In chromyl chloride, the number of d-electrons present on chromium is same as in (Given at no. of Ti: 22, V: 23, Cr: 24, Mn: 25, Fe: 26)

In chromyl chloride (CrO₂Cl₂), we are asked to identify which species has the same number of d-electrons on its central metal.

To determine the oxidation state of chromium in CrO₂Cl₂, let x represent the oxidation state of Cr. Each oxygen atom contributes -2 and each chloride contributes -1, so that $$x + 2(-2) + 2(-1) = 0 \implies x = +6$$.

Next, the electron configuration of Cr⁶⁺ follows from that of neutral Cr: [Ar] 3d⁵ 4s¹. Removing six electrons from the 3d and 4s orbitals gives Cr⁶⁺ = [Ar], which corresponds to 0 d-electrons.

We then compare this with each option. For Option A - V(IV), vanadium has [Ar] 3d³ 4s², and V⁴⁺ becomes [Ar] 3d¹, giving 1 d-electron. For Option B - Mn(VII), manganese has [Ar] 3d⁵ 4s², and Mn⁷⁺ becomes [Ar], giving 0 d-electrons ✓. For Option C - Fe(III), iron is [Ar] 3d⁶ 4s², and Fe³⁺ is [Ar] 3d⁵, giving 5 d-electrons. For Option D - Ti(III), titanium is [Ar] 3d² 4s², and Ti³⁺ is [Ar] 3d¹, giving 1 d-electron.

Since Mn(VII) has 0 d-electrons, the same number as Cr(VI) in chromyl chloride, the correct answer is Option B: Mn(VII).