The magnetic moment of ${\left[Mn{\left(CN\right)}_6\right]}^{3-}$ is 2.8 BM and that of  ${\left[MnB{r}_4\right]}^{2-}$ is 5.9 BM. Find the hybridization and geometries of these complex ions.

$\Rightarrow$ The oxidation state of  Mn is +3 in this complex.
The outer electronic configuration of  $Mn\Rightarrow \left|Ar\right|3d^{4}4s^{\circ }4p^{\circ }$
$\Rightarrow$ When ${\left(\text{CN}\right)}_{\text{6}}$ comes it needs 6 orbitals but the available orbitals are 5 only ( one S, 1 orbital of d as rest are filled with single electrons and 3 orbitals of
P) CN being strong ligand will make one electron of d orbital to pair with other d orbital to pair with other d orbital hence vacating one more orbitals in d,s and p will be 6 which will be taken by ${\left(\text{CN}\right)}_{\text{6}}$ 
So hybridization is $d^{2}s{p}^3$ and unpaired electrons is 2. 
($\because C{N}^{-}$ is a strong ligand, electrons forced to pair against Hund’s rule )
$\Rightarrow$${{\left[Mn{\left(CN\right)}_6\right]}^3}^{-}{d}^{2}s{p}^3,$  octahedral and ${{\left[MnB{r}_4\right]}^2}^{-} s{p}^3,$ tetrahedral