# An octahedral complex of ${\mathrm{Co}}^{3+}$ is diamagnetic. The hybridisation involved in the formation of the complex is:

1. A

${\mathrm{sp}}^{3}{d}^{2}$

2. B

$ds{p}^{2}$

3. C

${d}^{2}s{p}^{3}$

4. D

$ds{p}^{3}d$

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### Solution:

The outer electronic configuration of  is $\left(3\mathrm{d}{\right)}^{6}$, i.e. free ;

Since the octahedral complex of ${\mathrm{Co}}^{3+}$ is diamagnetic, it does not possess any unpaired electron. The configuration of ${\mathrm{Co}}^{3+}$ ion in the complex will be

Thus, ${\mathrm{Co}}^{3+}$ion in the complex involves ${\mathrm{d}}^{2}{\mathrm{sp}}^{3}$ hybridization.

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