Periodic Trends in Properties: Radius, Ionization Enthalpy & Electronegativity — Class 11 Chemistry Notes (JEE & NEET)

01

Why Properties Repeat — Complete Theory

Every trend in this section is a tug-of-war between two forces. Force one: the nucleus — every added proton pulls electrons in harder. Force two: shielding and distance — every added shell pushes the outermost electrons out and screens them from the nucleus. What the outermost electron actually feels is the effective nuclear charge, Z_eff ≈ Z − σ (σ = shielding by inner electrons). Across a period, electrons enter the same shell: shielding barely grows, so Z_eff climbs — pull wins. Down a group, a fresh shell opens each step: distance and shielding dominate — push wins. That single sentence generates every trend below.

1 · Atomic radius. Three flavours exist, and exams quote them precisely: covalent radius (half the internuclear distance between two bonded identical atoms), van der Waals radius (half the closest approach of non-bonded atoms — always larger), and metallic radius (in a metal lattice). Across a period, radius decreases (Na 186 → Cl 99 pm) — Z_eff tightens the same shell. Down a group it increases (Li 152 → Cs 262 pm) — shells stack. Two side rules: cation < parent atom < anion (fewer electrons, less repulsion / more electrons, more repulsion), and noble gases jump up because only their van der Waals radius is quoted.

Across periodr ↓ · Z_eff ↑ Down groupr ↑ · shells ↑ Ionscation < atom < anion

2 · Ionization enthalpy (IE). The energy to remove an electron from an isolated gaseous atom in its ground state — IE₁ for the first electron, always positive (you must pay). Successive removals cost more: IE₁ < IE₂ < IE₃, with a giant jump once a noble-gas configuration is reached. Across a period IE₁ generally increases (harder to pull a tightly-held electron); down a group it decreases (outer electron is farther, shielded). The exceptions — Be > B (2s² penetration beats 2p¹) and N > O (half-filled 2p³ symmetry beats 2p⁴ pairing) — are the most-asked whys in this chapter.

3 · Electron gain enthalpy (EA). The energy change when an isolated gaseous atom gains an electron. Negative = energy released = the atom wants the electron. Across a period it generally turns more negative; down a group less negative. Two deliberate wrinkles: chlorine (−349 kJ/mol) beats fluorine (−328) because F’s tiny 2p is too crowded to accept an electron comfortably; and half-filled N (and noble gases) have positive (endothermic) values — they refuse the electron.

4 · Electronegativity (EN). The tendency of a bonded atom to attract the shared pair toward itself — a relative, unitless number (Pauling scale), not an energy. F = 4.0 is the ceiling; O 3.5, N 3.0, Cl 3.0. Trends mirror IE: increases across, decreases down. Unlike EA, EN has no exception at fluorine — it stays the champion of both rankings students most often confuse.

5 · Metallic and non-metallic character. Metals are IE-low, EA-low electron donors; non-metals are IE-high, EA-high electron acceptors. So metallic character decreases across (Na → Cl) and increases down (Li → Cs; francium aside, Cs is the most reactive metal). The periodic bridge: oxides turn from basic (Na₂O, MgO) → amphoteric (Al₂O₃) → acidic (SiO₂, P₄O₁₀, SO₃, Cl₂O₇) across a period. Add the diagonal relationship — Li↔Mg, Be↔Al, B↔Si behave alike because their radii and Z_eff per charge coincide — and this chapter’s conceptual toolkit is complete. (The configuration engine (§ 3.5) behind all of it is built next.)

Table 1 — Period 3 census: the five properties, element by element (memorise the pattern, not just numbers)
Property	Na	Mg	Al	Si	P	S	Cl	Ar
Atomic radius (pm, covalent)	186	160	143	118	110	104	99	188 (vdW)
IE₁ (kJ/mol)	496	738	578	786	1012	1000	1256	1521
EA (kJ/mol)	−53	> 0	−43	−134	−72	−200	−349	> 0
EN (Pauling)	0.9	1.2	1.5	1.8	2.1	2.5	3.0	—
Metallic character	strong	strong	metal	metalloid	non-metal	non-metal	non-metal	noble

Table 2 — The exception ledger: where the trend bends and why (JEE’s favourite section)
Exception	Data	Reason
Be > B (IE₁)	899 vs 801 kJ/mol	Be loses a penetrated 2s² electron; B’s lone 2p¹ sits higher energy and leaves easily
N > O (IE₁)	1402 vs 1314 kJ/mol	N’s half-filled 2p³ is symmetric and stable; O’s 2p⁴ pair suffers interelectronic repulsion
Cl < F reversed for EA	−349 vs −328 kJ/mol	F’s crowded 2p repels the incoming electron; Cl’s spacious 3p accepts it freely
N, noble gases: positive EA	> 0	Half-filled 2p³ (N) and closed shells (noble gases) resist an extra electron; P (−72) is only weakly negative
Al ≈ Ga (IE₁, group 13)	578 vs 579 kJ/mol	Poor shielding by 3d electrons in Ga cancels its larger Z — d-block contraction effect

02

Visualising the Trends & Charting Them Live

Ek naksha, ek bench — the trend map first, then chart any property across any series with exceptions flagged.

