CHEM 121 INTEGRATED RESOURCE PACKAGE

Chapter 2 Exercises Answers

1

Page #

Exercise #

Answer

9 1

10 1

H atomic radius is: °

°

Α=Α

37.02

74.0

C atomic radius = (C─H bond distance) − (H atomic radius)

= °°°

Α=Α−Α 73.037.010.1

13 1 a) Although S and S2- have the same estimated value for Zeff = Z – S = 16 – 10 = +6, this simple approximation of Zeff neglects the repulsion of electrons within the valence shell (which can be viewed as partially screening one another from the nucleus – see table on page 2-9). Thus, the actual Zeff acting on the electrons in the outer shell is greater in S than in neutral S2- Therefore, the radius of S is smaller. In general terms: The nuclear charge is the same in S and S2-, but there are two more electrons in S2-. Therefore, there is more electron – electron repulsion in the anion and the size of the anion is greater. b) Na+ has one less electron, but the same nuclear charge Z. Therefore there is less electron – electron repulsion and electrons are held more tightly in Na+ than in Na. In addition, the effective nuclear charge acting on the outer shell electrons in Na+ is about +9 and that acting on the outer shell electron in neutral Na is about + 1. Thus, the radius of Na+ is smaller than that of Na.

13 2 F < N < B < Li < Rb

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Chapter 2 Exercises Answers

2

15 1 You can rationalize this observation using Coulomb’s law:

r

qqE

−+

∝ :

When a neutral atom of Be is ionized, the electron is separated from a singly charged cation: Be(g) ���� Be+(g) + e- IE1 is thus the energy required to separate two charges of magnitude 1. When Be+ is ionized, the electron is separated from a doubly charged cation: Be+(g) ���� Be2+(g) + e- IE2 is thus the energy required to separate a charge of magnitude +2 and a charge of magnitude -1. Therefore, IE2 is about twice as big as IE1.

15 2 Using Coulomb’s law to compare IE2 and IE3: When Be2+ is ionized, the electron is separated from a triply charged ion: Be2+(g) ���� Be3+(g) + e- IE3 is thus the energy required to separate a charge of magnitude +3 and a charge of magnitude -1. Therefore, one would expect IE3 to be the factor 3 bigger than IE1. However, as we can see, IE3 is greater than IE1 by the factor 16.5. This is because the third electron is removed from the first electron shell, which is much closer to the nucleus than the valence shell of Be and Be+. The fact that IE3 of Be is 16.5 greater than IE2, as opposed to 3 times greater, is an experimental evidence of the shell structure of atoms.

15 3 Ca < Mg < S < Cl

16 1 Oxygen

CHEM 121 INTEGRATED RESOURCE PACKAGE

Chapter 2 Exercises Answers

3

19 1 a)

b)

c)

d)

23 1 a)

b)

c)

23 2