First slide

Aplicaitons of VBT

Question

Pick out the correct statement with respect to ${[Mn (CN)_{6}]}^{3 -}$

Moderate

A

$It is {sp}^{3} d^{2} hybridised and tetrahedral .$
B

$It is d^{2} {sp}^{3} hybridised and octahedral .$
C

$It is {dsp}^{2} hybridised and square planar .$
D

$It is {sp}^{3} d^{2} hybridised and octahedral .$

Solution

${[Mn (CN)_{6}]}^{3 -} Let oxidation state of Mn be x . \begin{array}{l} x + 6 \times (- 1) = - 3 \\ x = + 3 \end{array} Electronic configuration of Mn : [Ar] 4 s^{2} 3 d^{5} Electronic configuration of {Mn}^{3 +} : [Ar] 3 d^{4} {CN}^{-} is a strong field ligand thus, it causes pairing of electrons in 3 d -orbital.$

Thus, ${[Mn (CN)_{6}]}^{3 -} has d^{2} {sp}^{3}$ hybridisation and has octahedral geometry.

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