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The Complete Chem 120

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by: Leslie Pike

The Complete Chem 120 CHEM 120

Leslie Pike
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This is an entire semester's worth of lecture notes. It is divided into four sections; this will (roughly) be the material that is covered in Exam 1, 2, 3, and 4, respectively. Exams will vary by p...
College Chemistry I
Dr. Darwin Dahl
Chemistry, Chem 120, chem120
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Date Created: 01/19/16
Chem 120 Complete Study Guide Section I SI units and conversions between them There are seven SI base units: • Meter: length • Kilogram: mass, NOT WEIGHT, the two mean two very different things • Candela: brightness of light • Ampere: current flow • Mole: the number 6.022*10 23 • Second: time • Kelvin: heat Important note: the liter is NOT one of the seven base units! One milliliter is equal to 1 cubic centimeter. (Cubic centimeter can be abbreviated cc.) The prefixes for SI units: • Tera • Giga • Mega • Kilo • (none) • Milli • Micro • Nano • Pico • Femto • Atto 3 0 3 -3 Each of these differ by 10 , with (none) being 10 , kilo being 10 , etc. Milli is 10 , Micro is 10 , et cetera. Exceptions to this rule are hecto, deka, deci, and centi (100, 10, 0.1, 0.01). Nomenclature For diatomic compounds: For nonmetal-nonmetal compounds, and metalloid-nonmetal compounds, USE the prefixes mono-, di-, tri-, etc. to indicate number of atoms. Example: N O2wo4ld be dinitrogen tetroxide. Use a periodic table to determine which elements are nonmetals or metalloids. For metal-nonmetal compounds, do NOT use the prefixes mono-, di- tri-, etc. There are two different ways to name metal compounds, depending on whether or not the metal is a fixed oxidation-state metal or is not fixed. How to determine the difference:  All metals in the first column of the periodic table (ignore hydrogen) have a fixed oxidation state of +1, no exceptions.  All metals in the second column of the periodic table have a fixed oxidation state of +2, no exceptions.  The metals aluminum, cadmium, silver, and zinc have fixed oxidation states of +3, +2, +1, and +2, respectively.  ALL OTHER metals have variable oxidation states. When naming a compound containing a metal with variable oxidation state, indicate the oxidation state with roman numerals. Example: FeO would be iron(II) oxide. Fe2O 3ould be iron(III) oxide. Determining the oxidation state requires knowledge of the charge of the anion to which the metal is bonded (in the previous example, oxygen has a charge of -2). The halogens column (fluorine, chlorine, bromine, iodine) will have a charge of -1. The chalcogen column (the one right before the halogen column) has a charge of -2. The charge of polyatomic anions varies depending on the particular anion; these just have to be memorized. Fixed oxidation state metals are named the same way, except you can leave off the roman numerals because the charge is known. Example: Zn(C H O ) is written zinc 2 3 2 2 acetate. There is no need to write zinc(II) acetate because zinc always has a charge of +2. For compounds containing polyatomic anions: Name the metal first, list its oxidation state if necessary, then write the name of the polyatomic anion. Dr. Dahl has posted a list of common polyatomic anions to Blackboard. (I am not allowed to copy and paste it into this document because that would be considered plagiarism; you will have to download it from Blackboard. I strongly suggest you do so, if you have not done so already.) Names of Important People  Democritus: coined the name “atom”; proposed that matter was composed of small, indivisible particles  Lavoisier: studied conservation of mass; burned things in jars and weighed the jar before and after to prove that no mass had been lost  Mendeleev: Periodic table  Stoney: discovered electrons  Rontgen: X-rays  JJ Thomson: charge to mass ratio of electron. IMPORTANT!!!  Rutherford: alpha and beta radiation; gold foil experiment, discovery of the atomic nucleus  Curie: radioactivity  Millikan: charged oil drop experiment, determined the weight of the electron  Moseley: modern-day periodic table  Aston: Mass spectrometer  Bohr: electron shells  De Broglie: electrons behave as particles and waves both Unit Conversions to Memorize  1.00 pounds = 454 grams  1.00 quarts = 946 mL (approximately 1 L, but 1qt=1L is usually not precise enough)  1.00 cubic centimeters = 1.00 mL (this is an EXACT relationship)  1 mile = 1.609 kilometers (this will save you lots of arithmetic and paper space!) 