Chemistry 121 Problem set V Solutions - 1
Chem 121 Problem Set V Lewis Structures, VSEPR and Polarity
ANSWERS
1.
Species
Elecronegativity difference in bond
Bond Polarity
Mp
very weakly polar covalent
< -40°C
NCl3
∆E = 3.0 - 3.0 = 0 for N-Cl
AlCl3
strongly polar covalent
190°C at 2.5 atm
∆E = 3.0 - 1.5 = 1.5 for Cl-Al
SO3
polar
covalent
-73°C
∆E = 3.5 - 2.5 = 1.0 for O-S
KI
ionic
(barely)
1330°C
∆E = 2.5 - 0.8 = 1.7 for I-K
CH4
weakly polar covalent
-182°C
∆E = 2.5 - 2.1 = 0.4 for C-H
CF4
strongly polar covalent
-150°C
∆E = 4.0 - 2.5 = 1.5 for F-C
Why does CF4 have a similar mp to CH4 when the bond polarities are very different?
2.
N2
a) valence electrons = 10
b) first structure gives bp 2 + lp 12 = 14,
c) second structure gives bp 6 + lp 4 = 10
d) Formal charge(N) = 5 - 3 - 2 = 0
N
N
10 - 14 = -4 or two extra bonds
N N
NO+ a) valence electrons = 5 + 6 - 1 = 10
b) first structure gives bp 2 + lp 12 = 14,
10 - 14 = -4 or two extra bonds
c) second structure gives bp 6 + lp 4 = 10
d) FC(N) = 5 - 3- 2 = 0 ,
FC(O) = 6 - 3 - 2 = +1
N
O2
O
a) valence electrons = 12
b) first structure gives bp 2 + lp 12 = 14,
c) second structure gives bp 4 + lp 8 = 12
d) FC(O) = 6 - 2 - 4 = 0
O
ClO3-
N
O
O
O
O
a) valence electrons = 7 + 18 + 1 = 26
b) first structure gives bp 6 + lp 20 = 26,
26 - 26 = 0 ok
c) second structure has FC (O) = 6 - 1 - 6 = -1 FC (Cl) = 7 - 3 - 2 = +2
O
O
Cl
Cl2
O
O
O
O
O
O
Cl
Cl
O
O
a) valence electrons = 14
b) first structure gives bp 2 + lp 12 = 14,
c) FC(Cl) = 7 - 1 - 6 = 0
Cl
O
O
Cl
O
Cl
O
O
Cl2
12 - 14 = -2 or one extra bond
O
14 - 14 = 0 ok
Cl
XeO3 a) valence electrons = 8 + 18 = 26
b) first structure gives bp 6 + lp 20 = 26,
26 - 26 = 0 ok
O
O
Chemistry 121 Problem set V Solutions - 2
c) second structure has FC (Xe) = 8 - 2 - 3 = +3
FC (O) = 6 - 1 - 6 = -1
O
O
Xe
Xe
O
O
3
O
O
Xe
O
O
O
NF3 a) valence electrons = 5 + 21 = 26
b) first structure gives bp 6 + lp 20 = 26
26 - 26 = 0 ok
c) first structure has FC(N) = 5 - 3 - 2 = 0 FC(F) = 7 - 1 - 6 = 0
F
N
F
XeF4
F
a) valence electrons = 8 + 28 = 36
b) first structure gives bp 8 + lp 24 = 32 36 - 32 = 4 need 4 more electrons on central atom
c) second structure has FC(Xe) = 8 - 4 - 4 = 0 FC(F) = 7 - 1 - 6 = 0
F
F
F
Xe
F
F
F
F
F
BF3
Xe
a) valence electrons = 3 + 21 = 24
b) first structure gives
bp 6 + lp 20 = 26
24 - 26 = -2 need to lose two electrons
but do not use multiple bonds in this case as boron is electron deficient and only 6 electrons on boron
