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  4. In aqueous solution, [Co(H2O)6]2+ (X), reacts with molecular oxygen in the presence of excess liquor NH3 to give a new complex Y. The number of unpaired electrons in X and Y are, respectively

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Q. In aqueous solution, $[Co(H_2O)_6]^{2+} (X)$, reacts with molecular oxygen in the presence of excess liquor $NH_3$ to give a new complex $Y$. The number of unpaired electrons in $X$ and $Y$ are, respectively

KVPYKVPY 2013The p-Block Elements

Solution:

In aqueous solution $[Co(H_2O)_6]^{2+}X$ reacts with molecular oxygen in the presence of excess liquor $NH_3$ to give a new complex $[Co(NH_3)_6]$
$\underset{(X)}{[Co(H_2O)_6]_2 } + 6NH_3 \to$
$[Co(NH_3)_6]_{2+} \xrightarrow{O_2} \underset{(Y)}{[Co(NH_3)_6]^{3+}}$
In $[Co(H_2O)_6]^{2+}, Co$ is in $+2$ oxidation $H_2O$ being weak ligand will not pair up the electrons, thus its electronic configuration will be $t^{5}_{2g} e^{2}_{g}.$. So, it will have three unpaired electrons.
In $[Co(NH_3)_6]^{3+} Co$ is in $+3$ oxidation, $NH_3$ being strong ligand will pair up the electrons, thus the electronic
configuration will be $t^{6}_{2g} e^{0}_{g}.$ So, it will have no unpaired electron.