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Pick out the correct statement with respect $[Mn(CN)_6]^{3-}$:
Explanation
Strong-field $CN^-$ pairs electrons leaving two inner d orbitals for $d^2sp^3$.
Detailed Solution
In $[Mn(CN)_6]^{3-}$, Mn(III) $= [Ar]\,3d^4$
$CN^-$, being a strong field ligand, forces pairing of electrons, giving $t_{2g}^4e_g^0$.
$\because$ Coordination number of Mn = 6, the hybridisation is $d^2sp^3$ and the structure is octahedral.
$CN^-$, being a strong field ligand, forces pairing of electrons, giving $t_{2g}^4e_g^0$.
$\because$ Coordination number of Mn = 6, the hybridisation is $d^2sp^3$ and the structure is octahedral.