kr]5s14d10 Ag+ [Kr]4d10 Compound: AgCl
Cd: [Kr]5s24d10 Cd2+ [Kr]4d10 Compound: CdS
The valence electrons do not adhere to the "octet rule" in this case (a limitation of the usefulness of this rule)
Note: The silver and cadmium atoms lost the 5s electrons in achieving the ionic state
When a positive ion is formed from an atom, electrons are always lost first from the subshell with the largest principle quantum number
A transition metal always loses electrons first from the higher 's' subshell, before losing from the underlying 'd' subshell.
Iron will not have a noble gas core (iron salts will have color)
Fe: [Ar]4s23d6 Fe2+ [Ar] 3d6
Fe: [Ar]4s23d6 Fe3+ [Ar] 3d5
Polyatomic ions
In polyatomic ions, two or more atoms are bound together by covalent (chemical) bonds. They form a stable grouping which carries a charge (positive or negative). The group of atoms as a whole acts as a charged species in forming an ionic compound with an oppositely charged ion.
________________________________________
1996 Michael Blaber


Basic Concepts of Chemical Bonding
Sizes of Ions
________________________________________
Sizes of Ions
Sizes of ions influence:
• packing of ions in ionic lattices, and therefore, the lattice energy
• biological recognition - some ions can pass through certain membrane channels, others may be too large
The size of an ion is influenced by:
• nuclear charge
• number of electrons
• valence orbitals
Cations
• formed by removing one or more valence electrons
• vacates the most spatially extended orbitals
• decreases the total electron-electron repulsion in the outer orbital
Cations are therefore smaller than the parent atom
Anions
• formed by addition of one or more valence electrons
• fills in the most spacially extended orbitals
• increases electron-electron repulsion in outer orbital
Anions are therefore larger than the parent atom
For ions of the same charge (e.g. in the same group) the size increases as we go down a group in the periodic table
As the principle quantum increases the size of both the parent atom and the ion will increase


How does the nuclear charge affect ion size?
Consider the following collection of ions:
ion electrons protons
O2- 10 8
F- 10 9
Na+ 10 11
Mg2+ 10 12
Al3+ 10 13
Each ion:
• contains the same number of electrons (10; with configuration 1s22s22p6) and are thus termed a collection of isoelectronic ions
• varies in the nuclear charge
The radius of each ion decreases with an increase in nuclear charge:

Oxygen and fluorine precede neon and are nonmetals, sodium, magnesium and aluminum come after neon and are metals.
________________________________________
1996 Michael Blaber

Basic Concepts of Chemical Bonding
Covalent Bonding
________________________________________
Covalent Bonding
Ionic substances:
• usually brittle
• high melting point
• organized into an ordered lattice of atoms, which can be cleaved along a smooth line
the electrostatic forces organize the ions of ionic substances into a rigid, organized three-dimensional arrangement
The vast majority of chemical substances are not ionic in nature
• gases and liquids, in addition to solids
• low melting temperatures
G.N. Lewis
reasoned that an atom might attain a noble gas electron configuration by sharing electrons
A chemical bond formed by sharing a pair of electrons is called a covalent bond
The diatomic hydrogen molecule (H2) is the simplest model of a covalent bond, and is represented in Lewis structures as:

The shared pair of electrons provides each hydrogen atom with two electrons in its valence shell (the 1s) orbital.
In a sense, it has the electron configuration of the noble gas helium
When two chlorine atoms covalently bond to form Cl2, the following sharing of electrons occurs:

Each chlorine atom shared the bonding pair of electrons and achieves the electron configuration of the noble gas argon.
In Lewis structures the bonding pair of electrons is usually displayed as a line, and the unshared electrons as dots:

The shared electrons are not located in a fixed position between the nuclei. In the case of the H2 compound, the electron density is concentrated between the two nuclei:

The two atoms are bound into the H2 molecule mainly due to the attraction of the positively charged nuclei for the negatively charged electron cloud located between them
For the nonmetals (and the 's' block metals) the number of valence electrons is equal to the group number:
Element Group Valence
electrons Bonds needed to form valence octet
F 7A 7 1
O 6A 6 2
N 5A 5 3
C 4A 4 4
Examples of hydride compounds of the above elements (covalent bonds with hydrogen:

Thus, the Lewis bonds successfully describe the covalent interactions between various nonmetal elements
Multiple bonds
The sharing of a pair of electrons represents a single covalent bond, usually just referred to as a single bond
In many molecules atoms attain complete octets by sharing more than one pair of electrons between them.
Two electron pairs shared a double bond
Three electron pairs shared a triple bond

Because each nitrogen contains 5 valence electrons, they need to share 3 pairs to each achieve a valence octet.
• N2 is fairly inert, due to the strong triple bond between the two nitrogens
• The N - N bond distance in N2 is 1.10 Å (fairly short)
From a study of various Nitrogen containing compounds bond distance as a function of bond type can be summarized as follows:
• N-N 1.47Å
• N=N 1.24Å
• N=N 1.10Å
As a general rule, the distance between bonded atoms decreases as the number of shared electron pairs increases
________________________________________
1996 Michael Blaber

