A correct statement is

Option (1) \(\left[\mathrm{MnBr}_{4}\right]^{2-}\)
\(\mathrm{Br}^{-}\)weak ligand does not allow pairing of \(e^{-} s\), therefore, hybridization of the Mn in complex is sp3 and geometry is
tetrahedral.
Option \((2)\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\)
\(\mathrm{NH}_{3}\) is a strong field ligand, therefore, pairing of electrons will take place. But due to insufficiency of vacant orbital,
pairing will not take place
Electronic configuration of \(\mathrm{Ni}(\mathrm{Z}=8)[\mathrm{Ar}] 3 d^{8} 4 s^{2}\)
Electronic configuration of \(N i^{2+}\) is \([\mathrm{Ar}] 3 d^{8}\)
\(N i^{2+}\) in the presence of strong field ligand
\begin{array}{|l|l|l|l|l|}
\hline 1 L & 1 L & 1 L & 1 & 1 \\
\hline
\end{array}


It is outer orbital complex and not an inner orbital complex Option (3) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\)
\(\mathrm{NH}_{3}\) is a strong field ligand, therefore, pairing will take place Electronic configuration of \(C_{0}(Z=27)\) is \([A r] 3 d^{74} S^{2}\) Electronic configuration of \(\mathrm{Co}^{2+}\) is \([\mathrm{Ar}] 3 d^{7}\) \(\mathrm{Co}^{2+}\) in the presence of strong field ligand.

There is one unpaired \(e^{-}\)therefore, it is paramagnetic
Option (4) \(\left[\mathrm{CoBr}_{2}(e n)_{2}\right]^{-}\)does not have ambidentate ligands, therefore, is does not exhibits linkage isomerism.