FIG. 1 — The whole chapter in one map: across the period (Z_eff wins) radius shrinks while IE, EN and non-metallic character climb; down the group (shells win) everything reverses. The red flags mark the bends — Be>B, N>O, and chlorine beating fluorine at electron capture.

Try it live

Trend Bench

Property

Series

Red bars = trend exceptions — exactly the bars NTA asks about. Data follows exam convention (covalent radii, Pauling EN, kJ/mol).

03

Solved Examples (Step-by-Step)

Data → force analysis → verdict. Jo reasoning yahan chalti hai, wahi Trend Bench ke explanation box me live chalti hai.

EXAMPLE 01Foundation · Radius order

Four atoms, one shrinking shell

Arrange Na, Mg, Al and Si in decreasing order of atomic radius, and justify without a data table. Also place Na⁺ and Mg²⁺ relative to their parents.

DirectionAll four are period-3 neighbours — same n = 3 shell, Z_eff rising left → right.
OrderNa (186) > Mg (160) > Al (143) > Si (118) pm — pull wins across.
IonsNa⁺ < Na (a shell’s worth of repulsion gone) and Mg²⁺ < Mg — cations always shrink; anions always swell.
Bonus chainMg²⁺ < Na⁺ < Mg < Na — Mg²⁺ holds more protons over the same electron count as Na⁺. Isoelectronic series rule: more Z, smaller r.

Na > Mg > Al > Si · Mg²⁺ < Na⁺ (isoelectronic)

EXAMPLE 02JEE Main · Exception why

The N > O interrogation

The first ionization enthalpy of nitrogen (1402 kJ/mol) is higher than that of oxygen (1314 kJ/mol), although oxygen has a higher nuclear charge. Explain. Which other pair shows the same behaviour?

ConfigsN: 1s² 2s² 2p³ — half-filled, symmetric · O: 1s² 2s² 2p⁴ — one paired set.
Force analysisRemoving from O’s 2p⁴ breaks into a half-filled set — repulsion in the pair makes that electron loose; removing from N destroys a stable half-filled shell — extra cost.
VerdictSymmetry beats charge: N > O.
Twin caseBe (899) > B (801) — same script: filled 2s² penetration vs lone 2p¹.

Half-filled stability wins — N > O, and Be > B by the same logic

EXAMPLE 03JEE Main · Successive IE

Reading IE tables like a fingerprint

Element X has IE₁ = 496 and IE₂ = 4562 kJ/mol; element Y has IE₁ = 738 and IE₂ = 1451 kJ/mol. Identify the group of each and explain the gap.

X’s jumpIE₂/IE₁ ≈ 9× — after losing one electron X hits a noble-gas core → X is group 1 (Na: Na⁺ = Ne config).
Y’s jumpModest IE₂/IE₁ (~2×); a big jump would only arrive at IE₃ → Y holds two valence electrons → group 2 (Mg).
VerifyNa⁺ (2,8) vs Na (2,8,1) — pulling from the closed Ne core costs 4562 kJ/mol ✓. Rule: jump position = valence count.

X = group 1 · Y = group 2 — the jump marks the core

04

Practice Questions (With Solutions)

Attempt first — options lock after one shot, exactly like the real exam. Then read the working, chahe galti ho ya na ho.

Attempted 0/4 · Correct 0

Q1NEET · Radius order

The correct decreasing order of atomic radius is:

Solution

Period 3, same shell, Z_eff rising → radius falls: Al (143) > Si (118) > P (110) > Cl (99).
Option C is the direction trap (reading the trend backwards); D scrambles neighbours. Verify on the Trend Bench.

(A) Al > Si > P > Cl

Q2JEE Main · EA anomaly

The element with the most negative electron gain enthalpy is:

Solution

Cl: −349 kJ/mol beats F’s −328 — F’s crowded 2p charges a repulsion tax on the incoming electron.
A is the EN-reflex answer — the single most exploited trap in this chapter. D refuses electrons outright (positive EA).

