Consider 'n' is the number of lone pair of electrons present in the equatorial position of the most stable structure of…

🧠 First find $n$ (LP in equatorial positions of ClF₃), then find which ions have $n$ unpaired electrons.

🗺️ Step 1 — Find n for ClF₃:

$\ce{ClF3}$: Cl has 3 bonds + 2 LP = 5 e-pairs → trigonal bipyramidal electron geometry.

Most stable structure: both lone pairs in equatorial positions (T-shaped) — equatorial LP minimizes 90° LP–BP repulsions.

$n = 2$ lone pairs in equatorial positions.

Step 2 — Ions with 2 unpaired electrons:

| Ion | Configuration | Unpaired e⁻ | |-----|---------------|------------| | $\ce{V^{3+}}$ | [Ar]$3d^2$ | 2 ✓ | | $\ce{Ti^{3+}}$ | [Ar]$3d^1$ | 1 ✗ | | $\ce{Cu^{2+}}$ | [Ar]$3d^9$ | 1 ✗ | | $\ce{Ni^{2+}}$ | [Ar]$3d^8$ | 2 ✓ | | $\ce{Ti^{2+}}$ | [Ar]$3d^2$ | 2 ✓ |

Ions with $n = 2$ unpaired electrons: A (V³⁺), D (Ni²⁺), E (Ti²⁺)

$\boxed{\text{Answer: A, D and E only}}$