S1.3 Electron configurations

Count the total electrons in each option: option A has 23 electrons (vanadium), option B has 24 electrons (chromium), option C has 23 electrons (vanadium), option D has 29 electrons (copper).

Chromium ($Z=24$) has the actual configuration $[\text{Ar}]4s^{1}3d^5$, not $4s^{2}3d^4$. Copper ($Z=29$) has the actual configuration $[\text{Ar}]4s^{1}3d^{10}$, not $4s^{2}3d^9$ or $4s^{1}3d^9$.

Vanadium ($Z=23$) follows the Aufbau principle normally with no exceptions. Its ground-state configuration is $[\text{Ar}]4s^{2}3d^3$, which matches option A exactly.

Option A represents the correct ground-state configuration for vanadium. Options B, C, and D all give incorrect configurations for the respective elements — B and C both correspond to chromium but neither is correct (B has wrong distribution; C has wrong total), and D is an incorrect configuration for copper.

The question asks which configuration is valid for a ground-state atom. We need to check each option against the known filling rules and the two exceptions (Cr and Cu).

Option B has 24 electrons, so it represents chromium. The ground-state configuration of Cr is $[\text{Ar}]4s^{1}3d^5$ — not $4s^{2}3d^4$. This option is incorrect.

Option C also has 23 electrons (vanadium). The correct ground-state configuration for vanadium is $4s^{2}3d^3$, not $4s^{1}3d^4$. Vanadium is not one of the exceptions, so moving an electron from 4s to 3d is not justified.

Option D has 29 electrons, representing copper. The actual ground-state configuration of copper is $[\text{Ar}]4s^{1}3d^{10}$, not $4s^{1}3d^9$ (which would only account for 28 electrons — incorrect).

Option A represents vanadium ($Z=23$) with $4s^{2}3d^3$, which correctly follows the Aufbau principle. This is the only valid ground-state configuration.

The key to this question is locating the large jump between successive ionization energies. The 1st IE is 496 kJ mol${}^{-1}$, and the 2nd IE jumps dramatically to 4562 kJ mol${}^{-1}$ — an approximately ninefold increase.

A large jump between the $n$th and $(n+1)$th ionization energies indicates that the $n$th electron was in the outer shell, and the $(n+1)$th electron must be removed from a full inner shell that is much closer to the nucleus. Here, the jump occurs after the 1st electron, meaning this element has only one electron in its outermost shell.

An element in Period 3 with one electron in its outermost shell is sodium ($Z=11$), which has a single $3s^1$ electron. After removing this electron, the next electron must come from the full $2p$ sublevel, which requires considerably more energy.

The ground-state electron configuration of sodium is $1s^{2}2s^{2}2p^{6}3s^1$, corresponding to option C.

The enormous jump between the 1st IE (496 kJ mol${}^{-1}$) and 2nd IE (4562 kJ mol${}^{-1}$) tells us this element has exactly one electron in its outermost energy level.

Option A is aluminium ($Z=13$) with configuration $3s^{2}3p^1$. For aluminium, the large jump occurs between the 3rd and 4th IEs (after removing 3 outer electrons), not after the 1st.

Option B is magnesium ($Z=12$) with configuration $3s^2$. For magnesium, the large jump occurs between the 2nd and 3rd IEs, because both $3s$ electrons can be removed before hitting the inner shell.

Option D is silicon ($Z=14$) with configuration $3s^{2}3p^2$. Silicon has four outer electrons, so the large jump would appear between the 4th and 5th IEs.

Option C ($1s^{2}2s^{2}2p^{6}3s^1$) represents sodium, which has a single outer electron. Removing this one $3s$ electron is relatively easy; the next electron must come from the inner $2p$ shell, causing the dramatic jump.