Chapter 21, Problem 28E

Interpretation Introduction

Interpretation: The electron configuration of metal ion in each given complex compound is to be stated.

Concept introduction: Valence electrons are filled in four principal orbitals according to the energy levels. The four orbitals are $s,p,d$ and $f$.

The metals from group $\text{3-12}$ are known as first row transition metals. The general form of valence electron configuration of these metals is,

$\left[\text{Nobel}\text{gas}\right]n{f}^{\text{x}}n{d}^{\text{y}}n{s}^{\text{z}}$

The distribution of electrons among different orbitals is given by electron configuration. When an element loses an electron, a positive charge is accumulated on it, whereas when an element gains an electron, negative charge is accumulated on it.

Answer to Problem 28E

Answer

1. (a) The electron configuration of metal ion ${\text{Fe}}^{2+}$ in ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\left[\text{Fe}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{2}}{\text{Cl}}_{\text{4}}\right]$ is $\left[\text{Ar}\right]{\text{3d}}^{\text{6}}$.
2. (b) The electron configuration of metal ion ${\text{Co}}^{2+}$ in $\left[\text{Co}{\left({\text{NH}}_3\right)}_{\text{2}}\left({\text{NH}}_2{\text{CH}}_2{\text{CH}}_2{\text{NH}}_2\right)\right]{\text{I}}_2$ is $\left[\text{Ar}\right]{\text{3d}}^{\text{7}}$.
3. (c) The electron configuration of metal ion ${\text{Ta}}^{5+}$ in ${\text{Na}}_2\left[{\text{TaF}}_{\text{7}}\right]$ is $\left[\text{Xe}\right]{\text{4f}}^{\text{14}}$.
4. (d) The electron configuration of metal ion ${\text{Pt}}^{2+}$ in $\left[\text{Pt}{\left({\text{NH}}_3\right)}_{\text{2}}{\text{I}}_{\text{2}}\right]\left[{\text{PtI}}_{\text{4}}\right]$ is $\left[\text{Xe}\right]{\text{4f}}^{\text{14}}{\text{5d}}^{\text{8}}$. The electron configuration of metal ion ${\text{Pt}}^{4+}$ in $\left[\text{Pt}{\left({\text{NH}}_3\right)}_{\text{2}}{\text{I}}_{\text{2}}\right]\left[{\text{PtI}}_{\text{4}}\right]$ is $\left[\text{Xe}\right]{\text{4f}}^{\text{14}}{\text{5d}}^{\text{6}}$.

Explanation of Solution

EXPLANATION

(a)

To write: The electron configuration of metal ion in ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\left[\text{Fe}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{2}}{\text{Cl}}_{\text{4}}\right]$.

The metals from group $\text{3-12}$ are known as first row transition metals. The general form of valence electron configuration of these metals is,

$\left[\text{Nobel}\text{gas}\right]n{f}^{\text{x}}n{d}^{\text{y}}n{s}^{\text{z}}$

Where,

• $n$ is the shell of principal orbital.
• $\text{x}$ is the number of electrons in $f$ orbital.
• $\text{y}$ is the number of electrons in $d$ orbital.
• $\text{z}$ is the number of electrons in $s$ orbital.

In the complex compound ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\left[\text{Fe}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{2}}{\text{Cl}}_{\text{4}}\right]$, the metal ion is ${\text{Fe}}^{2+}$.

Iron is a group $8$ element. The atomic number of iron $\left(\text{Fe}\right)$ is $\text{26}$ and its electronic configuration is,

$\left[\text{Ar}\right]{\text{3d}}^{\text{6}}{\text{4s}}^{\text{2}}$

It loses two electrons to acquire two positive charges and becomes ${\text{Fe}}^{2+}$. The electronic configuration of ${\text{Fe}}^{2+}$ is,

${\text{Fe}}^{2+}=\text{Fe}-2{\text{e}}^{-}$

Substitute the electron configuration of $\text{Fe}$ in the above equation.

$\begin{array}{c}{\text{Fe}}^{2+}=\text{Fe}-2{\text{e}}^{-}\\ =\left[\text{Ar}\right]{\text{3d}}^{\text{6}}{\text{4s}}^{\text{2}}-2{\text{e}}^{-}\\ =\left[\text{Ar}\right]{\text{3d}}^{\text{6}}\end{array}$

(b)

To write: The electron configuration of metal ion in $\left[\text{Co}{\left({\text{NH}}_3\right)}_{\text{2}}\left({\text{NH}}_2{\text{CH}}_2{\text{CH}}_2{\text{NH}}_2\right)\right]{\text{I}}_2$.

In the complex compound $\left[\text{Ag}{\left({\text{NH}}_3\right)}_{\text{2}}\right]\text{Cl}$, metal ion is ${\text{Co}}^{2+}$.

