1. Find oxidation state and electronic configuration of metal in the reactant

In aqueous solution, $NiCl_2$ contains $Ni^{2+}$.

Nickel atom: $Z=28$

$Ni: [Ar]3d^8 4s^2$

So,

$Ni^{2+}: [Ar]3d^8$

For free $Ni^{2+}$, the number of unpaired electrons in $3d^8$ is 2.

2. Determine oxidation state of Ni in the product

Given complex: $[Ni(CN)_6]^{2-}$

Let oxidation state of Ni be $x$.

Since each $CN^-$ has charge $-1$,

$x + 6(-1) = -2$ $x - 6 = -2$ $x = +4$

So metal centre is $Ni^{4+}$.

Now electronic configuration:

$Ni^{4+}: [Ar]3d^6$

3. Nature of ligand and spin state

$CN^-$ is a strong field ligand. Hence in octahedral complex $[Ni(CN)_6]^{2-}$, electrons pair up to give a low-spin arrangement.

For octahedral low-spin $d^6$:

$t_{2g}^6 e_g^0$

All electrons are paired.

So number of unpaired electrons in $[Ni(CN)_6]^{2-}$ is 0.

4. Calculate total change in number of unpaired electrons

Initial unpaired electrons on metal centre = $2$

Final unpaired electrons on metal centre = $0$

Therefore,

$\text{change} = 2 - 0 = 2$

5. Final answer

The total change in number of unpaired electrons is:

$\boxed{2}$

6. Comparison with stored correct answer

Stored correct answer = $2$

This matches the derived answer.