Chapter 22, Problem 5QP

Write the electron configurations of the following ions: ${\text{V}}^{\text{5+}}{\text{,Cr}}^{\text{3+}}{\text{,Mn}}^{\text{2+}}{\text{,Fe}}^{\text{3+}}{\text{,Cu}}^{\text{2+}}{\text{,Sc}}^{\text{3+}}{\text{,Ti}}^{\text{+}}\text{.}$

Interpretation Introduction

Interpretation:

The electronic configuration of the given transition metal ions is to be determined.

Concept introduction:

The transition metals are those that form at least one stable ion with incomplete d subshell.

The transition metals show a variable oxidation state.

The transition metals have partially filled d subshell.

The energy of electrons present in the d subshell is similar.

The nearest inert gas to the third transition series is argon with atomic number 18.

Answer to Problem 5QP

Solution: $\begin{array}{l}{\text{V}}^{\text{5+}}\text{: }\left[\text{Ar}\right]{\text{; Cr}}^{\text{3+}}\text{: }\left[\text{Ar}\right]{\text{3d}}^{\text{3}}{\text{; Mn}}^{\text{2+}}\text{: }\left[\text{Ar}\right]{\text{3d}}^{\text{5}}\text{; }\\ {\text{Fe}}^{\text{3+}}\text{: }\left[\text{Ar}\right]{\text{3d}}^{\text{5}}{\text{; Cu}}^{\text{2+}}\text{: }\left[\text{Ar}\right]{\text{3d}}^{\text{9}}{\text{; Sc}}^{\text{3+}}\text{: }\left[\text{Ar}\right]{\text{; Ti}}^{\text{4+}}\text{: }\left[\text{Ar}\right]\end{array}$

Explanation of Solution

The transition metals have unpaired electrons in a d subshell and the species containing unpaired electrons are unstable and readily lose an electron to form a bond with another chemical species.

Atomic number of vanadium atom is $\text{23}$.

The electronic configuration of vanadium atom is as follows:

${\text{V :1s}}^{\text{2}}{\text{ 2s}}^{\text{2}}{\text{ 2p}}^{\text{6}}{\text{ 3s}}^{\text{2}}{\text{ 3p}}^{\text{6}}{\text{ 4s}}^{\text{2}}{\text{ 3d}}^{\text{3}}$

When a vanadium atom loses five electrons from its valence shell, it forms a vanadium ion with $+5$ oxidation state.

Therefore, the electronic configuration of vanadium ion is as follows:

${\text{V}}^{5+}{\text{: 1s}}^{\text{2}}{\text{ 2s}}^{\text{2}}{\text{ 2p}}^{\text{6}}{\text{ 3s}}^{\text{2}}{\text{ 3p}}^{\text{6}}$

The electronic configuration of vanadium ion in terms of the nearest inert gas is as follows:

${\text{V}}^{\text{5+}}\text{: }\left[\text{Ar}\right]$

Atomic number of a chromium atom is $\text{24}$.

The electronic configuration of a chromium atom is as follows:

${\text{Cr :1s}}^{\text{2}}{\text{ 2s}}^{\text{2}}{\text{ 2p}}^{\text{6}}{\text{ 3s}}^{\text{2}}{\text{ 3p}}^{\text{6}}{\text{ 4s}}^{\text{1}}{\text{ 3d}}^{\text{5}}$

When chromium atom loses three electrons from its valence shell, it forms a chromium ion with $+3$ oxidation state.

Therefore, the electronic configuration of the chromium ion is as follows:

${\text{Cr}}^{3+}{\text{: 1s}}^{\text{2}}{\text{ 2s}}^{\text{2}}{\text{ 2p}}^{\text{6}}{\text{ 3s}}^{\text{2}}{\text{ 3p}}^{\text{6}}{\text{3d}}^3$

The electronic configuration of a chromium ion in terms of the nearest inert gas is as follows:

${\text{Cr}}^{\text{3+}}\text{: }\left[\text{Ar}\right]{\text{3d}}^{\text{3}}$

Atomic Number of manganese atom is $\text{25}$.

The electronic configuration of a manganese atom is as follows:

${\text{Mn :1s}}^{\text{2}}{\text{ 2s}}^{\text{2}}{\text{ 2p}}^{\text{6}}{\text{ 3s}}^{\text{2}}{\text{ 3p}}^{\text{6}}{\text{ 4s}}^{\text{2}}{\text{ 3d}}^{\text{5}}$

When manganese atom loses two electrons from its valence shell, it forms a manganese ion with $+2$ oxidation state.

