Chapter 26, Problem 26.48P

(a)

Interpretation Introduction

Interpretation:

Electronic configuration of $d$-orbital has to be written for high-spin $\text{Ni(II)}$ complex.

Explanation of Solution

Given complex contains $\text{Ni(II)}$ as metal ion.  The Roman numeral that is present in the parenthesis indicates the oxidation state of the metal ion.  Atomic number of nickel is $28$.  Therefore, $\text{Ni(II)}$ will have $26$ electrons.

In the octahedral complex, the $d$ orbitals splits into two energy levels and they are $t_{2g}$ and $e_g$.  In this $t_{2g}$ is the lower energy degenerate orbital and $e_g$ is the higher energy degenerate orbital.

Electronic configuration of nickel is given as shown below;

  $\text{Ni}:[\text{Ar}]4s^2 3d^8$

Nickel loses two electrons to form $\text{Ni(II)}$ ion.  The electronic configuration of $\text{Ni(II)}$ is given as shown below.

    $\text{Ni}:[\text{Ar}]4s^2 3d^8\to \text{Ni(II)}:[\text{Ar}]4s^0 3d^8+2e^{-}$

Therefore, there are eight electrons in $d$-orbital.  As it is given as a high-spin complex, the electrons enter into the higher energy $e_g$ orbital before pairing starts in $t_{2g}$ orbital.  Therefore, the $d$-orbital electronic configuration of $\text{Ni(II)}$ is $t_{2g}^6{e}_g^2$.

(b)

Interpretation Introduction

Interpretation:

Electronic configuration of $d$-orbital has to be written for high-spin $\text{Mn(II)}$ complex.

Explanation of Solution

Given complex contains $\text{Mn(II)}$ as metal ion.  The Roman numeral that is present in the parenthesis indicates the oxidation state of the metal ion.  Atomic number of manganese is $25$.  Therefore, $\text{Mn(II)}$ will have $23$ electrons.

In the octahedral complex, the $d$ orbitals splits into two energy levels and they are $t_{2g}$ and $e_g$.  In this $t_{2g}$ is the lower energy degenerate orbital and $e_g$ is the higher energy degenerate orbital.

Electronic configuration of manganese is given as shown below;

  $\text{Mn}:[\text{Ar}]4s^2 3d^5$

Manganese loses two electrons to form $\text{Mn(II)}$ ion.  The electronic configuration of $\text{Mn(II)}$ is given as shown below.

    $\text{Mn}:[\text{Ar}]4s^2 3d^5\to \text{Mn(II)}:[\text{Ar}]4s^0 3d^5+2e^{-}$

Therefore, there are five electrons in $d$-orbital.  As it is given as a high-spin complex, the electrons enter into the higher energy $e_g$ orbital before pairing starts in $t_{2g}$ orbital.  Therefore, the $d$-orbital electronic configuration of $\text{Mn(II)}$ is $t_{2g}^3{e}_g^2$.

(c)

Interpretation Introduction

Interpretation:

Electronic configuration of $d$-orbital has to be written for low-spin $\text{Fe(III)}$ complex.

Explanation of Solution

Given complex contains $\text{Fe(III)}$ as metal ion.  The Roman numeral that is present in the parenthesis indicates the oxidation state of the metal ion.  Atomic number of iron is $26$.  Therefore, $\text{Fe(III)}$ will have $23$ electrons.

In the octahedral complex, the $d$ orbitals splits into two energy levels and they are $t_{2g}$ and $e_g$.  In this $t_{2g}$ is the lower energy degenerate orbital and $e_g$ is the higher energy degenerate orbital.

Electronic configuration of iron is given as shown below;

  $\text{Fe}:[\text{Ar}]4s^2 3d^6$

Iron loses three electrons to form $\text{Fe(III)}$ ion.  The electronic configuration of $\text{Fe(III)}$ is given as shown below.

    $\text{Fe}:[\text{Ar}]4s^2 3d^6\to \text{Fe(III)}:[\text{Ar}]4s^0 3d^5+3e^{-}$

Therefore, there are five electrons in $d$-orbital.  As it is given as a low-spin complex, the electrons starts to pair in $t_{2g}$ orbital before entering the $e_g$ orbital.  Therefore, the $d$-orbital electronic configuration of $\text{Fe(III)}$ is $t_{2g}^5{e}_g^0$.

(d)

Interpretation Introduction

Interpretation:

Electronic configuration of $d$-orbital has to be written for $\text{Ti(IV)}$ complex.

Explanation of Solution

Given complex contains $\text{Ti(IV)}$ as metal ion.  The Roman numeral that is present in the parenthesis indicates the oxidation state of the metal ion.  Atomic number of titanium is $22$.  Therefore, $\text{Ti(IV)}$ will have $18$ electrons.

In the octahedral complex, the $d$ orbitals splits into two energy levels and they are $t_{2g}$  and $e_g$.  In this $t_{2g}$ is the lower energy degenerate orbital and $e_g$ is the higher energy degenerate orbital.

Electronic configuration of titanium is given as shown below;

  $\text{Ti}:[\text{Ar}]4s^2 3d^2$

Titanium loses four electrons to form $\text{Ti(IV)}$ ion.  The electronic configuration of $\text{Ti(IV)}$ is given as shown below.

    $\text{Ti}:[\text{Ar}]4s^2 3d^2\to \text{Ti(IV)}:[\text{Ar}]4s^0 3d^0+4e^{-}$

Therefore, there are no electrons in $d$-orbital.  Therefore, the $d$-orbital electronic configuration of $\text{Ti(IV)}$ is $t_{2g}^0{e}_g^0$.

(e)

Interpretation Introduction

Interpretation:

Electronic configuration of $d$-orbital has to be written for $\text{Ni(II)}$ complex.

Explanation of Solution

Given complex contains $\text{Ni(II)}$ as metal ion.  The Roman numeral that is present in the parenthesis indicates the oxidation state of the metal ion.  Atomic number of nickel is $28$.  Therefore, $\text{Ni(II)}$ will have $26$ electrons.

In the octahedral complex, the $d$ orbitals splits into two energy levels and they are $t_{2g}$  and $e_g$.  In this $t_{2g}$ is the lower energy degenerate orbital and $e_g$ is the higher energy degenerate orbital.

Electronic configuration of nickel is given as shown below;

  $\text{Ni}:[\text{Ar}]4s^2 3d^8$

Nickel loses two electrons to form $\text{Ni(II)}$ ion.  The electronic configuration of $\text{Ni(II)}$ is given as shown below.

    $\text{Ni}:[\text{Ar}]4s^2 3d^8\to \text{Ni(II)}:[\text{Ar}]4s^0 3d^8+2e^{-}$

Therefore, there are eight electrons in $d$-orbital.

In high-spin complex, the electrons enter into the higher energy $e_g$ orbital before pairing starts in $t_{2g}$ orbital.  Therefore, the $d$-orbital electronic configuration of $\text{Ni(II)}$ in high-spin complex is $t_{2g}^6{e}_g^2$.

In low-spin complex, the electrons pairing starts in $t_{2g}$ orbital before entering into $e_g$ orbital.  Therefore, the $d$-orbital electronic configuration of $\text{Ni(II)}$ in low-spin complex is $t_{2g}^6{e}_g^2$.

It is general that for all the transition metal ions that has more than six electrons in their $d$-orbital will have the same $d$ orbital configuration for high-spin and low-spin complexes.