Chapter 7, Problem 60QP

Write the ground-state electron configurations of the following transition metal ions: $\left(\text{a}\right){\text{ Sc}}^{\text{3+}}\text{,}\left(\text{b}\right){\text{Ti}}^{\text{4+}}\text{,}\left(\text{c}\right){\text{ V}}^{\text{5+}}\text{, }\left(\text{d}\right){\text{ Cr}}^{\text{3+}}\text{, }\left(\text{e}\right){\text{ Bln}}^{\text{2+}}\text{, }\left(\text{f}\right){\text{ Fe}}^{\text{2+}}\text{, }\left(\text{g}\right){\text{ Fe}}^{\text{3+}}\text{, }\left(\text{h}\right){\text{Co}}^{\text{2+}}\text{,}\left(\text{i}\right){\text{Ni}}^{\text{2+}}{\text{, (j)Cu}}^{\text{+}}\text{,}\left(\text{k}\right){\text{Cu}}^{\text{2+}}\text{,}\left(\text{1}\right){\text{ Ag}}^{\text{+}}\text{, }\left(\text{m}\right){\text{Au}}^{\text{+}}\text{,}\left(\text{n}\right){\text{Au}}^{\text{3+}}\text{,}\left(\text{o}\right){\text{ Pt}}^{\text{2+}}\text{,}$

Interpretation Introduction

Interpretation:

The ground-state electronic configuration of the given ions is to be written.

Concept introduction:

Elements that lose electrons are called cations. They have a lesser number of electrons than that of their neutral states.

Elements that gain electrons are called anions. They have a greater number of electrons than that of their neutral states.

Answer to Problem 60QP

Solution:

a) $[\text{Ar}]3d^{0}4s^0$

b) $[\text{Ar}]3d^{0}4s^0$

c) $[\text{Ar}]3d^{0}4s^0$

d) $[\text{Ar}]3d^{3}4s^0$

e) $[\text{Ar}]3d^{5}4s^0$

f) $[\text{Ar}]3d^{6}4s^0$

g) $[\text{Ar}]3d^{5}4s^0$

h) $[\text{Ar}]3d^{7}4s^0$

i) $[\text{Ar}]3d^{8}4s^0$

j) $[\text{Ar}]3d^{10}4s^0$

k) $[\text{Ar}]3d^{9}4s^0$

l) $[\text{Kr}]4d^{10}5s^0$

m) $[\text{Xe}]5d^{10}6s^0$

n) $[\text{Xe}]5d^{8}6s^0$

o) $[\text{Xe}]5d^{8}6s^0$

Explanation of Solution

a)Ion ${\text{Sc}}^{\text{3+}}$

The neutral state of scandium is ${\text{Sc}}_{21}$ with the electronic configuration $[\text{Ar}]3d^{1}4s^2$. ${\text{Sc}}^{\text{3+}}$ has eighteen electrons. The ground state electronic configuration of ${\text{Sc}}^{\text{3+}}$ is $[\text{Ar}]3d^{0}4s^0$. It is isoelectronic with ${\text{Ar}}_{18}$.

So, its ground-state electronic configuration is $[\text{Ar}]$.

b)Ion ${\text{Ti}}^{4+}$

The neutral state of titanium is ${\text{Ti}}_{22}$ with the electronic configuration $[\text{Ar}]3d^{2}4s^2$. ${\text{Ti}}^{4+}$ has 18 electrons. Electronic configuration of ${\text{Ti}}^{4+}$ is $[\text{Ar}]$. It is isoelectronic with ${\text{Ar}}_{18}$ and ${\text{Sc}}^{\text{3+}}$. However, titanium does not prefer ${\text{Ti}}^{4+}$ state as there is a lot of energy involved in this configuration.

Its ground-state electronic configuration is $[\text{Ar}]$.

c)Ion ${\text{V}}^{5+}$

The neutral state of vanadium is ${\text{V}}_{23}$, having theelectronic configuration $[\text{Ar}]3d^{3}4s^2$. ${\text{V}}^{5+}$ has eighteen electrons. It is isoelectronic with ${\text{Ar}}_{18}$, ${\text{Sc}}^{\text{3+}}$, and ${\text{Ti}}^{4+}$. However, vanadium does not prefer ${\text{V}}^{5+}$ state as there is a lot of energy involved in this configuration.

So, its ground-state electronic configuration is $[\text{Ar}]$.

d)Ion ${\text{Cr}}^{3+}$

The neutral state of chromium is ${\text{Cr}}_{24}$, having theelectronic configuration $[\text{Ar}]3d^{5}4s^1$. To attain the half-filled stability, chromium exists in this configuration. ${\text{Cr}}^{3+}$ has twenty-one electrons. Electronic configuration of ${\text{Cr}}^{3+}$ is $[\text{Ar}]3d^3$.

So, its ground-state electronic configuration is $[\text{Ar}]3d^3$.

e)Ion ${\text{Mn}}^{2+}$

The neutral state of manganese is ${\text{Mn}}_{25}$, having a very stable electronic configuration $[\text{Ar}]3d^{5}4s^2$. ${\text{Mn}}^{2+}$ has twenty-three electrons. Electronic configuration of ${\text{Mn}}^{2+}$ is $[\text{Ar}]3d^5$. Manganese is stable in this configuration because of the half-filled stability of its $d-\text{orbital}$.