Basic Concepts of Chemical Bonding
Bond Polarity and Electronegativity
________________________________________
Bond Polarity and Electronegativity
The electron pairs shared between two atoms are not necessarily shared equally
Extreme examples:
1. In Cl2 the shared electron pairs is shared equally
2. In NaCl the 3s electron is stripped from the Na atom and is incorporated into the electronic structure of the Cl atom - and the compound is most accurately described as consisting of individual Na+ and Cl- ions
For most covalent substances, their bond character falls between these two extremes
Bond polarity is a useful concept for describing the sharing of electrons between atoms
• A nonpolar covalent bond is one in which the electrons are shared equally between two atoms
• A polar covalent bond is one in which one atom has a greater attraction for the electrons than the other atom. If this relative attraction is great enough, then the bond is an ionic bond
Electronegativity
A quantity termed 'electronegativity' is used to determine whether a given bond will be nonpolar covalent, polar covalent, or ionic.
Electronegativity is defined as the ability of an atom in a particular molecule to attract electrons to itself
(the greater the value, the greater the attractiveness for electrons)
Electronegativity is a function of:
• the atom's ionization energy (how strongly the atom holds on to its own electrons)
• the atom's electron affinity (how strongly the atom attracts other electrons)
(Note that both of these are properties of the isolated atom)
For example, an element which has:
• A large (negative) electron affinity
• A high ionization energy (always endothermic, or positive for neutral atoms)
Will:
• Attract electrons from other atoms
• Resist having its own electrons attracted away
Such an atom will be highly electronegative
Fluorine is the most electronegative element (electronegativity = 4.0), the least electronegative is Cesium (notice that are at diagonal corners of the periodic chart)

General trends:
• Electronegativity increases from left to right along a period
• For the representative elements (s and p block) the electronegativity decreases as you go down a group
• The transition metal group is not as predictable as far as electronegativity
Electronegativity and bond polarity
We can use the difference in electronegativity between two atoms to gauge the polarity of the bonding between them
Compound F2 HF LiF
Electronegativity Difference 4.0 - 4.0 = 0 4.0 - 2.1 = 1.9 4.0 - 1.0 = 3.0
Type of Bond Nonpolar covalent Polar covalent Ionic (non-covalent)
• In F2 the electrons are shared equally between the atoms, the bond is nonpolar covalent
• In HF the fluorine atom has greater electronegativity than the hydrogen atom.
The sharing of electrons in HF is unequal: the fluorine atom attracts electron density away from the hydrogen (the bond is thus a polar covalent bond)
The H-F bond can thus be represented as:

• The '+' and '-' symbols indicate partial positive and negative charges.
• The arrow indicates the "pull" of electrons off the hydrogen and towards the more electronegative atom
• In lithium fluoride the much greater relative electronegativity of the fluorine atom completely strips the electron from the lithium and the result is an ionic bond (no sharing of the electron)
A general rule of thumb for predicting the type of bond based upon electronegativity differences:
• If the electronegativities are equal (i.e. if the electronegativity difference is 0), the bond is non-polar covalent
• If the difference in electronegativities between the two atoms is greater than 0, but less than 2.0, the bond is polar covalent
• If the difference in electronegativities between the two atoms is 2.0, or greater, the bond is ionic
________________________________________
1996 Michael Blaber

Basic Concepts of Chemical Bonding
Drawing Lewis Structures
________________________________________
Drawing Lewis Structures
The general procedure...
1. Sum the valence electrons from all atoms
• Use the periodic table for reference
• Add an electron for each indicated negative charge, subtract an electron for each indicated positive charge
2. Write the symbols for the atoms to show which atoms are attached to which, and connect them with a single bond
• You may need some additional evidence to decide bonding interactions
• If a central atom has various groups bonded to it, it is usually listed first: CO32-, SF4
• Often atoms are written in the order of their connections: HCN
3. Complete the octets of the atoms bonded to the central atom (H only has two)
4. Place any leftover electrons on the central atom (even if it results in more than an octet)
5. If there are not enough electrons to give the central atom an octet, try multiple bonds (use one or more of the unshared pairs of electrons on the atoms bonded to the central atom to form double or triple bonds)
Draw the Lewis structure of phosphorous trichloride (PCl3)
This is an example of a central atom, P, surrounded by chlorine atoms
1. We will have 5(P) plus 21 (3*7, for Cl), or 26 total valence electrons
2. The general symbol, starting with only single bonds, would be:

3. Completing the octets of the Cl atoms bonded to the central P atom:

4. This gives us a total of (18 electrons) plus the 6 in the three single bonds, or 24 electrons total. Thus we have 2 extra valence electrons which are not accounted for. We will place them on the central element:

5. The central atom now has an octect, and there is no need to invoke any double or triple bonds to achieve an octet for the central atom. We are finished.
Draw the Lewis structure for the NO+ ion
1. We will have 5 (N) plus 6 (O) minus 1 (1+ ion), or 10 valence electrons
2. The general structure starting only with single bonds would be:

3. Completing the octet of the O bonded to N:

4. This gives us a total of 6 plus 2 for the single bond, or 8 electrons. There are 2 unaccounted for electrons and we will place them on the N:

5. There are only 4 atoms on the N atom, not enough for an octet, so lets try a double bond between the N and O:

The oxygen still has an octet, but the N only has 6 valence electrons, so lets try a triple bond:

Each atom now has a valence octet. We are finished.
The brackets with the + symbol are used to indicate that this is an ion with a net charge of 1+

Formal Charge
In some cases we can draw several different Lewis structures which fulfill the octet rule for a compound. Which one is the most reasonable?
One method is to tabulate the valence electrons around each atom in a Lewis structure to determine the formal charge. The formal charge is the charge that an atom in a molecule would have if we considered each atom to have the same electronegativity in a compound.

