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Q.
The spin only magnetic moment $ (\mu_s) $ of a complex $ [Mn(Br_4)]^{4-} $ is $ 5.9\, BM $ . The geometry of the complex will be
AMUAMU 2016Coordination Compounds
Solution:
Since, the coordination number of $Mn^{2+}$ ion in the complex ion is $4$, it will be either tetrahedral ($sp^3$-hybridisation) or square planar ($dsp^2$ hybridisation). But the fact that the magnetic moment of the complex ion is $5.9 \,BM$
$\mu_s = \sqrt{n(n+2)} $
$\Rightarrow 5.9 = \sqrt{n(n+2)}$
$\Rightarrow n = 5$
$\therefore $ The complex has $5$ unpaired $e^-$ suggests that it should be tetrahedral in shape rather the square planar because of the presence of five impaired electrons in the $d$-orbitals.