1 Angstrom = 10 -1meters Miscellaneous WHERE TO ROUND: There are two different types of numbers: ratios and measurements. A ratio would be 60 minutes per hour, or 5280 feet per mile. These numbers are EXACT values, and they are considered to have an infinite number of significant figures. Measurements would be an iron ingot which weighs 3.52 grams. This number is NOT exact, because a less sensitive scale may read 4 grams and a more sensitive one 3.523681 grams. These numbers do NOT have an infinite number of significant figures. These numbers are usually written in scientific notation, and the number of -14 digits is consider23 the number of significant figures. Ex: 2.33*10 has 3 significant figures. 6.022*10 has four significant figures. When doing multiplication and division problems, round your answer to the same number of significant figures that are in the number with the fewest number of significant figures. When doing addition and subtraction problems, round to however many decimal places in the number with the fewest decimal places. HOW TO ROUND: (in these cases, we are rounding to 1 decimal place)  Less than 5, round down. 5.42 rounds to 5.4. 7.6327 rounds to 7.6  Greater than 5, round up. 5.57 rounds to 5.6. 3.285 rounds to 3.3  Equal to five, round down if number to the left is even, and round up if number to the left is odd. 3.65 rounds DOWN to 3.6. 3.35 rounds UP to 3.4 An extensive property can be changed by adding more mass. Weight, volume, length are extensive An intensive property cannot be changed by adding more mass. Color, density are extensive A physical property can be determined WITHOUT changing the nature of the substance, i.e. color. A chemical property CANNOT be determined without changing the nature of the substance, i.e. reactivity. An empirical formula tells RATIOS rather than actual number of atoms (although, depending on the molecule, it may tell that as well). The empirical formula for glucose (and most other sugars) is CH O2 because most sugars contain carbon, hydrogen, and oxygen in a 1-2-1 ratio. Section II Moles An atomic mass unit is one-twelfth of the mass of one atom of carbon-12. T23 masses of atoms are measured in atomic mass units. There are 6.022*10 amus in one gram. The molar mass is the mass of one mole of a substance. One mole is 6.022*10 23 units of a substance. A mole is a number, like a pair or a dozen or a gross. You can have a dozen toothbrushes or a gross of toothbrushes or a mole of toothbrushes (technically speaking, you would have nowhere to store one mole of toothbrushes, but in theory you could have that many). The mole is special because one mole of a substance weighs the same number in grams as one molecule weighs in amus. This is very convenient when measuring proportions for a chemical reaction. For instance, diatomic hydrogen weighs 2 amus. One mole of diatomic hydrogen weighs 2 grams. Diatomic oxygen weighs 32 amus. One mole of diatomic oxygen weighs 32 grams. In order to make one mole of water, you would have to weigh out 2 grams of hydrogen and 16 grams of oxygen. (Remember, hydrogen and oxygen combine in a 2:1 ratio to make water.) As a general rule, the molecular mass listed on the periodic table is used for the molar mass of the substance. EXCEPTIONS TO GENERAL RULE: hydrogen, nitrogen, oxygen, fluorine, chlorine, bromine, and iodine. These seven elements exist in gaseous form at room temperature. As gases, they form diatomic molecules, UNLIKE noble gases which exist as single atoms. Therefore, a mole of nitrogen gas will weigh 28 grams, NOT 14 grams. However, for percent composition problems, you will NOT use the molecular weight; you will use the atomic weight. (Compounds are composed of atoms, not molecules.) Be sure to pay very close attention to which molar mass (molecular or atomic) the problem asks you for! Example mole problem: You are performing a chemical reaction which requires two moles of sucrose (the scientific name for ordinary table sugar). Obviously, you 23 cannot count out 12.044*10 sucrose molecules. You will have to weigh the sucrose to determine the amount you need. To determine what one mole of sucrose weighs, first determine what one molecule of sucrose weighs. The formula for sucrose is C12 O22 11 has twelve carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms. Add up the masses of these atoms to get the weight of the entire molecule, 342 amu. Therefore, one mole of sucrose will weigh 342 grams, and two moles will weigh 684 grams. Conversely, you buy a 5-pound bag of table sugar. You want to know how many molecules of sucrose you have in the bag. You are doing a conversion problem; you have pounds and you want to end up with molecules. First, convert pounds to grams. There are 454 grams in one pound, so in 5 pounds there are 2270 grams. Then, convert grams to moles. One mole of sucrose weighs 342 grams, so divide 2270 grams substance by 342 grams per mole to get the number of moles, 6.64. 