The spin only magnetic moment ${[M n {(B r)}_{4}]}^{2 -}$ is 5.9BM. Then possible hybridization of Mn in the complex is

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$s p^{3} d$

$s p^{3}$

$d^{2} s p^{3}$

$d s p^{2}$

answer is D.

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Detailed Solution

Since, the coordination number of Mn2+ ion in the complex ion is 4,it will be either tetrahedral
(sp3hybridization) or square planar (dsp2hybridization) in geometry.
But, due to the fact that magnetic moment of the complex ion is 5.9BM.
So, it should be tetrahedral in shape rather than square planar because of the presence of five unpaired electrons in the d orbitals which will ultimately lead to
sp3hybridization. So, Mn2+ will have Mn will be
3d54s0in outermost shell.

Now, 4sand 4porbital of Mn will participate in hybridization and the hybridization will be
sp3 and the geometry will be tetrahedral.

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