Intuition The one-line pictureAn atom is a positive nucleus surrounded by electrons. When we push one extra electron onto a neutral gaseous atom, the atom either hugs it (energy released) or rejects it (energy needed). Electron gain enthalpy measures how much energy is released (usually) when this happens.
Definition Electron gain enthalpy (Δ e g H \Delta_{eg}H Δ e g H enthalpy change when one mole of electrons is added to one mole of gaseous atoms (or ions) to form gaseous negative ions:
X ( g ) + e − ⟶ X − ( g ) Δ e g H X(g) + e^- \longrightarrow X^-(g) \qquad \Delta_{eg}H X ( g ) + e − ⟶ X − ( g ) Δ e g H
If energy is released → Δ e g H \Delta_{eg}H Δ e g H negative (exothermic, atom likes the electron).
If energy must be supplied → Δ e g H \Delta_{eg}H Δ e g H positive (endothermic, atom resists).
Definition Electron affinity (A e A_e A e energy released for the same process, defined with the opposite sign convention :
A e = − Δ e g H ( at 0 K , ignoring the small 5 2 R T term ) A_e = -\Delta_{eg}H \quad(\text{at } 0\,K,\text{ ignoring the small } \tfrac{5}{2}RT \text{ term}) A e = − Δ e g H ( at 0 K , ignoring the small 2 5 R T term ) large positive electron affinity = a large negative electron gain enthalpy = atom strongly wants the electron. They are the same idea, opposite sign .
WHY two quantities? Chemists historically spoke of "affinity" (how much an atom loves electrons → positive = good). Thermodynamics prefers enthalpy where exothermic = negative. Don't mix the signs.
Intuition Feel the physicsThe incoming electron sits far away with essentially zero energy. As it falls into a partly-empty orbital, the nucleus' positive charge attracts it → the system drops to lower energy → that energy leaves as heat. The stronger the effective nuclear pull and the more room in the orbital, the more energy released.
Two competing effects decide the magnitude:
Effect
Pushes Δ e g H \Delta_{eg}H Δ e g H
Reason
Strong effective nuclear charge Z e f f Z_{eff} Z e f f
more negative
tighter attraction for the new electron
Small atom / compact orbital
more negative ...
closer to nucleus
...but too small → electron–electron repulsion
more positive
the new electron is crammed into a crowded, tiny orbital
That tension (attraction vs. crowding) is the whole story behind the anomalies.
Intuition Full and half-full shells push electronsaway
Noble gases (Ne, Ar): shell complete → new electron must enter the next shell → strongly positive Δ e g H \Delta_{eg}H Δ e g H Group 2 (Be, Mg) and Group 15 (N, P) : filled n s 2 ns^2 n s 2 n p 3 np^3 n p 3 resist an extra electron → Δ e g H \Delta_{eg}H Δ e g H
Intuition Why the tiny F atom loses to ClYou'd expect F (smaller, higher Z e f f Z_{eff} Z e f f 2p orbital is tiny and already packed with electrons. Adding one more forces it into a cramped space → huge electron–electron repulsion partly cancels the nuclear attraction. Cl's 3p orbital is larger and roomier , so the new electron fits comfortably → net energy released is greater .
Δ e g H ( F ) ≈ − 328 kJ/mol , Δ e g H ( Cl ) ≈ − 349 kJ/mol \Delta_{eg}H(\text{F}) \approx -328\ \text{kJ/mol}, \qquad \Delta_{eg}H(\text{Cl}) \approx -349\ \text{kJ/mol} Δ e g H ( F ) ≈ − 328 kJ/mol , Δ e g H ( Cl ) ≈ − 349 kJ/mol
Cl is more negative → Cl has the higher electron affinity. Same reason: O < S and N < P for the second-row-being-smaller anomaly.
"Chlorine has the bigger house, so it welcomes the guest." Small F = crowded room, rejects part of the guest's stay. Compact 2p ⇒ repulsion ⇒ anomaly.
Worked example 1. Predict the sign for NeonQ: Is Δ e g H ( Ne ) \Delta_{eg}H(\text{Ne}) Δ e g H ( Ne ) Step 1 — config: Ne is 1 s 2 2 s 2 2 p 6 1s^2 2s^2 2p^6 1 s 2 2 s 2 2 p 6 complete shell. Why? Determines whether there's room.
Step 2 — where does the new electron go? Into the 3 s 3s 3 s Why? No space in n=2.
Step 3 — verdict: weak attraction + must occupy high-energy shell → energy must be supplied → Δ e g H \Delta_{eg}H Δ e g H positive (≈ +116 kJ/mol). ✅
Δ e g H \Delta_{eg}H Δ e g H Step 1 — configs: C 2 p 2 2p^2 2 p 2 2 p 3 2p^3 2 p 3 2 p 4 2p^4 2 p 4 2 p 5 2p^5 2 p 5 Step 2 — apply trend: across period → more negative, except N's half-filled stability. Why? Stable configs resist electrons.
Step 3 — result: F < O < C < N \text{F} < \text{O} < \text{C} < \text{N} F < O < C < N F most negative , N least (even positive-ish). Order of magnitude: F ≈ −328, O ≈ −141, C ≈ −122, N ≈ +0 to +7 . ✅
Why N breaks rank? half-filled 2 p 3 2p^3 2 p 3
Worked example 3. Explain whyΔ e g H 2 \Delta_{eg}H_2 Δ e g H 2 second electron) is always positive
Q: O − ( g ) + e − → O 2 − ( g ) O^-(g) + e^- \to O^{2-}(g) O − ( g ) + e − → O 2 − ( g ) Step 1: the species O − O^- O − already negative . Why matters? Like charges repel.
