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

We are chasing one central result: $Δ_{e g} H (F) \approx - 328 kJ/mol, Δ_{e g} H (Cl) \approx - 349 kJ/mol$ Cl is more negative → Cl releases more energy. Let's earn that, symbol by symbol.

Prerequisites we lean on: Effective Nuclear Charge (Z_eff), Atomic Radius, and the parent topic note.

Step 1 — What "grabbing an electron" even looks like

WHAT. Start with a neutral gaseous atom and one lonely electron sitting infinitely far away, doing nothing. We slide that electron in until it settles into the atom.

WHY. Before we can ask "how much energy comes out?", we must fix the picture of the event. Every symbol later ( $Δ_{e g} H$ , $Z_{e f f}$ , repulsion) refers to something happening in this picture.

PICTURE. Look at the figure. On the left, the free electron (yellow dot) has energy we define as zero — it feels no pull. On the right it has fallen into the atom and the whole system has dropped to a lower energy. That drop is the energy released.

The number line at the bottom is our scoreboard: the deeper the final level sits below zero, the more negative $Δ_{e g} H$ , the happier the atom.

Step 2 — The pull: why energy comes out at all

WHAT. Zoom into why the electron falls downhill. The nucleus has positive charge; the incoming electron is negative; opposites attract, so the electron is pulled inward, and pulling something into an attractive well releases energy.

WHY. We need one clean force to call "the good guy" before we introduce the "bad guy" (repulsion) in Step 3. This attractive pull is the entire reason $Δ_{e g} H$ is usually negative.

PICTURE. The green arrow is the nuclear pull on the new electron. But the nucleus is not naked — the atom's own inner electrons (blue) sit between the nucleus and the newcomer and block part of the pull. What's left over is called the effective nuclear charge.

So far the story is simple: big pull = big energy out. If that were the whole story, tiny high- $Z_{e f f}$ fluorine would crush chlorine. It doesn't. Enter the bad guy.

Step 3 — The shove: electron–electron repulsion

WHAT. The orbital we're dropping the electron into is not empty — it already holds other electrons. All electrons are negative, so they repel the newcomer. This repulsion pushes the energy back up, working against the release.

WHY. This is the missing force that breaks the naive "smallest wins" rule. Without it you can never explain Cl > F. It is the second of exactly two competing effects.

PICTURE. Red arrows = the existing electrons shoving the new one away. In a small, packed orbital the electrons sit close together, so these red arrows are large. In a roomy orbital they're spread out, so the red arrows are weak.

Step 4 — The two contenders side by side: F vs Cl

WHAT. Line up fluorine and chlorine. F is smaller with a higher $Z_{e f f}$ → bigger green (stronger pull). But F's $2 p$ orbital is tiny and crammed → bigger red (stronger repulsion) too. Cl's $3 p$ orbital is larger → smaller green and smaller red.

WHY. The anomaly is a comparison, so we must draw both atoms at once and watch which arrow shrinks faster as we go from F to Cl.

PICTURE. F on the left: small tight ball, huge green pull and huge red shove. Cl on the right: bigger ball, gentler green, much gentler red. The key insight the picture shows: going F → Cl, the red arrow shrinks more dramatically than the green does, because spreading out relieves cramming faster than it relieves attraction.

Step 5 — Adding up the arrows: the net winner

WHAT. Take the green (negative) and red (positive) contributions for each atom and add them on the energy scoreboard from Step 1.

WHY. Individual arrows don't settle the question — only the net (green minus red) does. This is the arithmetic that produces the actual numbers.

PICTURE. Two vertical bars. For F: a long green bar down, then a fat red bar back up → the net lands at $- 328$ . For Cl: a shorter green bar down, but a much shorter red bar back up → the net lands lower, at $- 349$ . Cl wins because it kept more of its release.

Step 6 — Edge case: the second electron (always uphill)

WHAT. Now push a second electron onto an already-negative ion, e.g. $O^{-} (g) + e^{-} \to O^{2 -} (g)$ .

WHY. We must cover the degenerate case where our "attraction" hero flips sign. The parent note claims this is always positive — let's see it.

PICTURE. The target is now a red ball (net negative charge). There is no green pull toward a negative object at all — instead a red repulsive arrow greets the incoming electron from far away. Energy must be supplied; the level goes up.

Step 7 — Edge case: full and half-full shells slam the door

WHAT. Try noble gases (Ne: full $2 p^{6}$ ) and half-filled shells (N: $2 p^{3}$ ). Here the orbital has no comfortable room — either it's complete (Ne) or it's in a specially stable half-filled arrangement (N) that resists disruption.

WHY. These are the exceptions to "across a period gets more negative." A complete picture must show when the trend breaks.

PICTURE. Ne: the $n = 2$ shell is a closed ring; the newcomer is forced out to a distant, shielded $n = 3$ orbital → weak pull, energy must be supplied → positive. N: its three $2 p$ electrons sit one-per-box (half-filled, extra stable); adding one forces pairing in a box → repulsion → near-zero or positive.

The one-picture summary

Everything above compressed into a single scoreboard: for each atom, the green pull drags the energy down, the red shove pushes it back up, and where the dust settles is $Δ_{e g} H$ . Watch F lose to Cl because its red bar is fat; watch Ne, N sit above zero because they have no room.

Recall Feynman retelling of the whole walkthrough

Picture an atom as a house and the extra electron as a guest arriving from far away with empty hands (zero energy). The doorway — the nucleus — pulls the guest in, and the house rewards you with energy for filling it: that's the green pull, and that's why grabbing an electron usually releases heat. But the rooms already have people in them, and they shove the newcomer — that's the red repulsion, and it costs energy. The real "welcome energy" is the pull minus the shove. Fluorine's tug is fierce, but its house is so cramped the shove is even fiercer, so it only throws off $- 328$ . Chlorine tugs a little softer but has roomy halls, so its shove is tiny — net it releases more, $- 349$ . That's the whole anomaly. Two edge cases finish the story: shove a second electron onto an already-negative ion and there's no pull at all, only shove, so it always costs energy ( $+ 744$ for oxygen); and a house that's already full (neon) or perfectly half-arranged (nitrogen) simply won't take a guest without a fight, so their numbers sit above zero.