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CHEM 1030 Exam 3 Study Guide

by: Alyssa Anderson

CHEM 1030 Exam 3 Study Guide CHEM 1030

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This is the study guide for Exam 3, which covers everything from molecular geometry, hybridization, chemical reactions, aqueous solutions, acids/bases, oxidation, thermodynamics, and so on. It incl...
Fundamentals Chemistry I
Dr. Streit
Study Guide
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This 22 page Study Guide was uploaded by Alyssa Anderson on Wednesday April 20, 2016. The Study Guide belongs to CHEM 1030 at a university taught by Dr. Streit in Spring 2016. Since its upload, it has received 78 views.

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Date Created: 04/20/16
1 CHEM 1030 EXAM 3 STUDY GUIDE Molecular Geometry A. Molecular shape can be predicted by using the valence-shell electron-pair repulsion (VSEPR) model (AB ) X B. A is the central atom surrounded by X B atoms C. X can have integer values of 2 to 6 D. The basis of the VSEPR model is that electrons repel each other E. Electrons are found in various domains, either in lone pairs, single bonds, double bonds, or triple bonds F. Electrons will arrange themselves to be as far apart as possible, so as to minimize repulsion interactions G. The electron domain geometry is the arrangement of electron domains around the central atom —> 2 electron domains is linear —> 3 electron domains is trigonal planar —> 4 electron domains is tetrahedral —> 5 election domains is trigonal bipyramidal —> 6 electron domains is octahedral H. The molecular geometry is the arrangement of bonded atoms I. In an ABXmolecule, a bond angle is the angle between two adjacent A-B bonds J. AB Xolecules contain two different bond angles between adjacent bonds 1. Axial positions are perpendicular to the original plane 2. Equatorial positions are the bonds on the same plane K. When the central atom in a AB Xmolecule bears one of more lone pairs, the electron- domain geometry and the molecular geometry are no longer the same 2 3 The steps to determine the electron-domain and molecular geometrics are as follows: Step 1: Draw the lewis structure Step 2: Count the number of electron domains on the central atom Step 3: Determine the electron density geometry by applying the VSEPR Step 4: Determine the geometry by considering the positions of the atoms only Worked Example 7.1 Problem: Determine the shapes of SO3 and ICl4- Solution: With SO3, there are two single bonds and one double bond, so there are 3 domains, which should be arranged as a trigonal plane. With ICl4-, even though there are 4 outside atoms, there are two lone pairs so it is square planar, not tetrahedral. Deviation from ideal bond angles A. Some electron domains are better than others at repelling neighboring domains B. Lone pairs take up more space than bonded pairs C. Triple bonds are stronger than double which are stronger than single bonds Molecular Geometry and Polarity A. Molecular polarity is one of the most important consequences of molecular geometry B. A diatomic molecule is polar when the electronegativities of the two atoms are different C. The polarity of a molecule made up of three or more atoms depends on: 1. The polarity of the individual bonds 2. The molecular geometry D. The bonds in H2O are polar and the molecule is polar E. The bonds in CCL4 are polar but the molecule is non-polar F. Thebonds in CHCl3 are polar and the molecule is polar G. Dipole moments can be used to distinguish between structural isomers Note: Trans- and cis- can be determined by knowing if it’s non-polar and polar 4 Geometry of Molecules with More Than One Central Atom A. The geometry of more complex molecules can be determined by treating them as though they have multiple central atoms B. Write out the information of each central atom (number of domains, shape, etc.) Intermolecular Forces A. Intermolecular forces are attractive forces between neighboring molecules B. These faces are known collectively as van der Waals forces Dipole-Dipole Interactions A. Dipole interactions are attractive forces that act between polar molecules B. They are arranged differently according to their physical state Hydrogen Bonding A. Hydrogen bonding is a special type of dipole-dipole interaction B. Hydrogen bonding only occurs in molecules that contain H bonded to a small, highly electronegatively charged atom Disperson Forces A. Also called London Dispersion Forces B. Result from the