Electronic configuration

Aufbau principle and the $(n+l)$ rule. In the ground state, orbitals fill in order of increasing energy. That order is fixed by the $(n+l)$ rule: the orbital with the smaller value of $n+l$ fills first; if two orbitals tie on $n+l$, the one with the smaller $n$ fills first. This is why $4s$ ($n+l = 4+0=4$) fills before $3d$ ($n+l=3+2=5$), even though $3d$ has the smaller principal quantum number.

Pauli's exclusion principle. No two electrons in the same atom share all four quantum numbers, so each orbital holds at most two electrons, and those two must have opposite spins ($m_s = +\tfrac12$ and $-\tfrac12$).

Hund's rule of maximum multiplicity. Within a set of degenerate orbitals (the three $p$, five $d$, or seven $f$ orbitals of a subshell), electrons occupy separate orbitals with parallel spins first; pairing starts only once every orbital in that subshell already holds one electron. This keeps like-spin electrons apart, lowering their mutual repulsion.

Counting capacity. Each subshell of azimuthal quantum number $l$ has $(2l+1)$ orbitals, so its maximum electron count is $2(2l+1)$: an $s$ subshell ($l=0$) holds 2, a $p$ subshell ($l=1$) holds 6, a $d$ subshell ($l=2$) holds 10, an $f$ subshell ($l=3$) holds 14. Summing all subshells of a given principal shell $n$ gives the maximum electrons that shell can carry:

where: $N_{\max}(n)$ is the maximum number of electrons the $n$-th shell can hold, and $n$ is the principal quantum number ($n=1,2,3,\ldots$). This is derived below.

Removing electrons to form cations. For a positive ion, electrons are removed first from the subshell with the highest principal quantum number $n$ present in the atom — not necessarily the subshell filled last by the $(n+l)$ rule. That is why $4s$ electrons leave before $3d$ electrons for every first-row transition metal cation, even though $4s$ filled first.

Question. Write the ground-state electronic configuration of iron ($Z=26$). Then write the configuration of $\text{Fe}^{2+}$ and $\text{Fe}^{3+}$, and state which ion has more unpaired electrons in its $3d$ subshell.

Step 1. Fill 26 electrons in $(n+l)$ order: $1s, 2s, 2p, 3s, 3p, 4s, 3d$. Assigning maximum capacities ($s{:}2$, $p{:}6$, $d{:}10$) until 26 electrons are used gives

Check: $2+2+6+2+6+2+6=26$ ✓. In noble-gas shorthand, iron is $[\text{Ar}]\,3d^6\,4s^2$. Unlike chromium or copper, iron's $3d^64s^2$ is the actual ground state — $d^6$ is neither half-filled nor fully-filled, so there is no extra-stability shift here.

Step 2. For $\text{Fe}^{2+}$, remove 2 electrons. Electrons leave the highest-$n$ subshell first, so both come from $4s$ (not $3d$):

Step 3. For $\text{Fe}^{3+}$, remove 3 electrons. The two $4s$ electrons go first, accounting for 2; the third electron must then come from $3d^6$:

Step 4. Apply Hund's rule to compare unpaired electrons. Distributing $3d^6$ across five degenerate $d$ orbitals gives four singly-occupied orbitals and one doubly-occupied orbital, so $\text{Fe}^{2+}$ ($3d^6$) has 4 unpaired electrons. Distributing $3d^5$ gives all five orbitals singly occupied with parallel spins, so $\text{Fe}^{3+}$ ($3d^5$) has 5 unpaired electrons — more than $\text{Fe}^{2+}$, because $d^5$ is a half-filled subshell.

Final answer: Fe is $[\text{Ar}]\,3d^6\,4s^2$; Fe$^{2+}$ is $[\text{Ar}]\,3d^6$ (4 unpaired electrons); Fe$^{3+}$ is $[\text{Ar}]\,3d^5$ (5 unpaired electrons, the more stable half-filled configuration).

Competitive stretch

Entrance exams often ask for the number of unpaired electrons directly, without asking for the full configuration first — always write the configuration, then draw the orbital boxes for just the outermost subshell before counting arrows, rather than guessing from the superscript alone.

• Filling $3d$ before $4s$ because $3<4$ — the $(n+l)$ rule governs filling order, and $4s$ ($n+l=4$) fills before $3d$ ($n+l=5$).
• Removing electrons from $3d$ before $4s$ when forming a transition-metal cation — cations always lose electrons from the highest-$n$ subshell first, here $4s$, regardless of fill order.
• Pairing electrons in one orbital of a subshell while another degenerate orbital of the same subshell is still empty — Hund's rule requires spreading electrons singly across all orbitals first.
• Mechanically writing $3d^4\,4s^2$ or $3d^9\,4s^2$ for chromium or copper — always check whether shifting one $4s$ electron creates a half-filled ($d^5$) or fully-filled ($d^{10}$) subshell.
• Using $2n^2$ for an ion's outer shell without first identifying that shell's actual subshells present — the formula gives the maximum capacity of a full shell, not the electron count of a partially filled or ionized atom.

Ask the tutor: Want to practise deriving the (n+l) filling order from the triangular diagram, or work through the configuration and unpaired-electron count of another transition-metal ion?