Question

Pick out the correct statement with respect to $[Mn(CN)_6]^{3-}$ :

• A. It is $sp^3d^2$ hybridised and tetrahedral
• B. It is $d^2sp^3$ hybridised and octahedral
• C. It is $dsp^2$ hybridised and square planar
• D. It is $sp^3d^2$ hybridised and octahedral

Answer 1

Answer: (B)

$\left[ Mn ( CN )_{6}\right]^{3-}$

$Mn ( III )=[ Ar ] 3 d^{4}$

$CN ^{-}$ being strong field ligand forces pairing of electrons

This gives $t_{2 g}^{4} e_{g}^{0}$

$\therefore Mn ( III )=[ Ar ]$



$\because$ Coordination number of $Mn =6$

$\therefore $ Structure $=$ octahedral

$\left[ Mn ( CN )_{6}\right]^{3-}=$