CHM 2045C Module 4 Part II Sample Exam Answers
Chapter
10 Sample Multiple Choice Answers
1. The
correct answer is:
Electrons simultaneously attracted by
more than one nucleus
Explanation:
A sharing of electrons
Solution Reference:
Page 439, Orbitals
and Bonding Theories
2. The
correct answer is:
BrCl
Explanation:
In covalent compounds, bromine and chlorine
would have the tendency to form one bond, as each has one half-filled orbital.
Solution Reference:
Page 439, Orbitals
and Bonding Theories
3.The
correct answer is:
trigonal planar.
Solution Reference:
Page 442, Hybridization Involving s and p
Atomic Orbitals
4.The
correct answer is:
4 sigma bonds.
Solution Reference:
Page 442, Hybridization Involving s and p
Atomic Orbitals
5.The
correct answer is:
H2NNH2.
Explanation:
There are four regions of electron density about each nitrogen, so nitrogen
atoms can be described as being sp3 hybridized.
Solution Reference:
Page 442, Hybridization Involving s and p
Atomic Orbitals
6.The correct answer is:
3 only
Explanation:
Solution Reference:
Page 442, Hybridization Involving s and p
Atomic Orbitals
7.
The
correct answer is:
2, 4, and 5
Explanation:
Phosphine has a pyramidal shape and the phosphorus
atom is hybridized sp3.
Solution Reference:
Page 442, Hybridization Involving s and p
Atomic Orbitals, and Page 423, Molecular Polarity
8.The
correct answer is:
NH3, NH4+
Explanation:
Each nitrogen atom has four pairs of electrons
surrounding it.
Solution Reference:
Page 442, Hybridization Involving s and p
Atomic Orbitals
9.The
correct answer is:
sp3d
Explanation:
The P atom has five regions of electron
density. Thus, there must be five hybrid orbitals.
Solution Reference:
Page 449, Hybridization Involving s, p, and d
Atomic Orbitals
10. The
correct answer is:
linear.
Solution Reference:
Page 442, Hybridization Involving s and p
Atomic Orbitals.
11.
The
correct answer is:
sp3, sp2
Explanation:
Draw the Lewis structures and find the number
of regions of electron density, or make an energy level diagram.
Solution Reference:
Page 492, Hybridization Involving s and p
Atomic Orbitals
12.
The
correct answer is:
1.
Explanation:
Solution Reference:
Page 452, Multiple Bonds
13.
The
correct answer is:
H2CCH2.
Solution Reference:
Page 452, Multiple Bonds
14.
The
correct answer is:
NO2+
Explanation:
Both NO2+ and N2O
are isoelectronic (16 valence electrons).
Solution Reference:
Page 452, Multiple Bonds
15.
The
correct answer is:
1.
Solution Reference:
Page 452, Multiple Bonds
16. The
correct answer is:
2, 2, 4
Explanation:
Solution Reference:
Page 452, Multiple Bonds
17. The
correct answer is:
O22- < O2
< O22+
Explanation:
Bond Orders:
O22- = 1
O2 = 2
O22+ = 3
Solution Reference:
Page 459, Molecular Orbital Theory
18. The
correct answer is:
2 only
Explanation:
Bond Orders:
F2, 1; F2+, 3/2
CN-, 3; CN+, 2
NO-, 2; NO+, 3
Solution Reference:
Page 459, Molecular Orbital Theory
19. The
correct answer is:
molecular orbital theory.
Solution Reference:
Page 459, Molecular Orbital Theory
20. The
correct answer is:
P.
Solution Reference:
Page 470, Metals and Semiconductors
21.
The
correct answer is:
BO and NO
Solution Reference:
Page 459, Molecular Orbital Theory
22. The
correct answer is:
O22+
Explanation:
Which species has 10 valence electrons?
Solution Reference:
Page 459, Molecular Orbital Theory
23. The
correct answer is:
[core electrons] (s2s)2 (s2s*)2 (p2p)4.
Solution Reference:
Page 459, Molecular Orbital Theory
24. The
correct answer is:
Molecular Orbital Theory
Solution Reference:
Page 459, Molecular Orbital Theory
25.The
correct answer is:
1 only
Solution Reference:
Page 459, Molecular Orbital Theory