Example 9.7 pg 250
The spin only magnetic moment on MnBr4 2- is 5.9 BM. predict the geometry of the complex ion?
ANSWER MY SPECIFIC DOUBT ONLY
: I DID NOT UNDERSTAND LAST LINE OF NCERT THAT ' IT SHOULD TETRAHEDRAL IN SAME RATHER THAN SQUARE PLANAR BECAUSE OF THE PRESENCE OF FIVE UNPAIRED ELECTRONS IN THE D ORBITALS. WHYYYYY? EXPLAIN CLEARLY WHYYYYYYYY

Hi,

The magnetic moment of ${\left[{\text{MnBr}}_4\right]}^{2-}$ is calculated as given below :

Oxidation state of Mn in ${\left[{\text{MnBr}}_4\right]}^{2-}$ is +2
x + 4(-1) = -2
x - 4 = -2
x = +2

So, ${\left[{\text{MnBr}}_4\right]}^{2-}$ contain 5 unpaired electrons.
Magnetic moment is 

$\text{μ = }\sqrt{\mathrm{n} \left(\mathrm{n} +2\right)}\text{ = }\sqrt{5 \left(5 +2\right)}$ = 5.9 BM

$$\begin{gathered} \text{Electronic Configuration } \\ {\text{Mn}}^{+2} = \left[\mathrm{Ar}\right] 3{\mathrm{d}}^5 4{\mathrm{s}}^0 4{\mathrm{p}}^0 \end{gathered}$$

As Br^- is a weak ligand so pairing of electrons will not takes place. Now 3d orbital of Mn^​+2 contain 5 unpaired electrons. So this orbital is not involved in hybridisation. 

Now 4s and 4p orbitals of Mn⁺² undergoes hybridisation.
sp³ hybridisation takes place due to which ${\left[{\text{MnBr}}_4\right]}^{2-}$ have tetrahedral geometry.
 
Regards