Chapter 19, Problem 20E

(a)

Interpretation Introduction

Interpretation:

The electronic configuration of the transition metal ion in ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\left[\text{Fe}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{2}}{\text{Cl}}_{\text{4}}\right]$ should be stated.

Concept Introduction:

The study of the compounds that have central atom that is surrounded by the anions or molecules is known as coordination chemistry. The anions or molecules that surrounds central atom are known as ligands. These ligands are attached to the central atom by the dative bonds known as coordinate bonds.

Answer to Problem 20E

The electronic configuration of ${\text{Fe}}^{2+}$ is $\left[\text{Ar}\right]3d^6$ .

Explanation of Solution

The dissociation of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\left[\text{Fe}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{2}}{\text{Cl}}_{\text{4}}\right]$ is shown below:

  ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\left[\text{Fe}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{2}}{\text{Cl}}_{\text{4}}\right]\to 2{\text{NH}}_{\text{4}}{}^{+}+{\left[\text{Fe}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{2}}{\text{Cl}}_{\text{4}}\right]}^{2-}$

The oxidation state of central metal is calculated as follows:

  $\begin{array}{c}\text{O}\text{.S}\left(\text{Fe}\right)+2\times \text{O}\text{.S}\left({\text{H}}_{\text{2}}\text{O}\right)+4\times \text{O}\text{.S}\left(\text{Cl}\right)=-2\\ \text{O}\text{.S}\left(\text{Fe}\right)+2\times \text{0}+4\times \left(-1\right)=-2\\ \text{O}\text{.S}\left(\text{Fe}\right)-4=-2\\ \text{O}\text{.S}\left(\text{Fe}\right)=-2+4\\ =+2\end{array}$

Therefore, the oxidation state of Fe is +2. The electronic configuration of Fe is $\left[\text{Ar}\right]3d^{6}4s^2$ . Therefore, ${\text{Fe}}^{2+}$ is formed by losing of 2 electrons by $\text{Fe}$ . Therfore, the electronic configuration of ${\text{Fe}}^{2+}$ is $\left[\text{Ar}\right]3d^6$ .

(b)

Interpretation Introduction

Interpretation:

The electronic configuration of the transition metal ion in $\left[\text{Co}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}{\left({\text{NH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\right)}_{\text{2}}\right]{\text{I}}_{\text{2}}$ should be stated.

Concept Introduction:

The study of the compounds that have central atom that is surrounded by the anions or molecules is known as coordination chemistry. The anions or molecules that surrounds central atom are known as ligands. These ligands are attached to the central atom by the dative bonds known as coordinate bonds.

Answer to Problem 20E

The electronic configuration of ${\text{Co}}^{2+}$ is $\left[\text{Ar}\right]3d^7$ .

Explanation of Solution

The dissociation of $\left[\text{Co}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}{\left({\text{NH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\right)}_{\text{2}}\right]{\text{I}}_{\text{2}}$ is shown below:

  $\left[\text{Co}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}{\left({\text{NH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\right)}_{\text{2}}\right]{\text{I}}_{\text{2}}\to {\left[\text{Co}{\left({\text{NH}}_{\text{3}}\right)}_{\text{2}}{\left({\text{NH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\right)}_{\text{2}}\right]}^{+2}+{\text{2I}}^{-}$

The oxidation state of central metal is calculated as follows:

  $\begin{array}{c}\text{O}\text{.S}\left(\text{Co}\right)+2\times \text{O}\text{.S}\left({\text{NH}}_{\text{3}}\right)+2\times \text{O}\text{.S}\left({\text{NH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\right)=+2\\ \text{O}\text{.S}\left(\text{Co}\right)+2\times \text{0}+4\times \left(0\right)=+2\\ \text{O}\text{.S}\left(\text{Co}\right)=+2\end{array}$

Therefore, the oxidation state of Co is +2. The electronic configuration of Co is $\left[\text{Ar}\right]3d^{7}4s^2$ . Therefore, ${\text{Co}}^{2+}$ is formed by losing of 2 electrons by $\text{Co}$ . Therfore, the electronic configuration of ${\text{Co}}^{2+}$ is $\left[\text{Ar}\right]3d^7$ .

(c)

Interpretation Introduction

Interpretation:

The electronic configuration of the transition metal ion in ${\text{Na}}_{\text{2}}\left[{\text{TaF}}_7\right]$ should be stated.

Concept Introduction:

The study of the compounds that have central atom that is surrounded by the anions or molecules is known as coordination chemistry. The anions or molecules that surrounds central atom are known as ligands. These ligands are attached to the central atom by the dative bonds known as coordinate bonds.

