Chapter 22, Problem 22.57QP

(a)

Interpretation Introduction

Interpretation:

Using crystal field theory, the energy level diagram of d-orbitals of the complex ion ${\left[V{\left(CN\right)}_6\right]}^{3-}$ in octahedral field has to be drawn.

Concept Introduction:

There are five d-orbitals in a metal ion. They have similar energy levels that they are degenerated. Under the influence of ligands during complex formation, the degeneracy in d-orbitals is destroyed that they are split into two sets of orbitals - one set having lower energy and the another set is higher in energy.  Crystal field splitting refers to the difference in energy levels between these two sets of d-orbitals.

When ligands approach the metal ion the degeneracy in d-orbitals of the metal ion is destroyed and they split into two different energy levels. In case of octahedral complex, the $d_{x^2-y^2}and{d}_{z^2}$ orbitals occupy higher energy level and the $d_{xy},d_{xz},d_{yz}$ orbitals occupy lower energy level.  The energy gap between these two set of orbitals is termed as crystal field splitting energy.

Answer to Problem 22.57QP

The energy level diagram of d-orbitals of the complex ion ${\left[V{\left(CN\right)}_6\right]}^{3-}$ is drawn as,

Two unpaired electrons are present in this complex ion.

Explanation of Solution

Atomic number of Vanadium is $23$ and its electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{3}4s^2$.  In the complex ${\left[V{\left(CN\right)}_6\right]}^{3-}$ Vanadium is in $+3$ oxidation state as follows –

$\begin{array}{l}Oxidation\text{ state of Vanadium}\text{= charge on complex - charge of ligands}\\ \text{= -3-}\left[6\left(-1\right)\right]\\ =\text{ -3+6 = +3}\end{array}$

The electronic configuration of $V^{3+}$ is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^2$. Thus there two d-electrons and they are unpaired and distributed in octahedral field as –

(b)

Interpretation Introduction

Interpretation:

Using crystal field theory, the energy level diagram of d-orbitals of the high-spin complex ion ${\left[Co{\left(C_2{O}_4\right)}_3\right]}^{4-}$ in octahedral field has to be drawn.

Concept Introduction:

There are five d-orbitals in a metal ion. They have similar energy levels that they are degenerated. Under the influence of ligands during complex formation, the degeneracy in d-orbitals is destroyed that they are split into two sets of orbitals - one set having lower energy and the another set is higher in energy.  Crystal field splitting refers to the difference in energy levels between these two sets of d-orbitals.

When ligands approach the metal ion the degeneracy in d-orbitals of the metal ion is destroyed and they split into two different energy levels. In case of octahedral complex, the $d_{x^2-y^2}and{d}_{z^2}$ orbitals occupy higher energy level and the $d_{xy},d_{xz},d_{yz}$ orbitals occupy lower energy level.  The energy gap between these two set of orbitals is termed as crystal field splitting energy.

Answer to Problem 22.57QP

The energy level diagram of d-orbitals of the high-spin complex ion ${\left[Co{\left(C_2{O}_4\right)}_3\right]}^{4-}$ is drawn as,

There are three unpaired electrons in the complex ion ${\left[Co{\left(C_2{O}_4\right)}_3\right]}^{4-}$.

Explanation of Solution

Atomic number of Cobalt is $27$ and its electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{7}4s^2$.  In the complex ${\left[Co{\left(C_2{O}_4\right)}_3\right]}^{4-}$ Cobalt is in $+2$ oxidation state as follows –

$\begin{array}{l}Oxidation\text{ state of Cobalt}\text{= charge on complex - charge of ligands}\\ \text{= -4-}\left[3\left(-2\right)\right]\\ =\text{ -4+6 = +2}\end{array}$

The electronic configuration of $C{o}^{2+}$ is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^7$. Given that the complex is high-spin which means it has unpaired electrons. Thus the seven d-electrons are distributed in octahedral field as –

From the above arrangement we could say there are three unpaired electrons in the complex ion ${\left[Co{\left(C_2{O}_4\right)}_3\right]}^{4-}$.

(c)

Interpretation Introduction

Interpretation:

Using crystal field theory, the energy level diagram of d-orbitals of the low-spin complex ion ${\left[Mn{\left(CN\right)}_6\right]}^{3-}$ in octahedral field has to be drawn.

Concept Introduction:

There are five d-orbitals in a metal ion. They have similar energy levels that they are degenerated. Under the influence of ligands during complex formation, the degeneracy in d-orbitals is destroyed that they are split into two sets of orbitals - one set having lower energy and the another set is higher in energy.  Crystal field splitting refers to the difference in energy levels between these two sets of d-orbitals.

When ligands approach the metal ion the degeneracy in d-orbitals of the metal ion is destroyed and they split into two different energy levels. In case of octahedral complex, the $d_{x^2-y^2}and{d}_{z^2}$ orbitals occupy higher energy level and the $d_{xy},d_{xz},d_{yz}$ orbitals occupy lower energy level.  The energy gap between these two set of orbitals is termed as crystal field splitting energy.

Answer to Problem 22.57QP

The energy level diagram of d-orbitals of the low-spin complex ion ${\left[Mn{\left(CN\right)}_6\right]}^{3-}$ is drawn as,

There are two unpaired electrons in the complex ion ${\left[Mn{\left(CN\right)}_6\right]}^{3-}$.

Explanation of Solution

Atomic number of Manganese is $25$ and its electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}4s^2$.  In the complex ${\left[Mn{\left(CN\right)}_6\right]}^{3-}$ Manganese is in $+3$ oxidation state as follows –

$\begin{array}{l}Oxidation\text{ state of Manganese = charge on complex - charge of ligands}\\ \text{= -3-}\left[6\left(-1\right)\right]\\ =\text{ -3+6 = +3}\end{array}$

The electronic configuration of $M{n}^{3+}$ is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^4$. Given that the complex is low-spin which means the electrons are paired up to maximum possibility. Thus the four d-electrons are distributed in octahedral field as –

From the above arrangement we could say there are two unpaired electrons in the complex ion ${\left[Mn{\left(CN\right)}_6\right]}^{3-}$.