Periodic table trends — atomic radius, ionization energy across Period 3

Question

How do atomic radius and ionization energy vary across Period 3 (Na to Ar)? Explain the reason for the trend. Why does ionization energy show an anomalous dip at Al and S?

(NCERT Class 11 — Classification of Elements)

Solution — Step by Step

Atomic radius decreases from left to right across Period 3:

Na (186 pm) > Mg (160) > Al (143) > Si (117) > P (110) > S (104) > Cl (99)

Why? As we move across, the nuclear charge increases (+1 proton each time), but electrons are added to the same shell (3rd shell). The increased nuclear pull draws the electron cloud tighter, shrinking the atom.

Ionization energy (IE) generally increases from left to right:

Na (496) < Mg (738) < Al (577) < Si (786) < P (1012) < S (1000) < Cl (1251) kJ/mol

Smaller atomic radius means outer electrons are closer to the nucleus and harder to remove, so more energy is needed.

IE drops from Mg to Al despite increasing nuclear charge. Why?

Mg has configuration $[Ne] 3s^2$ — the $3s$ subshell is fully filled (extra stability). Al has $[Ne] 3s^2 3p^1$ — the single $3p$ electron is easier to remove because:

The $3p$ orbital is higher in energy than $3s$
The $3p$ electron is shielded by the $3s^2$ electrons

So less energy is needed to remove Al’s outer electron despite Al having one more proton.

IE drops slightly from P to S. Why?

P has $[Ne] 3s^2 3p^3$ — all three $3p$ orbitals are half-filled (extra stability due to exchange energy). S has $[Ne] 3s^2 3p^4$ — one $3p$ orbital now has a pair of electrons. The electron-electron repulsion in that paired orbital makes it easier to remove one electron.

So P’s half-filled stability gives it a higher IE than expected.

Why This Works

Periodic trends arise from the competition between nuclear charge (pulls electrons in) and electron shielding (pushes them out). Across a period, nuclear charge wins because new electrons are added to the same shell — they do not shield each other effectively from the increasing nuclear pull.

The anomalies at Al and S are due to subshell effects: fully-filled ( $s^2$ ) and half-filled ( $p^3$ ) configurations have extra stability. These exceptions appear in every period and are a favourite exam question.

Alternative Method

Instead of memorising trends, think of it as: “Would you rather be an atom holding your electrons tightly (high IE) or loosely (low IE)?” Atoms with full/half-filled subshells hold on tighter. Atoms with one electron in a new subshell hold on loosely. This intuition covers all the anomalies.

For CBSE 11th and JEE, the Al and S anomalies are very frequently tested. JEE Main often gives a list of elements and asks you to arrange them in order of IE. If you forget the anomalies, you will get the wrong order for adjacent elements.

Common Mistake

Students often say “IE increases across a period” as a blanket rule and forget the dips at Al and S (or analogous positions in other periods). In MCQs, the question specifically tests whether you know IE(Mg) > IE(Al) and IE(P) > IE(S). Always check for full/half-filled subshell exceptions before answering.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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