c) second structure has FC(B) = 3 - 3 - 0 = 0 FC(F) = 7 - 1 - 6 = 0
F
F
B
F
MnO4-
F
B
F
F
a) valence electrons = 7 + 24 + 1 = 32
b) first structure gives bp 8 + lp 24 = 32
c) first structure has FC(Mn) = 7 - 4 - 0 = +3
d) resonance
O Mn O
O
3
O Mn O
O Mn O
O
O
O
O
O
O
O Mn O
O
32 - 32 = 0
FC(O) = 6 - 1 - 6 = -1
O
Mn O
O
O
O
O Mn O
O
O
O
Mn O
O
SO2 a) valence electrons = 6 + 12 = 18
b) first structure has bp 4 + lp 16 = 20 18 - 20 = -2 need to lose 2 electrons as double bonds
c) second structure has bp 6 + lp 12 = 18
d) second structure has FC(S) = 6 - 3 - 2 = +1 FC(-O) = 6 - 1 - 6 = -1 FC(=O) = 6 - 2 - 4 = 0
Chemistry 121 Problem set V Solutions - 3
e) resonance (the 3d obitals on S are close in energy to the 3p orbitals and S can take more than 8
electrons)
O
S
S
O
S
O
O
S
O
O3
O
S
O
O
S
O
O
O
a) valence electrons = 18
b) first structure has bp 4 + lp 16 = 20 18 - 20 = -2 need to lose 2 electrons as double bonds
c) second structure has bp 6 + lp 12 = 18
d) second structure has FC(-O=) = 6 - 3 - 2 = +1 FC(-O) = 6 - 1 - 6 = -1 FC(=O) = 6 -2 - 4 = 0
e) resonance (no two double bond hybrid as the lowest energy empty orbitals are too high in energy and
so oxygen can only take 8 electrons, no 2d orbitals.
O
3.
XeF4
O
O
O
O
O
O
O
O
O
O
O
O
a) valence electrons = 8 + 28 = 36
b) first structure has bp 8 + lp 28 = 36 (problem 12)
c) VSEPR (central atom) bp 4 + lp 2 = 6
VSEPR shape is octahedral (second structure)
d) Molecular shape is square planar
e) third structure: ∆EN (F-Xe) = 1.38 and all 4 ∆EN (F-Xe) vectors cancel; ∆EN (lp-Xe) = 1.1 and the
two lp ∆EN vectors cancel
f) nonpolar
F
F
Xe
F
F
Xe
F
F
F
F
SF6
a) valence electrons = 6 + 42 = 48
b) first structure has bp 12 + lp 36 = 48
c) VSEPR (central atom) bp 6 + lp 0 = 6
VSEPR shape is octahedral (second structure)
d) Molecular shape is octahedral
e) third structure: ∆EN (F-S) = 1.4 and all 6 ∆EN (F-S) vectors cancel
f) nonpolar
F
F
F
F
F
F
F
F
F
S
S
F
F
F
OF2
a) valence electrons = 6 + 14 = 20
b) first structure has bp 4 + lp 16 = 20
c) VSEPR (central atom) bp 2 + lp 2 = 4
d) Molecular shape is bent
VSEPR shape is tetrahedral (second structure)
Chemistry 121 Problem set V Solutions - 4
e) third structure: ∆EN (F-O) = 0.54 and resultant ∆EN (F-O) vector between F-O bonds; ∆EN (lp-O) =
0.22 and resultant ∆EN (lp-O) vector between lp-O bonds; both resultant vectors come close to cancelling
(0.54 − 0.26 = 0.28).