Step 2: forcing another electron onto a negative ion means fighting Coulomb repulsion → energy must be supplied.
Step 3: Δ e g H 2 \Delta_{eg}H_2 Δ e g H 2 always positive (endothermic), e.g. O: +744 kJ/mol. This is why oxide formation overall needs lattice energy to be favourable. ✅
Common mistake "Fluorine must have the highest electron affinity because it's smallest and most electronegative."Why it feels right: F is the most electronegative element and the smallest halogen; small size usually = strong attraction. The flaw: electronegativity (bonding tendency) ≠ electron gain enthalpy (isolated-atom energetics). F's 2p is so compact that electron–electron repulsion cancels part of the attraction. Fix: Cl > F. Remember size can be too small .
Δ e g H \Delta_{eg}H Δ e g H Why it feels right: "positive = big/good" in everyday language. The flaw: sign conventions are opposite. Positive Δ e g H \Delta_{eg}H Δ e g H Fix: A e = − Δ e g H A_e = -\Delta_{eg}H A e = − Δ e g H Δ e g H \Delta_{eg}H Δ e g H
Common mistake Reporting the second electron gain as exothermic.Why it feels right: first one released energy, so surely the next does too. The flaw: the ion is now negative , repelling the electron. Fix: Δ e g H 2 > 0 \Delta_{eg}H_2 > 0 Δ e g H 2 > 0
Recall Feynman: explain to a 12-year-oldImagine an atom is a house and electrons are guests. Most atoms happily let one more guest in and even give you a little "thank-you" energy. Chlorine is a big comfy house — plenty of room — so it welcomes the extra guest warmly and releases a lot of energy. Fluorine is a tiny cramped house: it wants the guest but its rooms are so packed that the guest bumps into everyone, so it can only give back a little energy — less than roomy chlorine! And a full house (neon) simply won't let anyone in without a fight.
Which sign of Δ e g H \Delta_{eg}H Δ e g H likes the electron? ::: Negative (exothermic).
Why is Cl's Δ e g H \Delta_{eg}H Δ e g H
Define electron gain enthalpy Δ e g H \Delta_{eg}H Δ e g H Enthalpy change when 1 mole of electrons is added to 1 mole of gaseous atoms:
X ( g ) + e − → X − ( g ) X(g)+e^-\to X^-(g) X ( g ) + e − → X − ( g ) .
Relation between electron affinity A e A_e A e Δ e g H \Delta_{eg}H Δ e g H A e = − Δ e g H A_e = -\Delta_{eg}H A e = − Δ e g H (same process, opposite sign).
What sign of Δ e g H \Delta_{eg}H Δ e g H Negative (exothermic).
Trend of Δ e g H \Delta_{eg}H Δ e g H Becomes more negative (rising
Z e f f Z_{eff} Z e f f , atom nearer full shell).
Trend of Δ e g H \Delta_{eg}H Δ e g H Generally becomes less negative (larger atom, weaker attraction).
Why is Cl's Δ e g H \Delta_{eg}H Δ e g H F's compact 2p orbital causes large e⁻–e⁻ repulsion, cancelling part of the nuclear attraction; Cl's larger 3p accommodates the electron better.
Sign of Δ e g H \Delta_{eg}H Δ e g H Positive (complete shell; electron must enter next shell).
Why do N and P have near-zero/positive Δ e g H \Delta_{eg}H Δ e g H Half-filled
n p 3 np^3 n p 3 is stable and resists an extra electron.
Sign of the second electron gain enthalpy Δ e g H 2 \Delta_{eg}H_2 Δ e g H 2 Always positive (adding e⁻ to an already negative ion → repulsion).
Which pairs show the "second-period-too-small" anomaly? F<Cl, O<S, N<P (period-3 more negative than period-2).
Approx Δ e g H \Delta_{eg}H Δ e g H F ≈ −328, Cl ≈ −349.
attraction, more negative
Electron gain enthalpy ΔegH
Energy released, exothermic
Energy needed, endothermic
Effective nuclear charge Zeff
Electron-electron repulsion
Intuition Hinglish mein samjhoDekho, electron gain enthalpy ka matlab simple hai: ek neutral gaseous atom ko ek extra electron do, to kitni energy release ya absorb hoti hai. Agar energy release hoti hai to Δ e g H \Delta_{eg}H Δ e g H negative hota hai (atom ko electron pasand hai), aur agar energy deni padti hai to positive (atom mana kar raha hai). Electron affinity bilkul yahi cheez hai bas sign ulta — A e = − Δ e g H A_e = -\Delta_{eg}H A e = − Δ e g H Δ e g H \Delta_{eg}H Δ e g H
Trend yaad rakho: period me left se right jao to Z e f f Z_{eff} Z e f f
Ab star anomaly: Cl > F . Logic ke hisaab se F (chhota, high Z e f f Z_{eff} Z e f f electron-electron repulsion attraction ko kaat deta hai. Cl ka 3p bada aur khula hai, electron aaram se fit ho jaata hai — isliye Cl zyada energy release karta hai. Yahi kahani O<S aur N<P me bhi hai.
Do exam-trap yaad rakho: (1) N aur noble gases ka Δ e g H \Delta_{eg}H Δ e g H Second electron gain hamesha positive hota hai, kyunki ion already negative hai aur electron ko repel karta hai. In do points pe questions pakke aate hain — 80/20 rule!
