Ch. 9 Review Packet Answers - 1
Chapter 9 Review Packet
1.
The hybridization of the central nitrogen of N2O is:
a)
not hybridized.
b)
sp
c)
sp2
d)
sp3
e)
dsp3
2.
The hybridization of the central atom, sulfur, in SF4 is:
a)
sp
b)
sp2
c)
sp3
d)
dsp3
e)
d2sp3
3.
The hybridization of sulfur in the sulfate anion, SO42, is:
a)
sp3
b)
sp2
c)
sp
d)
s2p3
e)
not hybridized
4.
In the BeF2 molecule the Be valence orbitals are:
a)
sp hybrids
b)
sp2 hybrids
•• •
• ••
• F — Be — F •
c)
sp3 hybrids
••
••
d)
dsp2 hybrids
e)
none of the above
5.
The hybridization of the centra1 atom, Al, in AlBr3 is
a)
sp
b)
sp2
c)
sp3
d)
dsp3
e)
d2sp3
6.
The hybridization of Se in SeF6 is
a)
sp
b)
sp2
c)
sp3
d)
dsp3
e)
d2sp3
••
•Br
•
•
•
•
•
Al
•
•
Br
•
•
••
Br••
Ch. 9 Review Packet Answers - 2
7.
The hybridization of Br in BrF3 is
a)
sp
b)
sp2
c)
sp3
d)
dsp3
e)
d2sp3
8.
What is the bond order of Ne2?
a)
0
(σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)4(σ2p*)2
b)
½
c)
1
B.O. = (bonding  antibonding)/2 = (8  8)/2 = 0
d)
1½
e)
2
9.
What is the bond order of C2+1?
a)
0
(σ2s)2(σ2s*)2(π2p)3
b)
½
c)
1
d)
1½
B.O. = (bonding  antibonding)/2 = (5  2)/2 = 1½
e)
2
10.
H
H
|
|
Consider the molecule
H—N—N—H
••
••
The hybridization of each N atom is:
a)
sp
b)
sp2
c)
sp3
d)
dsp3
e)
d2sp3
•
B ≡ N•
11.
Which of the following statements about the molecule BN is false?
a)
It is paramagnetic.
b)
Its bond order is 2.
(σ2s)2(σ2s*)2(π2p)4
c)
The total number of electrons is 12. B.O. = (bonding  antibonding)/2 = (6  2)/2 = 2
d)
It has two pi bonds.
All electrons are paired, so it is diamagnetic.
e)
All of these are true.
Atomic numbers of B is 5 and N is 7 for a total of 12.
12.
Which of the following molecules or ions is not paramagnetic in its ground state?
a)
O2
O2 (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)2 unpaired electrons = paramagnetic
b)
O2+1
O2+1 (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)1 unpaired electrons = paramagnetic
c)
B2
B2 (σ2s)2(σ2s*)2(π2p)2
unpaired electrons = paramagnetic
d)
NO
2
* 2
4
2
* 1
NO
(σ
)
(σ
)
(π
)
(σ
)
(π
)
unpaired
electrons = paramagnetic
2s
2s
2p
2p
2p
e)
F2
2
* 2
2
4
* 4
F2 (σ2s) (σ2s ) (σ2p) (π2p) (π2p ) paired electrons = diamagnetic
Ch. 9 Review Packet Answers - 3
For questions 13 and 14, consider the skeletal structure shown below:
N—C—C—N
Draw the Lewis structure and answer the following:
13.
How many of the atoms are sp hybridized?
a)
0
b)
1
c)
2
•
•
• N≡C—C≡N•
d)
3
e)
4
14.
How many pi bonds does the molecule contain?
a)
b)
c)
d)
e)
0
2
4
6
7
15.
For which of the following diatomic molecules would the bond order become greater if an
electron is removed, i.e., if the molecule is converted to the positive ion in its ground state?
a)
B2
Valence
Valence
B.O. before:
B.O. after:
b)
C2
electrons before: electrons after:
c)
P2
B2 = 6
B2+1 = 5
(4  2)/2 = 1
(3  2)/2 = ½
+1
d)
F2
C2 = 8
C2 = 7
(6  2)/2 = 2
(5  2)/2 = 1½
e)
Na2
+1
P2 = 10
P2 = 9
(8  2)/2 = 3
(7  2)/2 = 2½
+1
F2 = 14
F2 = 13
(8  6)/2 = 1
(8  5)/2 = 1½
Na2 = 2
Na2+1 = 1
(2  0)/2 = 1
(1  0)/2 = ½
16.
