Ionization energy — first, second, …; trends and anomalies (e.g. B - Be): Worked examples

The scenario matrix

Every ionization-energy question is one of these cells. The examples below are labelled with the cell they hit, so together they cover the whole grid.

Cell	Case class	The question it tests	Example
A	Plain period trend	Order two neighbours, no anomaly	Ex 1
B	Anomaly type 1 (s vs p)	B < Be, Al < Mg	Ex 2
C	Anomaly type 2 (half vs paired)	O < N, S < P	Ex 3
D	Group trend (down)	Order two elements in a column	Ex 4
E	Successive IE jump	Read the group number off a data row	Ex 5
F	Coulomb estimate (numeric)	Predict a number from $Z_{eff}$ , $n$	Ex 6
G	Degenerate / limiting input	$I E_{2}$ of H, noble-gas core, single electron	Ex 7
H	Real-world word problem	Flame colours / which metal ionizes in a lamp	Ex 8
I	Exam twist (combined effects)	Both an anomaly AND a trend fight each other	Ex 9

Nine cells → nine worked examples. Let's go.

Cell A — the plain period trend

Worked example Ex 1 — Order

$I E_{1}$ : Li, Be, C These three sit in period 2 with no anomaly between them (the anomalies are at B and O). Put them in increasing order of $I E_{1}$ .

Forecast: guess the order before reading. Which of Li, Be, C do you think holds its electron tightest?

Write $Z$ : Li $Z = 3$ , Be $Z = 4$ , C $Z = 6$ . Why this step? Across a period $n$ (the shell) is fixed at $2$ , so the only thing changing our machine's output is $Z_{eff}$ , which tracks $Z$ .
Same shell, so shielding is small. Each added electron goes into the same $n = 2$ shell and barely blocks the nucleus, so $Z_{eff}$ climbs steadily Li → Be → C. Why this step? It justifies " $I E$ rises left-to-right" — the parent's period rule.
Order: $I E_{1} (Li) < I E_{1} (Be) < I E_{1} (C)$ .

Verify: experimental values (kJ/mol): Li $= 520$ , Be $= 899$ , C $= 1086$ . Strictly increasing — matches. ✅

Cell B — anomaly type 1 (full-s beats lone-p)

Worked example Ex 2 — Which is larger,

$I E_{1} (Be)$ or $I E_{1} (B)$ ? B is one proton more than Be, so the naive rule screams " $I E (B)$ bigger." Test that.

Forecast: trust the extra proton, or suspect a trap?

Configurations: Be $= 1 s^{2} 2 s^{2}$ , B $= 1 s^{2} 2 s^{2} 2 p^{1}$ . Why this step? The removed electron's orbital decides the answer, not the proton count alone.
Where does the electron leave from? In B it leaves a $2 p$ orbital, which is higher in energy than $2 s$ and is shielded by the filled $2 s^{2}$ pair. See the energy-ladder figure below. Why this step? A shielded, higher orbital means smaller effective pull → easier removal, per Half-filled and Fully-filled Stability.
Conclusion: the orbital effect beats the one extra proton, so $I E_{1} (B) < I E_{1} (Be)$ — a dip.

Verify: Be $= 899$ , B $= 801$ kJ/mol. Indeed $801 < 899$ , the dip is real. ✅ (Same story: Al $= 577 <$ Mg $= 738$ .)

Cell C — anomaly type 2 (half-filled beats paired)

Worked example Ex 3 — Which is larger,

$I E_{1} (N)$ or $I E_{1} (O)$ ? O has one proton more than N — naive rule says O bigger. Test it.

Forecast: guess whether N or O clings harder.

Configurations of the $2 p$ subshell: N $= 2 p^{3}$ (three orbitals, one electron each), O $= 2 p^{4}$ (one orbital now doubly occupied). Why this step? The difference is pairing, not shielding — a different mechanism from Ex 2.
Repulsion in O. In O two electrons share one $2 p$ orbital and repel each other. Removing one relieves that repulsion, so it leaves easily. N's half-filled $2 p^{3}$ is symmetric and exchange-stabilised, so it clings. Why this step? The removal energy is lowered by whatever destabilises the starting atom; O starts destabilised by pairing.
Conclusion: $I E_{1} (O) < I E_{1} (N)$ — the second period dip.

