$\left[ Co \left( NH _{3}\right)_{6}\right]^{2+}$ In this ion, Co is present as $Co ^{2+} . NH _{3}$, being strong field ligand, pair up the unpaired electrons.
Electronic configuration of $= Co ^{2+}=3 d ^{7}, 4 s^{\circ}, 4 p^{0}$
Electronic configuration of $CO ^{2+}$ in $=$
$\left[ Co \left( NH _{3}\right)_{6}\right]^{3+}$
Due to the presence of one unpaired electron, $\left[ Co \left( NH _{3}\right)_{6}\right]^{+}$ is paramagnetic.
$\left[ MnBr _{4}\right]^{2-}$ In this ion, Mn is present as $Mn ^{2+}$. Br, being weak field ligand, leave unpaired electrons of the metal ion as such.
Electronic configuration of $Mn ^{2+}=3 d^{5} 4 s^{0} 4 p^{0}$
Electronic configuration of $Mn ^{2+}$ in $\left[ MnBr _{4}\right]^{2}$
Due to $s p^{3}$ hybridisation, shape of $\left[ MnBr _{4}\right]^{2-}$ is tetrahedral.
(c) Linkage isomerism occurs when two or more atoms in a monodentate ligand may function as a donor.
Therefore, $\left[ CoB _{2}( en )_{2}\right]^{-}$ does not exhibit linkage isomerism.
(d) $\left. Ni \left( NH _{3}\right)_{6}\right]^{2+}$ In this ion, $Ni$ is present as $Ni ^{2+}$.
Electronic configuration of $Ni ^{2+}=[ Ar ] 3 d^{8} 4 s^{\circ}$
$NH _{3}$ is a strong field ligand, it can pair up electrons only if they are more than 4 and less than $8 .$
Electronic configuration of nine $\left[ Ni \left( NH _{3}\right)_{6}\right]^{2+}$
Since. $(n-1) d$, i.e., $3 d$ orbital does not take part in hybridisation, it is not an inner orbital complex. It is an outer orbital complex.
