JEE Chem — Periodicity

Chapter Overview & Weightage

Periodicity is foundational for both Inorganic and Physical Chemistry in JEE. While direct periodic-property questions appear sparingly (1 question per JEE Main paper on average), the concepts here underpin every Inorganic Chemistry question — making it indirectly responsible for 20–25% of the chemistry marks.

Key Concepts You Must Know

Atomic radius: Decreases across a period, increases down a group. Anomalies in Group 13 (Al ≈ Ga due to poor shielding by 3d).

Ionisation energy (IE): Energy needed to remove an electron. Increases across, decreases down. Half-filled and fully-filled stability anomalies (e.g., $IE_1$ of N > O).

Electron affinity (EA): Energy released when an atom gains an electron. Generally increases across, decreases down. Anomalies: $EA$ of F < Cl due to compact 2p repulsion.

Electronegativity: Tendency to attract bonding electrons. Pauling and Mulliken scales. Increases across, decreases down. Fluorine has highest EN (3.98 Pauling).

Effective nuclear charge ($Z_{\text{eff}}$): $Z_{\text{eff}} = Z - \sigma$, where $\sigma$ is shielding. Slater’s rules give a quantitative measure.

Diagonal relationship: Li-Mg, Be-Al, B-Si due to similar charge-to-radius ratios.

Important Trends and Formulas

Solved Previous Year Questions

PYQ 1 — JEE Main 2024

Arrange $N$, $O$, $F$, $Ne$ in order of increasing first ionisation energy.

$N(2p^3)$ has half-filled stability. $O(2p^4)$ has one paired electron — easier to remove. So $IE(N) > IE(O)$. Across the period: $C < N$, $O < F$, $F < Ne$.

Order: $O < N < F < Ne$ — wait, that’s wrong. Let me redo: $C < O < N < F < Ne$. From the data: $IE_1$ values (kJ/mol): C = 1086, N = 1402, O = 1314, F = 1681, Ne = 2080.

So increasing: $\boxed{C < O < N < F < Ne}$.

PYQ 2 — JEE Main 2023

Among Cl, F, Br, I, which has the highest electron affinity?

Despite F being most electronegative, $EA(F) < EA(Cl)$ due to compact 2p orbitals causing electron-electron repulsion when adding an extra electron. So Cl has the highest electron affinity.

PYQ 3 — JEE Advanced 2022

Compute the effective nuclear charge experienced by the outermost electron in K (atomic number 19) using Slater’s rules.

K: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1$. Outermost electron is $4s^1$.

Shielding from same group: 0 (only itself). Shielding from (n-1) = 3rd shell ($3s^2 3p^6$ = 8 electrons): $8 \times 0.85 = 6.8$. Shielding from (n-2) and lower (10 electrons in $1s^2 2s^2 2p^6$): $10 \times 1.00 = 10$.

Total $\sigma = 0 + 6.8 + 10 = 16.8$.

$Z_{\text{eff}} = 19 - 16.8 = 2.2$.

Difficulty Distribution

Expert Strategy

Week 1 — Master the trends. Sketch atomic radius, IE, EA, and EN graphs across a period and down a group. Internalise the direction of each trend before memorising values.

Week 2 — Anomalies. $IE(N) > IE(O)$, $EA(Cl) > EA(F)$, Al ≈ Ga, $Be(IE) > B(IE)$. These anomalies appear directly in JEE questions.

Week 3 — Slater’s rules and effective nuclear charge. Practice computing $Z_{\text{eff}}$ for at least 5 different elements. JEE Advanced asks this as a 4-mark numerical.

Common Traps