Electron gain enthalpy (ΔegH) is the enthalpy change when an isolated gaseous atom gains one electron to form a monovalent anion:
X(g) + e⁻ → X⁻(g) ; ΔegH
The sign convention matters: a negative ΔegH means energy is released (exothermic, favourable); a positive value means energy must be supplied (endothermic, unfavourable).
General periodic trends:
Key anomalies you must know:
Fluorine vs Chlorine: F has a less negative ΔegH than Cl. Fluorine's tiny 2p orbital is already crowded — severe electron–electron repulsion destabilises the incoming electron. Chlorine's larger 3p orbital accommodates the extra electron more comfortably. So the order is Cl > F (more negative for Cl).
Noble gases and alkaline earth metals: Atoms with completely filled subshells (ns², ns²np⁶) have positive or near-zero ΔegH — they resist gaining an electron because the added electron must enter a higher-energy shell.
Nitrogen (half-filled 2p³): Has near-zero or slightly positive ΔegH because the extra stability of a half-filled p subshell makes electron addition unfavourable.
Oxygen vs Sulphur: O has a less negative ΔegH than S, paralleling the F vs Cl anomaly — small compact 2p orbitals create electron–electron repulsion.
NEET watch-out: Questions frequently present four elements and ask which has the most negative ΔegH. The distractor is F when the answer is Cl, exploiting the assumption that "smallest halogen = highest affinity."
(Reference: NCERT Class 11 Chemistry Chapter 3, page 91)
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Electron gain enthalpy is defined as the enthalpy change when:
Answer: D. By definition (NCERT Class 11 Chemistry Chapter 3, page 91), electron gain enthalpy is the enthalpy change when an isolated gaseous atom accepts one electron to form a monovalent gaseous anion.
Why A is wrong: A describes ionization enthalpy (removal of electron), not electron gain enthalpy.
Why B is wrong: B is wrong because the definition requires the atom to be in the gaseous state (isolated), not aqueous solution. Hydration effects would alter the energy.
Why C is wrong: C describes an ion gaining an electron — electron gain enthalpy is defined for a neutral gaseous atom, not a cation.
Which of the following has the most negative electron gain enthalpy?
Answer: A. Chlorine has the most negative ΔegH among halogens. Despite fluorine being smaller with higher Z_eff, its compact 2p orbital causes severe inter-electron repulsion, making Cl's ΔegH more negative than F's. Down the group from Cl, Br and I have decreasing |ΔegH| due to increasing size and shielding.
Why B is wrong: B exploits the common assumption that smallest atom = most negative ΔegH. Fluorine's tiny 2p orbital has extreme e⁻–e⁻ repulsion, making its ΔegH less negative than Cl's.
Why C is wrong: C is wrong — Br is larger than Cl with more shielding, so the added electron is less tightly held. ΔegH of Br is less negative than Cl.
Why D is wrong: D is wrong — I is the largest halogen; its ΔegH is the least negative among halogens due to maximum atomic size and electron shielding.
The electron gain enthalpy of nitrogen is approximately zero (or slightly positive). This is because:
Answer: B. Nitrogen's 2p³ (one electron in each p orbital) configuration has exchange energy stabilisation. Adding an electron would disrupt this half-filled stability, making the process energetically unfavourable — hence near-zero or slightly positive ΔegH.
Why A is wrong: A — small radius would actually increase nuclear attraction for the incoming electron, which would make ΔegH more negative, not less. The dominant factor here is half-filled stability, not size.
Why C is wrong: C — high electronegativity generally correlates with more negative ΔegH, so this would predict a favourable electron gain, contradicting the observation. The half-filled p³ stability overrides this.
Why D is wrong: D — this is factually wrong. Phosphorus (also group 15) has a clearly negative ΔegH. The near-zero value is specific to nitrogen's compact 2p subshell, not universal to group 15.
Among O, S, Se, Te — the element with the most negative electron gain enthalpy is:
Answer: D. Sulphur has the most negative ΔegH in Group 16, paralleling the F vs Cl anomaly. Oxygen's compact 2p orbital creates significant electron–electron repulsion despite its higher Z_eff. Below S, increasing size and shielding make Se and Te less negative.
Why A is wrong: A exploits the misconception that the smallest atom at the top of the group always has the most negative ΔegH. Oxygen's 2p orbital is too compact — inter-electron repulsion dominates, making its ΔegH less negative than S.
Why B is wrong: B is wrong — Te is even larger than Se, with more shielding. It has the least negative ΔegH in this set.
Why C is wrong: C is wrong — Se is below S with greater atomic size and shielding, resulting in weaker attraction for the incoming electron. Its ΔegH is less negative than S.
The electron gain enthalpy of Be, Mg, and Ne is positive. The common reason is:
Answer: C. Be (2s²), Mg (3s²), and Ne (2s²2p⁶) all have completely filled subshells. An added electron must enter a higher-energy orbital with poor nuclear shielding, requiring energy input — hence positive ΔegH.
Why A is wrong: A — high IE is a separate property (energy to remove an electron). While these atoms do have relatively high IE, that doesn't directly explain why gaining an electron is endothermic. The filled-subshell stability is the correct reason.
Why B is wrong: B — Be and Mg have small atomic radii, not large ones. Ne's van der Waals radius is irrelevant to ΔegH. The positive value is due to electronic configuration stability, not size.
