Chapter 18, Problem 18.9P

Interpretation Introduction

(a)

Interpretation:

The $d^n$ for the metal in ${\left[\text{W}{\left(\text{CO}\right)}_{\text{5}}\right]}^{2-}$ complex is to be stated.

Concept introduction:

The number of unshared valence electrons on the metal is the number $n$ in the $d^n$ notation. For example, if the metal in a complex has $6$ unshared valence electrons, the complex is said to have $d^6$ configuration.

Answer to Problem 18.9P

The $d^n$ notation for the ${\left[\text{W}{\left(\text{CO}\right)}_{\text{5}}\right]}^{2-}$ complex is $d^8$.

Explanation of Solution

The formula for calculation of $n$ in the $d^n$ notation is given below.

$n=\left(\text{Valence}\text{electrons}\text{in}\text{neutral}\text{M}-\text{Oxidation}\text{state}\text{of}\text{M}\right)$

The formula for the calculation of oxidation state is given below

${\text{X}}_{\text{C}}={\text{X}}_{\text{M}}+{\text{X}}_{\text{L}}\times \text{n}$

• n is the number of ligands.

• ${\text{X}}_{\text{C}}$ is the charge on the complex.

• ${\text{X}}_{\text{M}}$ is the charge on the metal.

• ${\text{X}}_{\text{L}}$ is the charge on the ligand.

Therefore, the oxidation state of $\text{W}$ in ${\left[\text{W}{\left(\text{CO}\right)}_{\text{5}}\right]}^{2-}$ is calculated below.

$\begin{array}{rll}{\text{X}}_{\text{C}}& =& {\text{X}}_{\text{M}}+\left({\text{X}}_{\text{CO}}\times {\text{n}}_{\text{CO}}\right)\\ {\text{X}}_{\text{M}}-0\left(0\right)& =& -2\\ {\text{X}}_{\text{M}}& =& -2\end{array}$

Now, valence electronic configuration of $\text{W}$ is ${\text{5d}}^{\text{4}}{\text{6s}}^2$, so, valence electrons in tungsten are $6$.

L-type ligands have no effect on the $d^n$ notation. Since carbonyl ligand is a L- type ligand it would not have any effect in the $d^n$.

Value of $n$ for the metal complex is calculated below.

$\begin{array}{rll}n& =& \left(\text{Valence}\text{electrons}\text{in}\text{neutral}\text{M}-\text{Oxidation}\text{state}\text{of}\text{M}\right)\\ & =& \text{6}-\left(-\text{2}\right)\\ & =& \text{8}\end{array}$

Therefore, the given complex is a $d^8$ complex

Conclusion

The value of $n$ in the ${\left[\text{W}{\left(\text{CO}\right)}_{\text{5}}\right]}^{2-}$ complex is $8$.

Interpretation Introduction

(b)

Interpretation:

The $d^n$ for the metal in $\text{Pd}{\left({\text{PPh}}_3\right)}_{\text{4}}$ complex is to be stated.

Concept introduction:

The number of unshared valence electrons on the metal is the number $\text{n}$ in the $d^n$ notation. For example, if the metal in a complex has $6$ unshared valence electron the complex is said to have $d^6$ configuration.

Answer to Problem 18.9P

The $d^n$ notation for the $\text{Pd}{\left({\text{PPh}}_3\right)}_{\text{4}}$ complex is $d^{10}$.

Explanation of Solution

The formula for calculation of $\text{n}$ in the $d^n$ notation is given below.

$n=\left(\text{Valence}\text{electrons}\text{in}\text{neutral}\text{M}-\text{Oxidation}\text{state}\text{of}\text{M}\right)$

The oxidation state of the metal is calculated by the formula given below.

$\text{Oxidation}\text{state}\text{of}\text{M}=\left(\begin{array}{l}\text{Net}\text{charge}\text{present over the complex}-\text{Sum of charges of X-type}\\ \text{ ligand in dissosciated state}\end{array}\right)$

Since phosphine is a neutral ligand and there is no overall charge on the complex.

The formula for the calculation of oxidation state is given below

${\text{X}}_{\text{C}}={\text{X}}_{\text{M}}+{\text{X}}_{\text{L}}\times \text{n}$

• n is the number of ligands

• ${\text{X}}_{\text{C}}$ is the charge on the complex

• ${\text{X}}_{\text{M}}$ is the charge on the metal

• ${\text{X}}_{\text{L}}$ is the charge on the ligand

So, the oxidation state of Pd in the complex is calculated as shown below.

