Geometrical shapes of the complexes formed by the reaction of $$N{i^{2 + }}$$  with $$C{l^ - },\,C{N^ - }\,$$  and $${H_2}O,$$  respectively,are

Solution :
$$N{i^{ + 2}} + 4C{l^ - } \to \mathop {{{\left[ {NiC{l_4}} \right]}^{2 - }}}\limits_{s{p^3}} $$
$${\left[ {NiC{l_4}} \right]^{2 - }} = 3{d^8}$$    configuration with nickel in $$+ 2$$  oxidation state, $$C{l^ - }$$  being weak field ligand does not compel for pairing of electrons.
So,
$${\left[ {NiC{l_4}} \right]^{2 - }}$$

Hence, complex has tetrahedral geometry

$$N{i^{ + 2}} + 4C{N^ - } \to {\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$
$${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }} = 3{d^8}$$    configuration with nickel in $$ + \,\,\,2$$  oxidation state, $$C{N^ - }$$  being strong field ligand compels for pairing of electrons.
So,
$${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{ - 2}}$$

Hence, complex has square planar geometry.

$$N{i^{ + 2}} + 6{H_2}O \to {\left[ {Ni{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }}$$
$$\left[ {Ni{{\left( {{H_2}O} \right)}_6}} \right] = 3{d^8}$$    configuration with nickel in $$ + 2$$  oxidation state. As with $$3{d^8}$$   configuration two $$d - $$  orbitals are not available for $${d^2}s{p^3}$$  hybridisation. So, hybridisation of $$Ni\left( {{\text{II}}} \right)$$  is $$s{p^3}{d^2}$$  and $$Ni\left( {{\text{II}}} \right)$$  with six co-ordination will have octahedral geometry.

Note : With water as ligand, $$Ni\left( {{\text{II}}} \right)$$  forms octahedral complexes.