To solve this, we determine the specific electron count for the reference ion and compare it with the options:

1. Reference: Fe$^{2+}$ (Z=26): Electronic configuration of Fe atom: $[Ar] 3d^6 4s^2$ . For Fe$^{2+}$, 2 electrons are removed from the 4s orbital: $[Ar] 3d^6$.

• Number of d-electrons = 6.

2. Compare with Options:

• Option A: Mg (Z=12): Configuration is $1s^2 2s^2 2p^6 3s^2$. Total s-electrons = $2+2+2 = 6$. (Equal)
• Option B: Cl (Z=17): Configuration is $1s^2 2s^2 2p^6 3s^2 3p^5$. Total p-electrons = $6+5 = 11$. (Not Equal)
• Option C: Fe (Z=26): Configuration is $[Ar] 3d^6 4s^2$. Total d-electrons = 6. (Equal)
• Option D: Ne (Z=10): Configuration is $1s^2 2s^2 2p^6$. Total p-electrons = 6. (Equal)

Thus, the number of d-electrons in Fe$^{2+}$ (6) is not equal to the number of p-electrons in Cl (11).