Columbia attractions between instantaneous dipoles of non-polar molecules Worked Example 7.4 Ion-Dipole Interactions A. Ion-dipole interactions are Coulombic attractions between ions (either positive or negative) and polar molecules B. There are stronger and weaker attractions 5 Valence Bond Theory A. According to valence bond theory, atoms share electrons when atomic orbitals overlap B. The shape is just a region around the nucleus where you can find electrons C. Bonds happen when they OVERLAP D. Rules 1. A bond forms when single occupied atomic orbitals on two atoms overlap (occupy the same region of space) 2. The two electrons shared in the region of orbital must be of the opposite spin F. The H-H bond in H2 forms when the singly occupied 1s orbitals of the two H atoms overlap G. The F-F bond in F2 forms when the singly occupied 2p orbitals of the two F atoms overlap H. A covalent bond will form if the potential energy of the molecule is lower than the combined potential energies of the isolated atoms I. Note: The two electrons shared between the atoms must have a positive counterpart Hybridization of Atomic Orbitals A. Hybridization or mixing of atomic orbitals can account for observed bond angles in molecules that could not be described by the direct overlap of atomic orbitals B. Example: BeCl2. 1. Lewis structure and VSEPR theory predict Cl-Be-Cl (linear geometry) with a bond angle of 180* 2. Beryllium cannot form two bonds because it has a full 2s2 set 3. The two bonds formed would not be equivalent because it would become 2s1/2p1 4. Experimentally, the bonds are identical in length and strength. 5. Mixing one s orbital and one p orbital to yield two sp orbitals 6. The hybrid orbitals on Be each overlap with a singly occupied 3p orbital on a Cl atom 6 7. The mathematical combination of the quantum mechanical wave functions for an s orbital and a p orbital has one small lobe and one large lobe, and like any two electron domains on an atom, they re oriented in opposite directions with a 180* angle between them. 8. With two sp hybrid orbitals, each containing an unpaired electron, we can see how the Be atom is able to form two identical bonds with two Cl atoms. 9. Each of the singly occupied sp hybrid orbitals on the Be atom overlaps with the singly occupied 3p atomic orbitals on the Cl atom. C. Example: BF3. 1. Draw the Lewis diagram and count the number of electron domains. 2. Combine the necessary number of atomic orbitals to generate the required number of hybrid orbitals 3. Place electrons in the hybrid orbitals, putting one election in each orbital before pairing any electrons 4. The ground-state electron configuration of the B atom is [He] 2s2/2p1, containing just ONE unpaired electrons needed to explain the formation of three bonds 5. The ground-state and excited-state electron configurations can be represented as 2s2/2p1 —>(promoted to) 2s1/2p2 6. Because the three ones in BF3 are identical, we must hybridize the three singly occupied atomic orbitals (the one s and 2 p orbitals) to give the singly occupied hybrid orbitals: 2s1/2p2 —> (hybridization) sp2/sp2/sp2 7. Mixing of one s orbital and two p orbitals to yield three sp orbitals D. Example: CH4. 1. Draw the ground state orbital diagram for the central atom 2. Combine the necessary number of atomic orbitals, putting one election in each orbital before pairing any electrons. 3. Promote electrons from 2s2/2p2 —> 2s1/2p3 4. Hybridize the carbon atom from 2s1/2p3 —> sp3/sp3/sp3/sp3 5. Mixing of one s orbital and three p orbitals to yield four spy orbitals 6. The set of four spy orbitals on carbon, like any four electron domains on a central atom, assumes a tetrahedral arrangement 3. Rather than 2s1, 2p3, make it 2s4. (four electrons in the s block) 7 E. NOTE 1. Recall that elements in the third period of the periodic table and beyond do not necessarily obey the octet rule because they have d orbitals that can hold additional electrons 2. In order to explain the bonding and geometry of molecules in which there are more than four electron domains, on the central atom, we must include d orbitals in our hybridization scheme F.Example: PCl5. 1. Draw the ground state electron orbital diagram for the central atom 2. Maximize the number of unpaired electrons by promotion 3. Hybrid orbitals on phosphorus overlap with the 3p orbitals on chlorine 4. Promote from 3s2/3p3 —> 3s1/3p3/3d1 5. Hybridize Hybridization In Molecules Containing Multiple Bonds A. The remaining unhybridized p orbital is perpendicular to the plane in which the atoms of the molecule lie B. Pi bonds restrict free rotation around the bond axis C. The acetylene molecule is linear with sp hybridized carbons D. Promotion of an electron maximizes the number of unpaired electrons E. Formation of the C-C pi bond in acetylene 1. 