A) It is \[{{\operatorname{sp}}^{3}}{{d}^{2}}\] hybridised and tetrahedral.
B) It is \[{{d}^{2}}{{\operatorname{sp}}^{3}}\] hybridised and octahedral.
C) It is \[{{\operatorname{dsp}}^{2}}\] hybridised and square planar.
D) It is \[{{\operatorname{sp}}^{3}}{{d}^{2}}\] hybridised and octahedral.
Correct Answer: B
Solution :
In the complex \[{{\left[ Mn{{(CN)}_{6}} \right]}^{3-}}\], O.S. of Mn is +3 B.C. of \[M{{n}^{3+}}\to 3{{d}^{4}}\] The presence of a strong field ligand \[C{{N}^{-}}\] causes pairing of electrons. As, coordination number of \[\operatorname{Mn} = 6\], so it will form an octahedral complex. \[\therefore \,\,\, {{[Mn{{\left( CN \right)}_{6}}]}^{3-}}\]You need to login to perform this action.
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