Question

The spin only magnetic moment of ${\left[Mn{Br}_4\right]}^{2-}$ is $5.9B.M$ Geometry of the complex ion is:

• A. $Tetrahedral$
• B. $Octahedral$
• C. $Squareplanae$
• D. $Pentagonalpyramidal$

Answer 1

The correct option is A $Tetrahedral$

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.