To calculate formal charge, assign electrons to their respective atoms as follows:
1. All of the unshared electrons are assigned to the atom on which they are found
2. The bonding electrons are divided up equally between each atom involved in the bond
3. The number of valence electrons expected in the isolated atom is compared to the actual number of electrons assigned from the Lewis structure:
The formal charge equals the number of valence electrons in the isolated atom, minus the number of electrons assigned in the Lewis structure
Example: Carbon Dioxide (CO2)
Carbon has 4 valence electrons
Each oxygen has 6 valence electrons, therefore our Lewis structure of CO2 will have 16 electrons:

One way we could draw the Lewis structure is:

Another way we could draw the Lewis structure is:

Both structures fulfill the octet rule. But what are the formal charges?


Which structure is correct? In general, when several Lewis structures can be drawn the most stable structure is the one in which:
• The formal charges are the smallest
• Any negative charge is found on the most electronegative atom

In the above case, the second structure is the one with the smallest formal charges (i.e. 0 on all the atoms).
• Furthermore, in the first possible Lewis structure the carbon has a formal charge of 0 and one of the oxygens it is bonded to has a formal charge of +1.
• Oxygen is more electronegative than Carbon, so this situation would seem unlikely.
It is important to remember that formal charges do not represent the actual charges on the atoms. Actual charges are determined by the electronegativity of the atoms involved.

________________________________________
1996 Michael Blaber

Basic Concepts of Chemical Bonding
Resonance Structures
________________________________________
Resonance Structures
The Lewis structure of ozone (O3)
1. Sum of valence electrons = (6*3) = 18
2. Drawing the bond connectivities:

3. Complete the octets of the atoms bonded to the central atom:

4. Place any leftover electrons (18-16 = 2) on the central atom:

5. Does the central atom have an octet?
• NO, it has 6 electrons
• Add a multiple bond (first try a double bond) to see if the central atom can achieve an octet:

6. Does the central atom have an octet?
• YES, we are done
• Ozone would appear to have one single bond, and one double bond
However... known facts about the structure of ozone
• The bond lengths between the central oxygen and the other two oxygens are identical:

• We would expect that if one bond was a double bond that it should be shorter than the other (single) bond
• Since all the atoms are identical (oxygens) which atom is chosen for the double bond?

These Lewis structures are equivalent except for the placement of the electrons (i.e. the location of the double bond)
Equivalent Lewis structures are called resonance structures, or resonance forms
The correct way to describe ozone as a Lewis structure would be:

This indicates that the ozone molecule is described by an average of the two Lewis structures (i.e. the resonance forms)
The important points to remember about resonance forms are:
• The molecule is not rapidly oscillating between different discrete forms
• There is only one form of the ozone molecule, and the bond lengths between the oxygens are intermediate between characteristic single and double bond lengths between a pair of oxygens
• We draw two Lewis structures (in this case) because a single structure is insufficient to describe the real structure
The Nitrate (NO3-) ion:
1. Count up the valence electrons: (1*5) + (3*6) + 1(ion) = 24 electrons
2. Draw the bond connectivities:

3. Add octet electrons to the atoms bonded to the center atom:

4. Place any leftover electrons (24-24 = 0) on the center atom:

5. Does the central atom have an octet?
• NO, it has 6 electrons
• Add a multiple bond (first try a double bond) to see if the central atom can achieve an octet:

6. Does the central atom have an octet?
• YES
• Are there possible resonance structures? YES

Note: We would expect that the bond lengths in the NO3- ion to be somewhat shorter than a single bond
________________________________________
1996 Michael Blaber

Basic Concepts of Chemical Bonding
Exceptions to the Octet Rule
________________________________________
Exceptions to the Octet Rule
There are three general ways in which the octet rule breaks down:
1. Molecules with an odd number of electrons
2. Molecules in which an atom has less than an octet
3. Molecules in which an atom has more than an octet
Odd number of electrons
Draw the Lewis structure for the molecule nitrous oxide (NO):
1. Total electrons: 6+5=11
2. Bonding structure:

3. Octet on "outer" element:

4. Remainder of electrons (11-8 = 3) on "central" atom:

5. There are currently 5 valence electrons around the nitrogen. A double bond would place 7 around the nitrogen, and a triple bond would place 9 around the nitrogen.
We appear unable to get an octet around each atom
Less than an octet (most often encountered with elements of Boron and Beryllium)
Draw the Lewis structure for boron trifluoride (BF3):
1. Add electrons (3*7) + 3 = 24
2. Draw connectivities:

3. Add octets to outer atoms:

4. Add extra electrons (24-24=0) to central atom:

5. Does central electron have octet?
• NO. It has 6 electrons
• Add a multiple bond (double bond) to see if central atom can achieve an octet:

6. The central Boron now has an octet (there would be three resonance Lewis structures)
However...
• In this structure with a double bond the fluorine atom is sharing extra electrons with the boron.
• The fluorine would have a '+' partial charge, and the boron a '-' partial charge, this is inconsistent with the electronegativities of fluorine and boron.
• Thus, the structure of BF3, with single bonds, and 6 valence electrons around the central boron is the most likely structure
BF3 reacts strongly with compounds which have an unshared pair of electrons which can be used to form a bond with the boron:

More than an octet (most common example of exceptions to the octet rule)
PCl5 is a legitimate compound, whereas NCl5 is not.

