Atomic Structure: Question 3

Worked solution

Step 1: Predict the configuration from the simple Aufbau order

Following the normal orbital-filling (Aufbau) order alone, with 24 electrons to place, you would fill up to $3p^6$ (18 electrons), then add 2 more to $4s$ and the remaining 4 to $3d$:

$1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^4\,4s^2 \quad (\text{option A, total } = 2+2+6+2+6+4+2 = 24)$

This is a perfectly reasonable prediction, but it is not what is actually observed for chromium.

Step 2: Apply the chromium exception

Chromium is one of two Period 4 exceptions (with copper) where an electron is promoted from $4s$ into $3d$ relative to the naive prediction. This happens because a half-filled $3d$ sub-shell ($3d^5$), with one electron in each of the five $3d$ orbitals, is an unusually stable, lower-energy arrangement, the extra exchange stability from five parallel-spin electrons outweighs the small energy cost of promoting one electron out of $4s$.

The observed ground-state configuration of chromium is therefore:

$1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^5\,4s^1$

Total electrons: $2+2+6+2+6+5+1 = 24$ ✓, matching chromium’s proton number. This is option B.

Step 3: Why the other options are wrong

• A ($3d^4\,4s^2$, total 24): the Aufbau-predicted but experimentally incorrect configuration for chromium. It does not account for the extra stability of a half-filled $3d$ sub-shell.
• C ($3d^6$, total 24): has the correct total number of electrons but no $4s$ electrons at all, which never happens for a Period 4 element in the ground state. The $4s$ sub-shell always holds at least one electron here.
• D ($3d^5\,4s^2$, total 25): has the correct $3d^5$ split but one electron too many overall (this is the configuration of the $\text{Cr}^-$ ion or the total electron count for manganese, not neutral chromium).

Final answer

• The correct configuration is $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,\boxed{3d^5\,4s^1}$, option B.