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.
$(I)$ સિસ $- [Co(NH_3)_2(en)_2]^{3+}$ $(II)$ ટ્રાન્સ $-[IrCl_2(C_2O_4)_2]^{3-}$
$(III)\, [Rh(en)_3]^{3+}$ $(IV)$ સિસ $-[Ir(H_2O)_3Cl_3$