The configuration (2s)2(2s*)2(π2py)1(π2px)1 is the molecular orbital description for the
ground state of:
a)
Li2+1
The configuration has six electrons.
b)
Be2
B2 is the only answer choice that
c)
B2
has six valence electrons.
d)
B22
e)
C2
17.
Which of the following species is paramagnetic?
a)
C2
C2 (σ2s)2(σ2s*)2(π2p)4
b)
B2
B2 (σ2s)2(σ2s*)2(π2p)2
c)
N2
N2 (σ2s)2(σ2s*)2(π2p)4(σ2p)2
d)
H2
H2 (σ2s)2
e)
none of these
paired electrons = diamagnetic
unpaired electrons = paramagnetic
paired electrons = diamagnetic
paired electrons = diamagnetic
Ch. 9 Review Packet Answers - 4
18.
19.
Order the following from shortest to longest bond: C2, B2, H2, N2
a)
H2, N2, C2, B2
Bond Order
b)
N2, C2, B2, H2
As B.O. ↑, bond length ↓.
C2
(6  2)/2 = 2
c)
C2, N2, H2, B2
Bond Length:
B2
(4  2)/2 = 1
d)
C2, B2, H2, N2
N2 < C2 < B2 = H2
H2
(2  0)/2 = 1
e)
none of these
N2
(8  2)/2 = 3
Which charge(s) of O2 would give you a bond order of 2.5?
a)
2
O2 (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)2 B.O. = (bonding  antibonding)/2 = (8  4)/2 = 2
b)
1
O2+1 (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)1 B.O. = (bonding  antibonding)/2 = (8  3)/2 = 2½
c)
+1
1
2
* 2
2
4
* 3
d)
two of these O2 (σ2s) (σ2s ) (σ2p) (π2p) (π2p ) B.O. = (bonding  antibonding)/2 = (8  5)/2 = 1½
e)
none of these
20.
For how many of the following does bond order decrease if you take away one electron
from the neutral molecule? B2, C2, P2, F2
a)
0
Valence
Valence
B.O. before:
B.O. after:
b)
1
electrons
before:
electrons
after:
c)
2
B2 = 6
B2+1 = 5
(4  2)/2 = 1
(3  2)/2 = ½
d)
3
+1
e)
4
C2 = 8
C2 = 7
(6  2)/2 = 2
(5  2)/2 = 1½
P2 = 10
P2+1 = 9
(8  2)/2 = 3
(7  2)/2 = 2½
F2 = 14
F2+1 = 13
(8  6)/2 = 1
(8  5)/2 = 1½
21.
Which of the following e1ectron distributions among the molecular orbitals best describes
the NO molecule?
π2py = π2px
π2py* = π2px*
2s
2s*
2pz
2pz*
a)
2
2
4
2
4
2
b)
2
2
4
2
4
1
c)
2
2
4
1
3
0
d)
2
2
4
2
2
0
e)
2
2
4
2
1
0
22.
The C — C — H bond angles in ethylene, C2H4, are 120. What is the hybridization of the
carbon orbitals?
a)
sp
b)
sp2
c)
sp3
d)
dsp3
e)
d2sp3
Ch. 9 Review Packet Answers - 5
23.
What is the bond order of He2+1?
a)
0
b)
½
He2+1 (σ1s)2(σ1s*)1 B.O. = (bonding  antibonding)/2 = (2  1)/2 = ½
c)
1
d)
1½
e)
2
24.
Which of the following has the shortest bond length?
a)
O22
O2 (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)2
b)
O2
O2+1 (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)1
c)
O21
O2 1 (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)3
d)
O2+1
= (8  4)/2 = 2
B.O. = (bonding  antibonding)/2 = (8  3)/2 = 2½
B.O. = (bonding  antibonding)/2 = (8  5)/2 = 1½
B.O. = (bonding  antibonding)/2
As B.O. ↑, bond length ↓.
