The magnetic moment of MnCN63− is 2.8 BM and that of  MnBr42− is 5.9 BM. Find the hybridization and geometries of these complex ions.

⇒ The oxidation state of  Mn is +3 in this complex.The outer electronic configuration of  Mn⇒Ar3d44s∘4p∘⇒ When CN6 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 ofP) 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 CN6 So hybridization is d2sp3 and unpaired electrons is 2. (∵ CN− is a strong ligand, electrons forced to pair against Hund’s rule )⇒MnCN63−d2sp3,  octahedral and MnBr42− sp3, tetrahedral