Coordination Compounds - Result Question 43
####46. Pick out the correct statement with respect to $[Mn(CN)_6]^{3-}$
[2017]
(a) It is $s p^{3} d^{2}$ hybridised and tetrahedral
(b) It is $d^{2} s p^{3}$ hybridised and octahedral
(c) It is $d s p^{2}$ hybridised and square planar
(d) It is $s p^{3} d^{2}$ hybridised and octahedral
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Solution:
- (b) In the complex $[Mn(CN)_6]^{3-}$, O.S. of $Mn$ is +3 E.C. of $Mn^{+3} \to 3 d^{4}$
The presence of a strong field ligand $CN^{-}$causes pairing of electrons.
As, coordination number of $Mn=6$, so it will form an octahedral complex.
$\therefore[Mn(CN)_6]^{3-}=$
