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

(a) Cl: a halogen, one electron short of the Ar shell → it eagerly grabs an electron → energy released → ${\Delta }_{eg}H$ is negative ($\approx -349\text{kJ/mol}$). (b) Ne: its shell is complete ($1s^{2}2s^{2}2p^6$). The new electron must start a brand-new shell ($3s$) far out and shielded → the atom fights it → energy must be supplied → ${\Delta }_{eg}H$ is positive. (c) $O^{-}+e^{-}\to O^{2-}$: the target is already negative. Like charges repel, so we must push against Coulomb repulsion → energy absorbed → ${\Delta }_{eg}{H}_2$ is positive (always).

Use $A_e=-{\Delta }_{eg}H$. $A_e=-(-349)=+349\text{kJ/mol}$ The affinity is positive and large — consistent with "this atom loves electrons." Same physics, flipped sign.

(a) Na vs. Cl — same period (period 3). Moving right, $Z_{eff}$ rises and the atom nears a full shell → Cl is far more negative. (Na barely wants an electron; Cl is desperate for one.) (b) F vs. I — same group (17). Down a group, the atom grows, the incoming electron sits farther from the nucleus, attraction weakens → the heavier one is less negative → F is more negative than I by the plain trend. (This is the plain-trend answer; the F-vs-Cl subtlety appears in Level 3.) (c) O vs. Ne — same period. O ($2p^4$) still has room and high $Z_{eff}$ → negative. Ne has a complete shell → positive. So O is more negative.

${\Delta }_{eg}H(\text{S})-{\Delta }_{eg}H(\text{O})=(-200)-(-141)=-59\text{kJ/mol}$ S is 59 kJ/mol more negative, so sulphur releases more energy. This is the same "second-row atom is too compact" anomaly as Cl > F: O's tiny $2p$ orbital is crowded, S's roomier $3p$ accepts the electron more comfortably.

Configurations of the added-to $2p$ subshell: C $2p^2$, N $2p^3$ (half-filled), O $2p^4$, F $2p^5$. > Plain across-period trend would give F < O < N < C (F most negative). But N's half-filled $2p^3$ is extra stable, so it resists the extra electron (which must pair up) → N is pushed way up (near zero or positive). Final order, most negative → least: $\text{F}<\text{O}<\text{C}<\text{N}$ Approximate values: F $\approx -328$, O $\approx -141$, C $\approx -122$, N $\approx +7\text{kJ/mol}$. N is the odd one out because pairing into a stable half-filled shell costs energy.

Gap: ${\Delta }_{eg}H(\text{Cl})-{\Delta }_{eg}H(\text{F})=(-349)-(-328)=-21\text{kJ/mol}$. Cl is 21 kJ/mol more negative. The tug-of-war (see figure): F has higher $Z_{eff}$ and smaller size, so its attraction pull is genuinely stronger. But its $2p$ orbital is tiny and already holds 5 electrons; forcing a 6th in raises the repulsion term steeply. In Cl the $3p$ orbital is larger — attraction is a bit weaker, but repulsion is much weaker. Net: $\underset{\text{F wins}}{\underbrace{(\text{attraction})}}-\underset{\text{F loses big}}{\underbrace{(\text{repulsion})}}\Rightarrow \text{Cl comes out ahead.}$ The repulsion penalty on the compact F orbital more than erases F's attraction advantage.

(a) Enthalpies of sequential steps add (Hess's law): ${\Delta }_{eg}{H}_{\text{total}}={\Delta }_{eg}{H}_1+{\Delta }_{eg}{H}_2=(-141)+(+744)=+603\text{kJ/mol}$ (b) Positive ⇒ endothermic. Making a bare $O^{2-}$ ion from an O atom costs energy overall, because the second electron has to be forced onto an already-negative ion (Coulomb repulsion dominates). (c) The huge lattice energy released when $M{g}^{2+}$ and $O^{2-}$ pack into the MgO crystal more than pays back this +603 kJ/mol, so the overall solid-formation is still favourable.

To ionize Na we pay its ionization enthalpy; when Cl gains the electron we get back its electron-gain energy: $\Delta H=\underset{\text{Na}\to {\text{Na}}^{+}}{\underbrace{(+496)}}+\underset{\text{Cl}+e^{-}\to {\text{Cl}}^{-}}{\underbrace{(-349)}}=+147\text{kJ/mol}$ Positive ⇒ the isolated gas-phase transfer is endothermic. Cl's electron affinity does not fully repay Na's ionization cost. (As in Exercise 4.1, the ionic lattice energy of solid NaCl is what makes the real reaction exothermic — electron gain enthalpy alone never tells the whole story.)

Reason from most-wanted to least-wanted electron:

• Cl ($3p^5$, one short of full shell, highest $Z_{eff}$) → wants the electron most → the most negative: $-349$.
• S ($3p^4$, still roomy, high $Z_{eff}$ but one electron farther from completion than Cl) → strongly negative but less than Cl: $-200$.
• Na ($3s^1$; adding an electron gives the modestly stable $3s^2$) → mildly negative: $-53$.
• Mg ($3s^2$, a fully-filled $n{s}^2$ that resists a new electron entering $3p$) → positive: $+67$.

Assignment: Cl $=-349$, S $=-200$, Na $=-53$, Mg $=+67$. The one positive value must go to the filled-subshell element (Mg), and the two strongly negative ones to the near-complete-shell nonmetals (Cl then S).

Successive changes going down the group: $\text{F}\to \text{Cl}:(-349)-(-328)=-21(\text{more negative — anomaly})$ $\text{Cl}\to \text{Br}:(-325)-(-349)=+24(\text{less negative — normal})$ $\text{Br}\to \text{I}:(-295)-(-325)=+30(\text{less negative — normal})$ Only the first step (F→Cl) goes the "wrong" way; from Cl downward the trend is perfectly normal (values become less negative). So the student is half right: the rule fails only at the top, because F's compact $2p$ orbital carries an unusually large electron–electron repulsion penalty. From Cl onward, orbitals are roomy and ordinary size-based weakening dominates. Verdict: the group rule is not "simply wrong" — it holds for Cl→Br→I and breaks only at the F→Cl boundary.