Chapter 26, Problem 26.47P

(a)

Interpretation Introduction

Interpretation:

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

Explanation of Solution

Given complex contains $\text{Nb(III)}$ as metal ion.  The Roman numeral that is present in the parenthesis indicates the oxidation state of the metal ion.  Atomic number of niobium is $41$.  Therefore, $\text{Nb(III)}$ will have $38$ 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 niobium is given as shown below;

  $\text{Nb}:[\text{Kr}]5s^2 4d^3$

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

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

Therefore, there are two electrons in $d$-orbital.  The $d$-orbital electronic configuration of $\text{Nb(III)}$ is $t_{2g}^2{e}_g^0$.

(b)

Interpretation Introduction

Interpretation:

Electronic configuration of $d$-orbital has to be written for $\text{Mo(II)}$ complex that has ${\Delta }_0$ greater than electron-pairing energy..

Explanation of Solution

Given complex contains $\text{Mo(II)}$ as metal ion.  The Roman numeral that is present in the parenthesis indicates the oxidation state of the metal ion.  Atomic number of molybdenum is $42$.  Therefore, $\text{Mo(II)}$ will have $40$ 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 molybdenum is given as shown below;

  $\text{Mo}:[\text{Kr}]5s^2 4d^4$

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

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

Therefore, there are four electrons in $d$-orbital.  As in the problem it is said that the ${\Delta }_0$ is greater than the electron-pairing energy, the electrons will get paired in $t_{2g}$ orbital itself.  Therefore, the $d$-orbital electronic configuration of $\text{Mo(II)}$ is $t_{2g}^4{e}_g^0$.

(c)

Interpretation Introduction

Interpretation:

Electronic configuration of $d$-orbital has to be written for $\text{Mn(II)}$ complex that has ${\Delta }_0$ lesser than electron-pairing energy..

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 in the problem it is said that the ${\Delta }_0$ is lesser than the electron-pairing energy, the electrons will get be filled in all the degenerate orbital before getting paired.  Therefore, the $d$-orbital electronic configuration of $\text{Mn(II)}$ is $t_{2g}^3{e}_g^2$.

(d)

Interpretation Introduction

Interpretation:

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

Explanation of Solution

Given complex contains $\text{Au(I)}$ as metal ion.  The Roman numeral that is present in the parenthesis indicates the oxidation state of the metal ion.  Atomic number of gold is $79$.  Therefore, $\text{Au(I)}$ will have $78$ 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 gold is given as shown below;

  $\text{Au}:[\text{Xe}]6s^1 4f^{14} 5d^{10}$

Gold loses one electron to form $\text{Au(I)}$ ion.  The electronic configuration of $\text{Au(I)}$ is given as shown below.

    $\text{Au}:[\text{Xe}]6s^1 4f^{14} 5d^{10}\to \text{Au(I)}:[\text{Xe}]6s^0 4f^{14} 5d^{10}+1e^{-}$

Therefore, there are ten electrons in $d$-orbital.  The $d$-orbital electronic configuration of $\text{Au(I)}$ is $t_{2g}^6{e}_g^4$.

(e)

Interpretation Introduction

Interpretation:

Electronic configuration of $d$-orbital has to be written for $\text{Ir(III)}$ complex that has ${\Delta }_0$ greater than electron-pairing energy..

Explanation of Solution

Given complex contains $\text{Ir(III)}$ as metal ion.  The Roman numeral that is present in the parenthesis indicates the oxidation state of the metal ion.  Atomic number of iridium is $77$.  Therefore, $\text{Ir(III)}$ will have $74$ 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 iridium is given as shown below;

  $\text{Ir}:[\text{Xe}]6s^2 4f^{14} 5d^7$

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

    $\text{Ir}:[\text{Xe}]6s^2 4f^{14} 5d^7\to \text{Ir(III)}:[\text{Xe}]6s^0 4f^{14} 5d^6+3e^{-}$

Therefore, there are six electrons in $d$-orbital.  As in the problem it is said that the ${\Delta }_0$ is greater than the electron-pairing energy, the electrons will get paired in $t_{2g}$ orbital itself before entering $e_g$ orbital.  Therefore, the $d$-orbital electronic configuration of $\text{Ir(III)}$ is $t_{2g}^6{e}_g^0$.