Question

The spin only magnetic moment of $[MnB{r}_4{]}^{2–}$ is $5.9BM.$ Predict the geometry of the complex ion?

Answer 1

$\text{Magnetic moment}$

Magnetic moment of complex is the combination of spin and orbital magnetic moment.

$\text{Spin only magnetic moment:}$

It generates due to the precession of unpaired electron about their own axes.

Spin only magnetic moment can be calculated by the following formula

Magnetic moment $(M.M)=\sqrt{n(n+2)}\dots \dots \dots \dots ..\left(i\right)$

Where, $n$ is number of unpaired electrons

$\text{Calculation of unpaired electron}$

Given: Magnetic moment of complex = $5.9BM$

Now from the equation $\left(i\right)$

$\sqrt{n(n+2)}=5.9$

$\Rightarrow n=5$

$\text{Geometry of the complex}$

Since, the coordination number of $M{n}^{2+}$ ion in the complex ion is $4,$ it will be either tetrahedral $\left(s{p}^3\text{hybridization}\right)$ or square planar $\left(ds{p}^2\text{hybridization}\right)$ in geometry.

But, due to the fact that magnetic moment of the complex ion is $5.9BM.$ So, it should be tetrahedral in shape rather than square planar because of the presence of five unpaired electrons in the d orbitals which will ultimately lead to $s{p}^3$ hybridization. So, $M{n}^{2+}$ will have Mn will be $3d^{5}4s^0$ in outermost shell.

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