Question

In aqueous solution, $[Co(H_{2}O{)}_6{]}^{2+}(X)$, reacts with molecular oxygen in the presence of excess liquor $N{H}_3$ to give a new complex $Y$. The number of unpaired electrons in $X$ and $Y$ are, respectively

• A. 3, 1
• B. 3, 0
• C. 3, 3
• D. 7, 0

Answer 1

Answer: (B)

In aqueous solution $[Co(H_{2}O{)}_6{]}^{2+}X$ reacts with molecular oxygen in the presence of excess liquor $N{H}_3$ to give a new complex $[Co(N{H}_3{)}_6]$
$\underset{(X)}{[Co(H_{2}O{)}_6{]}_2}+6N{H}_3\to$
$[Co(N{H}_3{)}_6{]}_{2+}\stackrel{O_2}{\to }\underset{(Y)}{[Co(N{H}_3{)}_6{]}^{3+}}$
In $[Co(H_{2}O{)}_6{]}^{2+},Co$ is in $+2$ oxidation $H_{2}O$ being weak ligand will not pair up the electrons, thus its electronic configuration will be $t_{2g}^5{e}_g^2.$. So, it will have three unpaired electrons.
In $[Co(N{H}_3{)}_6{]}^{3+}Co$ is in $+3$ oxidation, $N{H}_3$ being strong ligand will pair up the electrons, thus the electronic
configuration will be $t_{2g}^6{e}_g^0.$ So, it will have no unpaired electron.