Cobalt is a group $9$ element. The atomic number of cobalt $\left(\text{Co}\right)$ is $\text{27}$ and its electronic configuration is,

$\left[\text{Ar}\right]{\text{3d}}^{\text{7}}{\text{4s}}^{\text{2}}$

It loses two electrons to acquire two positive charges and becomes ${\text{Co}}^{2+}$. The electronic configuration of ${\text{Co}}^{2+}$ is,

${\text{Co}}^{2+}=\text{Co}-2{\text{e}}^{-}$

Substitute the electron configuration of $\text{Co}$ in the above equation.

$\begin{array}{c}{\text{Co}}^{2+}=\text{Co}-2{\text{e}}^{-}\\ =\left[\text{Ar}\right]{\text{3d}}^{\text{7}}{\text{4s}}^{\text{2}}-2{\text{e}}^{-}\\ =\left[\text{Ar}\right]{\text{3d}}^{\text{7}}\end{array}$

 (c)

To write: The electron configuration of metal ion in ${\text{Na}}_2\left[{\text{TaF}}_{\text{7}}\right]$.

In the complex compound ${\text{Na}}_2\left[{\text{TaF}}_{\text{7}}\right]$, metal ion is ${\text{Ta}}^{5+}$.

Tantalum is a group $5$ element. The atomic number of tantalum $\left(\text{Ta}\right)$ is $\text{73}$ and its electronic configuration is,

$\left[\text{Xe}\right]{\text{4f}}^{\text{14}}{\text{5d}}^{\text{3}}{\text{6s}}^{\text{2}}$

It loses five electrons to acquire five positive charges and becomes ${\text{Ta}}^{5+}$. The electronic configuration of ${\text{Ta}}^{5+}$ is,

${\text{Ta}}^{5+}=\text{Ta}-5{\text{e}}^{-}$

Substitute the electron configuration of $\text{Ta}$ in the above equation.

$\begin{array}{c}{\text{Ta}}^{5+}=\text{Ta}-5{\text{e}}^{-}\\ =\left[\text{Xe}\right]{\text{4f}}^{\text{14}}-2{\text{e}}^{-}\\ =\left[\text{Xe}\right]{\text{4f}}^{\text{14}}\end{array}$

(d)

To write: The electron configuration of metal ion in $\left[\text{Pt}{\left({\text{NH}}_3\right)}_{\text{2}}{\text{I}}_{\text{2}}\right]\left[{\text{PtI}}_{\text{4}}\right]$.

In the complex compound $\left[\text{Pt}{\left({\text{NH}}_3\right)}_{\text{2}}{\text{I}}_{\text{2}}\right]\left[{\text{PtI}}_{\text{4}}\right]$, metal ion is ${\text{Pt}}^{2+}$ and ${\text{Pt}}^{4+}$.

Platinum is a group $10$ element. The atomic number of platinum $\left(\text{Pt}\right)$ is $\text{78}$ and its electronic configuration is,

$\left[\text{Xe}\right]{\text{4f}}^{\text{14}}{\text{5d}}^{\text{9}}{\text{6s}}^{\text{1}}$

It loses two electrons to acquire two positive charges and becomes ${\text{Pt}}^{2+}$. The electronic configuration of ${\text{Pt}}^{2+}$ is,

${\text{Pt}}^{2+}=\text{Pt}-2{\text{e}}^{-}$

Substitute the electron configuration of $\text{Pt}$ in the above equation.

$\begin{array}{c}{\text{Pt}}^{2+}=\text{Pt}-2{\text{e}}^{-}\\ =\left[\text{Xe}\right]4{\text{f}}^{\text{14}}{\text{5d}}^{\text{9}}{\text{6s}}^{\text{1}}-2{\text{e}}^{-}\\ =\left[\text{Xe}\right]{\text{4f}}^{\text{14}}{\text{5d}}^{\text{8}}\end{array}$

Platinum loses four electrons to acquire four positive charges and becomes ${\text{Pt}}^{4+}$. The electronic configuration of ${\text{Pt}}^{4+}$ is,

${\text{Pt}}^{4+}=\text{Pt}-4{\text{e}}^{-}$

Substitute the electron configuration of $\text{Pt}$ in the above equation.

$\begin{array}{c}{\text{Pt}}^{4+}=\text{Pt}-4{\text{e}}^{-}\\ =\left[\text{Xe}\right]4{\text{f}}^{\text{14}}{\text{5d}}^{\text{9}}{\text{6s}}^{\text{1}}-4{\text{e}}^{-}\\ =\left[\text{Xe}\right]4{\text{f}}^{\text{14}}{\text{5d}}^{\text{6}}\end{array}$

Conclusion

Conclusion

The electron configuration is the representation of distribution of electrons among different orbitals.