23 Now, to get the number of molecules. There are 6.022*10 molecules in one mole. Therefore, in 6.64 moles, you will have about 4*10 24molecules. (This problem should be more hideous than anything you will see on a Chem 120 exam.) Percent Composition Say you are given an unknown compound that was analyzed in a lab and found to contain 2% hydrogen, 33% sulfur, and 65% oxygen. You will need to find the moles- to-moles ratio of the atoms to identify the compound. Convert these percentages to moles. To do this, assume you have 100 grams of the substance. This means that two grams will be hydrogen, 33 grams will be sulfur, and 65 grams will be oxygen. Convert grams to atomic moles (NOT molecular moles, this is where percent composition problems and stoichiometry problems differ). Atomic hydrogen weighs one gram per mole, so two grams of hydrogen is two moles. Sulfur weighs 32 grams per mole; you have 33 grams in our sample, giving you just a little more than a mole. Atomic oxygen weighs 16 grams per mole; you have 65 grams; which is just a little more than four moles. Our empirical formula would be H SO2, w4ich is sulfuric acid. This problem is made a little bit easier by the fact that one mole of sulfuric acid weighs 98 grams, almost exactly the 100 grams assumed for the calculation; however, the concept is the same for all percent composition problems. Conversely, to determine percent composition from given ratios, first determine the molar mass of the compound. Then, divide the weight of the atom in the compound by the weight of the entire compound. For example, iron(III) oxide weighs 158 grams/mole. There are two moles of iron in every iron(III) oxide mole. Iron has a molar mass of 55 g/mol, so divide 110/158 to get 69.6% iron by weight in iron(III) oxide. (NOTE: Due to rounding, the percent values may not exactly total 100%, and the ratios may not quite be whole numbers. If you get close, and your answer makes sense, you’ve done it correctly.) One way to determine percent composition of organic compounds (compounds containing only carbon, hydrogen, and possibly oxygen) is by combustion analysis. Combustion analysis burns an organic compound in an atmosphere full of oxygen. The resulting water and carbon dioxide are weighed. To calculate the molecular formula of the compound, first take the weight of carbon dioxide and water, and convert this into moles. (Water weighs 18 grams per mole, and carbon dioxide weighs 46 grams per mole.) Then, do moles-to-moles conversion (two moles of hydrogen in one mole of water, one mole of carbon in one mole of carbon dioxide). You now have your ratio of carbon to hydrogen. IMPORTANT NOTE: YOU CANNOT USE THIS SAME PROCEDURE TO CALCULATE THE PERCENTAGE OF OXYGEN!!!! THE COMPOUND IS BURNED IN AN ATMOSPHERE OF OXYGEN. THIS MEANS THE PRODUCTS WILL HAVE MORE OXYGEN THAN THE ORIGINAL COMPOUND DID, AND YOU HAVE NO MEANS OF TELLING HOW MUCH MORE JUST BY LOOKING AT THE NUMBERS. To calculate the amount of oxygen, convert your carbon and hydrogen back to grams. If this total weight is less than the weight of the original sample, the remainder of the weight will be oxygen. Convert this weight to moles (atomic oxygen weighs 16 grams per mole). Stoichiometry Stoichiometry can be defined as the relationships between reactants and products in a chemical equation. Stoichiometry problems start with a chemical equation (usually it will not be balanced and you will be expected to balance it). For example, the decomposition of potassium chlorate into potassium chloride and oxygen gas: KClO 3 KCl + O 2 This equation is unbalanced; there are three oxygen moles on the left and two on the right. Dr. Dahl did not explain this in class, but there is a shortcut to balancing equations (and because the exam is multiple choice, you will not be penalized for using the shortcut): KClO 3 KCl + 3/2O 2 Ta-da, there are now three oxygen