1. Find the oxidation state and electronic configuration of Ni in each complex

Nickel has atomic number $28$.

Its ground-state configuration is: $Ni: [Ar] \, 3d^8 4s^2$

For coordination compounds, the metal ion configuration is determined from its oxidation state.

2. Recall the VBT hybridization patterns for coordination number 4

For 4-coordinate complexes:

• Tetrahedral geometry $\Rightarrow sp^3$
• Square planar geometry $\Rightarrow dsp^2$

Thus, we need the compound in which Ni forms a square planar complex.

3. Check each option

Option A: $NiCl_2\cdot 6H_2O$

This is hydrated nickel chloride, effectively containing the hexaaqua complex in solution/crystal form, i.e. nickel is typically present as $[Ni(H_2O)_6]^{2+}$.

• Oxidation state of Ni = $+2$
• Configuration of $Ni^{2+}$: $Ni^{2+}: [Ar] \, 3d^8$
• Coordination number = $6$

For 6-coordinate $Ni^{2+}$ with weak ligand $H_2O$, the complex is octahedral, not 4-coordinate square planar.

So this is not $dsp^2$.

Option B: $K_2[Ni(CN)_4]$

Here the complex ion is $[Ni(CN)_4]^{2-}$.

Let oxidation state of Ni be $x$: $x + 4(-1) = -2$ $x = +2$

So Ni is $Ni^{2+}$ with configuration: $Ni^{2+}: [Ar] \, 3d^8$

Now, $CN^-$ is a strong-field ligand. It causes pairing of $3d$ electrons, making one $3d$ orbital available for hybridization.

Thus hybridization is: $dsp^2$ which gives square planar geometry.

So Option B is correct.

Option C: $[Ni(CO)_4]$

Here CO is neutral, so oxidation state of Ni is $0$.

Configuration of $Ni^0$: $Ni: [Ar] \, 3d^8 4s^2$

In $[Ni(CO)_4]$, nickel forms a tetrahedral complex with hybridization: $sp^3$

So this is not $dsp^2$.

Option D: $Na_2[NiCl_4]$

The complex ion is $[NiCl_4]^{2-}$.

Let oxidation state of Ni be $x$: $x + 4(-1) = -2$ $x = +2$

So Ni is again $Ni^{2+}$: $[Ar] \, 3d^8$

But $Cl^-$ is a weak-field ligand, so pairing does not occur in the $3d$ orbitals. Hence the complex uses outer orbitals and forms a tetrahedral complex with: $sp^3$

So this is not $dsp^2$.

4. Final conclusion

The compound having $dsp^2$ hybridization according to valence bond theory is: $K_2[Ni(CN)_4]$

So the correct option is B.