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.
(પરમાણુ ક્રમાંક $Co=27$)