moles on both sides. (Some professors do not like fractions; some are fine with them. To be safe, multiply everything by the denominator to remove the fraction.) Now you have dealt with the equation. On to the rest of the problem. Stoichiometry problems give you the mass of one reactant, and ask you how many grams of product can be produced from this reactant. Alternatively, you may be given the mass of the product, and be asked how many grams of reactant are needed. (If you know how to convert grams to moles, good. If you don’t, go back to the previous section and read it until you know how to convert grams to moles. This is important knowledge; I will be very surprised if it DOESN’T show up on the exam.) Sometimes, the product will be in aqueous solution, and you will be dealing with molars instead of grams. Regardless of which you are given, CONVERT IT TO MOLES. Converting molars to moles is actually simpler than converting grams to moles. Multiply the molarity of the solution by the liters of solution. The volume will cancel, leaving you with the moles. (You may be given the volume in milliliters instead of liters. In this case, just move the decimal point three places to the left. No big deal.) Specific Gravity Specific gravity is measured in units of grams per cubic centimeter (or grams per mL). By definition, water has a specific gravity of 1. Everything else is based off of that. massof solution(g) SpecificGravity= volumeof solution(mL) Calculating molarity from specific gravity is yet another factor-label problem. Sample problem: what is the molarity of a 35% hydrochloric acid solution with a specific gravity of 1.2? (Hint: in a 100 g sample that is 35% HCl by weight, 35 g will be HCl and 100 g will be the weight of the entire solution.) 1.2gsolution∗35gHCl mL solution 100gsolution ∗mol HCl ∗1000mL 36.45gHCl L =11.5M HCl Parts per Million (ppm) The concentration of very dilute solutions is measured in parts per million (ppm). Definition as follows: massof solute(mg) ppm= mass of solution(kg) Note: the solute weighs so little that “mass of solvent” and “mass of solution” can be used interchangeably here. If the solvent is water, we can take a shortcut, because by definition water weighs one kilogram per liter: ppm= massof solute(mg) volumeof solution(L) Sample problem: You need to prepare a 300 mL solution of 50ppm Cl. All you have available is barium chloride and water. How much barium chloride (in grams) do you need? 50mgCl ∗0.3L L ∗gCl 1 1000mgCl ∗molCl ∗mol BaCl 2 34.54gCl 2molCl ∗207.9gBaCl 2 =0.044gBaCl 2 mol BaCl 2 This gives you the weight of barium chloride. Dilutions The relationship of dilutions is defined as follows: M i iM V f f This is a plug-and-chug type of problem. Plug in the knowns, and use algebra to solve for the unknown. Acids, Bases, Salts, and Precipitates An acid is a molecule that can lose a proton. A base is a molecule that can gain a proton (or lose a hydroxyl group). Strong acids and bases dissociate completely in water; weak acids and bases only dissociate a small percent in water. The six strong acids are:  Hydrochloric acid, HCl  Hydrobromic acid, HBr  Hydroiodic acid, HI  Sulfuric acid, H2SO 4  Nitric acid, HNO 3  Perchloric acid, HClO 4 Most bases are strong, with the exception of ammonia, NH . Note: AMM3NIA is different from AMMONIUM. The latter has four hydrogens and a charge of +1. An acid-base reaction involves a proton transfer. A neutralization reaction turns an acid and a base into water and a salt. A precipitation reaction (as seen in chem lab) involves the exchange of anions and the formation of a non-soluble product, which precipitates out of solution. Most compounds containing silver, lead, and mercury are insoluble. All compounds containing alkali metals, alkali earth metals, and nitrates are soluble. The types of equations we have seen so far, such as the potassium chlorate one in a previous section, are molecular equations. Ionic equations are just like molecular equations, but with all aqueous compounds (unless they are weak electrolytes) broken into ions. (A weak electrolyte is an acid, base, or salt that does not dissociate fully in water. A strong electrolyte is an acid, base, or salt that dissociates fully in water.) A net ionic equation is an ionic equation, with the spectator ions removed. They can be computed by working “backwards” in many cases: find the product that is a solid, liquid, gas, or weak electrolyte, and remove everything else from the equation except