Structure of Atom • Topic 3 of 3

Electronic Configuration

Once we know that electrons occupy orbitals labelled by quantum numbers, the next question is practical: in a real atom, which orbitals get filled, and in what order? The arrangement of electrons in the orbitals of an atom is its electronic configuration, written by listing each occupied sub-shell with the number of electrons as a superscript, for example carbon as $1s^2\,2s^2\,2p^2$. Three rules govern the filling.

1. Aufbau principle. "Aufbau" is German for "building up". Electrons enter the lowest-energy orbital available first, then fill upwards. The energy order is set by the $(n+l)$ rule (Madelung rule): an orbital with a lower value of $n+l$ fills first; if two orbitals have the same $n+l$, the one with the smaller $n$ fills first. So $4s$ (with $n+l=4+0=4$) fills before $3d$ ($n+l=3+2=5$), even though $3d$ has a smaller $n$. The resulting order is $1s,2s,2p,3s,3p,4s,3d,4p,5s,4d,5p,6s,4f,5d,6p,7s$.

2. Pauli exclusion principle. No two electrons in an atom can have all four quantum numbers identical. Since an orbital is fixed by $n,l,m_l$, the two electrons sharing it must differ in $m_s$ — one spin up ($+\tfrac{1}{2}$), one spin down ($-\tfrac{1}{2}$). Hence an orbital holds at most two electrons, and they must be paired with opposite spins. This is why a sub-shell of $2l+1$ orbitals holds at most $2(2l+1)$ electrons.

3. Hund's rule of maximum multiplicity. When filling degenerate (equal-energy) orbitals of a sub-shell, electrons occupy them singly first, all with parallel spins, before any pairing begins. Pairing two electrons in the same small orbital costs energy (electron-electron repulsion), so spreading out is favoured. For nitrogen ($2p^3$) all three $p$ electrons are unpaired in separate orbitals; only at oxygen ($2p^4$) does pairing start.

To write the configuration of an ion, first write the neutral atom, then add electrons (for an anion) into the next orbital, or remove electrons (for a cation) — and here is a key subtlety: electrons are removed from the orbital with the highest $n$ first, not simply in the reverse of the Aufbau order. So for iron, $\text{Fe}=[\text{Ar}]\,3d^6\,4s^2$, but $\text{Fe}^{2+}=[\text{Ar}]\,3d^6$ — the $4s$ electrons leave before the $3d$ electrons, because $4s$ has the higher principal quantum number.

Extra stability of half-filled and fully-filled sub-shells. Configurations in which a sub-shell is exactly half-filled ($p^3,d^5,f^7$) or completely filled ($p^6,d^{10},f^{14}$) are unusually stable, thanks to their symmetrical charge distribution and greater exchange energy (the stabilising energy gained when electrons of parallel spin swap places). This stability is large enough to override the simple Aufbau order in a few elements. Chromium is $[\text{Ar}]\,3d^5\,4s^1$ (not $3d^4\,4s^2$) and copper is $[\text{Ar}]\,3d^{10}\,4s^1$ (not $3d^9\,4s^2$): one $4s$ electron shifts into $3d$ to reach the extra-stable half-filled or fully-filled $d$ sub-shell. These two are the classic NCERT exceptions you must remember.

From the configuration we can read off the number of unpaired electrons (which governs magnetic behaviour — paramagnetic if any are unpaired, diamagnetic if all are paired) and the valence-shell arrangement that drives the element's chemistry. The configuration is therefore the bridge between the abstract quantum numbers and the real periodic table.

Aufbau filling order arrows through the sub-shellsAufbau Order (follow the diagonal arrows)1s2s2p3s3p3d4s4p4d4f5s5p5d5f6s6p6dLower (n+l) fills first; ties broken by lower n
Selected configurations — note the Cr and Cu exceptions
Element (Z)ConfigurationUnpaired e−
N (7)1s² 2s² 2p³3
Fe (26)[Ar] 3d⁶ 4s²4
Cr (24)[Ar] 3d⁵ 4s¹6
Cu (29)[Ar] 3d¹⁰ 4s¹1
1
Worked Example
Write the complete electronic configuration of phosphorus ($Z=15$) and state the number of unpaired electrons.
Solution
  1. Fill in Aufbau order: $1s^2\,2s^2\,2p^6\,3s^2\,3p^3$ (total $2+2+6+2+3=15$).
  2. The three $3p$ electrons occupy $p_x,p_y,p_z$ singly (Hund's rule).
  3. So all three $3p$ electrons are unpaired.

Answer: $1s^2\,2s^2\,2p^6\,3s^2\,3p^3$; $3$ unpaired electrons.

