Question
Describe how ionization energy, electronegativity, and electron affinity vary across Period 3 (Na to Ar). Explain any anomalies observed.
(NCERT Class 11, Chapter 3)
Solution — Step by Step
Moving left to right across Period 3 (Na → Mg → Al → Si → P → S → Cl → Ar):
- Nuclear charge increases (more protons)
- Atomic size decreases (electrons pulled closer)
- Therefore: ionization energy, electronegativity, and electron affinity generally increase
The outer electrons are held more tightly as we move across.
The general trend is increasing IE across the period, but two dips occur:
| Element | Na | Mg | Al | Si | P | S | Cl | Ar |
|---|---|---|---|---|---|---|---|---|
| IE (kJ/mol) | 496 | 738 | 577 | 786 | 1012 | 1000 | 1251 | 1521 |
Anomaly 1: Al has lower IE than Mg. Reason: Al’s outermost electron is in 3p (easier to remove) while Mg’s is in 3s (which is more stable and closer to the nucleus).
Anomaly 2: S has slightly lower IE than P. Reason: P has a half-filled 3p configuration () which is extra stable. S has one paired electron in 3p (), and the electron-electron repulsion in the paired orbital makes it easier to remove.
Electronegativity increases steadily across the period: Na (0.93) → Cl (3.16). Noble gases are not assigned electronegativity values (they don’t form bonds under normal conditions).
No significant anomalies here — the trend is smooth because electronegativity is a bonding property that correlates with both IE and electron affinity.
Electron affinity generally becomes more negative (more energy released) across the period. But there are exceptions:
- Mg and Ar have very low (near zero or positive) electron affinity because their subshells are fully filled ( for Mg, for Ar). Adding an electron would destabilise the configuration.
- P has lower electron affinity than Si because P has a stable half-filled configuration. Adding an electron disrupts this stability.
Why This Works
All three properties are controlled by the balance between nuclear charge (attraction) and electron-electron repulsion (shielding). Across a period, nuclear charge wins — so these properties generally increase. But the extra stability of half-filled and fully filled subshells creates dips. These dips are the most tested anomalies in exams.
Alternative Method — Remember the Anomaly Pattern
The anomaly pairs to memorise for Period 3:
- IE: Mg > Al (s vs p) and P > S (half-filled stability)
- EA: Si > P (half-filled stability) and Ar has near-zero EA (filled shell)
For JEE/NEET, the most frequently asked comparison is: “Why is IE of N greater than O?” (same logic as P > S in Period 3). Both test the extra stability of half-filled subshells. If you understand the Period 3 trends, you automatically handle Period 2 as well.
Common Mistake
Students memorise the general trend (IE increases across a period) and apply it blindly without accounting for exceptions. In MCQs, the answer is often the exception: “Which has higher IE — Al or Mg?” The answer is Mg, not Al, despite Mg being to the left. Ignoring the s-to-p transition anomaly is one of the costliest mistakes in this chapter.