Question

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

• A. tetrahedral
• B. square planar
• C. square pyramidal
• D. tetragonal

Answer 1

Answer: (A)

tetrahedral

Answer 2

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.