Expanded valence shells are observed only for elements in period 3 (i.e. n=3) and beyond
• The 'octet' rule is based upon available ns and np orbitals for valence electrons (2 electrons in the s orbitals, and 6 in the p orbitals)
• Beginning with the n=3 principle quantum number, the d orbitals become available (l=2)
The orbital diagram for the valence shell of phosphorous is:

Third period elements occasionally exceed the octet rule by using their empty d orbitals to accommodate additional electrons
Size is also an important consideration:
• The larger the central atom, the larger the number of electrons which can surround it
• Expanded valence shells occur most often when the central atom is bonded to small electronegative atoms, such as F, Cl and O.
Draw the Lewis structure for ICl4-
1. Count up the valence electrons: 7+(4*7)+1 = 36 electrons
2. Draw the connectivities:

3. Add octet of electrons to outer atoms:

4. Add extra electrons (36-32=4) to central atom:

5. The ICl4- ion thus has 12 valence electrons around the central Iodine (in the 5d orbitals)

________________________________________
1996 Michael Blaber

Basic Concepts of Chemical Bonding
Strengths of Covalent Bonds
________________________________________
Strengths of Covalent Bonds
The stability of a molecule is a function of the strength of the covalent bonds holding the atoms together.
How do we measure the strength of a covalent bond?
Bond-dissociation energy (i.e. "bond energy")
Bond energy is the enthalpy change (H, heat input) required to break a bond (in 1 mole of a gaseous substance)

What about when we have a compound which is not a diatomic molecule? Consider the dissociation of methane:

There are four equivalent C-H bonds, thus we can that the dissociation energy for a single C-H bond would be:
D(C-H) = (1660/4) kJ/mol = 415 kJ/mol
Note:
• The bond energy for a given bond is influenced by the rest of the molecule.
• However, this is a relatively small effect (suggesting that bonding electrons are localized between the bonding atoms).
• Thus, the bond energy for most bonds varies little from the average bonding energy for that type of bond
Bond energy is always a positive value - it takes energy to break a covalent bond (conversely energy is released during bond formation)
Average bond energies:
Bond (kJ/mol)
C-F 485
C-Cl 328
C-Br 276
C-I 240

C-C 348
C-N 293
C-O 358
C-F 485

C-C 348
C=C 614
C=C 839
The more stable a molecule (i.e. the stronger the bonds) the less likely the molecule is to undergo a chemical reaction
Bond Energies and the Enthalpy of Reactions
If we know which bonds are broken and which bonds are made during a chemical reaction, we can estimate the enthalpy change of the reaction (Hrxn) even if we don't know the enthalpies of formation (Hf°)for the reactants and products:
H = (bond energies of broken bonds) - (bond energies of formed bonds)
Example: The reaction between 1 mol of chlorine and 1 mol methane

Bonds broken: 1 mol of Cl-Cl bonds, 1 mol of C-H bonds
Bonds formed: 1 mol of H-Cl bonds, 1 mol of C-Cl bonds
H = [D(Cl-Cl) + D(C-H)] - [D(H-Cl)+D(C-Cl)]
[242 kJ + 413 kJ] - [431 kJ + 328 kJ]
-104 kJ
Thus, the reaction is exothermic (because the bonds in the products are stronger than the bonds in the reactants)
Example: The combustion of 1 mol of ethane

bonds broken: 6 moles C-H bonds, 1 mol C-C bonds, 7/2 moles of O=O bonds
bonds formed: 4 moles C=O bonds, 6 moles O-H bonds
H = [(6*413) + (348) + (7/2*495)] - [(4*799) + (6*463)]
= 4558 - 5974
= -1416 kJ (the reaction is exothermic)
Bond strength and bond length
Bond Bond Energy
(kJ/mol) Bond Length
(Å)
C-C 348 1.54
C=C 614 1.34
C=C 839 1.20

As the number of bonds between two atoms increases, the bond grows shorter and stronger
________________________________________
1996 Michael Blaber



Basic Concepts of Chemical Bonding
Oxidation Numbers
________________________________________
Oxidation Numbers
When a covalent bond forms between two atoms with different electronegativities the shared electrons in the bond lie closer to the more electronegative atom:

