Chapter 19, Problem 7E

(a)

Interpretation Introduction

Interpretation: The electronic configuration of ${\text{Cr, Cr}}^{\text{2+}}{\text{, Cr}}^{\text{3+ }}$ ions needs to be determined.

Concept Introduction: The electronic configuration of an atom or ion represents the distribution of electrons in the atom or ion. It represents the sequence of atomic orbitals in the increasing order of their energy.

In each orbital, the electrons must be filled according to their maximum capacity. For example; s-orbital can accommodate 2 electrons, p-orbital can accommodate 6 electrons, d-orbital can accommodate 10 electrons and f-orbital can accommodate 14 electrons maximum.

Answer to Problem 7E

  $\begin{array}{l}{\text{Cr = 1s}}^{\text{2}}{\text{,2s}}^{\text{2}}{\text{,2p}}^{\text{6}}{\text{,3s}}^{\text{2}}{\text{,3p}}^{\text{6}}{\text{,3d}}^5{\text{,4s}}^1\\ {\text{Cr}}^{\text{2+}}{\text{= 1s}}^{\text{2}}{\text{,2s}}^{\text{2}}{\text{,2p}}^{\text{6}}{\text{,3s}}^{\text{2}}{\text{,3p}}^{\text{6}}{\text{,3d}}^{\text{4}}\\ {\text{Cr}}^{\text{3+ }}={\text{1s}}^{\text{2}}{\text{,2s}}^{\text{2}}{\text{,2p}}^{\text{6}}{\text{,3s}}^{\text{2}}{\text{,3p}}^{\text{6}}{\text{,3d}}^3\end{array}$

Explanation of Solution

The atomic number of Cr is 24. It is placed in the group-VIB of the periodic table. It is a transition element therefore the 3d orbitals must be filled before 4s orbital. But to form the cation, the electrons must be removed from the outermost shell of the atom. Therefore the electronic configuration of ${\text{Cr, Cr}}^{\text{2+}}{\text{, Cr}}^{\text{3+ }}$ must be:

  $\begin{array}{l}{\text{Cr = 1s}}^{\text{2}}{\text{,2s}}^{\text{2}}{\text{,2p}}^{\text{6}}{\text{,3s}}^{\text{2}}{\text{,3p}}^{\text{6}}{\text{,3d}}^5{\text{,4s}}^1\\ {\text{Cr}}^{\text{2+}}{\text{= 1s}}^{\text{2}}{\text{,2s}}^{\text{2}}{\text{,2p}}^{\text{6}}{\text{,3s}}^{\text{2}}{\text{,3p}}^{\text{6}}{\text{,3d}}^{\text{4}}\\ {\text{Cr}}^{\text{3+ }}={\text{1s}}^{\text{2}}{\text{,2s}}^{\text{2}}{\text{,2p}}^{\text{6}}{\text{,3s}}^{\text{2}}{\text{,3p}}^{\text{6}}{\text{,3d}}^3\end{array}$

(b)

Interpretation Introduction

Interpretation: The electronic configuration of ${\text{Cu, Cu}}^{\text{+}}{\text{, Cu}}^{\text{2+ }}$ ions needs to be determined.

Concept Introduction: The electronic configuration of an atom or ion represents the distribution of electrons in the atom or ion. It represents the sequence of atomic orbitals in the increasing order of their energy.

In each orbital, the electrons must be filled according to their maximum capacity. For example; s-orbital can accommodate 2 electrons, p-orbital can accommodate 6 electrons, d-orbital can accommodate 10 electrons and f-orbital can accommodate 14 electrons maximum.

Answer to Problem 7E

  $\begin{array}{l}{\text{Cu = 1s}}^{\text{2}}{\text{,2s}}^{\text{2}}{\text{,2p}}^{\text{6}}{\text{,3s}}^{\text{2}}{\text{,3p}}^{\text{6}}{\text{,3d}}^{10}{\text{,4s}}^1\\ {\text{Cu}}^{\text{+}}{\text{= 1s}}^{\text{2}}{\text{,2s}}^{\text{2}}{\text{,2p}}^{\text{6}}{\text{,3s}}^{\text{2}}{\text{,3p}}^{\text{6}}{\text{,3d}}^{10}\\ {\text{Cu}}^{\text{2+ }}={\text{1s}}^{\text{2}}{\text{,2s}}^{\text{2}}{\text{,2p}}^{\text{6}}{\text{,3s}}^{\text{2}}{\text{,3p}}^{\text{6}}{\text{,3d}}^9\end{array}$

