The magnetic moment of ${[M n {(C N)}_{6}]}^{3 -}$ is 2.8 BM and that of ${[M n B r_{4}]}^{2 -}$ is 5.9 BM. Find the hybridization and geometries of these complex ions.

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$s p^{3} d^{2}, octahedral and d s p^{2},tetrahedral$

$s p^{3} d^{2}, octahedral and s p^{3}, square planar$

$d^{2} s p^{3}, octahedral and d s p^{2} square planar$

$d^{2} s p^{3}, octahedral and s p^{3}, tetrahedral$

answer is C.

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Detailed Solution

$\Rightarrow$ The oxidation state of Mn is +3 in this complex.
The outer electronic configuration of $M n \Rightarrow |A r| 3 d^{4} 4 s^{\circ} 4 p^{\circ}$
$\Rightarrow$ When ${(CN)}_{6}$ comes it needs 6 orbitals but the available orbitals are 5 only ( one S, 1 orbital of d as rest are filled with single electrons and 3 orbitals of
P) CN being strong ligand will make one electron of d orbital to pair with other d orbital to pair with other d orbital hence vacating one more orbitals in d,s and p will be 6 which will be taken by ${(CN)}_{6}$
So hybridization is $d^{2} s p^{3}$ and unpaired electrons is 2.
( $∵ C N^{-}$ is a strong ligand, electrons forced to pair against Hund’s rule )
$\Rightarrow$ ${[M n {(C N)}_{6}]}^{3}^{-} d^{2} s p^{3},$ octahedral and ${[M n B r_{4}]}^{2}^{-} s p^{3},$ tetrahedral