• The oxidation number of an atom is the charge that results when the electrons in a covalent bond are assigned to the more electronegative atom
• It is the charge an atom would possess if the bonding were ionic
In HCl (above) the oxidation number for the hydrogen would be +1 and that of the Cl would be -1
in oxidation numbers we write the sign first to distinguish them from ionic (electronic) charges
Oxidation numbers do not refer to real charges on the atoms, except in the case of actual ionic substances.
Oxidation numbers can be determined using the following rules:
1. The oxidation number for an element in its elemental form is 0 (holds true for isolated atoms and elemental substances which bond identical atoms: e.g. Cl2, etc)
2. The oxidation number of a monoatomic ion is the same as its charge (e.g. oxidation number of Na+ = +1, and that of S2- is -2)
3. In binary compounds (two different elements) the element with greater electronegativity is assigned a negative oxidation number equal to its charge in simple ionic compounds of the element (e.g. in the compound PCl3 the chlorine is more electronegative than the phosphorous. In simple ionic compounds Cl has an ionic charge of 1-, thus, its oxidation state is -1)
4. The sum of the oxidation numbers is zero for an electrically neutral compound and equals the overall charge for an ionic species.
5. Alkali metals exhibit only an oxidation state of +1 in compounds
6. Alkaline earth metals exhibit only an oxidation state of +2 in compounds
PCl3
The chlorine is more electronegative and so its oxidation number is set to -1. The overall molecule is neutral, so the oxidation number of P, in this case, is +3.
CO32-
The oxygen is more electronegative and receives an oxidation number of -2. The overall molecule has a net charge of 2- (an overall oxidation number of ¬2), therefore, the C must have an oxidation state of +4, i.e. (3*-2) + 'C' = -2.
Examples of Sulfur
H2S
Sulfur (2.5) is more electronegative than hydrogen (2.1), thus it has an oxidation number of -2. The hydrogen will have an oxidation number of +1.
S8
This is an elemental form of sulfur, and thus would have an oxidation number of 0.
SCl2
Chlorine (3.0) is more electronegative than sulfur (2.5), thus it has an oxidation number of -1. The sulfur thus has an oxidation number of +2.
Na2SO3
Sodium (alkali metal) always has an oxidation number of +1. The oxygen (3.5) is more electronegative than sulfur (2.5), thus the oxygen would have an oxidation number of -2. The sulfur would therefore have an oxidation number of +4.
SO42-
The oxygen is more electronegative and thus has an oxidation number of -2. The sulfur thus has an oxidation number of +6.
• Sulfur exhibits a variety of oxidation numbers (-2 to +6)
• In general the most negative oxidation number corresponds to the number of electrons which must be added to give an octet of valence electrons
• The most positive oxidation number corresponds to a loss of all valence electrons
Oxidation Numbers and Nomenclature
Compounds of the alkali (oxidation number +1) and alkaline earth metals (oxidation number +2) are typically ionic in nature.
Compounds of metals with higher oxidation numbers (e.g. tin +4) tend to form molecular compounds
• In ionic and covalent molecular compounds usually the less electronegative element is given first.
• In ionic compounds the names are given which refer to the oxidation (ionic) state
• In molecular compounds the names are given which refer to the number of molecules present in the compound
Ionic Molecular
MgH2 magnesium hydride H2S dihydrogen sulfide
FeF2 iron(II) fluoride OF2 oxygen difluoride
Mn2O3 manganese(III) oxide Cl2O3 dichlorine trioxide
________________________________________
1996 Michael Blaber

Molecular Geometry and Bonding Theories
Molecular Geometries
________________________________________
Molecular shapes, or geometries, are critical to molecular recognition and function.
Molecular Geometries
The Lewis structure of carbon tetrachloride:

• Provides information about connectivities
• Provides information about valence orbitals
• Provides information about bond character
However, the Lewis structure provides no information about the shape of the molecule
The structure of a molecule is defined by:
• The bond angles
• The bond lengths
In carbon tetrachloride:
• Each C-Cl bond length is 1.78Å
• Each Cl-C-Cl bond angle is 109.5°
Carbon tetrachloride is tetrahedral in structure:

Molecular Geometries of ABn molecules
A central atom A is bonded to two or more B atoms


These structures can generally be predicted, when A is a nonmetal, using the "valence-shell electron-pair repulsion model (VSEPR)
________________________________________
1996 Michael Blaber






Molecular Geometry and Bonding Theories
Valence Shell Electron Pair Repulsion (VSEPR) Model
________________________________________
The Valence Shell Electron Pair Repulsion Model

Balloons tied together adopt arrangements which minimize steric clashes between neighbors:

• Atoms are bonded together by electron pairs in valence orbitals
• Electrons are all negatively charged and tend to repel other electrons
• Bonding pairs of shared electrons tend to repel other bonding pairs of electrons in the valence orbital
The best spatial arrangement of the bonding pairs of electrons in the valence orbitals is one in which the repulsions are minimized
Like the balloon example:
• Two electron pairs in the valence orbital are arranged linearly
• Three electron pairs are organized in a trigonal planar arrangement
• Four electron pairs are organized in a tetrahedral arrangement
• Five electron pairs are arranged in a trigonal bipyramid
• Six electron pairs are organized in an octahedral arrangement

The shape of a molecule can be related to these five basic arrangements
Predicting Molecular Geometries
In Lewis structures there are two types of valence electron pairs:
• bonding pairs (shared by atoms in bonds)
• nonbonding pairs (also called lone pairs)
The Lewis structure of ammonia:

• Three bonding pairs of electrons
• One nonbonding pair of electrons
The electron shell repulsion between these four electron pairs is minimized in a tetrahedral arrangement (i.e. the "electron pair geometry" is tetrahedral)
This arrangement is for the valence electron pairs. What about the atoms in a compound?
• The molecular geometry is the location of the atoms of a compound in space
• We can predict the molecular geometry from the electron pair geometry
• In the above example (ammonia), we would predict that the three hydrogens would form the vertices of a tetrahedron, and the nonbonding electron pair the fourth. Thus, ammonia would have a trigonal pyramide arrangement of its H atoms
Steps involved in determining the VSEPR model:
1. Draw the Lewis structure
2. Count total number of electron pairs around the central atom. Arrange them to minimize the electron shell repulsion
3. Describe the molecular geometry in terms of the angular arrangement of the bonding pairs
Four or Fewer Valence-Shell Electron Pairs
Structural types for molecules or ions which obey the octet rule:

Note: a double or triple bond is counted as one bonding pair when predicting geometry
Using the VSEPR model predict the molecular geometries of a) SnCl3- and b) O3

The Effect of Nonbonding Electrons and Multiple Bonds on Bond Angles
The VSEPR model can be used to explain slight distortions from ideal bond geometries observed in some structures.
Methane, ammonia and water all have tetrahedral electron-pair geometries, but the bond angles of ammonia and water are slightly distorted from an ideal tetrahedron:

