1. Find the $d$-electron count of each metal ion

Water is a weak-field ligand, so all these octahedral complexes are high spin.

We first write the ground-state electronic configuration of each atom and then remove electrons for the given oxidation state.

2. Complex-wise analysis

(i) $[V(H_2O)_6]^{3+}$

• $V$ ($Z=23$): $[Ar]3d^34s^2$
• $V^{3+}$: remove $2$ electrons from $4s$ and $1$ from $3d$
• So, $V^{3+} = 3d^2$

In octahedral field: $t_{2g}^2e_g^0$ Number of unpaired electrons $=2$ (even)

(ii) $[Cr(H_2O)_6]^{2+}$

• $Cr$ ($Z=24$): $[Ar]3d^54s^1$
• $Cr^{2+}$: remove $1$ from $4s$ and $1$ from $3d$
• So, $Cr^{2+} = 3d^4$

High-spin octahedral arrangement: $t_{2g}^3e_g^1$ Number of unpaired electrons $=4$ (even)

(iii) $[Fe(H_2O)_6]^{3+}$

• $Fe$ ($Z=26$): $[Ar]3d^64s^2$
• $Fe^{3+}$: remove $2$ from $4s$ and $1$ from $3d$
• So, $Fe^{3+} = 3d^5$

High-spin octahedral arrangement: $t_{2g}^3e_g^2$ Number of unpaired electrons $=5$ (odd)

(iv) $[Ni(H_2O)_6]^{3+}$

• $Ni$ ($Z=28$): $[Ar]3d^84s^2$
• $Ni^{3+}$: remove $2$ from $4s$ and $1$ from $3d$
• So, $Ni^{3+} = 3d^7$

High-spin octahedral arrangement: $t_{2g}^5e_g^2$ Number of unpaired electrons $=3$ (odd)

(v) $[Cu(H_2O)_6]^{2+}$

• $Cu$ ($Z=29$): $[Ar]3d^{10}4s^1$
• $Cu^{2+}$: remove $1$ from $4s$ and $1$ from $3d$
• So, $Cu^{2+} = 3d^9$

Octahedral arrangement: $t_{2g}^6e_g^3$ Number of unpaired electrons $=1$ (odd)

3. Count complexes with even number of unpaired electrons

• $[V(H_2O)_6]^{3+}$: $2$ unpaired ✅
• $[Cr(H_2O)_6]^{2+}$: $4$ unpaired ✅
• $[Fe(H_2O)_6]^{3+}$: $5$ unpaired ❌
• $[Ni(H_2O)_6]^{3+}$: $3$ unpaired ❌
• $[Cu(H_2O)_6]^{2+}$: $1$ unpaired ❌

Therefore, the number of such complexes is $2$

4. Compare with stored correct answer

Stored correct answer is C, i.e. $2$.

So, the derived answer agrees with the stored answer.