2 pi bonds 2. 2 between the two carbon atoms F. Worked Example 7.8 Molecular Orbital Theory A. The atomic orbitals involved in bonding actually combine to form new orbitals that are the “property” of the entire molecu;e, rather than of the individual atoms forming the bonds, called molecular orbitals B. In molecular orbital theory electrons shared by atoms in a molecule reside in the molecular orbitals 8 *Chapter 8: Chemical Reactions* A chemical equation uses chemical symbols to denote what occurs in a chemical reaction A. NH3 + HCl —> NH4Cl B. Ammonia and Hydrochloride reaction to form ammonium chloride Labels (indicate the physical state) A. Gas (g) B. Liquid (l) C. Solid (s) D. Aqueous (aq)- dissolved in water E. Example: NH3(g)+ HCl(g)—> NH4Cl(g) Balancing Chemical Equations A. Chemical equations must be balanced so that the law od conservation of mass is obeyed B. Balancing is achieved by writing stoichiometric coefficients to the left of the chemical formulas C. Requires a trial-and-error approach D. Generally, it will facilitate the balancing process if you do the following: 1. Write the unbalanced equation 2. Balance each of the atoms, leaving 02 at the end 3. Continue process until it becomes even Calculations with Balanced Chemical Equations A. Consider the complete reaction of 3.82 moles of CO to form CO2 B. Calculate the number of moles of CO2 produced C. Moles you have x mole ratio = moles you should get 9 Determination of Empirical Formula A. In the combustion of 18.8 g of glucose (CxHyOz) 27.6 g of CO2 and 11.3 g of H2O are produced B. It is possible to determine the mass of carbon and hydrogen in the original sample by multiplying the (original mass) x (original molar mass) x (mole ratio) x (new molar mass) = new mass C. Divide by the smallest subscript to find the whole number ratio D. To get the empirical formula, follow these steps: 1. Multiply mass of sample by their molar masses to get moles of the samples, then obtaining the whole number ratio 2. Find the molar mass of the smallest ratio of the substance 3. Divide the molar mass by the empirical formula mass 4. Multiply the subscripts by the ratio Limiting Reactants A. Consider the reaction between 5 moles of CO and 8 moles of H2 to produce methanol B. CO + 2H2 —> CH2OH C. How many moles of H2 are necessary in order for all the CO to react? Moles H2 = 5 moles CO x mole ratio = 10 moles H2 D. How many moles of CO are necessary in order for al of the H2 to react? Moles CO = 8 moles H2 x mole ratio = 4 moles CO Reaction Yield A. The theoretical yield is the amount of product that arms when all the limiting reactant reacts to form the desired product B. The actual yield is the amount of product actually obtained from a reaction C. The practical yield tell what percentage the actual yield is of the theoretical yield D. % Yield = actual/theoretical x 100% 10 *Chapter 9* General Properties of Aqueous Solutions A. A solution is a homogenous mixture of two or more substances B. The substance present in the largest amount (moles) is referred to as the solvent C. The other substances are the solutes D. An electrolyte is a substance that dissolves in water to yield a solution that conducts electricity Strong Electrolytes and Weak Electrolytes A. An electron that disassociates completely is known as a strong electrolyte B. Water soluble ionic compounds/ strong acids/ strong bases C. Look up chart for strong acids D. A weak electrolyte is a compound that produces ions upon dissolving but exists in solution predominately as molecules that are NOT ionized 11 Precipitation Reactions A. An insoluble product that separates from a solution is called a precipitate B. A chemical reaction in which a precipitate forms is called a precipitation reaction Solubility Guidelines for Ionic Compounds in Water A. Water is a good solvent for ionic compounds because it is a polar molecule B. The polarity of water results from electron distributions within the molecules C. The oxygen atom has an attraction for the hydrogen atom’s electrons and is therefor partially negative compared to hydrogen D. Hydration occurs when water molecules remove the individual ions from an ionic solid surrounding them, preventing them from