Which of the following complexes is diamagnetic and square planar?

\(
\left[\mathrm{CoF}_6\right]^{3-} \Rightarrow \mathrm{Co}^{3+}[\mathrm{Ar}] 3 \mathrm{~d}^6 45^{\circ}
\)
\(\mathrm{Sp}^3 \mathrm{~d}^2\) Hybridisation, Octahedral shape
\(
\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+} \Rightarrow \mathrm{Co}^{3+}[\mathrm{Ar}] 3 \mathrm{~d}^6 45^{\circ}
\)
\(\mathrm{d}^2 \mathrm{sp}^3\) Hybridisation, pairing due to strong field ligand
In \(\left[\mathrm{NiCl}_4\right]^{2-} \mathrm{Ni}\) is in +2 oxidation state \(\mathrm{Ni}^{+2}-[\mathrm{Ar}] 3 \mathrm{~d}^8\)
Hybridization involved is \(\mathrm{sp}^3\) as participating orbitals are \(4 \mathrm{~S} \& 4 \mathrm{P}\).
In \(\left[\mathrm{NiCN}_4\right]^{2-} \mathrm{Ni}\) is in +2 oxidation state.
\(
\mathrm{Ni}^{+2}-[\mathrm{Ar}] 3 \mathrm{~d}^8
\)
\(\mathrm{CN}^{-}\)is strong field ligand. It causes pairing of \(3 \mathrm{~d}\) electrons.
Hybridization involved is \(\mathrm{dsp}^2\) as participating orbitals are \(3 \mathrm{~d}\), \(4 \mathrm{~S}\) and \(4 \mathrm{P}\). Compound is diamagnetic due to pairing of electrons and shape is square planar.