ATOMIC STRUCTURE AND INTERATOMIC BONDING

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CHAPTER 2

ATOMIC STRUCTURE AND INTERATOMIC BONDING

PROBLEM SOLUTIONS

Electrons in Atoms

Problem 2.1 Cite the difference between atomic mass and atomic weight.

Answer 2.1 Atomic mass is the mass of an individual atom, whereas atomic weight is the average (weighted) of the atomic masses of an atom's naturally occurring isotopes.

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Problem 2.2

Silicon has three naturally occurring isotopes: 92.23% of

28 Si, with an atomic weight of 27.9769 amu, 4.68% of 29 Si, with an atomic weight of 28.9765 amu, and 3.09% of 30 Si, with an atomic weight of 29.9738 amu. On the basis of these data, confirm that the average atomic weight of Si is 28.0854 amu.

Solution 2.2 The average atomic weight of silicon (A Si

• is computed by adding fraction-of-occurrence/atomic weight

products for the three isotopes—i.e., using Equation 2.2. (Remember: fraction of occurrence is equal to the percent of occurrence divided by 100.) Thus

A Si = f 28 Si A 28 Si

• f

29 Si A 29 Si

• f

30 Si A 30 Si

= (0.9223)(27.9769 amu) + (0.0468)(28.9765 amu) + (0.0309)(29.9738 amu) = 28.0854 amu

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Problem 2.3

Zinc has five naturally occurring isotopes: 48.63% of

64 Zn with an atomic weight of 63.929 amu; 27.90% of 66 Zn with an atomic weight of 65.926 amu; 4.10% of 67 Zn with an atomic weight of 66.927 amu; 18.75% of 68 Zn with an atomic weight of 67.925 amu; and 0.62% of 70 Zn with an atomic weight of 69.925 amu. Calculate the average atomic weight of Zn.

Solution 2.3 The average atomic weight of zinc A Zn is computed by adding fraction-of-occurrence—atomic weight products for the five isotopes—i.e., using Equation 2.2. (Remember: fraction of occurrence is equal to the percent of occurrence divided by 100.) Thus

A Zn =f 64 Zn A 64 Zn +f 66 Zn A 66 Zn + f 67 Zn A 67 Zn +f 68 Zn A 68 Zn +f 70 Zn A 70 Zn

Including data provided in the problem statement we solve for A Zn as

A Zn =(0.4863)(63.929 amu) + (0.2790)(65.926 amu)

• (0.0410)(66.927 amu) + (0.1875)(67.925 amu) + (0.0062)(69.925 amu)

= 65.400 amu

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Problem 2.4

Indium has two naturally occurring isotopes:

113 In with an atomic weight of 112.904 amu, and 115 In with an atomic weight of 114.904 amu. If the average atomic weight for In is 114.818 amu, calculate the fraction-of- occurrences of these two isotopes.

Solution 2.4 The average atomic weight of indium (A In

• is computed by adding fraction-of-occurrence—atomic weight

products for the two isotopes—i.e., using Equation 2.2, or

A In =f 113 In A 113 In +f 115 In A 115 In

Because there are just two isotopes, the sum of the fracture-of-occurrences will be 1.000; or

f 113 In +f 115 In

=1.000

which means that f 113 In = 1.000-f 115 In

Substituting into this expression the one noted above for f 113 In , and incorporating the atomic weight values provided in the problem statement yields

114.818 amu=f 113 In A 113 In +f 115 In A 115 In

114.818 amu=(1.000-f 115 In )A 113 In +f 115 In A 115 In

114.818 amu=(1.000-f 115 In )(112.904 amu)+f 115 In (114.904 amu)

114.818 amu=112.904 amu-f 115 In (112.904 amu)+f 115 In (114.904 amu)

Solving this expression for f 115 In yields f 115 In =0.957 . Furthermore, because

f 113 In = 1.000-f 115 In

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