the cation and anion that form the non-dissolved product. For neutralization reactions, the net ionic equation will almost always be the combining of a proton with hydroxide to form liquid water. Oxidation-reduction reactions These reactions involve transfer of electrons. When an element is oxidized, it loses electrons. When an element is reduced, it gains electrons. Remember this with the term “oil rig”—oxidation is losing, reduction is gaining. When an element is oxidized, the element taking its electrons is called the oxidizing agent. (Makes logical sense.) Same is true for reduction. Section III Chapter 5: Gases  Boyle’s law: PiVi=PfVf  Charles’ law: Vi/Ti=Vf/Tf Note: TEMPERATURE MUST BE IN KELVINS FOR THIS TO WORK!!!  Avogadro’s law: Vi/ni=Vf/nf  Combined gas law: PiVi/niTi=PfVf/nfTf  Ideal gas law: PV=nRT, where R=0.0821L*atm/(mol*K)  STP=0C and 1 atm  Volume of a gas at STP is 22.414L/mol.  Modified ideal gas law: Pmm=(rho)RT where mm=molar mas (g/mol) and rho=density (g/L)  Partial pressures: Pi=XiPt Example problem: sample of natural gas at 1.37 atm. Contains 8.24 mol methane, 0.421 mol ethane, 0.116 mol propane. What is the partial pressure of the propane? Xpropane0.116/(8.24+0.421+0.116)=0.0132 Ppropane.0132*1.37atm=0.0181atm Note: the ideal gas law is PV=nRT. At normal temperatures and pressures, ALL gases obey this law. Gases deviate from this law at extremely high pressures and low temperatures. Also, this law no longer applies once a gas becomes a liquid. We assume for all problems worked in class that a) the gas does not deviate from this law, and b) the gas does not become a liquid. Sample problem: You have a 1.5 g impure sample of potassium chlorate. You heat it and it decomposes into potassium chloride and oxygen. You collect 30 mL of oxygen gas, over water. The temperature is 20 C. The pressure is 730 mmHg. Calculate the percentage of potassium chlorate in the sample. Balanced reaction: 2KClO  3KCl + 3O 2 First, realize that this is a stoichiometry problem. You need to find the number of moles of oxygen gas that you collected, so that you can find the number of moles of potassium chlorate that you originally had. Convert this to grams, and divide by the grams of sample to get the percentage. To find the number of moles of oxygen, use the ideal gas equation, PV=nRT. Solving for n gives us: n=PV /RT To determine the pressure of the oxygen (remember that your container will have a combination of oxygen and water vapor, because you collected over water), use the law of partial pressures. P =P + P t oxygen water Pwateran be found on the table that Dr. Dahl will provide you with. 730=P oxygen 17.5. Poxygen712.5 mmHg. Now, we can plug and chug. Remember that pressure must be in atm, volume must be in L, and temperature must be in K. 712.5 ( 760 )(0.030) n= =0.001169molO 2 0.0821(273+20) The rest is stoichiometry. O ∗2molKClO 2 3∗122gKClO 3molO 2 3 0.00169mol =0.095gKClO 3 molKClO 3 Lastly, find the mass percentage. 0.095/1.5=6.3%. Gas equations Equation for velocity of a gas: v= √RT /M Here, R=8.314. M=molar mas, in KILOGRAMS per mole. NOT GRAMS!!! T=temperature in Kelvin. Kinetic energy The kinetic energy equation is 0.5mv . (This equation itself is not important for Chem 120.) From this is derived the equation used to compare the velocities of two gases: v= m 2 v2 √m 1 The kinetic energy of a gas is directly proportional to its temperature. KE 1=T 1 KE 2 T 2 Therefore, to get the relationship the kinetic energy of gas 1 and the kinetic energy of gas 2, divide the temperature of gas 1 by the temperature of gas 2. The rate-of-effusion equation that compares two different gases is similar to the velocity/mass relationship. Rate1= molar mass2 Rate2 √ molar mass1 For example, say it takes 2 liters of hydrogen gas (molar mass 2 g/mol) 5 minutes to escape from a container. How long would it take 3 liters of helium gas (molar mass 4 g/mol) to escape from the same container? The hydrogen escapes at a rate of 2L/5min or 0.4L/min. The helium is three liters, the number of minutes is unknown. Filling in the equation: (0.4L/min)/(3L/x min)=sqrt((4g/mol)/(2g/mol)) Three liters of helium would take 10.6 minutes to escape the container. Ch. 6: Thermochemistry The total heat change of the system is equal but opposite to the total change of the surroundings. For example, when a system gains 100 kJ, the surroundings lose 100 kJ and vice versa. The total energy change of a system is equal to the heat exchange plus the amount of