recombine so the substances dissolve E. Solubility is defined as the maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature G. There are many exceptions (see below) Molecular Equations A. In a molecular equation, compounds are represented by chemical formulas as though they exist in solution as molecules or formula units B. Na2SO4(aq) + Ba(OH)2(aq) —> 2NaOH(aq) + BaSO4(s) C. Reactions in which cations in two ionic compounds exchange anions are called metathesis or double replacement reactions Ionic Equations A. In an ionic equation, compounds that exist completely or predominately as ions in solution are represented as those ions B. In the reaction between aqueous sodium sulfate and barium hydroxide, the aqueous species are represented as ions which form precipitates Net Ionic Equations A. An equation that includes the species that are actually involved in the reaction is called a net ionic equation B. Ions that appear on both sides of the equation are called spectator ions C. Spectator ions do not participate in the reaction 12 Precipitation Reactions (to determine the molecular, ionic, and net ionic equations) A. Write and balance the molecular equation, predicting the products by assuming that the cations trade anions B. Write the ionic equation by separating strong electrolytes into their constituent ions C. Write the net ionic equation by identifying and cancelling spectator ions on both sides of the equation D. If both the reactants and products are all strong electrolytes, all the ions in solution are spectator ions. In this case, there is no net ionic equation and no reaction takes place Acid-Base Reactions A. Acids can either be strong or weak B. A strong acid is a strong electrolyte Strong Acids and Bases A. Strong bases are strong electrolytes (dissociate completely) B. Strong bases are the hydroxides of Group 1A and heavy Group 2A C. A weak acid is a weak electrolyte, it does not dissociate completely D. Most acids are weak acids An oxidation-reduction (or redox) reaction is a chemical reaction in which electrons are transferred from one reactant to another hy A. Oxidation is the loss of electrons B. Reduction is the gain of electrons 2+ C. Reducing agents (Zn loses two electrons and is oxidized to Zn ) 2+ D. Oxidizing agent (Cu gains two electrons and is reduced to Cu) E. A redox reaction is the sum of an oxidation half-reaction and a reception half- reaction 13 Bronsted Acids and Bases A. An Arrhenius acid is one that ionizes in water to produce H+ ions B. An Arrhenius base is one that dissociates in water to produce OH- ions C. A Bronsted acid is a proton donor D. A Bronsted base is a proton acceptor E. In these definitions, a proton refers to a hydrogen atom that has lost its electron, also known as a hydrogen ion (H+) F. Bronsted acids donate protons to water to form the hydronium ion (H3O+) G. A monoprotic acid has one proton to donate H. Hydrochloric acid is an example I. A polyprotic acid has more than one acidic hydrogen atom J. Sulfuric acid, H2SO4, is an example of a diuretic acid, there are two acidic hydrogen atoms K. Polyprotic acids lose protons in a stepwise fashion L. Bases that produce only one mole of hydroxide per mole of compound are called monobasic (sodium hydroxide is an example) K. Some strong bases produce more than one hydroxide per mole of compound (Barium hydroxide per mole of compound (Barium hydroxide is an example is an example of a dibasic base) The oxidation number is the charge an atom if electron were transferred completely A. H2(g) + F2(g) —> 2HF(g) B. The oxidation number of H i2 0 C. The oxidation number of F i2 0 D. The oxidation number of HF (+1)(-1) Oxidation Numbers (Assign the numbers to the elements in the compound KMnO4-) A. Step 1: Start with the oxidation numbers you know 1. K is +1 2. O4 is -8 (-2x4) 3. So Mn has to be -7 B. Step 2: The numbers in the boxes (total contribution to charge) must sum to the original charge 14 To assign oxidation numbers: A. The oxidation number of an element, in it elemental form, is 0 B. Figure out the charges through the periodic table C. Know the elements that nearly always have the same oxidation number Oxidation of Metals in Aqueous Solutions A. In a displacement reaction, an atom or an ion in a compound is replaced by an atom of another element B. Zinc displaces (replaces) copper in the dissolved salt 2+ C. Zn is oxidized to Zn 2+ D. Cu is reduced to Cu Concentration of Solutions A. Molarity (M), or molar concentration, is defined as the number of moles of solute per liter of solution B. Molarity (M) = moles solute (moles) / liters solution (L) C. Can also be written as L = mol/M D. Can also be written as mol = M x L Dilution A. Dilution is the process of preparing a less concentrated solution from a more concentrated one B. Moles of solute before dilution = moles of solute after dilution C. Example: In an experiment, a student needs 1.00 L of a 4.00 M KMnO4 solution. A stock solution of 1.00 M KMnO4 is available. How much of the stock is needed? 