We compare the number of electrons in the specific orbitals for each species:

1. Analyze Fe²⁺ (Z=26): Neutral Fe configuration: $[Ar] 3d^6 4s^2$. Fe²⁺ configuration (remove 2 electrons from 4s): $[Ar] 3d^6$.

• Number of d-electrons = 6.

2. Evaluate Options:

• Option A (Mg, Z=12): Configuration is $1s^2 2s^2 2p^6 3s^2$. Total s-electrons = $2(1s) + 2(2s) + 2(3s)$ = 6. (Equal to Fe²⁺ d-electrons).
• Option B (Cl, Z=17): Configuration is $1s^2 2s^2 2p^6 3s^2 3p^5$. Total p-electrons = $6(2p) + 5(3p)$ = 11. (Different from Fe²⁺ d-electrons).
• Option C (Fe, Z=26): Configuration is $[Ar] 3d^6 4s^2$. Total d-electrons = 6. (Equal to Fe²⁺ d-electrons).
• Option D (Ne, Z=10): Configuration is $1s^2 2s^2 2p^6$. Total p-electrons = 6. (Equal to Fe²⁺ d-electrons).

Therefore, the count for Chlorine (11) is the only one that differs.