The number of d-electrons in Fe^{2+} (atomic | vedclass

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(D) The electronic configuration of $_{26}Fe$ is $[Ar] \, 3d^6 \, 4s^2$. For $Fe^{2+}$,two electrons are removed from the $4s$ orbital,so the configur

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$p$-electrons in $Cl^{-}$ (At. No. of $Cl = 17$)

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(D) The electronic configuration of $_{26}Fe$ is $[Ar] \, 3d^6 \, 4s^2$. For $Fe^{2+}$,two electrons are removed from the $4s$ orbital,so the configuration is $[Ar] \, 3d^6$. The number of $d$-electrons in $Fe^{2+}$ is $6$. Now,let us check the options: $A$: $Ne$ $(Z=10)$ is $1s^2 \, 2s^2 \, 2p^6$. Number of $p$-electrons $= 6$. $B$: $Mg$ $(Z=12)$ is $1s^2 \, 2s^2 \, 2p^6 \, 3s^2$. Number of $s$-electrons $= 1s^2 + 2s^2 + 3s^2 = 6$. $C$: $Fe$ $(Z=26)$ is $[Ar] \, 3d^6 \, 4s^2$. Number of $d$-electrons $= 6$. $D$: $Cl^{-}$ $(Z=17+1=18)$ is $1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6$. Number of $p$-electrons $= 2p^6 + 3p^6 = 12$. Therefore,the number of $d$-electrons in $Fe^{2+}$ is not equal to the number of $p$-electrons in $Cl^{-}$.

The number of $d$-electrons in $Fe^{2+}$ (atomic number of $Fe = 26$) is not equal to that of the

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