2.2.5 · D5Periodic Trends
Question bank — Electron gain enthalpy - electron affinity — trends, anomalies (e.g. Cl - F)
Before you start, lock in the sign rule — nearly half the traps hang on it:
True or false — justify
Fluorine, being the most electronegative element, must have the most negative of all elements.
False — electronegativity is a bonding-in-a-molecule tendency, while is isolated-atom energetics; F's tiny 2p is so crowded that electron–electron repulsion makes Cl the more negative one.
A more negative always corresponds to a larger electron affinity .
True — since , a more negative enthalpy flips to a larger positive affinity; they are the same fact stated with opposite sign.
Adding a second electron to an atom releases even more energy than the first, since the atom already showed it "likes" electrons.
False — the second electron approaches a species that is already negatively charged, so Coulomb repulsion dominates and is always positive (endothermic).
Down any group, becomes uniformly less negative without exception.
False — the general trend is less negative (bigger atom, weaker pull), but the second-period elements (F, O, N) are anomalously less negative than the third-period members below them due to their compact orbitals.
Noble gases have because they are chemically inert.
False — inertness is a consequence, not the cause; their is strongly positive because the valence shell is full, so any extra electron must enter a higher, shielded shell that barely attracts it.
Nitrogen has a near-zero or slightly positive while carbon (to its left) is more negative.
True — this reverses the "across a period gets more negative" rule because N's half-filled is extra stable, so it resists pairing an extra electron into an already-occupied 2p orbital.
Since Cl releases more energy than F on gaining an electron, Cl is also more electronegative than F.
False — they measure different things; Cl beats F on (roomier 3p) but F is still the more electronegative element in bonds.
The magnitude of is decided by effective nuclear charge alone.
False — it is a tug-of-war between (attracts the new electron, more negative) and electron–electron repulsion in a small/crowded orbital (pushes it back, more positive); the anomalies come entirely from the second effect.
Spot the error
"Ne is , so the extra electron enters an empty 2p orbital, giving a small negative ."
The error is "empty 2p" — the 2p subshell is completely full (), so the electron must go into the next shell (3s), making positive, not negative.
"Be resists an electron because its 2p is half-filled."
The subshell reason is wrong: Be is , a fully filled subshell; that filled-s stability is why it resists, not a half-filled p (Be has no p electrons at all).
"F is smaller than Cl and has higher , therefore F must release more energy."
The reasoning ignores crowding: F's 2p orbital is so compact that the added electron suffers strong e⁻–e⁻ repulsion, cancelling part of the attraction — so Cl, with a roomier 3p, releases more.
"O has a positive because repels electrons."
Confuses first and second gain: the first gain is exothermic (negative); it is the second gain that is positive because of the already-negative ion.
"A positive means the atom has a large electron affinity."
Sign convention is inverted: positive = endothermic = the atom dislikes the electron = low affinity, since .
"Group 15 elements (N, P) resist electrons because their shells are completely full."
Not full — they are half-filled ; it is the special stability of a half-filled subshell (one electron in each p orbital) that makes them reluctant, not a complete shell.
Why questions
Why do halogens sit at the peak (most negative ) of their periods?
They are one electron short of a stable noble-gas configuration, so the incoming electron completes the shell and the atom strongly welcomes it, releasing a lot of energy.
Why does the "across a period → more negative" trend break at nitrogen and phosphorus?
Their half-filled configuration is unusually stable; adding an electron forces pairing in an already-occupied p orbital, and the extra repulsion cancels the expected nuclear-attraction gain.
Why is and (the third-period member more negative) despite the general down-a-group trend?
The second-period atoms are so small that electron–electron repulsion in their compact orbitals is severe; the slightly larger third-period orbitals ease that crowding, so they release more energy — the same reason Cl beats F.
Why must chemists distinguish "electron affinity" from "electron gain enthalpy" carefully?
They describe the identical process but with opposite sign conventions (), so mixing them flips whether "positive" means the atom loves or hates the electron.
Why does forming , which is common in oxides, require energy overall for the gas-phase ion?
Because is strongly positive (Coulomb repulsion against the already-negative ); the oxide is only stable in a solid because lattice energy more than repays that cost.
Why does a "too small" atom sometimes have a less negative than a larger one?
A very small orbital means the added electron is crammed close to the existing electrons, so repulsion rises faster than the nuclear attraction, reducing the net energy released.
Edge cases
What is the sign of for adding an electron to a cation like ?
Negative and large — pulling an electron onto a positive ion is strongly favourable (opposite of the second-electron case), essentially the reverse of ionization.
What happens to if the atom already has a completely filled valence shell (e.g. any noble gas)?
It is positive: the electron cannot fit in the full shell and must enter the next principal shell, which is far out and well-shielded, so energy must be supplied.
For (second electron gain), is there any atom for which it is negative?
No — every second gain adds an electron to an already-negative ion, so Coulomb repulsion makes positive for all elements.
Is defined for the gaseous state, and does it matter?
Yes, it is defined for gaseous atoms/ions; it matters because in solids or solutions, lattice/solvation energies dominate and would mask the pure atom's electron-accepting energetics.
For a Group 2 atom like Mg, where would the incoming electron go, and what is the expected sign?
Into a higher-energy p orbital (the subshell is filled and stable), so the atom resists and is near-zero or positive.
If two atoms had identical but different sizes, which would have the more negative ?
The larger one — same nuclear pull but more room means less electron–electron repulsion on the added electron, so more net energy is released.
Recall One-line summary of every trap
Almost all traps reduce to two ideas: (1) sign convention — negative means "wants it"; and (2) the attraction-vs-crowding tug-of-war that makes small orbitals (F, O, N) and stable configs (half/full shells) release less than you'd naively expect.