Question

The spin only magnetic moment $({\mu }_s)$ of a complex $[Mn(B{r}_4){]}^{4-}$ is $5.9BM$ . The geometry of the complex will be

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

Answer 1

Answer: (A)

Since, the coordination number of $M{n}^{2+}$ ion in the complex ion is $4$, it will be either tetrahedral ($s{p}^3$-hybridisation) or square planar ($ds{p}^2$ hybridisation). But the fact that the magnetic moment of the complex ion is $5.9BM$
${\mu }_s=\sqrt{n(n+2)}$
$\Rightarrow 5.9=\sqrt{n(n+2)}$
$\Rightarrow n=5$
$\therefore$ 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.