The bond angles decrease as the number of nonbonding electron pairs increases
Since the electron pairs of bonding atoms are somewhat delocalized from the individual atoms (i.e. they are shared by two atoms), whereas the nonbonding electron pairs are attracted to a single nucleus, the nonbonding pairs can be thought of as having a somewhat larger electron cloud near the parent atom (kind of like being a somewhat larger balloon in the balloon analogy). This "crowds" the bonding pairs and the geometry distortions reflect this.
Multiple bonds, which contain higher electron density than single bonds also distort geometry by crowing the bonding pairs of single bonds:

Electrons in multiple bonds, like nonbonding electrons, exert a greater repulsive force on adjacent electron pairs than do single bonds
Geometries of Molecules with Expanded Valence Shells
When the central atom has 'd' orbitals available (n = 3 and higher) then it may have more than 4 electron pairs around it. Such atoms exhibit a variety of molecular geometries:

The trigonal bipyramidal arrangement for atoms with 5 pairs of valence electrons contains two geometrically distinct types of electron pairs, axial and equitorial:

If there is a non-bonding pair of electrons (a "larger" electron cloud), it will go in the axial position to minimize electron repulsion
The octahedral structure contains 6 pairs of valence electrons. All positions are equivalent and at 90° from other electron pairs.
If there is one nonbonding pair of electrons, it makes no difference where we place them. However, if there are two nonbonding pairs of electrons, the second pair will be 180° from the first to minimize steric interactions
Molecules with no central atom
The VSEPR model can be used to determine the geometry of more complex molecules

• The first carbon has four pairs of valence electrons and will be tetrahedral
• The second carbon has "three" (multiple bonds count as one in VSEPR) and will be trigonal planar
• The oxygen on the right has four and will be tetrahedral (only two have bonds and thus it will appear as a "bent" conformation):

________________________________________
Examples of molecular geometry:


________________________________________
1996 Michael Blaber

Molecular Geometry and Bonding Theories
Polarity of Molecules
________________________________________
Polarity of Molecules
The "charge distribution" of a molecule is determined by
• The shape of the molecule
• The polarity of its bonds
A Polar Molecule:
• The center of the overall negative charge on the molecule does not coincide with the center of overall positive charge on the molecule
• The molecule can be oriented such that one end has a net negative charge and the other a net positive charge, i.e. the molecule is a dipole
A Nonpolar molecule
• Has no charges on the opposite ends of the molecule
• Or, has charges of the same sign on the opposite ends of the molecule
• Molecule is not a dipole
Any diatomic molecule with a polar bond is a polar molecule (dipole)

Polar molecules align themselves:
• in an electric field
• with respect to one another
• with respect to ions

The degree of polarity of a molecule is described by its dipole moment, = Q * r
where
• Q equals the charge on either end of the dipole
• r is the distance between the charges
the greater the distance or the higher the charge, the greater the magnitude of the dipole
Dipole moments are generally reported in Debye units
1 debye = 3.33 x 10-30 coulomb meters (C m)
Example: H-Cl a covalent polar compound
• The H-Cl bond distance is 1.27Å
• +1 and -1 charges in a dipole produce 1.60 x 10-19 C
 = Qr = (1.60 x 10-19 C)(1.27 x 10-10 m)
 = 2.03 x 10-29 C m
 = 2.03 x 10-29 C m (1 debye/3.33 x 10-30) = 6.10 debye
The actual dipole of H-Cl is 1.08 debye. The reason for this is that the compound is covalent and not ionic, thus the charges of the dipole are less that +1, and -1 (values expected for a fully ionic compound)
Compound Bond
Length (Å) Electronegativity
Difference Dipole
Moment (D)
HF 0.92 1.9 1.82
HCl 1.27 0.9 1.08
HBr 1.41 0.7 0.82
HI 1.61 0.4 0.44
Although the bond length is increasing, the dipole is decreasing as you move down the halogen group. The electronegativity decreases as we move down the group. Thus, the greater influence is the electronegativity of the two atoms (which influences the charge at the ends of the dipole).
The Polarity of Polyatomic Molecules
• Each polar bond in a polyatomic molecule will have an associated dipole
• The overall dipole of the molecule will be the sum of the individual dipoles

• Although in carbon dioxide the oxygens have a partial negative charge and the carbon a partial positive charge, the molecule has no dipole - it will not orient in an electrical field
• Water has a dipole and will orient in an electrical field
Although a polar bond is a prerequisite for a molecule to have a dipole, not all molecules with polar bonds exhibit dipoles
ABn molecules and non-polar geometries
For ABn molecules, where the central atom A is surrounded by identical atoms for B, there a certain molecular geometries which result in no effective dipole, regardless of how polar the individual bonds may be. These geometries are:

________________________________________
1996 Michael Blaber




Molecular Geometry and Bonding Theories
Covalent Bonding and Orbital Overlap
________________________________________
Covalent Bonding and Orbital Overlap
• The VSEPR model is a simple method which allows us to predict molecular geometries, but it does not explain why bonds exist between atoms.
• How can we explain molecular geometries and the basis of bonding at the same time?
Quantum mechanics and molecular orbitals
"Valence bond theory"
Combine Lewis' idea of electron pair bonds with electron orbitals (quantum mechanics)
Covalent bonding occurs when atoms share electrons (Lewis model)
• Concentrates electron density between nuclei
The buildup of electron density between two nuclei occurs when a valence atomic orbital of one atom merges with that of another atom (Valence bond theory)
• The orbitals share a region of space, i.e. they overlap
• The overlap of orbitals allows two electrons of opposite spin to share the common space between the nuclei, forming a covalent bond