f) weakly polar, ∆EN = 0.3 (the dipole moment is 0.3 D)
F
O
F
O
F
O
F
CH2Cl2
a) valence electrons = 4 + 14 + 2 = 20
b) first structure has bp 8 + lp 12 = 20
c) VSEPR (central atom) bp 4 + lp 0 = 4
VSEPR shape is tetrahedral (second structure)
d) Molecular shape is tetrahedral
e) third structure: ∆EN (Cl-C) = 0.61 and resultant ∆EN (Cl-C) vector between Cl-C bonds;
∆EN (C-H) = 0.35 and resultant ∆EN (C-H) vector between C-H bonds; resultant vectors are
additive (0.61+0.35 = 0.96).
f) polar, ∆EN = 0.96 (the dipole moment is 1.6 D)
H
Cl
C
H
C
Cl
Cl
NCl3
H
Cl
H
a) valence electrons = 5 + 21 = 26
b) first structure has bp 6 + lp 20 = 26
c) VSEPR (central atom) bp 3 + lp 1 = 4
VSEPR shape is tetrahedral (second structure)
d) Molecular shape is trigonal pyramidal
e) third structure: ∆EN (N-Cl) = 0.12 and resultant small ∆EN (N-Cl) vector between N-Cl bonds;
∆EN (lp-N) =0.66 and ∆EN (lp-N) vector overcomes the resultant from the ∆EN (N-Cl) vectors
(0.66 – 0.12 = 0.54).
f) polar, ∆EN = 0.54 (the dipole moment is 0.39 D)
Cl
Cl
N
N
Cl
Cl
Cl
N
Cl
CHF3 a) valence electrons = 4 + 21 + 1 = 26
b) first structure has bp 8 + lp 18 = 26
c) VSEPR (central atom) bp 4 + lp 0 = 4
VSEPR shape is tetrahedral (second structure)
d) Molecular shape is tetrahedral
e) third structure: ∆EN (F-C) = 1.43 and resultant ∆EN (F-C) vector between F-C bonds; ∆EN (C-H) =
0.34 and ∆EN (C-H) vector adds to the resultant from the ∆EN (F-C) vectors (0.35 + 1.43 = 1.78).
f) strongly polar, ∆EN = 1.78 (the dipole moment is 1.65 D)
H
H
F
C
F
C
F
F
F
F
CF4
a) valence electrons = 4 + 28 = 32
b) first structure has bp 8 + lp 24 = 32
c) VSEPR (central atom) bp 4 + lp 0 = 4
VSEPR shape is tetrahedral (second structure)
d) Molecular shape is tetrahedral
e) third structure: ∆EN (F-C) = 1.43 and all ∆EN (F-C) vectors cancel.
Chemistry 121 Problem set V Solutions - 5
f) nonpolar
F
F
F
C
C
F
F
4.
F
F
F
(a)
N
N
(b)
N
O
(c)
N
O
N
N
O
(c) is the least stable as it has the most formal charge (including two negatives on one nitrogen)
(b) is the most stable as the negative charge is on oxygen
(a) is less stable than (b) with the negative charge on nitrogen
Order of contribution to resonance hybrid is (b) > (a) >> (c)
The resonance hybrid will have the largest contribution from (b) and a smaller contribution from (a) so the
N-N bond will have an electron density between a triple and a double bond and thus its length will be
between 110 and 120 ppm, in fact it is 112 ppm. The N-O bond will have an electron density between a
single and a double bond and so its length will be between 115 and 147 ppm, in fact it is 119 ppm.
5a)
(i)
(ii)
H
(iii)
H
C
N
N
H
C
H
N
N
C
H
N
N
H
Lewis structure (iii) is not legitimate as carbon is forming 5 bonds and has 10 valence electrons. Carbon only
has orbitals for 8 valence electrons.
(i) is the most stable as the negative charge is on nitrogen
(ii) is less stable than (i) with the negative charge on carbon
Order of contribution to resonance hybrid is (i) > (ii)
b) Hybridization
H
H
C
N
N
H
sp2
C
N
N
H
sp2
sp
sp3
sp
sp
c) (i) H−C−N 120º, C−N−N 180º; (ii) H−C−N 109.5º, C−N−N 180º
6.