work done. Memorize this equation: Work = Force * Distance. In the case of an expanding gas: Work = -(Outside pressure * change in volume). Important Concept Chemical equations can be written to contain a heat change, along with the moles of products and reactants. Saying that a reaction requires 100 kJ of heat is NO DIFFERENT from saying that a reaction requires five moles of oxygen gas or 2 moles of sodium chloride. When doing stoichiometry problems, the heat is treated just like the compounds are treated. Additionally, the heat can be written outside of the equation, as a delta H at the end. If delta H is negative, this means that the heat belongs on the PRODUCTS side. If delta H is positive, the heat belongs on the REACTANTS side. When a chemistry problem says that heat is “evolved”, this means that the heat is given off, and therefore belongs on the REACTANTS side of the equation. The specific heat of a substance is the amount of heat required to raise one gram of that substance one degree Celsius (or Kelvin, it makes no difference). The specific heat of water is 4.184 J/(g * K). YOU MUST MEMORIZE THIS ONE. Dr. Dahl will provide all other specific heats. The heat capacity of a substance is the heat required to raise a given amount of a substance 1 degree C. The standard heat of formation of a substance is the delta H when one mole of that substance is formed from its elements at 1 atm. THE STANDARD HEAT OF FORMATION OF A SUBSTANCE IN ITS ELEMENTAL STATE IS ALWAYS ZERO. This is because no heat change takes place; the substance is already in the elemental state. The standard enthalpy of a reaction is calculated as follows: (sum of delta H of products)-(sum of delta H of reactants). Adding equations Chemical equations can be added. This is quite useful when determining the delta H of a reaction in which none of the reactants are in their elemental state. In a typical problem, you will be given two or three equations, and a “final” equation. Your job is to cancel terms in the equations so that they add to the “final” equation. Equations can be multiplied by a constant; in this case, delta H is also multiplied by a constant. (Remember, delta H is the SAME as the moles of a reactant or product!) Equations can be reversed. For instance, if heat was on the product side, it would now be on the reactant side, causing a sign change in delta H. Once you have multiplied and/or reversed the equations so that terms not included in the “final” equation cancel out, the delta Hs will add, to give you the final delta H. Section IV Chapter 7: Quantum Theory and Wavelengths Very important equation for calculating frequency and wavelength of light waves Speed=Frequency∗Wavelength For all light waves, speed is a constant, 3*10 m/s. Given speed and frequency, one can calculate wavelength, or given speed and wavelength, one can calculate frequency. Frequency is the number of wavelengths of that wave that pass a fixed point in one unit of time. Frequency is represented by the Greek letter nu, which looks like the English letter v. A high frequency means that more wavelengths pass the fixed point per time interval. High frequency is equivalent to high energy. The energy of a light wave is equal to Planck’s constant, 6.63*10 -3J*s, multiplied by frequency. The unit of frequency is Hz, or 1/s. If something has a frequency of 100,000 hertz, that means that 100,000 wavelengths pass through a point in one second. Wavelength is self-explanatory, the length of a wave from crest to crest or trough to trough. Wavelength is represented by the Greek letter lambda. Long wavelengths correspond to low energy, and short wavelengths correspond to high energy. Wavelength is often measured in angstroms. One angstrom is 10 -10meters. Niels Bohr came up with the energy shell model for the atom. This is NOT the Rutherford planetary model, and the electrons do NOT “orbit” the nucleus like planets orbiting the sun. Electrons float around in their set energy levels. Light is emitted when an electron jumps from a higher energy level to a lower energy level. Light is absorbed when an electron jumps from a lower energy level to a higher energy level. The energy of an electron is equal to the negative Rydberg constant (Dr. Dahl will provide this on exams) divided by the quantum number squared. The change in energy when an electron changes states is described by the following equation: 1 1 ∆ E=R (H − ) ni2 nf2 Note: THIS IS BACKWARDS from the way these normally work. In this case, we have INITIAL MINUS FINAL instead of the other way around. The Balmer series is when an electron jumps from a higher energy state to n=2. Light in the visible spectrum is emitted during this jump. The Schrodinger wave equation describes the probability of where an electron will be at any given time. This equation itself is beyond the scope of a 100-level class; however, you will be expected to know its four variables (n, l, m , and l ). s  N=principal quantum number. It can be any positive integer. It represents the energy level, or the shell.  