ch 7 8 9 10 review answer key

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Name __________________________________________ Period ______ Date _____________ AP Chemistry

Chapters 7, 8, 9, and 10 Review—answer key

1. Account for each of the following observations about pairs of substances. In your answers, use appropriate principles of chemical bonding and/or intermolecular forces. In each part, your answer must include references to both substances.

(a) Even though NH3 and CH4 have similar molecular masses, NH3 has a much higher normal

boiling point (-33°C) than CH4 (-164°C).

CH4 is non-polar. Its intermolecular force of attraction is London dispersion force which is the

weakest of all the IMFs causing a low BP. NH3 is polar and has hydrogen bonding, the strongest

IMF causing it to have a high BP.

(b) At 25°C and 1.0 atm, ethane (C2H6) is a gas and hexane (C6H14) is a liquid.

Both ethane and hexane are non-polar molecules with London dispersion forces. Since hexane has a larger polarizable electron cloud, it has a stronger force of attraction than ethane which has a smaller polarizable electron cloud causing hexane to have a higher BP than ethane.

(c) Si melts at a much higher temperature (1,410°C) than Cl2 (-101°C).

Silicon is a covalent network solid with strong covalent bonds between atoms while chlorine are discreet diatomic molecules with weak London dispersion forces between molecules. It takes more energy to break up covalent bonds between the Si atoms than the weak IMF between the chlorine molecules.

(d) MgO melts at a much higher temperature (2,852°C) than NaF (993°C).

Since both are ionic, the +2/-2 charges in MgO makes MgO have a higher attractive force between ions thus having a higher lattice energy than NaF whose charges are +1/-1. 2. Calculate the change in energy for each of the following reactions:

(a) K(s) + ½ F2(g)  KF(s) ∆Htotal= -565 KJ/mol

K(s)  K(g) ∆H= +89 KJ/mol (enthalpy of atomization)

K(g)  K+(g) + e- ∆H= +418 KJ/mol (ionization energy)

½ F2(g)  F(g) ∆H= +154(1/2) KJ/mol (bond energy)

F(g) + e-  F-(g) ∆H= -328 KJ/mol (electron affinity)

K+

(g) + F-(g)  KF(s) ∆H= -821 KJ/mol (lattice energy—exothermic!)

(b) Ethene + water  ethanol ∆H=Σ bonds broken – Σ bonds formed (Draw the Lewis structures!)

Need to break C=C, O-H; need to form C-C, C-O, C-H ∆H = (614+467) – (347+358+413) = -37 KJ/mol

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(Draw the Lewis structures!)

2 C8H10 + 21 O2  16 CO2 + 10 H2O

Need to break 6 C=C, 10 C-C, 20 C-H, 21 O=O; need to form 32 C=O, 20 O-H ∆H= [6(614) + 10(347) + 20(413) + 21(495)] – [32(799) + 20(467)] = -9099 KJ/mol (d) LiI(s)  Li(s) + ½ I2(g) ∆Htotal= +225 KJ/mol

LiI(s)  Li+(g) + I-(g) ∆H= +757 KJ/mol (lattice energy)

I

-(g)  I(g) + e- ∆H= +295 KJ/mol (electron affinity)

I(g)  ½ I2(g) ∆H= -149(1/2) KJ/mol (bond energy)

Li+

(g) + e- Li(g) ∆H= -519 KJ/mol (ionization energy)

Li(g)  Li(s) ∆H= -159 KJ/mol (enthalpy of atomization)

3. According to periodic trends, explain the following: (a) Atomic radii generally decrease across the periodic table.

The effective nuclear charge increases across the periodic table causing a stronger pull on the layers of electrons bringing them closer to the nucleus making the atoms smaller. (b) Ionization energy generally increases across the periodic table but oxygen has a lower ionization energy than nitrogen.

Oxygen’s last valence electron increases the amount of electron to electron repulsion as it pairs up with another electron in the same orbital causing it to require less energy to remove compared to nitrogen’s last valence electron which is spread out in its own orbital.

(c) Rank the following ions from smallest to largest: K+_{, Ca}+2_{, S}-2_{, Cl}

-Ca+2_{, K}+_{, Cl}-_{, S}-2

All are isoelectronic with 18 electrons in three energy levels. Calcium has the highest effective nuclear charge causing it to be the smallest.