$\begin{array}{rll}{\text{X}}_{\text{M}}+{\text{X}}_{{\text{PPh}}_{\text{3}}}\times {\text{n}}_{{\text{PPh}}_{\text{3}}}& =& {\text{X}}_{\text{C}}\\ {\text{X}}_{\text{M}}+0\left(4\right)& =& 0\\ {\text{X}}_{\text{M}}& =& 0\end{array}$

Now, valence electronic configuration of $\text{Pd}$ is ${\text{4d}}^{\text{10}}$ so, valence electrons $=10$

Value of $n$ for the metal complex is calculated below.

$\begin{array}{rll}n& =& \left(\text{Valence}\text{electron}\text{in}\text{neutral}\text{M}-\text{Oxidation}\text{state}\text{of}\text{M}\right)\\ & =& 10-0\\ & =& 10\end{array}$

Therefore, the given complex is a $d^{10}$ complex.

Conclusion

The value of $n$ in the $\text{Pd}{\left({\text{PPh}}_3\right)}_{\text{4}}$ complex is $10$.

Interpretation Introduction

(c)

Interpretation:

The $d^n$ for the metal in $\left[\text{Rh}{\left(\text{PPh}\right)}_{\text{3}}{\left(\text{H}\right)}_{\text{2}}\text{Cl}\right]$ is to be stated.

Concept introduction:

The number of unshared valence electrons on the metal is the number $\text{n}$ in the $d^n$ notation. For example, if the metal in a complex has $6$ unshared valence electron the complex is said to have $d^6$ configuration.

Answer to Problem 18.9P

The $d^n$ notation for the complex is $d^6$.

Explanation of Solution

The given complex is shown below.

Figure 1

The formula for calculation of $\text{n}$ in the $d^n$ notation is given below.

$n=\left(\text{Valence}\text{electrons}\text{in}\text{neutral}\text{M}-\text{Oxidation}\text{state}\text{of}\text{M}\right)$

The oxidation state of metal is calculated by the formula given below.

$\text{Oxidation}\text{state}\text{of}\text{M}=\left(\begin{array}{l}\text{Net}\text{charge}\text{present over the complex}-\text{Sum of charges of X-type}\\ \text{ ligand in dissosciated state}\end{array}\right)$

Since, phosphine is a neutral ligand.and there is no overall charge on the complex. The charge assosciated with chlorine and hydride ligands is $-1$.

The formula for the calculation of oxidation state is given below.

${\text{X}}_{\text{C}}={\text{X}}_{\text{M}}+{\text{X}}_{\text{L}}\times \text{n}$

• n is the number of ligands.

• ${\text{X}}_{\text{C}}$ is the charge on the complex.

• ${\text{X}}_{\text{M}}$ is the charge on the metal.

• ${\text{X}}_{\text{L}}$ is the charge on the ligand.

So, the oxidation state of Rh in the complex is calculated below.

$\begin{array}{rll}{\text{X}}_{\text{M}}+\left({\text{X}}_{\text{Cl}}\times {\text{n}}_{\text{Cl}}\right)+\left({\text{X}}_{\text{H}}\times {\text{n}}_{\text{H}}\right)& =& {\text{X}}_{\text{C}}\\ {\text{X}}_{\text{M}}+1\left(-1\right)+2\left(-1\right)& =& 0\\ {\text{X}}_{\text{M}}-1-2& =& 0\\ {\text{X}}_{\text{M}}& =& +3\end{array}$

The valence electronic configuration of Rh is ${\text{4d}}^{\text{8}}{\text{5s}}^{\text{1}}$ so, number of valence electrons is $9$.

Value of $n$ for the metal complex is calculated below.

$\begin{array}{rll}n& =& \left(\text{Valence}\text{electron}\text{in}\text{neutral}\text{M}-\text{Oxidation}\text{state}\text{of}\text{M}\right)\\ & =& \text{9}-\text{3}\\ & =& \text{6}\end{array}$

Therefore, the given complex is a $d^6$ complex.

Conclusion

The value of $n$ in the $\left[\text{Rh}{\left(\text{PPh}\right)}_{\text{3}}{\left(\text{H}\right)}_{\text{2}}\text{Cl}\right]$ complex is $6$.