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Standard XII

Chemistry

Ammonia

In aqueous so...

Question

In aqueous solution, ${[C o {(H_{2} O)}_{6}]}^{2 +}$ $(X)$ 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.

3, 1

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3, 0

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3, 3

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7, 0

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Solution

The correct option is D $3, 0$
${[C o {(H_{2} O)}_{6}]}^{2 +}$ has $C o^{2 +}$ ion with $3 d^{7}$ outer electronic configuration and has 3 unpaired electrons as it is high spin or spin free complex.
It reacts with molecular oxygen in the presence of excess liquor $N H_{3}$ to give a new complex $Y,$ ${[C o {(N H_{3})}_{6}]}^{3 +}$ . It has $C o^{3 +}$ ion with $3 d^{6}$ outer electronic configuration and has 0 unpaired electrons as it is low spin or spin paired complex.

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Addition of excess aqueous ammonia to a pink coloured aqueous solution of $M C l_{2} . H_{2} O (x) a n d N H_{4} C l$ gives an octahedral complex Y in the presence of air. In 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 coloured complex Z. The calculated spin only magnetic moment of X and Z is 3.87 B.M., whereas it is zero for complex Y.
Among the following options, which statemet(s) is (are) correct?

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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.

In aqueous solution, ${[C o {(H_{2} O)}_{6}]}^{2 +}$ $(X)$ 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.