An octahedral complex of Co3+ is diamagnetic. The hybridisation involved in the formation of the complex is:

A
${sp}^{3} d^{2}$
B
$d s p^{2}$
C
$d^{2} s p^{3}$
D
$d s p^{3} d$

Solution:

The outer electronic configuration of $_{27} {Co}^{3 +}$ is $(3 d)^{6}$ , i.e. free $_{27} {Co}^{3 +} ion$ ;

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

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

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

Solution:

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