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)

It is $d^2sp^3$ hybridised and octahedral

Answer 2

$\left[ Mn ( CN )_{6}\right]^{3-}$ $Mn ( III )=[ Ar ] 3 d^{4}$ $CN ^{-}$ 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 $\left[ Mn ( CN )_{6}\right]^{3-}=$