Problem Set 8 Solutions
1.
For each of the following polyatomic ions, provide a Lewis structure, including resonance and
formal charge. (2pts each, 8 pts total)
a) BH4–
H
H
B
H
H
tetrahedral
b) HCO2–
–
O
O
C
H
C
O
c) AsCl4–
Cl
As
Cl
Cl
Cl
d) ICl2–
Cl
I
Cl
H
O
2.
For each of the following molecules, provide a Lewis structure, including formal charge and
resonance. Using the table of bond energies from page 314 of your text, along with the
following information, estimate ∆H°f for each of these molecules in the gas phase at 25°C.
∆H°f (C(g)) = +717 kJ/mol
a) CH3OH (methanol)
H
C
H
Carbon: Tetrahedral
Oxygen: Bent
Molecule will be polar
O
H
H
ΔH°f will be: C (s) + 1/2 O2 (g) + 2 H2 (g)  CH3OH (g)
The equivalent gas-phase reaction is:
C (g) + 1/2 O2 (g) + 2 H2 (g)  CH3OH
which has:
ΔH = 1/2DO=O + 2 DH–H – 3 DC–H – DO–H – DC–O
= 1/2(498.7) + 2(436.4) – 3(414) – 460 – 351
= –931 kJ/mol
and add:
C (s)  C (g)
ΔH°f (C(g)) = +717 kJ/mol
to get the overall estimated "H°f = - 214 kJ/mol
(3 pts)
!
b) COCl2 (phosgene)
O
C
Cl
Cl
ΔH°f will be: C (s) + 1/2 O2 (g) + Cl2 (g)  COCl2 (g)
The equivalent gas-phase reaction is:
C (g) + 1/2 O2 (g) + Cl2 (g)  COCl2
which has:
ΔH = 1/2 DO=O + DCl–Cl – DC=O – 2 DC–Cl
= 1/2(498.7) + 242.7 – 745 – 2(338)
= –929 kJ/mol
and add:
C (s)  C (g)
ΔH°f (C(g)) = +717 kJ/mol
to get the overall estimated !H°f = –212 kJ/mol
(3 pts)
(or ∆H°f = –192 kJ/mol if you used D(C-Cl) = 328 kJ/mol)
(or ∆H°f = –246.5 kJ/mol if students used all values from Practice Problems.)
c) H2NOH (hydroxylamine)
H
N
H
Nitrogen: Trigonal pyramidal
Oxygen: Bent
Molecule will be polar
O
H
ΔH°f will be: 1/2 N2 (g) + 1/2 O2 (g) + 3/2 H2 (g)  H2NOH (g)
This is already in the gas phase.
so ΔH = 1/2DN≡N + 1/2DO=O + 3/2DH–H – 2 DN–H – DN–O – DO–H
= 1/2(941.4) + 1/2(498.7) + 3/2(436.4) – 2(393) – 176 – 460
and estimate "H°f = - 47 kJ / mol
(3 pts)
!
d) N2O4 (bonded like O2N–NO2)
O
O
N
O
O
N
O
N
O
O
O
N
O
N
O
O
O
N
O
N
O
O
N
O
Both nitrogens are trigonal planar. Molecule will be nonpolar (dipoles will cancel)
ΔH°f will be: N2 (g) + 2 O2 (g)  N2O4 (g)
This is already in the gas phase.
so ΔH = DN≡N + 2 DO=O – DN–N – 2 DN–O – 2 DN=O
= 941.4 + 2(498.7) – 193 – 2(176) – 2(607)
and estimate "H°f = +180 kJ / mol
(3 pts)
(This estimate is particularly bad because it does not account for resonance and delocalization.)
!
3.
a)
On the basis of electronegativity, predict whether the molecule ICl (iodine chloride) will be
polar. If it should be polar, which end is positive and which end is negative? (2 pts)
Electronegativity of I is 2.5, and of Cl is 3.0. Therefore, ICl should be polar, with the Cl as the
negative end and the I as the positive end:
δ+ δ–
I
b)
Cl
Draw a Lewis structure for the molecule CO, including formal charge. It turns out that, in the
CO molecule, the carbon is the negative end of the molecule. Is this what you would expect
based on electronegativity? What could explain this? (2 pts)
C
O
Oxygen is more electronegative, but in this case the formal charge wins. The oxygen (with the
positive formal charge) is actually the positive end of the molecule; the carbon (with the
negative formal charge) is actually the negative end. Note that if you try to draw a Lewis
structure with no formal charges, it does not have complete octets. Always obey the octet rule
before worrying about formal charge!
c)
Draw a Lewis structure for the molecule O3, including formal charge and resonance. Based on
electronegativity differences, would you expect the O–O bonds in ozone to be polar? It turns
out that these bonds are polar; what could explain this? (2 pts)
O
O
O
O
O
O
Since there is no difference between the electronegativity of two O atoms, the polar bond
character could be explained by formal charges. The central O has a +1 formal charge, while
the terminal O’s share a –1 formal charge.
4.
Construct a Born-Haber cycle and calculate the lattice energy of CaC2 (s). Note that this solid
contains the diatomic ion C22–. (4 pts)
Useful Information: ∆H°f (CaC2(s))
= –60 kJ/mol
∆Hsub (Ca (s))
= +178 kJ/mol
∆Hsub (C (s))
= +717 kJ/mol
Bond dissociation energy of C2 (g)
=
First ionization energy of Ca (g)
=
Second ionization energy of Ca (g)
=
First electron affinity of C2 (g)
=
Second electron affinity of C2 (g)
=
IE1
Ca (g)
C2 (g)
Ca+ (g)
EA1
C2- (g)
IE2
EA2
+614 kJ/mol
+590 kJ/mol
+1143 kJ/mol
–315 kJ/mol
+410 kJ/mol
Ca2+ (g)
C22- (g)
-DC2
-LE
!Hsub(Ca)
2C (g)
2 !Hsub (C)
Ca (s)
+
2C (s)
CaC2 (s)
∆H°f (CaC2(s))
= ∆Hsub(Ca (s)) + IE1 + IE2 + 2∆Hsub(C (s)) – DC2 + EA1 + EA2 – LE
–60
Solve for LE
= 178 + 590 + 1143 + 2(717) – 614 + (–315) + 410 – LE
= 2886 kJ/mol