According to the valence bond theory the hybridisation of central metal atom is dsp² for which one of the following compounds?

According to VBT i.e. valence bond theory,

Electronic configuration of $\mathrm{Ni}=\left[\mathrm{Ar}\right]3{\mathrm{d}}^{8}4{\mathrm{s}}^2$.

(a) ${\mathrm{NiCl}}_2\cdot 6{\mathrm{H}}_2\mathrm{O}$

${\mathrm{NiCl}}_2\cdot 6{\mathrm{H}}_2\mathrm{O}⟶{\mathrm{NiCl}}_2+6{\mathrm{H}}_2\mathrm{O}$

Oxidation number of Ni(x) = x + 2(-1) = 0; where x,-1 and 2 are the oxidation number of Ni, oxidation number of Cl and number of Cl atoms respectively.

${\mathrm{NiCl}}_2\Rightarrow \mathrm{x}+(-2)=0\Rightarrow \mathrm{x}=2$

Electronic configuration of ${\mathrm{Ni}}^{2+}=\left[\mathrm{Ar}\right]3{\mathrm{d}}^{8}4{\mathrm{s}}^0$

Cl^– is a weak field ligand. So, no pairing of electrons occurs.

For C.N.= 6

(b) ${\mathrm{K}}_2\left[\mathrm{Ni}(\mathrm{CN}{)}_4\right]$

${\mathrm{K}}_2\left[\mathrm{Ni}(\mathrm{CN}{)}_4\right]⟶2{\mathrm{K}}^{+}+{\left[\mathrm{Ni}(\mathrm{CN}{)}_4\right]}^{2-}$

x + 4(-1) - (-2) = 0; where x, 4,-1 and -2 are the oxidation number of Ni, number of CN ligands, charge on one CN and charge on complex.

${\left[\mathrm{Ni}(\mathrm{CN}{)}_4\right]}^{2-}\Rightarrow \mathrm{x}-4+2=0\Rightarrow \mathrm{x}=+2$

Electronic configuration of ${\mathrm{Ni}}^{2+}\Rightarrow \left[\mathrm{Ar}\right]3{\mathrm{d}}^{8}4{\mathrm{s}}^0$

CN^– is a strong field ligand. So, pairing of electrons occur.

For C.N. = 4

(c) $\mathrm{Ni}(\mathrm{CO}{)}_4$

CO is neutral and strong field ligand. So, pairing of electrons occur.

Oxidation number of Ni is zero

Electronic configuration of $\mathrm{Ni}=\left[\mathrm{Ar}\right]3{\mathrm{d}}^{10}4{\mathrm{s}}^0$

For C.N.= 4

(d) ${\mathrm{Na}}_2\left[{\mathrm{NiCl}}_4\right]$

${\mathrm{Na}}_2\left[{\mathrm{NiCl}}_4\right]⟶2\mathrm{Na}+{\left[{\mathrm{NiCl}}_4\right]}^{2-}$

x + 4(-1) - (-2) = 0; where x, 4,-1 and -2 are the oxidation number of Ni, number of CN ligands, charge on one CN and charge on complex.

${\left[{\mathrm{NiCl}}_4\right]}^{2-}\Rightarrow \mathrm{x}-4+2=0\Rightarrow \mathrm{x}=+2$

Electronic configuration of ${\mathrm{Ni}}^{2+}=\left[\mathrm{Ar}\right]3{\mathrm{d}}^{8}4{\mathrm{s}}^0$

For C.N.= 4

Hence, only ${\mathrm{K}}_2\left[\mathrm{Ni}(\mathrm{CN}{)}_4\right]\text{ has }{\mathrm{dsp}}^2$ hybridisation.