Chapter 22, Problem 22.60QP

(a)

Interpretation Introduction

Interpretation:

Distribution of d-electrons and the geometry of the complex ion ${\left[Pt{\left(N{H}_3\right)}_2{\left(N{O}_2\right)}_2\right]}^{2+}$ has to be determined.

Concept Introduction:

Complex compounds exist in following geometries - tetrahedral, square planar, octahedral etc.

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 tetrahedral complex, the $d_{x^2-y^2}and{d}_{z^2}$ orbitals occupy lower energy level and the $d_{xy},d_{xz},d_{yz}$ orbitals occupy higher energy level.  The energy gap between these two set of orbitals is termed as crystal field splitting energy.

In case of square planar complex, the 5 d-orbitals split into following pattern and it is given below with increasing order of energy.

$d_{xy}={\text{ d}}_{yz}<d_{z^2}<d_{xy}\ll d_{x^2-y^2}$

Answer to Problem 22.60QP

The d-electrons are distributed in the complex ion ${\left[Pt{\left(N{H}_3\right)}_2{\left(N{O}_2\right)}_2\right]}^{2+}$ as follows –

The complex has square planar geometry.

Explanation of Solution

In the complex ion ${\left[Pt{\left(N{H}_3\right)}_2{\left(N{O}_2\right)}_2\right]}^{2+}$ the central metal ion Platinum is in $+4$ oxidation state as follows –

$\begin{array}{l}oxidation\text{ state of Pt}\text{= charge on complex - charge of ligands}\\ \text{= +2-}\left[\text{2}\left(0\right)+2\left(-1\right)\right]\\ \text{= +2-}\left(-2\right)=+4\end{array}$

Atomic number of Platinum is $78.$  The electrons are filled on the basis of building-up rule. The electronic configuration of Platinum could be $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^{2}5p^{6}4f^{14}5d^{9}6s^1$ which can be simplified as $\left[Xe\right]4f^{14}5d^{9}6s^1.$  The electronic configuration of $P{t}^{+4}$ could be $\left[Xe\right]4f^{14}5d^{6}6s^0.$  Thus there are six electrons in d-orbital of $P{t}^{+4}$ and it has two unpaired electrons which are distributed as follows –

The distribution of d-electrons as shown above correlates to that of square planar geometry.  Hence the complex ion ${\left[Pt{\left(N{H}_3\right)}_2{\left(N{O}_2\right)}_2\right]}^{2+}$ exists in square planar geometry.

Conclusion

Splitting of d-orbitals determines the geometry of the complex.

(b)

Interpretation Introduction

Interpretation:

Distribution of d-electrons and the geometry of the complex ion ${\left[MnC{l}_4\right]}^{2-}$ has to be determined.

Concept Introduction:

Complex compounds exist in following geometries - tetrahedral, square planar, octahedral etc.

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 tetrahedral complex, the $d_{x^2-y^2}and{d}_{z^2}$ orbitals occupy lower energy level and the $d_{xy},d_{xz},d_{yz}$ orbitals occupy higher energy level.  The energy gap between these two set of orbitals is termed as crystal field splitting energy.

In case of square planar complex, the 5 d-orbitals split into following pattern and it is given below with increasing order of energy.

$d_{xy}={\text{ d}}_{yz}<d_{z^2}<d_{xy}\ll d_{x^2-y^2}$

Answer to Problem 22.60QP

The d-electrons are distributed in the complex ion ${\left[MnC{l}_4\right]}^{2-}$ as follows –

The complex has tetrahedral geometry.

Explanation of Solution

In the complex ion ${\left[MnC{l}_4\right]}^{2-}$ the central metal ion Manganese is in $+2$ oxidation state as follows –

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

Atomic number of Manganese is $25.$  The electrons are filled on the basis of building-up rule. The electronic configuration of Manganese could be $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}4s^2$ which can be simplified as $\left[Ar\right]3d^{5}4s^2.$  The electronic configuration of $M{n}^{+2}$ could be $\left[Ar\right]3d^5.$  Thus there are five electrons in d-orbital of $M{n}^{+2}$ and all of them are unpaired which are distributed as follows –

The distribution of d-electrons as shown above correlates to that of tetrahedral geometry.  Hence the complex ion ${\left[MnC{l}_4\right]}^{2-}$ exists in tetrahedral geometry.