(B) Chlorine, −349 kJ/mol

Q3JEE Main · IE dip

The first ionization enthalpy of aluminium (578 kJ/mol) is lower than that of magnesium (738 kJ/mol) because:

Solution

Mg’s 3s² electrons penetrate closer to the nucleus (lower energy, harder to remove); Al’s new 3p¹ electron sits outside that shielded pair.
Same family of reasoning as Be > B and N > O — sublevel energy and pairing, not raw nuclear charge, decides these dips.

(B) 3p¹ vs 3s² penetration

Q4NEET · EN vs EA

Which statement about electronegativity is correct?

Solution

A describes electron gain enthalpy, not EN — the definition swap. B is wrong twice: EN is unitless and relative.
D inverts the trend (EN decreases down a group). C is the definition, verbatim.

(C) Bonded-atom pull · F = 4.0

05

Key Rules & Takeaways

Rule card

Eight lines that solve this topic

Z_eff ≈ Z − σ — the engineAcross: Z_eff ↑ (pull wins) · Down: shells ↑ (push wins).

r: across ↓ · down ↑Na 186 → Cl 99 pm · Li 152 → Cs 262 pm; vdW > covalent.

Cation < parent < anionIsoelectronic series: higher Z → smaller radius (Mg²⁺ < Na⁺).

IE₁: across ↑ · down ↓Exceptions Be>B, N>O; Al≈Ga (3d poor shielding).

IE₁ < IE₂ < IE₃ · jump = coreNa: 496 → 4562 (Ne core). Jump position = valence count = group.

EA: Cl −349 > F −328 (anomaly)N (half-filled 2p³) and noble gases: positive EA; P only weakly negative.

EN: relative, unitless · F = 4.0Bonded atom pulls shared pair — not an energy; trends mirror IE.

Metallic ↓ across · ↑ downOxides basic → amphoteric → acidic; diagonals Li-Mg, Be-Al, B-Si.

Anchor numbers: Na→Cl 186→99 pm · IE₁ Na 496 / IE₂ 4562 · N 1402 > O 1314 · Be 899 > B 801 · EA Cl −349, F −328 · EN F 4.0, O 3.5, N 3.0

Two forces explain everything — Z_eff across, shells down; every trend and every exception is one of these winning locally.
Exceptions are the exam — Be>B, N>O (half-filled/penetration), Cl>F (2p crowding), positive EA for N and noble gases; each has a two-line reason to reproduce.
IE tables are fingerprints — the position of the big jump counts the valence electrons and names the group, no memorisation needed.
EN and EA are different instruments — bonded-atom pull (relative, unitless, F tops) vs free-atom energy (kJ/mol, Cl tops); the swap is the chapter’s most-asked trap.

06

FAQs

Why does atomic radius decrease across a period?

Across a period, electrons enter the same shell while protons are added one by one, so the effective nuclear charge felt by the outermost electrons keeps rising. The same shell is pulled in more tightly, and the radius shrinks — from 186 pm in sodium to 99 pm in chlorine. Down a group the opposite happens: a new shell is added at each step, so radius increases despite rising nuclear charge.

Why is the electron gain enthalpy of chlorine more negative than that of fluorine?

Fluorine's incoming electron enters the tiny, crowded 2p subshell, where existing electrons repel it strongly — some of the released energy is spent against this repulsion. Chlorine's 3p subshell is larger and more spacious, so the added electron is repelled less. Hence chlorine releases more energy: −349 kJ/mol against fluorine's −328 kJ/mol, making chlorine the record holder.

Why is the first ionization enthalpy of nitrogen higher than that of oxygen?

Nitrogen has a stable half-filled 2p³ configuration, and removing one electron destroys this symmetry. Oxygen's 2p⁴ holds one paired electron, and the interelectronic repulsion in that pair makes it easier to remove — so oxygen's first ionization enthalpy (1314 kJ/mol) is lower than nitrogen's (1402 kJ/mol). The same logic gives Be (2s²) a higher ionization enthalpy than B (2s²2p¹).

What is the difference between electronegativity and electron gain enthalpy?

Electron gain enthalpy measures the energy change when an isolated gaseous atom accepts an electron — it is an absolute energy term in kJ/mol. Electronegativity measures the tendency of a bonded atom to attract the shared electron pair towards itself — it is a relative, unitless number on a scale such as Pauling's (fluorine = 4.0, the highest). One concerns a free atom gaining an electron; the other concerns an atom already in a bond.