Species
C
O
C−O bond length
113 pm
O
C
120 pm
O
H
H C
H
O
H
143 pm
Chemistry 121 Problem set V Solutions - 6
O
O
C
O
C
O
O
143 −
O
C
O
O
(143 − 120)
= 143 − 8 = 135pm
3
O
Order with respect to C−O bond length (longest to shortest): CH3OH > CO32− > CO2 > CO
7. ∆EN (C−O) = 3.4 − 2.5 = 0.9
C
O
The electronegativity vector is toward oxygen, i.e. oxygen has a higher electron density than carbon,
but the formal charges indicate that in CO, oxygen is deficient in electrons and carbon has an excess, which
would lead to a lower separation of charge and a smaller dipole moment than expected from electronegativity
values.
8. Cyanate, NCO−
(i)
N
(ii)
C
N
O
(iii)
C
O
N
C
O
The most stable Lewis structure is (ii) with the negative charge on oxygen, though (i) with the negative charge
on nitrogen will contribute to the resonance hybrid, making cyanate a stable anion.
Fulminate, CNO−
(i)
C
N
(ii)
C
O
N
(iii)
O
C
N
O
Carbon forms three bonds, under special conditions it will form three bonds and carry a single negative charge,
but such species are very reactive. Lewis structure (ii) is the only significant contributor to the resonance hybrid,
structures (i) and (iii) with two and three negative charges on carbon will contribute insignificantly to the
resonance hybrid.
Thus the cyanate ion with two Lewis structures contributing to the resonance hybrid (negative charge on oxygen
and nitrogen) would be expected to be reasonably stable, while the fulminate ion with only one significant
structure contributing to the resonance hybrid and with the negative charge on carbon would be expected to be
quite unstable.
9.
a) HSO4−
O
O
H
O
S
HO
O
S
O
O
HO
O
O
S
O
O
ClO4−
O
O
Cl
O
H2 PO4−
O
O
Cl
O
O
O
O
O
O
Cl
O
O
O
Cl
O
O
Chemistry 121 Problem set V Solutions - 7
O
O
HO
P
HO
O
P
O
OH
OH
thus ClO4− has the most Lewis structures contributing to the resonance hybrid and the negative charge is
delocalized to the greatest extent, making it the weakest base (and the most stable anion). H2 PO4− has
the least contributors to the resonance hybrid and the negative charge is delocalized to the smallest
extent, making it the least stable and thus the most basic.
So the order of increasing basicity is: ClO4− < HSO4− < H2 PO4−
b) ClO2−
O
Cl
O
O
Cl
O
ClO3−
O
O
Cl
Cl
O
O
O
O
Cl
O
O
O
ClO−
Cl
O
−
thus ClO is the least stable with the negative charge localized on one oxygen, making it the most basic;
ClO3− is the most stable with the negative charge delocalized on three oxygens, making it the least basic.
So the order of increasing basicity is: ClO3− < ClO2− < ClO−
10.