Explanation of Solution

The atomic number of Cu is 29. It is placed in the group-IB of the periodic table. It is a transition element therefore the 3d orbitals must be filled before 4s orbital. But to form the cation, the electrons must be removed from the outermost shell of the atom. Therefore the electronic configuration of ${\text{Cu, Cu}}^{\text{+}}{\text{, Cu}}^{\text{2+ }}$ must be:

  $\begin{array}{l}{\text{Cu = 1s}}^{\text{2}}{\text{,2s}}^{\text{2}}{\text{,2p}}^{\text{6}}{\text{,3s}}^{\text{2}}{\text{,3p}}^{\text{6}}{\text{,3d}}^{10}{\text{,4s}}^1\\ {\text{Cu}}^{\text{+}}{\text{= 1s}}^{\text{2}}{\text{,2s}}^{\text{2}}{\text{,2p}}^{\text{6}}{\text{,3s}}^{\text{2}}{\text{,3p}}^{\text{6}}{\text{,3d}}^{10}\\ {\text{Cu}}^{\text{2+ }}={\text{1s}}^{\text{2}}{\text{,2s}}^{\text{2}}{\text{,2p}}^{\text{6}}{\text{,3s}}^{\text{2}}{\text{,3p}}^{\text{6}}{\text{,3d}}^9\end{array}$

(c)

Interpretation Introduction

Interpretation: The electronic configuration of ${\text{V, V}}^{\text{2+}}{\text{, V}}^{\text{3+ }}$ ions needs to be determined.

Concept Introduction: The electronic configuration of an atom or ion represents the distribution of electrons in the atom or ion. It represents the sequence of atomic orbitals in the increasing order of their energy.

In each orbital, the electrons must be filled according to their maximum capacity. For example; s-orbital can accommodate 2 electrons, p-orbital can accommodate 6 electrons, d-orbital can accommodate 10 electrons and f-orbital can accommodate 14 electrons maximum.

Answer to Problem 7E

  $\begin{array}{l}{\text{V = 1s}}^{\text{2}}{\text{,2s}}^{\text{2}}{\text{,2p}}^{\text{6}}{\text{,3s}}^{\text{2}}{\text{,3p}}^{\text{6}}{\text{,3d}}^3{\text{,4s}}^2\\ {\text{V}}^{\text{2+}}{\text{= 1s}}^{\text{2}}{\text{,2s}}^{\text{2}}{\text{,2p}}^{\text{6}}{\text{,3s}}^{\text{2}}{\text{,3p}}^{\text{6}}{\text{,3d}}^3\\ {\text{V}}^{\text{3+ }}={\text{1s}}^{\text{2}}{\text{,2s}}^{\text{2}}{\text{,2p}}^{\text{6}}{\text{,3s}}^{\text{2}}{\text{,3p}}^{\text{6}}{\text{,3d}}^2\end{array}$

Explanation of Solution

The atomic number of V is 23. It is placed in the group-VB of the periodic table. It is a transition element therefore the 3d orbitals must be filled before 4s orbital. But to form the cation, the electrons must be removed from the outermost shell of the atom. Therefore, the electronic configuration of ${\text{V, V}}^{\text{2+}}{\text{, V}}^{\text{3+ }}$ must be:

  $\begin{array}{l}{\text{V = 1s}}^{\text{2}}{\text{,2s}}^{\text{2}}{\text{,2p}}^{\text{6}}{\text{,3s}}^{\text{2}}{\text{,3p}}^{\text{6}}{\text{,3d}}^3{\text{,4s}}^2\\ {\text{V}}^{2+}{\text{= 1s}}^{\text{2}}{\text{,2s}}^{\text{2}}{\text{,2p}}^{\text{6}}{\text{,3s}}^{\text{2}}{\text{,3p}}^{\text{6}}{\text{,3d}}^3\\ {\text{V}}^{\text{3+ }}={\text{1s}}^{\text{2}}{\text{,2s}}^{\text{2}}{\text{,2p}}^{\text{6}}{\text{,3s}}^{\text{2}}{\text{,3p}}^{\text{6}}{\text{,3d}}^2\end{array}$