A neutral transition metal atom possesses the following shell-wise electron distribution in its grou — Atomic Structure Chemistry Question
Question
A neutral transition metal atom possesses the following shell-wise electron distribution in its ground state: $2$ electrons in the K shell, $8$ electrons in the L shell, $11$ electrons in the M shell, and $2$ electrons in the N shell. Calculate the total number of $p$ -electrons present in this specific atom.
💡 Solution & Explanation
The given shell distribution is K= $2$ ($n=1$), L= $8$ ($n=2$), M= $11$ ($n=3$), N= $2$ ($n=4$). The corresponding electronic configuration following the Aufbau sequence is $1s^2$ (K), $2s^2 2p^6$ (L), $3s^2 3p^6 3d^3$ (M), and $4s^2$ (N). This atom is Vanadium ($Z=23$). The $p$ -electrons are solely located in the fully occupied $2p$ and $3p$ subshells. The total number of $p$ -electrons is $6$ (from $2p^6$) + $6$ (from $3p^6$) = $12$.