Electron gain enthalpy - electron affinity — trends, anomalies (e.g. Cl - F): Worked examples

This is the drill page for the parent topic. The parent built the idea; here we hit every kind of question an exam can throw at you — every sign, every stable-shell exception, the Cl–F anomaly, second-electron additions, and the sign-flip between $Δ_{e g} H$ and electron affinity.

Before line one, one reminder built from scratch so no symbol is unearned:

The scenario matrix

Every question in this topic is really one of these cells. We will hit each one.

Cell	What varies	Expected outcome	Example
A. Normal halogen	one el. from full shell	strongly negative $Δ_{e g} H$	Ex 1 (Cl)
B. Cl > F anomaly	tiny 2p vs roomy 3p	Cl more negative than F	Ex 2
C. Noble gas (degenerate)	complete shell, no room	positive $Δ_{e g} H$	Ex 3 (Ne)
D. Half/full-filled stability	$n s^{2}$ , $n p^{3}$ resist	near-zero or positive	Ex 4 (N vs C, O)
E. Second electron (like-charge)	electron onto an anion	always positive	Ex 5 (O⁻→O²⁻)
F. Sign-flip trap	convert $Δ_{e g} H \leftrightarrow A_{e}$	flip the sign, keep magnitude	Ex 6
G. Word problem (limiting)	thermochem cycle	net sign from summing steps	Ex 7
H. Exam twist (rank a mixed set)	period + group + anomaly together	full ordering	Ex 8

Prerequisites you can lean on: Effective Nuclear Charge (Z_eff), Atomic Radius, Half-filled and Fully-filled Stability, and the big-picture map in Periodic Trends Overview.

Cell A — the normal halogen

Worked example Example 1 — Sign and rough size for chlorine

Q: Predict the sign of $Δ_{e g} H (Cl)$ and say why it is large.

Forecast: Guess before reading — will Cl release or absorb energy, and is it a big number or a small one?

Step 1 — write the configuration. Cl is $[Ne] 3 s^{2} 3 p^{5}$ : one electron short of the full $3 p^{6}$ . Why this step? The "how full is the outer shell" fact decides whether there is a comfortable vacancy for the new electron.

Step 2 — where does the electron land? Into the empty slot of the $3 p$ subshell, completing $3 p^{6}$ (an argon-like closed shell). Why this step? Completing a shell is a big energy payoff — the atom reaches a very stable state.

Step 3 — sign. Reaching a stable closed shell → energy is released → $Δ_{e g} H$ is negative, and large. Book value $\approx - 349 kJ/mol$ . Why this step? Released energy = exothermic = negative sign by our bank-statement rule.

Verify: Cl is a halogen, the class with the most negative $Δ_{e g} H$ in each period — consistent. Units kJ/mol ✔. Sign matches "atom hugs the electron."

Figure — Electron gain enthalpy - electron affinity — trends, anomalies (e.g. Cl - F)

Cell B — the star anomaly Cl > F

Worked example Example 2 — Why does Cl beat F?

Q: Given $Δ_{e g} H (F) \approx - 328 kJ/mol$ and $Δ_{e g} H (Cl) \approx - 349 kJ/mol$ , which atom has the larger electron affinity, and explain the anomaly.

Forecast: F is smaller with a bigger Z_eff — surely it wins? Guess, then read on.

Step 1 — compare the numbers. $- 349 < - 328$ , so Cl's value is more negative. Why this step? "More negative $Δ_{e g} H$ " and "larger electron affinity" mean the same physical thing — more energy released.

Step 2 — convert to affinity to double-check. $A_{e} = - Δ_{e g} H$ , so $A_{e} (F) = + 328$ , $A_{e} (Cl) = + 349$ . Cl's affinity is bigger. Why this step? Guards against the classic sign confusion (Cell F).

Step 3 — the physical cause. F's incoming electron enters the compact 2p orbital, already packed → strong electron–electron repulsion cancels part of the nuclear attraction. Cl's larger 3p (bigger radius) has room, so less repulsion, more net energy released. Why this step? The anomaly is not about $Z_{e f f}$ — it is about crowding. This same crowding gives O < S and N < P.

Verify: Magnitude difference $349 - 328 = 21 kJ/mol$ — small but real, exactly the size you'd expect from a modest repulsion correction, not a total reversal of the period trend. ✔

Cell C — the degenerate case: no room at all

Cell D — half-filled and full-filled stability

Worked example Example 4 — Rank C, N, O, F (most negative first)

Q: Order $Δ_{e g} H$ for carbon, nitrogen, oxygen, fluorine.

Forecast: Naively "across a period = steadily more negative" gives C > N > O > F. One of these atoms breaks the march — which?

Step 1 — configurations. C $2 p^{2}$ , N $2 p^{3}$ (half-filled), O $2 p^{4}$ , F $2 p^{5}$ . Why this step? Half-filled $2 p^{3}$ is an extra-stable arrangement (Half-filled and Fully-filled Stability).