Answer to Problem 20E

The electronic configuration of ${\text{Ta}}^{5+}$ is $\left[\text{Xe}\right]4f^{14}$ .

Explanation of Solution

The dissociation of ${\text{Na}}_{\text{2}}\left[{\text{TaF}}_7\right]$ is shown below:

  ${\text{Na}}_{\text{2}}\left[{\text{TaF}}_7\right]\to {\text{2Na}}^{+}+{\left[{\text{TaF}}_7\right]}^{-2}$

The oxidation state of central metal is calculated as follows:

  $\begin{array}{c}\text{O}\text{.S}\left(\text{Ta}\right)+7\times \text{O}\text{.S}\left(\text{F}\right)=-2\\ \text{O}\text{.S}\left(\text{Ta}\right)+7\times \left(-1\right)=-2\\ \text{O}\text{.S}\left(\text{Ta}\right)-7=-2\\ \text{O}\text{.S}\left(\text{Ta}\right)=-2+7\\ \text{O}\text{.S}\left(\text{Ta}\right)=+5\end{array}$

Therefore, the oxidation state of $\text{Ta}$ is +5. The electronic configuration of $\text{Ta}$ is $\left[\text{Xe}\right]4f^{14}5d^{3}6s^2$ . Therefore, ${\text{Ta}}^{5+}$ is formed by losing of 5 electrons by $\text{Ta}$ . Therfore, the electronic configuration of ${\text{Ta}}^{5+}$ is

  $\left[\text{Xe}\right]4f^{14}$ .

(d)

Interpretation Introduction

Interpretation:

The electronic configuration of the transition metal ion in $\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{\text{I}}_{\text{2}}\right]\left[{\text{PtI}}_{\text{4}}\right]$ should be stated.

Concept Introduction:

The study of the compounds that have central atom that is surrounded by the anions or molecules is known as coordination chemistry. The anions or molecules that surrounds central atom are known as ligands. These ligands are attached to the central atom by the dative bonds known as coordinate bonds.

Answer to Problem 20E

The electronic configuration of ${\text{Pt}}^{2+}$ in $\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{\text{I}}_{\text{2}}\right]$ and ${\text{Pt}}^{4+}$ in $\left[{\text{PtI}}_{\text{4}}\right]$ is $\left[\text{Xe}\right]4f^{14}5d^8$ and $\left[\text{Xe}\right]4f^{14}5d^6$ respectively.

Explanation of Solution

The dissociation of $\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{\text{I}}_{\text{2}}\right]\left[{\text{PtI}}_{\text{4}}\right]$ is shown below:

  $\begin{array}{c}\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{\text{I}}_{\text{2}}\right]\to {\text{2I}}^{-2}+{\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}\right]}^{+2}\\ \left[{\text{PtI}}_{\text{4}}\right]\to {\text{2I}}^{-2}+{\text{Pt}}^{+4}\end{array}$

The oxidation state of central metal in ${\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}\right]}^{+2}$ is calculated as follows:

  $\begin{array}{c}\text{O}\text{.S}\left(\text{Pt}\right)+4\times \text{O}\text{.S}\left({\text{NH}}_{\text{3}}\right)=+2\\ \text{O}\text{.S}\left(\text{Pt}\right)+4\times \left(0\right)=+2\\ \text{O}\text{.S}\left(\text{Pt}\right)+0=+2\\ \text{O}\text{.S}\left(\text{Pt}\right)=+2\end{array}$

Therefore, the oxidation state of $\text{Pt}$ is +2 in $\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{\text{I}}_{\text{2}}\right]$ . The electronic configuration of $\text{Pt}$ is $\left[\text{Xe}\right]4f^{14}5d^{9}6s^1$ . Therefore, ${\text{Pt}}^{2+}$ is formed by losing of 2 electrons by $\text{Pt}$ . Therfore, the electronic configuration of ${\text{Pt}}^{2+}$ is $\left[\text{Xe}\right]4f^{14}5d^8$ .

The oxidation state of $\text{Pt}$ is +4 in $\left[{\text{PtI}}_{\text{4}}\right]$ . The electronic configuration of $\text{Pt}$ is $\left[\text{Xe}\right]4f^{14}5d^{9}6s^1$ . Therefore, ${\text{Pt}}^{4+}$ is formed by losing of 4 electrons by $\text{Pt}$ . Therfore, the electronic configuration of ${\text{Pt}}^{4+}$ is $\left[\text{Xe}\right]4f^{14}5d^6$ .