To solve this problem, we need to determine the number of d-electrons in $$Fe$^{2+}$ and compare it with the number of electrons in other given options.$\newline$1. Calculate the number of d-electrons in $$Fe$^{2+}: \newline$• The atomic number of iron (Fe) is 26, so the electronic configuration of neutral Fe is:$\newline [$Ar$] \, 3d^6 \, 4s^2 \newline$• In $$Fe$^{2+},$ two electrons are removed. Electrons are removed first from the 4s orbital, then from the 3d orbital.$\newline$• Therefore, the electronic configuration of $$Fe$^{2+}$ is:$\newline [$Ar$] \, 3d^6 \newline$• Thus, the number of d-electrons in $$Fe$^{2+}$ is 6.$\newline$2. Calculate the number of electrons in each option:$\newline$• Option 1: p-electrons in Cl (Z = 17)$\newline$ - The electronic configuration of Cl is:$\newline [$Ne$] \, 3s^2 \, 3p^5 \newline$ - The number of p-electrons is 5.$\newline$• Option 2: d-electrons in Fe (Z = 26)$\newline$ - The electronic configuration of neutral Fe is:$\newline [$Ar$] \, 3d^6 \, 4s^2 \newline$ - The number of d-electrons is 6.$\newline$• Option 3: p-electrons in Ne (Z = 10)$\newline$ - The electronic configuration of Ne is:$\newline 1s^2 \, 2s^2 \, 2p^6 \newline$ - The number of p-electrons is 6.$\newline$• Option 4: s-electrons in Mg (Z = 12)$\newline$ - The electronic configuration of Mg is:$\newline [$Ne$] \, 3s^2 \newline$ - The number of s-electrons is 4 (2 from 1s, 2 from 2s, and 2 from 3s).$\newline$3. Compare the number of electrons:$\newline$• The number of d-electrons in $$Fe$^{2+}$ is 6.$\newline$• The number of p-electrons in Cl is 5, which is not equal to 6.$\newline$Therefore, the number of d-electrons in $$Fe$^{2+}$ is not equal to the number of p-electrons in Cl.$\newline$This corresponds to Option 1.