∆EN(F-S) = 1.40 and ∆EN(lp-S) = 1.08
SF2
SF3+
SF3−
SF4
F
F
F
S
S
F
F
S
F
F
F
F
S
F
F
F
F
F
F
F
F
F
F
F
F
F
F
S
S
F
S
S
net vector = 0.32
permanent dipole
SF5+
net vector = 0.32
permanent dipole
net vector = 0.32
permanent dipole
SF5−
net vector = 0.32
permanent dipole
SF6
Chemistry 121 Problem set V Solutions - 8
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
S
vectors cancel
no permanent dipole
net vector = 0.32
permanent dipole
vectors cancel
no permanent dipole
11a) To obtain a square planar arrangement, we need an octahedral VSEPR shape (6 electron pairs) with two
lone pairs on the central atom. Thus
F
F
I
F
F
FC (I) = 7 − 4 − 4 = −1 thus the formula is IF4−
b) To obtain a pyramidal arrangement, we need an tetrahedral VSEPR shape (4 electron pairs) with one
lone pair on the central atom. Thus
F
I
F
F
FC (I) = 7 − 2 − 3 = +2 thus the formula is IF32+
c) To obtain a T-shaped arrangement, we need an trigonal bipyramid VSEPR shape (5 electron pairs)
with two lone pairs on the central atom. Thus
F
I
F
F
FC (I) = 7 − 4 − 3 = 0 thus the formula is IF3
d) To obtain a linear arrangement, we need an trigonal bipyramid VSEPR shape (5 electron pairs) with
three lone pairs on the central atom. Thus
F
I
F
FC (I) = 7 − 6 − 2 = −1 thus the formula is IF2−
e) To obtain a square pyramid arrangement, we need an octahedral VSEPR shape (6 electron pairs) with
one lone pair on the central atom. Thus
F
F
I
F
F
F
FC (I) = 7 − 2 − 5 = 0 thus the formula is IF5
f) To obtain a octahedral arrangement, we need an octahedral VSEPR shape (6 electron pairs). Thus
F
F
F
F
I
F
F
FC (I) = 7 − 0 − 6 = +1 thus the formula is IF6+
Chemistry 121 Problem set V Solutions - 9
12.
Since Sb and F have different electronegativities, a nonpolar species will have all the electronegativity
vectors cancel. ∆EN(F-Sb) = 1.93 and ∆EN(lp-Sb) = 1.61
So, working through the possible species with an even number of valence electrons and zero or a single
charge we have:
SbF, the single vector cannot cancel so it is polar.
SbF2+
SbF2−
SbF4+
SbF3
F
F
F
Sb
F
F
Sb
Sb
F
F
F
Sb
F
F
F
F
F
F
F
Sb
net vector = 0.28
permanent dipole
SbF4−
F
Sb
F
F
F
net vector = 0.28
permanent dipole
SbF6−
F
F
Sb
F
I
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
F
vectors cancel
no permanent dipole
SbF6+
SbF5
F
F
F
F
Sb
Sb
net vector = 0.28
permanent dipole
F
F
F
F
F
F
F
F
F
F
Sb
net vector = 0.28
permanent dipole
vectors cancel
no permanent dipole
can not exist
vectors cancel
no permanent dipole
13. The only VSEPR shape where all adjacent F-X-F angles are 90º is the octahedron, so we want XF6. The
Lewis structure must have no lone pairs on the central atom.
Chemistry 121 Problem set V Solutions - 10
F
F
F
F
F
F
F
F
X
F
F
F
F
which gives 6(6) + 6(2) = 48 valence electrons
thus: (valence electrons on X) + 6(7) = 48 and so: (valence electrons on X) = 6
X can be S, Se, Te but not O.
The only VSEPR shape where all adjacent F-X-F angles are slightly less than 90º is the octahedron with
one lone pair on the central atom, so we want XF5.
F
F
F
F
F
F
F
F
F
F
which gives 5(6) + 2 + 5(2) = 42 valence electrons
thus: (valence electrons on X) + 5(7) = 42 and so: (valence electrons on X) = 7
X can be Cl, Br, I but not F.
14.
The VSEPR shape where XF4 has a permanent dipole can not be a tetrahedron (4 bp) or an octahedron
with two lone pairs (4 bp and 2 lp) as in each case all the electronegativity vectors will cancel. It can not
be a trigonal planar as this shape has only 3 regions of electron density and we have 4 bp. The VSEPR
shape can only be a trigonal bipyramid.
F
F
X
F
F
X
F
F
F
F
which gives 4(6) + 2 + 4(2) = 34 valence electrons
thus: (valence electrons on X) + 4(7) = 34 and so: (valence electrons on X) = 6
X can be S, Se, Te but not O (as oxygen can have a maximum of 8 electrons on the central atom).