Why D is wrong: D — Ne is a noble gas (not a metal), and Be/Mg are reactive metals (they readily form Be²⁺, Mg²⁺). This statement is factually incorrect.
Arrange the following in order of increasingly negative electron gain enthalpy: Na, Cl, Ar, S.
Answer: A. Ar (noble gas, filled 3s²3p⁶) has positive ΔegH → least negative. Na (3s¹, metallic) has a slightly negative value. S has a significantly negative value (Group 16 needs 2 electrons for octet). Cl has the most negative value (one electron short of noble gas configuration, 3p⁵). Order of increasingly negative: Ar < Na < S < Cl.
Why B is wrong: B is wrong — Ar has positive ΔegH (filled shell), so it should be less negative than Na. Ar must come before Na in the increasing order. This option places Na before Ar, implying Na is more positive than Ar, which is incorrect.
Why C is wrong: C is wrong — this reverses the trend completely. Cl has the most negative ΔegH (halogen), not the least. Ar has positive ΔegH, so it should be at the 'least negative' end.
Why D is wrong: D is wrong — it places Cl before S, suggesting S is more negative than Cl. In period 3, Cl (3p⁵) has a more negative ΔegH than S (3p⁴) because Cl is one electron away from a stable octet.
Which statement about electron gain enthalpy is INCORRECT?
Answer: C. Fluorine does NOT have the most negative ΔegH — chlorine does. F's compact 2p orbital creates severe e⁻–e⁻ repulsion upon electron addition, making Cl's value (−349 kJ/mol) more negative than F's (−328 kJ/mol). This is a high-frequency NEET distractor.
Why A is wrong: A is a correct statement — halogens (Group 17) need one electron for a stable octet, so they have large negative ΔegH values. This cannot be the incorrect statement.
Why B is wrong: B is a correct statement — increasing Z_eff across a period generally increases the attraction for an incoming electron (with known anomalies at filled/half-filled subshells). This cannot be the incorrect statement.
Why D is wrong: D is a correct statement — noble gases have completely filled s²p⁶ configurations; the incoming electron must enter a higher shell, requiring energy input. This cannot be the incorrect statement.
The second electron gain enthalpy of oxygen (O⁻ → O²⁻) is positive because:
Answer: B. O⁻ is already negatively charged. Adding a second electron requires overcoming the electrostatic repulsion between the incoming electron and the net negative charge on O⁻. This makes the second electron gain enthalpy endothermic (positive) — energy must be supplied.
Why A is wrong: A — small atomic radius alone doesn't explain why the second ΔegH is positive. The decisive factor is the coulombic repulsion from the existing negative charge on O⁻, not the size of the neutral atom.
Why C is wrong: C — ionization enthalpy describes electron removal, not addition. High IE of neutral oxygen is irrelevant to why the O⁻ → O²⁻ process is endothermic.
Why D is wrong: D — this is factually wrong. O²⁻ has configuration 2s²2p⁶ (fully filled), not half-filled. The 2p subshell is complete in the oxide ion.
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Pattern: Periodic trends comparison — electron gain enthalpy ordering (from NEET pattern: periodic trend bundle, which includes CHE.U09.ELECTRON_GAIN_ENTHALPY in its topic_codes).
Given
Four elements: F (2p⁵), Cl (3p⁵), O (2p⁴), S (3p⁴).
Required
Order from least negative to most negative ΔegH.
Concept
• Halogens (Group 17) have more negative ΔegH than chalcogens (Group 16) in the same period — one electron short of octet vs two.• Within a group, the second-period element (F, O) has less negative ΔegH than the third-period element (Cl, S) due to compact 2p orbital electron–electron repulsion.
Framework (no formula needed — trend-based reasoning)
• Compare within groups first: Cl > F (more negative) and S > O (more negative).• Compare across period: Cl > S (halogen vs chalcogen in period 3); F > O in period 2, but F may be close to S.• Known values: Cl (−349) > F (−328) > S (−200) > O (−141) kJ/mol.
Substitution
Applying the two principles:• O is least negative (compact 2p⁴ + only Group 16).• S is next (larger 3p⁴, less repulsion than O, but still Group 16).• F is next (compact 2p⁵ — halogen but repulsion penalty).• Cl is most negative (3p⁵ — halogen with spacious orbital).
Ordering
Least negative → most negative: O < S < F < Cl
Final answer
O (−141 kJ/mol) < S (−200 kJ/mol) < F (−328 kJ/mol) < Cl (−349 kJ/mol)
Common trap
Placing F as most negative by assuming "top of group = highest affinity." The F vs Cl anomaly (inter-electron repulsion in the compact 2p shell) is a high-frequency NEET distractor. Similarly, placing O above S ignores the same 2p repulsion effect.
Similar NEET-style question
"Which of the following represents the correct order of electron gain enthalpy (most negative first)? (a) Cl > F > S > O (b) F > Cl > O > S (c) Cl > F > O > S (d) F > Cl > S > O" — Answer: (a).
Energy released when neutral atom gains electron. Across period: more negative (favourable). Down group: becomes less negative. Cl > F (Cl has larger size, less e⁻-e⁻ repulsion).
-- NCERT Class 11 Chemistry, Ch. 3, p. 90More in Classification of Elements and Periodicity in Properties: 4 exam traps and mistakes · 1 formula from its other lessons.
1 question from NEET 2026. Answers verified against NTA official keys.
Among the following options, the correct trend in the electron gain enthalpy is
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Tempts surface-level recall.
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