The hybridization of Ni, shape, and number of unpaired electrons present in ${\left[{\mathrm{NiCl}}_4\right]}^{2–}$ are respectively

It is given to find out the hybridization of Ni, shape, and number of unpaired electrons present in ${\left[{\mathrm{NiCl}}_4\right]}^{2–}$. 

In ${\left[{\mathrm{NiCl}}_4\right]}^{2–}$, the oxidation state of Ni is +2. So, the electronic configuration is $3{\mathrm{d}}^8$. 

Explanation

Six electrons are paired and two remain unpaired in the ground state. 
Chloride ion is a weak ligand and thus is unable to pair up the electrons. 
4s and 4p orbitals undergo ${\mathrm{sp}}^3$ hybridization. 

Hence the correct answer is A) ${\mathrm{sp}}^3$, tetrahedral, 2