Bond length:
O2 +1 < O2 < O2 1
25.
Experiments on B2 find that the molecule is paramagnetic. Explain how this result is
consistent with the π2p molecular orbital being lower in energy than the σ2p molecular
orbital.
Ch. 9 Review Packet Answers - 6
26.
A compound composed of 2.1% hydrogen, 29.8 % nitrogen, and 68.1% oxygen has a molar
mass of approximately 50 g/mol.
a.
What is the molecular formula of the compound?
b.
What is its Lewis structure if H is bonded to O?
c.
What is the geometry of the molecule?
d.
What is the hybridization of the orbitals around the N atom?
e.
How many σ and how many π bonds are there in the molecule?
Ch. 9 Review Packet Answers - 7
27.
Consider the structure of glycine, the simplest amino acid:
H
H
O
|
|
||
H—N — C — C — O—H
••
|
H
a)
Indicate the hybridizations at each N and C atom in the molecule.
N
sp3
C1
sp3
C2
sp2
b)
What is the total number of bonds in the molecule?
c)
10 total bonds
What is the total number of π bonds in the molecule?
1 π bond
28.
Answer the following questions about N2 and N21.
a.
Write the complete electron configuration for N2 using molecular orbitals.
N2 (σ2s)2(σ2s*)2(π2p)4(σ2p)2
b.
Write the complete electron configuration for N21 using molecular orbitals.
N21 (σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)1
c.
Calculate the bond order of N2.
B.O. = (bonding  antibonding)/2 = (8  2)/2 = 3
d.
Calculate the bond order of N21.
B.O. = (bonding  antibonding)/2 = (8  3)/2 = 2½
e.
Explain why N2 will be paramagnetic or diamagnetic.
It will be diamagnetic because all electrons are paired.
f.
Explain why N21 will be paramagnetic or diamagnetic.
It will be paramagnetic because one electron is unpaired.
g.
Compare the bond energy of N2 to that of N21.
N2 will have a greater bond energy than N21 because
N2 has a greater bond order than N21.
Ch. 9 Review Packet Answers - 8
29.
Answer the following questions about C2, C21, and C22.
a.
Write the complete electron configuration for C2 using molecular orbitals.
C2 (σ2s)2(σ2s*)2(π2p)4
b.
Write the complete electron configuration for C21 using molecular orbitals.
C21 (σ2s)2(σ2s*)2(π2p)4(σ2p)1
c.
Write the complete electron configuration for C22 using molecular orbitals.
C22 (σ2s)2(σ2s*)2(π2p)4(σ2p)2
d.
Place the species in order of greatest to least bond order.
C2 (σ2s)2(σ2s*)2(π2p)4
C21 (σ2s)2(σ2s*)2(π2p)4(σ2p)1
C2 2 (σ2s)2(σ2s*)2(π2p)4(σ2p)2
= (6  2)/2 = 2
B.O. = (bonding  antibonding)/2 = (7  2)/2 = 2½
B.O. = (bonding  antibonding)/2 = (8  2)/2 = 3
B.O. = (bonding  antibonding)/2
C2 2 > C2 1 > C2
e.
Place the species in order of least to greatest bond energies.
As B.O. ↑, bond energy ↑.
Bond Energy:
C2 < C2 1 < C22
f.
Explain the expected magnetism in these species.
C2 and C22 will be diamagnetic because they have paired electrons.
C21 will be paramagnetic because it has an unpaired electron.
30.
Answer the following questions regarding caffeine. (Remember that carbon is placed at
every vertex where an element is not shown and hydrogens are not shown if needed to make
up the fourth bond. Lone pairs have not been added.)
Ch. 9 Review Packet Answers - 9
a.
How many  bonds are in the molecule?
25
b.
How may  bonds are in the molecule?
4
c.
What is the hybridization on each carbon shown bonded to the nitrogens?
d.
What is the hybridization on each carbon assumed to be at the vertices of the
hexagon?
sp2
e.
What is the hybridization on each nitrogen? sp3
f.
What is the hybridization on each oxygen? sp2
g.
What is the OCN bond angle?
120
sp3