L= angular momentum quantum number. It can be any whole number, up to N-1. It represents whether the subshell is s, p, d, or f. The s subshell has an angular momentum quantum number of 0, p is 1, d is 2, f is 3. IMPORTANT THING TO WATCH OUT FOR: the s subshell is ZERO, not ONE, p is 1, etc. Dr. Dahl will probably try to trip you up on this.  M=lagnetic quantum number. It ranges from –l to 0 to l, it can be any integer value between these values. It represents the orbital, for example, p , x p y and p z  M ss the spin. It can be +1/2 or -1/2. Chapter 8: Electron configurations The ground state of an atom means that the atom is not an ion; it has an equal number of protons and electrons. The ground state also means that there are no excited electrons. The ground state electron configuration for magnesium is 1s 2s 2p 3s . This can be abbreviated as [Ne]3s , because ground state magnesium has a neon core and then an extra 3s subshell. When subshells fill, one electron is put in each orbital first. Then, a second electron is put in each orbital. Subshells fill in order of energy level. The (n-1)d subshell fills before the np subshell; 4s fills, then 3d fills, then 4p fills. There are exceptions to this rule, namely, when the d subshell has 4 electrons and when the d subshell has 9 electrons. In these cases, one of the electrons in the s subshell will go back into that last d orbital. When subshells empty, they empty from the highest n-value first. 4s empties BEFORE 3d empties. NOTE: THE ORDER IN WHICH SUBSHELLS EMPTY DOES NOT NECESSARILY MATCH THE ORDER IN WHICH SUBSHELLS FILL!!!!!!! If n-values are the same, the atom loses from the highest l-value first. 3d disappears before 3p and 3s. A paramagnetic atom has at least 1 unpaired electron. A diamagnetic atom has no unpaired electrons. Noble gases are diamagnetic, as are ground-state alkali earth metals. A good way to remember this is to remember that “di” means two, and in a “diamagnetic” atom each orbital has two electrons in it. All elements want to achieve noble gas configuration. This means that they have all of their shells full. When alkali metals exist in the +1 state, they have the same configuration as the noble gas right behind them. Na has the same electron configuration as [Ne]. Some atoms achieve pseudo noble gas configuration instead; this means that they have the noble gas core and also a filled (n)d subshell. When halogens exist in the -1 state, they have the same electron configuration as the noble gas in front of them. F has the same electron configuration as [Ne]. Elements with the same electron configuration are known as isoelectronic. Ne, Na , + F , and O are all isoelectronic. The main group, aka representative group, consists of the s-group and the p-group, i.e. IA, IIA, IIIA, IVA, VA, VIA, VIIA, and VIIIA. The main group does NOT contain the d- block or the lanthanides or the actinides. Non-main-group elements may not necessarily obey the following periodic table trends. Periodic table trends, and why:  Atomic radius increases from top to bottom and right to left. Reason: a new shell is added each period, and each shell is bigger than the previous shell. Also, as atomic nuclei increase in charge, they draw the electrons in closer, causing shell size to shrink.  Metallic trend increases from top to bottom and right to left. This is an easy one to remember; the periodic table is labeled in terms of metals, metalloids, and nonmetals.  Electron affinity, or the amount of energy required to remove an added electron, increases from bottom to top and left to right. Reason: the fewer n- levels an atom has, the closer the valence shell is to the nucleus, and the more strongly the nucleus attracts those valence electrons. Also, the closer the shell is to full, the more strongly the atom feels its missing electrons and the worse it wants them.  