Chapter 22, Problem 22.58QP

(a)

Interpretation Introduction

Interpretation:

Using crystal field theory, the energy level diagram of d-orbitals of the complex ion ${\left[ZrC{l}_6\right]}^{4-}$ 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 22.58QP

The energy level diagram of d-orbitals of the complex ion ${\left[ZrC{l}_6\right]}^{4-}$ is drawn as,

Two unpaired electrons are present in this complex ion.

Explanation of Solution

Atomic number of Zirconium is $40$ and its electronic configuration is $\left[Kr\right]4d^{2}5s^2$.  In the complex ${\left[ZrC{l}_6\right]}^{4-}$ Zirconium is in $+2$ oxidation state as follows –

$\begin{array}{l}Oxidation\text{ state of Zr}\text{= charge on complex - charge of ligands}\\ \text{= -4-}\left[6\left(-1\right)\right]\\ =\text{ -4+6 = +2}\end{array}$

The electronic configuration of $Z{r}^{2+}$ is $\left[Kr\right]4d^2.$ 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 low-spin complex ion ${\left[OsC{l}_6\right]}^{2-}$ 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 22.58QP

The energy level diagram of d-orbitals of the low-spin complex ion ${\left[OsC{l}_6\right]}^{2-}$ is drawn as,

The low-spin complex ion ${\left[OsC{l}_6\right]}^{2-}$ has two unpaired electrons.

Explanation of Solution

Atomic number of Osmium is $76$ and its electronic configuration is $\left[Xe\right]4f^{14}5d^{6}6s^2$.  In the complex ${\left[OsC{l}_6\right]}^{2-}$ Osmium is in $+4$ oxidation state as follows –

$\begin{array}{l}Oxidation\text{ state of Os}\text{= charge on complex - charge of ligands}\\ \text{= -2-}\left[6\left(-1\right)\right]\\ =\text{ -2+6 = +4}\end{array}$

The electronic configuration of $O{s}^{4+}$ is $\left[Xe\right]4f^{14}5d^4$. 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 ${\left[OsC{l}_6\right]}^{2-}$.

(c)

Interpretation Introduction

Interpretation:

Using crystal field theory, the energy level diagram of d-orbitals of the high-spin complex ion ${\left[MnC{l}_6\right]}^{4-}$ 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 22.58QP

The energy level diagram of d-orbitals of the high-spin complex ion ${\left[MnC{l}_6\right]}^{4-}$ in octahedral field is drawn as,

There are five unpaired electrons in the complex ion ${\left[MnC{l}_6\right]}^{4-}$.

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 ${\left[MnC{l}_6\right]}^{4-}$ Manganese is in $+3$ oxidation state as follows –

$\begin{array}{l}Oxidation\text{ state of Manganese = charge on complex - charge of ligands}\\ \text{= -4-}\left[6\left(-1\right)\right]\\ =\text{ -4+6 = +2}\end{array}$

The electronic configuration of $M{n}^{2+}$ is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^5$. Given that the complex is high-spin which means the electrons are unpaired. Thus the five d-electrons are distributed in octahedral field as –

From the above arrangement we could say there are five unpaired electrons in the complex ion ${\left[MnC{l}_6\right]}^{4-}$.