question_answer

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

A)  It is \[ds{{p}^{2}}\]hybridised and square planar

B)  It is \[s{{p}^{3}}{{d}^{2}}\]hybridised and octahedral

C)  It is \[s{{p}^{3}}{{d}^{2}}\] hybridised and tetrahedral

D)  It is\[{{d}^{2}}s{{p}^{3}}\] hybridised and octahedral

Correct Answer: D

Solution :

                                \[{{[Mn{{(CN)}_{6}}]}^{3-}}\]                 \[Mn(III)=[Ar]3{{d}^{4}}\] \[C{{N}^{-}}\]being strong field ligand forces pairing of electrons This gives\[t_{2}^{4}e_{g}^{0}\] \[\therefore \]Mn(III) = [Ar] ∵ Coordination number of Mn = 6 \[\therefore \]Structure = octahedral                 \[{{[Mn{{(CN)}_{6}}]}^{3-}}=\]