Electronegativity is how much that atom hogs the shared electrons when it bonds with another atom. Same trend as electron affinity, and for the same reasons. NOTE: The electronegativity trend, for the first two periods of the p- block, is that the top period has higher electronegativity than the bottom period when the diagonal is drawn from left to right, top to bottom. It is the other way around the rest of the time.  Reactivity. This is a weird trend. For alkali metals and alkali earth metals, the reactivity increases down the column. This is because as the valence shell gets farther from the nucleus, it is less attracted by the nucleus and those outer shell electrons are more easily lost. For halogens and chalcogens (the fluorine and oxygen columns), reactivity decreases down the column. This is because, as the valence shell gets farther from the nucleus, it is less attracted by the nucleus and the atom has less of a tendency to steal other atoms’ electrons. Chapter 9: Ionic and Covalent bonding If two atoms participating in a bond have an electronegativity difference of 0, the bond is nonpolar covalent. The atoms share the electrons evenly. If the electronegativity difference is between 0 and 1.7, the bond is polar covalent (they technically share electrons, but the more electronegative atom hogs the electrons most of the time). If the electronegativity difference is greater than 1.7, the bond is ionic, meaning that the more electronegative element completely steals the electrons. The lattice energy is the energy required to completely separate one mole of solid ionic compound into gaseous ions. Smaller atoms have a greater lattice energy. Ions with a larger charge have a higher lattice energy. A Lewis structure is a diagram of bonds and electrons in a molecule. Molecules want to be symmetrical as much as possible; keep this in mind when drawing a Lewis structure. Any two bonding atoms can form single bonds. Carbon, oxygen, nitrogen, phosphorous, and sulfur can form double bonds. Carbon and nitrogen can form triple bonds. As the bond order increases (second-order bond is a double bond and third-order bond is a triple bond), the length of the bonds decreases. Polar covalent bonds are symbolized by an arrow (dipole). The arrow points toward the more electronegative element. Polar covalent bonds do not necessarily mean that the molecule is polar; often the dipoles cancel each other out. Resonance is what happens when more than one equivalent Lewis structure can be drawn for a molecule. Examples: benzene, ozone, carbonate. In these cases, the atoms “share” the extra bond(s). The symbol for resonance is  , an arrow with two ends. The formal charge of an atom is calculated by counting a lone pair for 2- and a bonding pair for 1-. For an atom with a formal charge of 0, the – charge equals the + charge from the nucleus. When multiple Lewis structures can be drawn for a molecule, the correct Lewis structure is the one in which the atoms have the smallest formal charges. When multiple Lewis structures with the same formal charges can be drawn for a molecule, the most electronegative atom should have the negative charge (or the most electropositive atom should have the positive charge, etc.) Exceptions to the octet rule: Hydrogen and helium only have two electrons in their outermost shell. Alkali and alkali earth metals do not have enough electrons in their valence shell to form enough bonds to fill an octet. Instead, they form bonds with the valence electrons that they have, and they do not have full octets. This same situation can happen for IIIA elements, such as boron. Molecules with a principle quantum number greater than 2 can form extra bonds, on account of the d-orbital. Example: sulfur hexafluoride. Molecules involving noble gases typically break the octet rule, on account of the noble gas already having a full octet. The noble gas will have extra electron pairs. If the noble gas has a d-subshell, the extra pairs are not a problem. Argon, krypton, xenon, and radon can form bonds. Helium and neon CANNOT form bonds because they do not have a d-shell. Molecules with odd numbers of electrons break the octet rule. Example: nitrogen monoxide. Nitrogen and oxygen are double-bonded together, oxygen has two pairs of nonbonding electrons, nitrogen has a nonbonding pair and a lone electron. The delta H for a reaction can be calculated by subtracting the energy of all of the bonds of the products from the energy of all of the bonds of the reactants.


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