Question

The oxidation states of ${\text{Cr}}$ in $\left[ {{\text{Cr}}{{\left( {{{\text{H}}_2}{\text{O}}} \right)}_6}} \right]{\text{C}}{{\text{l}}_3}$ , $\left[ {{\text{Cr}}{{\left( {{{\text{C}}_6}{{\text{H}}_6}} \right)}_2}} \right]$ , and ${{\text{K}}_2}\left[ {{\text{Cr}}{{\left( {{\text{CN}}} \right)}_2}{{\left( {\text{O}} \right)}_2}\left( {{{\text{O}}_2}} \right)\left( {{\text{N}}{{\text{H}}_3}} \right)} \right]$ respectively are:
A.$ + 3$ , $0$ , and $ + 6$
B.$ + 3$ , $0$ , and $ + 4$
C.$ + 3$ , $ + 4$ and $ + 6$
D.$ + 3$ , $ + 2$ and $ + 4$

Answer 1

Hint: In this question, we have been asked to find out the oxidation states of a set of coordination complexes formed by the element chromium. The atomic number of chromium is 24 and the outermost electronic configuration of ${\text{Cr}}$ is $\left[ {Ar} \right]3{d^5}4{s^1}$ .

Complete step by step answer:
The outermost electronic configuration of ${\text{Cr}}$is $\left[ {Ar} \right]3{d^5}4{s^1}$ .
We can calculate the oxidation state of ${\text{Cr}}$ in the following coordination complexes as follows:

$\left[ {{\text{Cr}}{{\left( {{{\text{H}}_2}{\text{O}}} \right)}_6}} \right]{\text{C}}{{\text{l}}_3}$
Let us take the oxidation state of ${\text{Cr}}$ as $x$.
We know that ${{\text{H}}_2}{\text{O}}$ is a neutral ligand. So, its charge is taken as zero.
As we observe the complex, we can understand that the complex is a cationic complex, since there are $3{\text{C}}{{\text{l}}^ - }$ ions outside the complex to neutralise the charge of the complex. The complex has an overall charge of $ + 3$ .
Therefore, in order to calculate the unknown value of $x$ , we can write an equation as,
$x + (0 \times 6) = + 3$
Therefore,
$x = + 3$
So, the oxidation state of ${\text{Cr}}$ in $\left[ {{\text{Cr}}{{\left( {{{\text{H}}_2}{\text{O}}} \right)}_6}} \right]{\text{C}}{{\text{l}}_3}$ is $ + 3$ .

$\left[ {{\text{Cr}}{{\left( {{{\text{C}}_6}{{\text{H}}_6}} \right)}_2}} \right]$
Let us take the oxidation state of ${\text{Cr}}$ as $x$.
We know that benzene( ${{\text{C}}_6}{{\text{H}}_6}$ ) is a neutral ligand. So, its charge is taken to be zero. The complex is a neutral complex. This means that the overall charge of the complex is zero.
Therefore, in order to calculate the unknown value of $x$ , we can write an equation as,
$x + (0 \times 2) = 0$
Therefore,
$x = 0$
So, the oxidation state of ${\text{Cr}}$ in $\left[ {{\text{Cr}}{{\left( {{{\text{C}}_6}{{\text{H}}_6}} \right)}_2}} \right]$ is $0$ .

${{\text{K}}_2}\left[ {{\text{Cr}}{{\left( {{\text{CN}}} \right)}_2}{{\left( {\text{O}} \right)}_2}\left( {{{\text{O}}_2}} \right)\left( {{\text{N}}{{\text{H}}_3}} \right)} \right]$
Let us take the oxidation state of ${\text{Cr}}$ as $x$.
The ligands in the complex are ${\text{C}}{{\text{N}}^ - }$ , ${{\text{O}}^{2 - }}$ , ${\text{O}}_2^ - $ , which are anionic ligands and ${\text{N}}{{\text{H}}_3}$ , which is a neutral ligand.
The given complex is an anionic complex having a charge of $ - 2$ which is neutralised by $2{{\text{K}}^ + }$ ions outside the coordination complex. This means that the charge of the complex is $ - 2$ .
Therefore, in order to calculate the unknown value of $x$ , we can write an equation as,
$x + ( - 1 \times 2) + ( - 2 \times 2) + ( - 1 \times 2) + 0 = - 2$
Solving the equation, we get,
$x + ( - 2) + ( - 4) + ( - 2) + 0 = - 2$
The equation can be further simplified as,
$x - 8 = - 2$
We get,
$x = - 2 + 8$
Therefore,
$x = + 6$
So, the oxidation state of ${\text{Cr}}$ in ${{\text{K}}_2}\left[ {{\text{Cr}}{{\left( {{\text{CN}}} \right)}_2}{{\left( {\text{O}} \right)}_2}\left( {{{\text{O}}_2}} \right)\left( {{\text{N}}{{\text{H}}_3}} \right)} \right]$ is $ + 6$ .

Hence, option (A) is the correct answer.

Note:
Knowing the outermost electronic configuration of $3d$ transition elements and the various types of ligands – cationic, anionic and neutral, involved in the formation of coordination complexes would be helpful in calculating the oxidation state of central transition metal ions in the complex.