Conclusion

Splitting of d-orbitals determines the geometry of the complex.

(c)

Interpretation Introduction

Interpretation:

Distribution of d-electrons and the geometry of the complex ion ${\left[NiC{l}_4\right]}^{2-}$ has to be determined.

Concept Introduction:

Complex compounds exist in following geometries - tetrahedral, square planar, octahedral etc.

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 tetrahedral complex, the $d_{x^2-y^2}and{d}_{z^2}$ orbitals occupy lower energy level and the $d_{xy},d_{xz},d_{yz}$ orbitals occupy higher energy level.  The energy gap between these two set of orbitals is termed as crystal field splitting energy.

In case of square planar complex, the 5 d-orbitals split into following pattern and it is given below with increasing order of energy.

$d_{xy}={\text{ d}}_{yz}<d_{z^2}<d_{xy}\ll d_{x^2-y^2}$

Answer to Problem 22.60QP

The d-electrons are distributed in the complex ion ${\left[NiC{l}_4\right]}^{2-}$ as follows –

The complex has tetrahedral geometry.

Explanation of Solution

In the complex ion ${\left[NiC{l}_4\right]}^{2-}$ the central metal ion Nickel is in $+2$ oxidation state as follows –

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

Atomic number of Nickel is $28.$  The electrons are filled on the basis of building-up rule. The electronic configuration of Nickel could be $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{8}4s^2$ which can be simplified as $\left[Ar\right]3d^{8}4s^2.$  The electronic configuration of $N{i}^{+2}$ could be $\left[Ar\right]3d^8.$  Thus there are five electrons in d-orbital of $N{i}^{+2}$ and given that there are two electrons unpaired which are distributed as follows –

The distribution of d-electrons as shown above correlates to that of tetrahedral geometry.  Hence the complex ion ${\left[NiC{l}_4\right]}^{2-}$ exists in tetrahedral geometry.

Conclusion

Splitting of d-orbitals determines the geometry of the complex.

(d)

Interpretation Introduction

Interpretation:

Distribution of d-electrons and the geometry of the complex ion ${\left[Au{F}_4\right]}^{-}$ has to be determined.

Concept Introduction:

Complex compounds exist in following geometries - tetrahedral, square planar, octahedral etc.

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 tetrahedral complex, the $d_{x^2-y^2}and{d}_{z^2}$ orbitals occupy lower energy level and the $d_{xy},d_{xz},d_{yz}$ orbitals occupy higher energy level.  The energy gap between these two set of orbitals is termed as crystal field splitting energy.

In case of square planar complex, the 5 d-orbitals split into following pattern and it is given below with increasing order of energy.

$d_{xy}={\text{ d}}_{yz}<d_{z^2}<d_{xy}\ll d_{x^2-y^2}$

Answer to Problem 22.60QP

The d-electrons are distributed in the complex ion ${\left[Au{F}_4\right]}^{-}$ as follows –

The complex has square planar geometry.

Explanation of Solution

In the complex ion ${\left[Au{F}_4\right]}^{-}$ the central metal ion Gold is in $+3$ oxidation state as follows –

$\begin{array}{l}oxidation\text{ state of Au}\text{= charge on complex - charge of ligands}\\ \text{= -1-}\left[4\left(-1\right)\right]\\ \text{= -1+4}=+3\end{array}$

Atomic number of Gold is $79.$ The electrons are filled on the basis of building-up rule. The electronic configuration of Gold could be $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^{2}5p^{6}4f^{14}5d^{10}6s^1$ which can be simplified as $\left[Xe\right]4f^{14}5d^{10}6s^1.$  The electronic configuration of $A{u}^{+3}$ could be $\left[Xe\right]4f^{14}5d^8.$  Thus there are eight electrons in d-orbital of $A{u}^{+3}$ and given that there are no electrons unpaired which are distributed as follows –

The distribution of d-electrons as shown above correlates to that of square planar geometry.  Hence the complex ion ${\left[Au{F}_4\right]}^{-}$ exists in square planar geometry.

Conclusion

Splitting of d-orbitals determines the geometry of the complex.