Question

The addition of excess aqueous ammonia to a pink coloured aqueous solution of $CoC{l}_2.6H_{2}O\left(X\right)$ and $N{H}_{4}Cl$ gives an octahedral complex $Y$ in the presence of air. In an aqueous solution, complex $Y$ behaves as $1:3$ electrolyte. The reaction of $X$ with excess $HCl$ at room temperature results in the formation of a blue colored complex $Z$$\left(\right[CoC{l}_4{]}^{2-})$. The calculated spin only magnetic moment of $X$ and $Z$ is $3.87\text{BM}$ whereas, it is zero for complex $Y$.
Choose the correct statement among the following.

• A. Hybridisation of $Y$ is $s{p}^3{d}^2$
• B. Both $Z$ and $Y$ are diamagnetic in nature
• C. Coordination number of $Y$ is $4$
• D. Oxidation state of the central metal ion in $Y$ is $+3$

Answer 1

The correct option is D Oxidation state of the central metal ion in $Y$ is $+3$

According to the question,
$\underset{\text{Pink (X)}}{\left[Co\right(H_{2}O{)}_6]C{l}_2}\underset{O_2\left(air\right)}{\overset{ExcessN{H}_{4}OH/N{H}_{4}Cl}{\to }}\underset{Y}{\left[Co\right(N{H}_3{)}_6]C{l}_3}$

$\underset{\text{Pink (X)}}{\left[Co\right(H_{2}O{)}_6]C{l}_2}+\underset{Excess}{4C{l}^{-}}\to \underset{Z}{[CoC{l}_4{]}^{2-}}$

(a) The complex $Y$ is $\left[Co\right(N{H}_3{)}_6]C{l}_3$. Here, $Co$ is in $+3$ oxidation state. As $N{H}_3$ is a strong field ligand, the hybridisation of $Y$ is $d^{2}s{p}^3$ i.e. it forms an inner orbital complex.

(b) As can be seen from the spin only magnetic moment given in the question, $Z$ is paramagnetic whereas $Y$ is diamagnetic.

(c) The coordination number of $Y$ is $6$.

(d) Let the oxidation state of the central metal ion in $Y$ $\left(\right[Co\left(N{H}_3{)}_6\right]C{l}_3)$ be ${}^{\prime }{a}^{\prime }$
$a+0\times 6-1\times 3=0$
$a=+3$
Hence, the oxidation state of the central metal ion in $Y$ is $+3$.