1. Use the relationship that moles of suction before dilution = moles of solution after dilution to find the missing part. 2. To make the solution: pipet 400 mL out of solution and into 1.00 L volumetric flask and carefully dilute to the calibration mark D. Because most volumes measured in the laboratory are in milliliters rather than liters, it is worth pointing out that the equation can be written as M xcmL = Mcx mLd d 15 E. A series of dilutions that may be used to prepare a number of increasingly dilute solutions is called serial dilution 1. Step 1: Prepare a dilute solution from the stock 2. Step 2: Dilute portion of the prepared solution to make a more dilute solution 3. Step 3: Repeat as needed *Chapter 10* Energy and Energy Changes A. The systems is a part of the universe that is of specific interest B. The surroundings constitute the rest of the universe outside the system C. Universe = System + Surroundings D. The system is usually defined as the substances involved in chemical and physical changes E. Thermochemistry is the study of heat (the transfer of thermal energy ) F. Heat is the transfer of thermal energy G. Heat is either absorbed or released during a process H. SI Unit is a Joule (J). Often calories are used and 1 calorie is the amount of heat required to raise 1 g of water by 1*C I. 1 cal = 4.184 J J. calorie is not the same as a nutritional calories (Cal) K. 1 Cal = 1000 cal L. An exothermic process occurs when heat is transferred from the system to the surroundings (“Feels hot!”) M. An endothermic process occurs when heat is transferred from the surroundings to the system (“Feels cold!”) Thermodynamics A. Thermodynamics is the study od the interconversion of heat and other kinds of energy B. In thermodynamics, there are three types of systems: 1. An open system can exchange mass and energy with the surroundings 2. A closed system allows the transfer of energy but not mass 3. An isolated system does not exchange either mass or energy with its surrounding 16 States and State Functions A. State function are properties that are determined by the state of the system, regardless of how the condition was achieved B. The magnitude of chmage depends only on the initial and final states of the system 1. Energy 2. Pressure 3. Volume 4. Temperature The First Law of Thermodynamics A. *triangle* U is the change in the internal energy B. “sys” and “surr” denote system and surroundings, respectively C. *triangle* U = U f U is the different of the erengies of the initial and final states Work and Heat A. The overall change in the systems initial energy is given by *triangle* U = q + w B. q is heat 1. q is positive for an endothermic process (heat absorbed by the system) 2. q is negative for an exothermic process (heat released by the system) C. w is work 1. w is positive for work done ON the system 2. w is negative for work done BY the system Enthalpy: Reactions Carried Out at Constant Volume or at Constant Pressure A. Sodium azide detonates to give a large quantity of nitrogen gas B. 2NaN3(s) —> 2Na(s) + 3N2(g) C. Under constant volume conditions, pressure increases D. Under constant pressure, volume increases E. Pressure-volume or PV work, is done when there is a volume change under constant pressure (w = -P*triangle*V) F. P is the external opposing pressure G. *triangle* V is the change in the volume of the container H. Under conditions of constant pressure, ΔU = q − PΔV AND ΔU = q + w 17 Enthalpy A. The thermodynamic function of a system called enthalpy (H) is defined by the equation H = U + PV B. Pressure: pascal; 1Pa = 1 kg/(m s ) . 