There is an optimum distance between two bonded nuclei in covalent bonds:

• As the 1s orbitals of the hydrogen start to overlap there is a reduction in the potential energy of the system (due to the increase in electron density between the two positively charged nuclei)
• As the distance decreases further, repulsion between the nuclei becomes significant at short distances
• The internuclear distance at the minimum potential energy corresponds to the observed bond length
Therefore:
The observed bond length is the distance at which the attractive forces (nuclei for bonding electrons) is balance by the repulsive force (nuclei vs. nuclei; and additionally, electron vs. electron)
________________________________________
1996 Michael Blaber


Molecular Geometry and Bonding Theories
Hybrid Orbitals
________________________________________
Hybrid Orbitals
For polyatomic molecules we would like to be able to explain:
• The number of bonds formed
• Their geometries
sp Hybrid Orbitals
Consider the Lewis structure of gaseous molecules of BeF2:

• The VSEPR model predicts this structure will be linear
• What would valence bond theory predict about the structure?
The fluorine atom electron configuration:
• 1s22s22p5

• There is an unpaired electron in a 2p orbital
• This unpaired 2p electron can be paired with an unpaired electron in the Be atom to form a covalent bond
The Be atom electron configuration:
• 1s22s2

• In the ground state, there are no unpaired electrons (the Be atom is incapable of forming a covalent bond with a fluorine atom
• However, the Be atom could obtain an unpaired electron by promoting an electron from the 2s orbital to the 2p orbital:

This would actually result in two unpaired electrons, one in a 2s orbital and another in a 2p orbital
• The Be atom can now form two covalent bonds with fluorine atoms
• We would not expect these bonds to be identical (one is with a 2s electron orbital, the other is with a 2p electron orbital)
However, the structure of BeF2 is linear and the bond lengths are identical
• We can combine wavefunctions for the 2s and 2p electrons to produce a "hybrid" orbital for both electrons
• This hybrid orbital is an "sp" hybrid orbital

• The orbital diagram for this hybridization would be represented as:


Note:
• The Be 2sp orbitals are identical and oriented 180° from one another (i.e. bond lengths will be identical and the molecule linear)
• The promotion of a Be 2s electron to a 2p orbital to allow sp hybrid orbital formation requires energy.
o The elongated sp hybrid orbitals have one large lobe which can overlap (bond) with another atom more effectively
o This produces a stronger bond (higher bond energy) which offsets the energy required to promote the 2s electron
sp2 and sp3 Hybrid Orbitals
Whenever orbitals are mixed (hybridized):
• The number of hybrid orbitals produced is equal to the sum of the orbitals being hybridized
• Each hybrid orbital is identical except that they are oriented in different directions
BF3
Boron electron configuration:

• The three sp2 hybrid orbitals have a trigonal planar arrangement to minimize electron repulsion

NOTE: sp2 refers to a hybrid orbital being constructed from one s orbital and two p orbitals. Although it looks like an electron configuration notation, the superscript '2' DOES NOT refer to the number of electrons in an orbital.
• An s orbital can also mix with all 3 p orbitals in the same subshell
CH4


• Thus, using valence bond theory, we would describe the bonds in methane as follows: each of the carbon sp3 hybrid orbitals can overlap with the 1s orbitals of a hydrogen atom to form a bonding pair of electrons
NOTE: sp3 refers to a hybrid orbital being constructed from one s orbital and three p orbitals. Although it looks like an electron configuration notation, the superscript '3' DOES NOT refer to the number of electrons in an orbital.
ANOTHER NOTE: the two steps often observed when constructing hybrid orbitals is to 1) promote a valence electron from the ground state configuration to a higher energy orbital, and then 2) hybridize the appropriate valence electron orbitals to achieve the desired valence electron geometry (i.e. the correct number of hybrid orbitals for the appropriate valence electron geometry)
H2O
Oxygen


Hybridization Involving d Orbitals
Atoms in the third period and higher can utilize d orbitals to form hybrid orbitals
PF5


Similarly hybridizing one s, three p and two d orbitals yields six identical hybrid sp3d2 orbitals. These would be oriented in an octahedral geometry.
• Hybrid orbitals allows us to use valence bond theory to describe covalent bonds (sharing of electrons in overlapping orbitals of two atoms)
• When we know the molecular geometry, we can use the concept of hybridization to describe the electronic orbitals used by the central atom in bonding
Steps in predicting the hybrid orbitals used by an atom in bonding:
1. Draw the Lewis structure
2. Determine the electron pair geometry using the VSEPR model
3. Specify the hybrid orbitals needed to accommodate the electron pairs in the geometric arrangement
NH3
1. Lewis structure

2. VSEPR indicates tetrahedral geometry with one non-bonding pair of electrons (structure itself will be trigonal pyramidal)
3. Tetrahedral arrangement indicates four equivalent electron orbitals


Valence Electron Pair Geometry Number of Orbitals Hybrid Orbitals
Linear 2 sp
Trigonal Planar 3 sp2
Tetrahedral 4 sp3
Trigonal Bipyramidal 5 sp3d
Octahedral 6 sp3d2


________________________________________
• 1996 Michael Blaber
Molecular Geometry and Bonding Theories
Multiple Bonds
________________________________________
Multiple Bonds
The "internuclear axis" is the imaginary axis that passes through the two nuclei in a bond:

