Chapter 23, Problem 23.15P

(a)

Interpretation Introduction

Interpretation:

The electronic configuration and the number of unpaired electrons for $C{r}^{3+}$ ion 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 $C{r}^{3+}ions$:

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

The electronic configuration of $C{r}^{\text{3+}}\text{: }\left[\text{Ar}\right]{\text{ 3d}}^{\text{3}}$

There are three unpaired electrons.

(b)

Interpretation Introduction

Interpretation:

The electronic configuration and the number of unpaired electrons for $T{i}^{4+}ions$ 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 $4s$ electron and two $3d$ electron are removed to form $T{i}^{4+}ions$:

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

The electronic configuration of $\text{Ti: }\left[\text{Ar}\right]\text{0}\text{.}$

There is no unpaired electron.

(c)

Interpretation Introduction

Interpretation:

The electronic configuration and the number of unpaired electrons for $C{o}^{3+}ions$ 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 $C{o}^{3+}ions$:

The electronic configuration of $\text{Co: }\left[\text{Ar}\right]{\text{4s}}^{\text{2}}{\text{3d}}^7$.

The electronic configuration of $C{o}^{\text{3+}}\text{:}\left[\text{Ar}\right]{\text{3d}}^6\text{. }$

There are four 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 $T{a}^{2+}ions$ 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.

Transition elements in Periods 6 and 7 have a general electron configuration of

  $\left[\text{noble gas}\right]{\text{ns}}^{\text{2}}\left(\text{n– 2}\right){\text{f}}^{\text{14}}\left(\text{n– 1}\right){\text{d}}^{\text{x}}\text{.}$

Explanation of Solution

The two $5s$ electrons and one $4d$ electron are removed to form $T{a}^{2+}ions$:

The electronic configuration of $\text{Ta: }\left[Xe\right]{\text{6s}}^{\text{2}}{\text{4f}}^{\text{14}}{\text{5d}}^{\text{3}}\text{.}$

The electronic configuration of $\text{Ta: }\left[Xe\right]{\text{4f}}^{\text{14}}{\text{5d}}^{\text{3}}\text{.}$

There are three unpaired electrons.