Question

If magnetic moment of $[MnB{r}_4{]}^{2-}$ is 5.92 BM, then the geometry of this complex ion will be:

• A. square planar
• B. tetrahedral
• C. octahedral
• D. trigonal planar

Answer 1

Answer: (B)

tetrahedral

The oxidation state of $Mn$ in $[MnB{r}_4{]}^{2-}$ is +2 and since its magnetic moment is 5.92 BM

This implies$\sqrt{n(n+2)}$ = 5.92 BM, $n=5$. Hence $Mn$ has 5 unpaired electrons.

As $Br$ is a weak field ligand, therefore, this also implies that the pairing of electrons will not take place and electronic configuration of $Mn$ will be $3d^{5}4s^0$.

Now, $4s$ and $4p$ orbital of $Mn$ participate in hybridization and the hybridization is $s{p}^3$ and the geometry will be tetrahedral.