2
Worked Example
Using the $(n+l)$ rule, explain why $4s$ is filled before $3d$.
Solution
  1. For $4s$: $n+l=4+0=4$.
  2. For $3d$: $n+l=3+2=5$.
  3. The orbital with the lower $n+l$ fills first, so $4s$ (value $4$) fills before $3d$ (value $5$).

Answer: $4s$ has the smaller $(n+l)$ value, so by the Madelung rule it is lower in energy and fills first.

3
Worked Example
Write the electronic configuration of the iron(II) ion, $\text{Fe}^{2+}$ ($Z$ of Fe $=26$).
Solution
  1. Neutral Fe: $[\text{Ar}]\,3d^6\,4s^2$.
  2. To form $\text{Fe}^{2+}$, remove $2$ electrons from the orbital of highest $n$ — that is $4s$.
  3. Removing both $4s$ electrons leaves $[\text{Ar}]\,3d^6$.

Answer: $\text{Fe}^{2+}=[\text{Ar}]\,3d^6$ (the $4s$ electrons leave before the $3d$ electrons).

4
Worked Example
Explain why chromium has the configuration $[\text{Ar}]\,3d^5\,4s^1$ rather than $[\text{Ar}]\,3d^4\,4s^2$.
Solution
  1. The expected Aufbau filling would give $3d^4\,4s^2$.
  2. Shifting one $4s$ electron into $3d$ gives $3d^5\,4s^1$, making both sub-shells half-filled.
  3. Half-filled $3d^5$ has a symmetrical distribution and high exchange energy, so it is extra stable.

Answer: The extra stability of the half-filled $3d^5$ (and $4s^1$) sub-shell makes $[\text{Ar}]\,3d^5\,4s^1$ lower in energy.

5
Worked Example
State the four quantum numbers for the last (highest-energy) electron of sodium ($Z=11$).
Solution
  1. Sodium: $1s^2\,2s^2\,2p^6\,3s^1$; the last electron is in $3s$.
  2. For $3s$: $n=3$, $l=0$ (so $m_l=0$).
  3. A single electron is taken as $m_s=+\tfrac{1}{2}$ by convention.

Answer: $n=3,\ l=0,\ m_l=0,\ m_s=+\tfrac{1}{2}$.

6
Worked Example
How many unpaired electrons are there in a gaseous $\text{Mn}^{2+}$ ion ($Z$ of Mn $=25$)? Comment on its magnetic behaviour.
Solution
  1. Neutral Mn: $[\text{Ar}]\,3d^5\,4s^2$.
  2. Remove the two $4s$ electrons (highest $n$): $\text{Mn}^{2+}=[\text{Ar}]\,3d^5$.
  3. Five $3d$ electrons occupy five orbitals singly (Hund's rule) — all unpaired.

Answer: $5$ unpaired electrons; $\text{Mn}^{2+}$ is strongly paramagnetic.

Key Points

  • Electronic configuration lists occupied sub-shells, e.g. $\text{C}=1s^2\,2s^2\,2p^2$; filling follows Aufbau, Pauli and Hund.
  • Aufbau / $(n+l)$ rule: lower $n+l$ fills first; ties broken by lower $n$ — so $4s$ fills before $3d$.
  • Pauli exclusion: an orbital holds at most two electrons with opposite spins; Hund's rule: fill degenerate orbitals singly with parallel spins before pairing.
  • Ions: remove electrons from the highest-$n$ orbital first, so $\text{Fe}^{2+}=[\text{Ar}]\,3d^6$ (not losing $3d$ first).
  • Half-filled ($d^5$) and fully-filled ($d^{10}$) sub-shells have extra stability — hence $\text{Cr}=[\text{Ar}]\,3d^5\,4s^1$ and $\text{Cu}=[\text{Ar}]\,3d^{10}\,4s^1$.
Tap an option to check your answer0 / 4
Q1.According to the $(n+l)$ rule, which orbital is filled immediately after $4s$?
Explanation: $3d$ has $n+l=5$ versus $4s$'s $4$, so $3d$ ($n+l=5$, lower $n$ than $4p$ at the same value) fills right after $4s$.
Q2.The maximum number of electrons that can share the same orbital is:
Explanation: By the Pauli exclusion principle an orbital holds two electrons, and they must have opposite spins.
Q3.Hund's rule of maximum multiplicity tells us that, within a sub-shell, electrons:
Explanation: Degenerate orbitals are each singly occupied with parallel spins before any pairing, minimising repulsion.
Q4.The correct ground-state configuration of copper ($Z=29$) is:
Explanation: A $4s$ electron shifts into $3d$ to give the extra-stable fully-filled $3d^{10}\,4s^1$ configuration.