Chapter 23, Problem 23.14P

(a)

Interpretation Introduction

Interpretation:

The electronic configuration and the number of unpaired electrons for $S{c}^{3+}ions$ have to be given.

Concept introduction:

The transition elements in Periods 4 and 5 have a general electron configuration of

  $\left[\text{noble gas}\right]{\text{ns}}^{\text{2}}\left(\text{n– 1}\right){\text{d}}^{\text{x}}\text{.}$

Transition metals lose their $ns$ orbital electrons first in forming cations.  After losing the $ns$ electrons, additional electrons may be lost from the $\left(n–1\right)d$ orbitals.

Explanation of Solution

The two $4s$ electrons and one $3d$ electron are removed to form $S{c}^{3+}ions$:

The electronic configuration of $\text{Sc: }\left[\text{Ar}\right]{\text{4s}}^{\text{2}}{\text{3d}}^{\text{1}}$.

The electronic configuration of ${\text{Sc}}^{\text{3+}}\text{: }\left[\text{Ar}\right]\text{ or}{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}\text{. }$

There are no unpaired electrons.

(b)

Interpretation Introduction

Interpretation:

The electronic configuration and the number of unpaired electrons for $C{u}^{2+}ions$ have to be given.

Concept introduction:

The transition elements in Periods 4 and 5 have a general electron configuration of

  $\left[\text{noble gas}\right]{\text{ns}}^{\text{2}}\left(\text{n– 1}\right){\text{d}}^{\text{x}}\text{.}$

Transition metals lose their $ns$ orbital electrons first in forming cations.  After losing the $ns$ electrons, additional electrons may be lost from the $\left(n–1\right)d$ orbitals.

Explanation of Solution

The single $4s$ electron and one $3d$ electron are removed to form $C{u}^{2+}ions$:

The electronic configuration of  $\text{Cu: }\left[\text{Ar}\right]{\text{4s}}^{\text{1}}{\text{3d}}^{\text{10}}\text{.}$

The electronic configuration of ${\text{Cu}}^{\text{2+}}\text{:}\left[\text{Ar}\right]{\text{3d}}^{\text{9}}\text{. }$

There is one unpaired electron.

(c)

Interpretation Introduction

Interpretation:

The electronic configuration and the number of unpaired electrons for $F{e}^{3+}ions$ have to be given.

Concept introduction:

The transition elements in Periods 4 and 5 have a general electron configuration of

  $\left[\text{noble gas}\right]{\text{ns}}^{\text{2}}\left(\text{n– 1}\right){\text{d}}^{\text{x}}\text{.}$

Transition metals lose their $ns$ orbital electrons first in forming cations.  After losing the $ns$ electrons, additional electrons may be lost from the $\left(n–1\right)d$ orbitals.

Explanation of Solution

The two $4s$ electrons and one $3d$ electron are removed to form $F{e}^{3+}ions$:

The electronic configuration of $\text{Fe: }\left[\text{Ar}\right]{\text{4s}}^{\text{2}}{\text{3d}}^{\text{6}}$.

The electronic configuration of ${\text{Fe}}^{\text{3+}}\text{:}\left[\text{Ar}\right]{\text{3d}}^{\text{5}}\text{. }$

There are five unpaired electrons since each of the five $d$ electrons occupies its own orbital.

(d)

Interpretation Introduction

Interpretation:

The electronic configuration and the number of unpaired electrons for $N{b}^{3+}ions$ have to be given.

Concept introduction:

The transition elements in Periods 4 and 5 have a general electron configuration of

  $\left[\text{noble gas}\right]{\text{ns}}^{\text{2}}\left(\text{n– 1}\right){\text{d}}^{\text{x}}\text{.}$

Transition metals lose their $ns$ orbital electrons first in forming cations.  After losing the $ns$ electrons, additional electrons may be lost from the $\left(n–1\right)d$ orbitals.

Explanation of Solution

The two $5s$ electrons and one $4d$ electron are removed to form $N{b}^{3+}ions$:

The electronic configuration of $\text{Nb: }\left[\text{Kr}\right]{\text{5s}}^{\text{2}}{\text{4d}}^{\text{3}}\text{.}$

The electronic configuration of ${\text{Nb}}^{\text{3+}}\text{:}\left[\text{Kr}\right]{\text{4d}}^{\text{2}}\text{. }$

There are two unpaired electrons.