In aqueous solution,[Co(H_2O)_6]^{2+} (X) rea | vedclass

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(B) The reaction is: $[Co(H_2O)_6]^{2+} + 6 NH_3 \xrightarrow{O_2} [Co(NH_3)_6]^{3+} + 6 H_2O$.In complex $X$,$[Co(H_2O)_6]^{2+}$,the oxidation state

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$3, 0$

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(B) The reaction is: $[Co(H_2O)_6]^{2+} + 6 NH_3 \xrightarrow{O_2} [Co(NH_3)_6]^{3+} + 6 H_2O$.In complex $X$,$[Co(H_2O)_6]^{2+}$,the oxidation state of $Co$ is $+2$ ($d^7$ configuration).Since $H_2O$ is a weak field ligand,the electrons do not pair up,resulting in $t_{2g}^5 e_g^2$ configuration,which gives $3$ unpaired electrons.In complex $Y$,$[Co(NH_3)_6]^{3+}$,the oxidation state of $Co$ is $+3$ ($d^6$ configuration).Since $NH_3$ is a strong field ligand,it causes pairing of electrons,resulting in $t_{2g}^6 e_g^0$ configuration,which gives $0$ unpaired electrons.Thus,the number of unpaired electrons in $X$ and $Y$ are $3$ and $0$ respectively.

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:

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