2 C. Volume: cubic meters; m 3 D.PV: 1kg/(m s ) x m = 1(kg m )/s = 1 J 2 E. Enthalpy: joules Equations A. For any process, the change in enthalpy is ΔH = ΔU + Δ(PV) B. If pressure is constant, ΔH = ΔU + PΔV C. Rearrange to solve for ΔU = ΔH + PΔV D. Remember, q = Δp + ΔV E. Substitute equation 3 into equation 4 and solve: q = (pH − PΔV) + PΔV F.q p ΔH for a constant-pressure process G. The enthalpy of reaction (ΔH) is the difference between enthalpies of the products and the enthalpies of the reactants H. ΔH = H(products) – H(reactants) I. Assumes reactions in the lab occur at constant pressure 1. ΔH > 0 (positive)—> endothermic process 2. ΔH < 0 (negative) —> exothermic process Thermochemical Equations A. The following guidelines are useful when considering thermochemical equations 1. Always specific the physical states of reactants and products because they help determine the actual enthalpy changes (different states have different enthalpies) 2. When multiplying an equation by a factor (n), multiply the delta H blue by the same factor 3. Reversing an equation changes the sign but not the magnitude of delta H B. Worked example 10.3 1. Calculate the solar energy required to produce 75.0 g of C6H12O6. 2. Calculate the number of moles needed, then multiply it by the entire equation 18 Calorimetry A. The measurement of heat changes B. Heat changes are measured i a device called a calorimeter C. The specific heat (s) of a substance is the amount of heat required to raise the temperature of 1 g of the substance by 1*C. D. The heat capacity (C) is the amount of heat required to raise the temperature of an object by 1*C E. The “object” may be a given quantity of a particular substance F. Heat capacity of water = 4.184 J / (1g x *C) x 1000g = 4184 J/*C G. Specific heat capacity has units of J/(g • °C) H. Heat capacity has units of J/°C Specific Heat and Heat Capacity A. The heat associated with a temperature hcnage ay be calculated B. q = smΔT C. q = CΔT D. Calculate the amount of heat required o heat 1.01 kg of water from 0.05*C to 35.81*C. Constant-Pressure Calorimetry A. Concepts to consider for coffee-cup calorimetry: qp= delta H B. In an exothermic reaction, the system loses heat C. Worked example 10.5 D. Constant volume calorimetry is carried out in a device known as a constant-volume bomb E. A constant-volume calorimeter is an isolated system. F. Bomb calorimeters are typically used to determine heats of combustion. G. qcal −q rxn 19 Hess’s Law A. Hess’s law states that the change in enthalpy for a stepwise process is the sum of the enthalpy changes for each of the steps B. When applying Hess’s Law: 1. Manipulate thermochemical equations in a manner that gives the overall desired equation 2. Remember the rules for manipulating thermochemical equations: a. Always specify the physical states of reactants and products because they help determine the actual enthalpy changes. b. When multiplying an equation by a factor (n), multiply the ΔH value by same factor. c. Reversing an equation changes the sign but not the magnitude of ΔH. 3. Add the ΔH for each step after proper manipulation. 4. Process is useful for calculating enthalpies that cannot be found directly. Standard Enthalpies of Formation A. The standard enthalpy of formation (ΔH° ) isfdefined as the heat change that results when 1 mole of a compound is formed from its constituent elements in their standard states B. The superscripted degree sign denotes standard conditions 1. 1 atm pressure for gases 2. 1 M concentration for solutions D. “f” stands for formation E. ΔH f° for an element in its most stable form is zero. F. ΔH f° for many substances are tabulated in Appendix 2 of the textbook G. The standard enthalpy of reaction (ΔH °rxn ) is defined as the enthalpy of a reaction carried out under standard conditions. H. ΔH °rxn = [cΔH f°(C) + dΔH f°(D) ] – [aΔH f°(A) + bΔH f°(B)] I. ΔH °rxn = ΣnΔH f°(products) – ΣmΔH f°(reactants) J. n and m are the stoichiometric coefficients for the reactants and products. 20 Bond Enthalpy and the Stability of Covalent Molecules A. The bond enthalpy is the enthalpy change associated with breaking a bond in 1 mole of gaseous molecule B. H2(g) → H(g) + H(g) // ΔH° = 436.4 kJ/mol C. The enthalpy for a gas phase reaction is given by: ΔH° = ΣBE(reactants) – ΣBE(products) D. ΔH° = total energy input – total energy released E. Bond enthalpy change in an exothermic reaction F. Bond enthalpy change in an endothermic reaction 21 Lattice Energy and the Stability of Ionic Compounds A. A Born-Haber cycle is a cycle that relates the lattice energy of an ionic compound to quantities that can be measured + - B. Na(s) + 1/2 Cl (g2 → Na (g) + Cl (g) 22 Comparison of Ionic and Covalent Compounds A. Ionic and covalent compounds differ in their general physical properties because of the differences in the nature of their bonds. B. Check out table 10.6


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