Chapter 3, Problem 3.01VP

Interpretation Introduction

Interpretation & Concept Introduction:

The elements highlighted for each of the questions.

Answer to Problem 3.01VP

Solution:

a) Purple (Na), Red (Cr), Orange (Au)

b) Blue (Ne)

c) Orange (Au)

d) Red (Cr)

e) Green (Cl), Blue (Ne)

Explanation of Solution

a)

• The electron configuration of the element highlighted in purple is (Na) is $\left[Ne\right] {3s}^1$. It has a single s electron in the outer most third shell.

• The electron configuration of the element highlighted in red (Cr) is $\left[Ar\right] {3d}^5{4s}^1$. It has a single s electron in the outer most fourth shell.

• The electron configuration of the element highlighted in orange (Au) is $\left[Xe\right] {4f}^{14 }{5d}^{10}{6s}^1$. It has a single s electron in the outer most sixth shell.

b)

• The electron configuration of the element highlighted in blue (Ne) is $\left[He\right]{2s}^2{2p}^6$. It has filled sets of 2s and 2p orbitals in the outer most shell.

c)

• The electron configuration of the element highlighted in orange (Au) is $\left[Xe\right] {4f}^{14 }{5d}^{10}{6s}^1$. It has filled set of 5d orbital.

d)

• The electron configuration of the element highlighted in red (Cr) is $\left[Ar\right] {3d}^5{4s}^1$. It has half-filled set of 5d orbital.

e)

• The electron configuration of the element highlighted in green (Cl) is $\left[Ne\right] {3s}^2{3p}^5$. It has two s electrons in the outer most third shell.

• The electron configuration of the element highlighted in blue (Ne) is $\left[He\right]{2s}^2{2p}^6$. It has two s electrons in the outer most second shell.

Conclusion

The elements highlighted for each question are identified on the basis of their location in the periodic table.