Shape of $\left[ NiCl _4\right]^{2-}=$ Tetrahedral
The oxidation state of $Ni$ in the complex is $+2$. It has outer electronic configuration of $3 d^8$.
Since, chloride ion is a weak field ligand, there will be no electron pairing. Hence, one $4 s$ orbital and three $4 p$ orbitals will hybridized to overlap with four $3 p$ orbitals of $Cl$. This will result in tetrahedral geometry.
(en $=$ ઇથેન $-1,2-$ ડાયએમાઈન)
$(i)\ XeF_4\ (ii)\ SF_4\ (iii)\ [Ni (Cl_4)]^{2-}\ (iv)\ [ptCl_4]^{2-}$