CHM 2045C Module 4 Part II Sample Exam Answers
Chapter 10 Sample Multiple Choice Answers
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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.

 

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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


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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

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