Step 2 — apply the period trend, then patch N. Rising $Z_{e f f}$ pushes F most negative, then O, then C. But N resists the extra electron because it must pair up inside the stable half-filled set → pairing repulsion. Why this step? The exception overrides the smooth trend for exactly one atom.

Step 3 — final order (most negative → least). $F < O < C < N$ . Book magnitudes: F $\approx - 328$ , O $\approx - 141$ , C $\approx - 122$ , N $\approx + 7$ (essentially zero, even slightly positive). Why this step? This is the sortable answer an exam wants.

Verify: F is the most negative (halogen ✔). N is the odd one out and lands positive/near-zero ✔. The other three descend monotonically ( $- 328 < - 141 < - 122 < + 7$ ) ✔.

Cell E — the second electron, a like-charge fight

Worked example Example 5 —

$O^{-} (g) + e^{-} \to O^{2 -} (g)$ Q: Sign of the second electron gain enthalpy $Δ_{e g} H_{2}$ of oxygen?

Forecast: The first electron on O released energy. Does the second do the same?

Step 1 — identify the charge on the target. We now add an electron to $O^{-}$ , which is already negative. Why this step? Like charges repel — the incoming (negative) electron meets a (negative) ion.

Step 2 — the energy consequence. To push a negative electron onto a negative ion, we must fight Coulomb repulsion → energy must be poured in. Why this step? Work done against repulsion always costs energy.

Step 3 — sign. $Δ_{e g} H_{2}$ is always positive. For oxygen, $Δ_{e g} H_{2} \approx + 744 kJ/mol$ . Why this step? Positive = endothermic, as required.

Verify: Net for $O \to O^{2 -}$ = $Δ_{e g} H_{1} + Δ_{e g} H_{2} = (- 141) + (+ 744) = + 603 kJ/mol$ — endothermic overall. That is exactly why forming oxides needs lattice energy to be favourable. ✔

Cell F — the sign-flip trap

Cell G — word problem with a limiting thermochemical sum

Worked example Example 7 — Is the gas-phase step

$N a^{+} + C l \to N a^{+} + C l^{-}$ energetically downhill? Q: For the electron transfer $C l (g) + e^{-} \to C l^{-} (g)$ occurring near a spectator $N a^{+}$ , use $Δ_{e g} H (Cl) = - 349 kJ/mol$ . Is the electron capture step exothermic, and by how much per mole?

Forecast: Guess the sign of the energy change for the chlorine half.

Step 1 — isolate the relevant step. Only the electron gain on Cl matters here: $C l + e^{-} \to C l^{-}$ , $Δ_{e g} H = - 349 kJ/mol$ . Why this step? The spectator $N a^{+}$ does not change in this half-reaction, so it contributes nothing to this step's enthalpy.

Step 2 — read the sign. $- 349 < 0$ → exothermic: $349 kJ$ released per mole of Cl. Why this step? Negative enthalpy = energy leaves the system.

Step 3 — limiting sanity. If Cl were replaced by a noble-gas-like species with $Δ_{e g} H > 0$ , this step would instead cost energy — showing the sign genuinely depends on the atom. Why this step? Testing the degenerate limit confirms our reasoning is not accidental.

Verify: $349 kJ/mol$ released, units correct, sign negative for a halogen — all consistent with Cell A. ✔

Cell H — the mixed-set exam twist

Worked example Example 8 — Rank F, Cl, S, Ne (most negative

$Δ_{e g} H$ first) Q: Order these four using period trends, the Cl–F anomaly, and stable-shell logic together.

Forecast: Four different cells in one question — write your guess before reading.

Step 1 — classify each. F = halogen but tiny (anomaly), Cl = roomy halogen (Cell B winner among halogens), S = group-16 (one short of Cl's row, less negative than Cl), Ne = noble gas (Cell C, positive). Why this step? Sorting by type first prevents blind application of one trend.

Step 2 — order the negatives. Among the electron-lovers: Cl $(- 349)$ is most negative (roomy halogen), then F $(- 328)$ (anomaly keeps it just behind Cl), then S $(- 200)$ (group 16, less than the halogen). Why this step? Anomaly first, then column position.

Step 3 — place Ne last. Ne is $+ 116$ (positive), so it sits at the least negative end. Why this step? Positive always ranks below every negative value.

Final order (most negative → least): $Cl < F < S < Ne$ , i.e. $- 349, - 328, - 200, + 116 kJ/mol$ .

Verify: Strictly increasing sequence $- 349 < - 328 < - 200 < + 116$ ✔. Cl beats F (anomaly) ✔. Ne positive and last ✔.

Recall One-line summary of the matrix

Halogen → big negative; Cl > F because F's 2p is crowded; noble gas / half-full → positive; second electron → always positive; and $A_{e} = - Δ_{e g} H$ so never trust the word "positive" without checking which quantity you hold.