Correct Answer - C
`[Co(NH_3)_6]^(2+)` in this ion , Co is present as `Co^(2+).NH_3`, being strong field ligand, pair up the unpaired electrons. Electronic configuration of `Co^(2+)=3d^(7),4s^(0),4p^(0)`
Due to the presence of one unpaired electron ,`[Co(NH_3)_6]^(+)` is paramagnetic.
(b) `[MnBr_(4)]^(2-)` in this ion , Mn is present as `Mn^(2+)`. Br, being weak field ligand , leave unpaired electron of the metal ion as such , Electronic configuration of `Mn^(2+)=3d^(6),4s^(0),4p^(0)`.
Electronic configuration of `Mn^2+=3d^(5),4s^(0),4p^(0)`
Electronic configuration of `Mn^2+` in `[MnBr_4]^(2-)`
Due to `sp^3` hybridiation , shape of `[MnBr_(4)]^(2-)` -is tetrahedral.
(c) Linkage isomerism occurs when two or more atoms in a monodentate ligand may function as a donor.
Therefore , `[CoBr_2(en)_2]^(-)` does not exhibit linkage isomerism.
(d) `[Ni(NH_3)_6]^(2+)` in this ion , Ni is present as `Ni^(2+)`.
Electronic configuration of `Ni^(2+)=[Ar]3d^(8)4s^(0)`
`NH_3` is a strong field ligand. but cannot pair up electrons as if it pair up electron then hybridisation comes out to be `dsp^3d` which is not possible. Hence , here no pairing of `e^(-)` takes places.
Electronic configuration of Ni in `[Ni(NH_3)_6]^(2+)`
Since , `(n-1)`d , i.e., 3d orbital does not take part in hyidsation. it is not an inner orbital comples. it is an outer orbital complex.