Therefore, the electronic configuration of manganese ion is as follows:

${\text{Mn}}^{2+}{\text{: 1s}}^{\text{2}}{\text{ 2s}}^{\text{2}}{\text{ 2p}}^{\text{6}}{\text{ 3s}}^{\text{2}}{\text{ 3p}}^{\text{6}}{\text{3d}}^5$

The electronic configuration of a manganese ion in terms of the nearest inert gas is as follows:

${\text{Mn}}^{\text{2+}}\text{: }\left[\text{Ar}\right]{\text{3d}}^{\text{5}}$

Atomic number of an iron atom is $\text{26}$.

The electronic configuration of an iron atom is as follows:

${\text{Fe :1s}}^{\text{2}}{\text{ 2s}}^{\text{2}}{\text{ 2p}}^{\text{6}}{\text{ 3s}}^{\text{2}}{\text{ 3p}}^{\text{6}}{\text{ 4s}}^{\text{2}}{\text{ 3d}}^{\text{6}}$

When an iron atom loses three electrons from its valence shell, it forms an iron ion with $+3$ oxidation state.

Therefore, the electronic configuration of an iron ion is as follows:

${\text{Fe}}^{3+}{\text{: 1s}}^{\text{2}}{\text{ 2s}}^{\text{2}}{\text{ 2p}}^{\text{6}}{\text{ 3s}}^{\text{2}}{\text{ 3p}}^{\text{6}}{\text{3d}}^5$

The electronic configuration of an iron ion in terms of the nearest inert gas is as follows:

${\text{Fe}}^{\text{3+}}\text{: }\left[\text{Ar}\right]{\text{3d}}^{\text{5}}$

Atomic number of copper atom is $\text{29}$.

The electronic configuration of a copper atom is as follows:

${\text{Cu :1s}}^{\text{2}}{\text{ 2s}}^{\text{2}}{\text{ 2p}}^{\text{6}}{\text{ 3s}}^{\text{2}}{\text{ 3p}}^{\text{6}}{\text{ 4s}}^{\text{1}}{\text{ 3d}}^{\text{10}}$

When a copper atom loses two electrons from its valence shell, it forms a copper ion with $+2$ oxidation state.

Therefore, the electronic configuration of a copper ion is as follows:

${\text{Cu}}^{2+}{\text{: 1s}}^{\text{2}}{\text{ 2s}}^{\text{2}}{\text{ 2p}}^{\text{6}}{\text{ 3s}}^{\text{2}}{\text{ 3p}}^{\text{6}}{\text{3d}}^9$

The electronic configuration of a copperion in terms of the nearest inert gas is as follows:

${\text{Cu}}^{2+}\text{: }\left[\text{Ar}\right]{\text{3d}}^{\text{9}}$

Atomic number of a scandium atom is $\text{21}$.

The electronic configuration of a scandium atom is as follows:

${\text{Sc :1s}}^{\text{2}}{\text{ 2s}}^{\text{2}}{\text{ 2p}}^{\text{6}}{\text{ 3s}}^{\text{2}}{\text{ 3p}}^{\text{6}}{\text{ 4s}}^{\text{2}}{\text{ 3d}}^{\text{1}}$

When a scandiumatom loses three electrons from its valence shell, it forms a scandium ion with $+3$ oxidation state.

Therefore, the electronic configuration of a scandium ion is as follows:

${\text{Sc}}^{3+}{\text{: 1s}}^{\text{2}}{\text{ 2s}}^{\text{2}}{\text{ 2p}}^{\text{6}}{\text{ 3s}}^{\text{2}}{\text{ 3p}}^{\text{6}}$

The electronic configuration of a scandium ion in terms of the nearest inert gas is as follows:

${\text{Sc}}^{3+}\text{: }\left[\text{Ar}\right]$

Atomic number of a titanium atom is $\text{22}$.

The electronic configuration of atitanium atom is as follows:

${\text{Ti :1s}}^{\text{2}}{\text{ 2s}}^{\text{2}}{\text{ 2p}}^{\text{6}}{\text{ 3s}}^{\text{2}}{\text{ 3p}}^{\text{6}}{\text{ 4s}}^{\text{2}}{\text{ 3d}}^{\text{2}}$

When titanium atom loses four electrons from its valence shell, it forms a titanium ion with $+4$ oxidation state.

Therefore, the electronic configuration of a titanium ion is as follows:

${\text{Ti}}^{4+}{\text{: 1s}}^{\text{2}}{\text{ 2s}}^{\text{2}}{\text{ 2p}}^{\text{6}}{\text{ 3s}}^{\text{2}}{\text{ 3p}}^{\text{6}}$

The electronic configuration of a titanium ion in terms of the nearest inert gas is as follows:

${\text{Ti}}^{4+}\text{: }\left[\text{Ar}\right]$