(d) Fluorine is the most electronegative and Francium is the least.

Fluorine has the highest effective nuclear charge with the least number of energy levels of electrons shielding the nucleus from shared electron pairs while Francium has the lowest effective nuclear charge with the most number of energy levels of electrons shielding it from shared electron pairs.

4. Explain why PI5 exists but NI5 does not.

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Nitrogen is in the second period containing only s and p orbitals so it cannot accommodate the expanded octet.

5. Write the electron configuration for Sulfur and answer the following: 1s2_2s2_2p6_3s2_3p4

(a) Predict the charge on sulfide ion and justify your answer.

The charge should be -2 because sulfur was only 2 electrons away from a complete octet. (b) What are the four quantum numbers for the 8th_{electron in sulfur? 2, 1, -1, -1/2}

For the 13th_{? 3, 1, -1, +1/2}

(c) Which is larger, sulfur atom or sulfide ion? Explain.

Sulfide ion is larger than the sulfur atom due to the extra repulsion caused by the added electrons.

6. Which has the largest lattice energy? Justify. (a) NaF, LiF, KF

LiF will have the largest lattice energy since it has the smallest radius. Since lattice energy = k (Q1Q2/r), when divided by a smaller r, the quotient is larger. The Q1Q2 is not a

factor since the charges are the same on all three. (b) NaF, MgO, AlN

AlN will have the largest lattice energy since it has the largest charges (+3/-3 compared to +2/-2 and +1/-1). Since lattice energy = k (Q1Q2/r), when the product of Q1Q2 is large,

the lattice energy is large. The size is not a factor since they are all relatively the same. 7. What is the wavelength in nm emitted from a hydrogen atom when the electron falls from energy level 5 to energy level 2?

∆E = - 2.178 x 10-18_{J (1/2}2_{– 1/5}2_{) = - 4.574 x 10}-19_J

4.574 x 10-19 _{J = (6.626.x 10}-34_{Js) (ν)}

ν=6.903 x 1014_Hz

3.00 x 108_{m/s = (λ) (6.903 x 10}14 _Hz)

λ = 4.346 x 10-7_{m (10}9_{nm/m) = 434.6 nm}

8. Using the molecular orbital theory, predict which of the following are paramagnetic and diamagnetic: N2-2, O2-2, Be2+2, C2-, O2+, Li2+. Calculate the bond order for each.

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_ σp*

_ _

_ _ _ πp* πp* _ _ _

p _ p σp

_ _ πp πp

_

_ σs* _

s _ s σs

N2-2 paramagnetic, BO=2

O2-2 diamagnetic, BO=1

Be2+2 diamagnetic, BO=1

C2- paramagnetic, BO=2 1/2

O2+ paramagnetic, BO=2 1/2

Li2+ paramagnetic, BO=1/2

9. Draw the Lewis structures for the following and include: i. shape

ii. bond angle

iii. bond dipoles (and electronegativity differences) iv. polarity of molecule

v. IMF

vi. hybridization of central atom

vii. formal charges (different Lewis structure if necessary)

viii. types and number of covalent bonds present ( bonds or  bonds)

(a) SO3-2

Trigonal pyramid, 107° SO; ∆En = 1.0 Polar, dipole-dipole sp3

S = +1, O = -1; redraw to include 1 double bond 2  bonds only, 1 with  and  bonds

(b) XeOF2

T-shape, <90°

XeF, ∆En=4.0; XeO, ∆En=3.5 Polar, dipole-dipole

sp3_d

Xe = +1, O=-1, F=0; redraw with double bond to O 2  bonds only, 1 with  and  bonds

(c)OCl2

Bent, 104.5° ClO, ∆En=0.5 Polar, dipole-dipole sp3

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(d) C2H2

Linear, 180°

HC, ∆En=0.4, C-C, ∆En=0

Non-polar, London dispersion forces sp

Formal charges all = 0

C-H  bond; C-C triple bond has 1  and 2 

(e) Br3

-Linear, 180° ∆En=0

Non-polar, London dispersion forces sp3_d