So, its ground-state electronic configuration is $[\text{Ar}]3d^5$.

f)Ion ${\text{Fe}}^{\text{2+}}$

The neutral state of ferrous (iron) is ${\text{Fe}}_{26}$, having theelectronic configuration $[\text{Ar}]3d^{6}4s^2$. ${\text{Fe}}^{\text{2+}}$(ferrous ion) has twenty-four electrons. Electronic configuration of ${\text{Fe}}^{\text{2+}}$ is $[\text{Ar}]3d^6$.

So, its ground-state electronic configuration is $[\text{Ar}]3d^6$.

g)Ion ${\text{Fe}}^{\text{3+}}$

The neutral state of ferrous (iron) is ${\text{Fe}}_{26}$, having theelectronic configuration $[\text{Ar}]3d^{6}4s^2$. ${\text{Fe}}^{\text{3+}}$(ferric ion) has twenty-three electrons. Electronic configuration of ${\text{Fe}}^{\text{2+}}$ is $[\text{Ar}]3d^5$. Ferric ion is more stable than that of ferrous ion because of the half-filled stability of its $d-\text{orbital}$, and hence, ferrous readily changes from $+2$ to $+3$ oxidation state.

So, its ground-state electronic configuration is $[\text{Ar}]3d^5$.

h) Ion ${\text{Co}}^{2+}$

The neutral state of cobalt is ${\text{Co}}_{27}$, having theelectronic configuration $[\text{Ar}]3d^{7}4s^2$. ${\text{Co}}^{2+}$ has twenty-five electrons. Electronic configuration of ${\text{Co}}^{2+}$ is $[\text{Ar}]3d^7$.

So, its ground-state electronic configuration is $[\text{Ar}]3d^7$.

i) Ion ${\text{Ni}}^{2+}$

The neutral state of nickel is ${\text{Ni}}_{28}$, having theelectronic configuration $[\text{Ar}]3d^{8}4s^2$. ${\text{Ni}}^{2+}$ has twenty-six electrons, with the electronic configuration $[\text{Ar}]3d^8$.

So, its ground-state electronic configuration is $[\text{Ar}]3d^8$.

j)Ion ${\text{Cu}}^{+}$

The neutral state of copper is ${\text{Cu}}_{29}$, having theelectronic configuration $[\text{Ar}]3d^{10}4s^1$. ${\text{Cu}}^{+}$(cuprous ion) has twenty-eight electrons, with the electronic configuration $[\text{Ar}]3d^{10}$. Cuprous ion is very stable because of the fully-filled stability of its $d-\text{orbital}$.

So, its ground-state electronic configuration is $[\text{Ar}]3d^{10}$.

k) Ion ${\text{Cu}}^{2+}$

The neutral state of copper is ${\text{Cu}}_{29}$, having theelectronic configuration $[\text{Ar}]3d^{10}4s^1$. The ${\text{Cu}}^{2+}$(cupric ion) has twenty-seven electrons, with the electronic configuration $[\text{Ar}]3d^9$. Cupric ion is less stable than that of the cuprous ion.

So, its ground-state electronic configuration is $[\text{Ar}]3d^9$.

l)Ion ${\text{Ag}}^{\text{+}}$

The neutral state of silver is ${\text{Ag}}_{47}$, having theelectronic configuration $[\text{Kr}]4d^{10}5s^1$. The ${\text{Ag}}^{\text{+}}$ ion has forty-six electrons, with the electronic configuration $[\text{Kr}]4d^{10}$. ${\text{Ag}}^{\text{+}}$ ion is very stable because of the fully-filled stability of its $d-\text{orbital}$.

So, its ground-state electronic configuration is $[\text{Kr}]4d^{10}$.

m)Ion ${\text{Au}}^{+}$

The neutral state of gold is ${\text{Au}}_{79}$, having theelectronic configuration of $[\text{Xe}]5d^{10}6s^1$. The ${\text{Au}}^{+}$ ion has seventy-eight electrons, with the electronic configuration $[\text{Xe}]5d^{10}$. To attain fully-filled stability of its $d-\text{orbital}$, gold exists in this electronic configuration.

So, its ground-state electronic configuration is $[\text{Xe}]5d^{10}$.

n)Ion ${\text{Au}}^{3+}$

The neutral state of gold is ${\text{Au}}_{79}$, having theelectronic configuration $[\text{Xe}]5d^{10}6s^1$. The ${\text{Au}}^{3+}$ ion has seventy-six electrons, with the electronic configuration $[\text{Xe}]5d^8$. ${\text{Au}}^{3+}$ is less stable than that of ${\text{Au}}^{+}$ because of its incomplete $d-\text{orbital}$.

So, its ground-state electronic configuration is $[\text{Xe}]5d^8$.

o)Ion ${\text{Pt}}^{2+}$

The neutral state of platinum is ${\text{Au}}_{79}$, having theelectronic configuration $[\text{Xe}]5d^{9}6s^1$. ${\text{Pt}}^{2+}$ ion has seventy-seven electrons, with the electronic configuration $[\text{Xe}]5d^8$. It is isoelectronic with ${\text{Au}}^{3+}$.

So, its ground-state electronic configuration is $[\text{Xe}]5d^8$.