The covalent bonds we have been considering so far exhibit bonding orbitals which are symmetrical about the internuclear axis (either an s orbital - which is symmetric in all directions, or a p orbital that is pointing along the bond towards the other atom, or a hybrid orbital that is pointing along the axis towards the other atom)
Bonds in which the electron density is symmetrical about the internuclear axis are termed "sigma" or "" bonds
In multiple bonds, the bonding orbitals arise from a different type arrangement:
• Multiple bonds involve the overlap between two p orbitals
• These p orbitals are oriented perpendicular to the internuclear (bond) axis

This type of overlap of two p orbitals is called a "pi" or "" bond. Note that this is a single  bond (which is made up of the overlap of two p orbitals)
In  bonds:
• The overlapping regions of the bonding orbitals lie above and below the internuclear axis (there is no probability of finding the electron in that region)
• The size of the overlap is smaller than a  bond, and thus the bond strength is typically less than that of a  bond
________________________________________
Generally speaking:
• A single bond is composed of a  bond
• A double bond is composed of one  bond and one  bond
• A triple bond is composed of one  bond and two  bonds

________________________________________
C2H4 (ethylene; see structure above)
• The arrangement of bonds suggests that the geometry of the bonds around each carbon is trigonal planar
• Trigonal planar suggests sp2 hybrid orbitals are being used (these would be  bonds)
What about the electron configuration?
Carbon: 1s2 2s2 2p2

• Thus, we have an extra unpaired electron in a p orbital available for bonding
• This extra p electron orbital is oriented perpendicular to the plane of the three sp2 orbitals (to minimize repulsion):

• The unpaired electrons in the p orbitals can overlap one another above and below the internuclear axis to form a covalent bond

• This interaction above and below the internuclear axis represents the single  bond between the two p orbitals
Experimentally:
• we know that the 6 atoms of ethylene lie in the same plane.
• If there was a single  bond between the two carbons, there would be nothing stopping the atoms from rotating around the C-C bond.
• But, the atoms are held rigid in a planar orientation.
• This orientation allows the overlap of the two p orbitals, with formation of a bond.
• In addition to this rigidity, the C-C bond length is shorter than that expected for a single bond.
• Thus, extra electrons (from the  bond) must be situated between the two C-C nuclei.
C2H2 (acetylene)

• The linear bond arrangement suggests that the carbon atoms are utilizing sp hybrid orbitals for bonding

• This leaves two unpaired electrons in p orbitals
• To minimize electron replusion, these p orbitals are at right angles to each other, and to the internuclear axis:

• These p orbitals can overlap two form two  bonds in addition to the single  bond (forming a triple bond)

Delocalized Bonding
localized electrons are electrons which are associated completely with the atoms forming the bond in question
In some molecules, particularly with resonance structures, we cannot associate bonding electrons with specific atoms
C6H6 (Benzene)

Benzene has two resonance forms
• The six carbon - carbon bonds are of equal length, intermediate between a single bond and double bond
• The molecule is planar
• The bond angle around each carbon is approximately 120°
The apparent hybridization orbital consistent with the geometry would be sp2 (trigonal planar arrangement)
• This would leave a single p orbital associated with each carbon (perpendicular to the plane of the ring)

With six p electrons we could form three discrete  bonds
• However, this would result in three double bonds in the ring, and three single bonds
• This would cause the bond lengths to be different around the ring (which they are not)
• This would also result in one resonance structure being the only possible structure
The best model is one in which the  electrons are "smeared" around the ring, and not localized to a particular atom

• Because we cannot say that the electrons in the  bonds are localized to a particular atom they are described as being delocalized among the six carbon atoms
Benzene is typically drawn in two different ways:

• The circle indicates the delocalization of the p bonds
________________________________________
Structure of NO3-

The Lewis structure of NO3- ion suggests that three resonance structures describe the molecular structure
• For any individual Lewis resonance structure the electronic structure for the central N atom is predicted to be sp2 hybrid orbitals participating in  bonds with each of the O atoms, and an electron in a p orbital participating in a  bond with one oxygen (forming a double bond)
• Two of the O atoms are predicted to have sp3 hybrid orbitals, with one orbital participating in a  bond with the central N atom and the other orbitals filled with non-bonding electron pairs. The other O atom is predicted to have sp2 hybrid orbitals, with one orbital participating in a  bond with the central N and two orbitals filled with non-bonding pairs of electrons. Furthermore, this last O atom is participating in a double bond with the central N atom and therefore should have an electron in a p orbital to participate in a  bond with the central N
How will this arrangement look as far as the orbital diagrams?

• There are 24 valence electrons in the expected valence orbitals above
• Summing the valence electrons from the formula gives: (3 x 6) for O, plus 5 for N, plus 1 for ionic charge = 24
What might we expect for the electron configuration if we just started with the N atom?

• We would predict that the N can only make two  bonds, it would have one pair of non-bonding electrons, and a p electron left over to participate in a  bond with one of the  bonds
• This is different from what the Lewis structure shows, and from our prediction of hybrid orbitals from the expected geometry
• If we look at the sp3 O atoms above we see that they actually have 7 electrons (1 more than expected), while the sp2 O atom has the expected 6. Furthermore, the N atom (in the correct sp2 configuration) has 4 electrons (1 less than expected)
• The "extra" electron from the ionic charge is correctly accounted for in the summation of electrons
Thus, the correct way to determine electron configurations appears to be:
• begin by predicting the hybridization orbitals
• then determine lone pair arrangements and s and p bonding electrons for each atom
• confirm that all bonding electrons are correct and that the total of electrons is correct
________________________________________
1996 Michael Blaber