Verify: N $= 1402$ , O $= 1314$ kJ/mol. $1314 < 1402$ . ✅ (Same for S $= 1000 <$ P $= 1012$ .)

Cell D — the group trend (going down)

Cell E — reading a group number from IE jumps

Worked example Ex 5 — An unknown element X has successive IEs (kJ/mol):

$577, 1817, 2745, 11577, 14842, \dots$ What group is it in? Forecast: count how many electrons come off "cheaply" before the wall.

List the jump ratios: $1817/577 \approx 3.1$ , $2745/1817 \approx 1.5$ , then $11577/2745 \approx 4.2$ . Why this step? A big sudden jump flags that the next electron comes from a lower, noble-gas core shell — very hard to reach.
Locate the wall. The huge jump happens going from $I E_{3}$ to $I E_{4}$ . So 3 electrons came off before we hit the core. Why this step? The number of "cheap" removals before the wall = number of valence electrons = group number.
Conclusion: 3 valence electrons ⇒ Group 13. (This matches aluminium: $[Ne] 3 s^{2} 3 p^{1}$ .)

Verify: Al's real data (kJ/mol) is $577, 1817, 2745, 11577, 14842$ — the jump after the 3rd electron confirms Group 13. ✅

Cell F — a Coulomb numerical estimate

Worked example Ex 6 — Estimate

$I E_{1}$ of boron with the $Z_{eff}$ model B $= 1 s^{2} 2 s^{2} 2 p^{1}$ ; remove the $2 p$ electron. Predict a number in eV, then compare.

Forecast: will the estimate land near the true $\approx 8.3$ eV, or wildly off?

Slater's $S$ for the $2 p$ electron. Same-shell ( $2 s, 2 p$ ) electrons shield $0.35$ each; there are 4 of them (two $2 s$ , one $2 p$ ... careful: 4 other $n = 2$ electrons? No — count electrons other than the one we remove). B has $2 s^{2} 2 p^{1}$ → 2 other valence electrons at $0.35$ ; the two $1 s$ at $0.85$ . $S = 2 (0.85) + 2 (0.35) = 1.70 + 0.70 = 2.40$ Why this step? Slater's Rules give $S$ so we can turn "shielding" into a number.
Effective charge: $Z_{eff} = Z - S = 5 - 2.40 = 2.60$ . Why this step? This is the net pull our electron actually feels.
Plug into the machine with $n = 2$ : $I E \approx 13.6 \times \frac{2.6 0 ^{2}}{2 ^{2}} = 13.6 \times \frac{6.76}{4} \approx 22.98 eV$ Why this step? Direct application of $I E = 13.6 Z_{eff}^{2} / n^{2}$ .

Verify: experimental $I E_{1} (B) = 8.30$ eV. Our crude number ( $\approx 23$ eV) is too high — Slater over-counts the true pull for a diffuse $2 p$ electron. Right order of magnitude, wrong precision; the model tells us direction, not exact values. ✅ (limitation, honestly reported)

Cell G — degenerate & limiting inputs

Worked example Ex 7 — Three edge cases in one

(a) What is $I E_{2}$ of hydrogen? (b) Why is $I E (He)$ the largest of all neutral atoms? (c) What happens to $I E$ as $n \to \infty$ ?

Forecast: which of these is a "trick — undefined" case?

(a) $I E_{2}$ of H. H has only one electron. After $I E_{1}$ you have a bare proton $H^{+}$ — no electron left to remove. So $I E_{2} (H)$ is undefined / does not exist. Why this step? Always check the electron count before computing a successive IE — the degenerate case is "there is nothing there."
(b) He $= 1 s^{2}$ . It is a tiny, unscreened $n = 1$ shell with $Z = 2$ . Using the machine with $Z_{eff} \approx 2 - 0.30 = 1.70$ , $n = 1$ : $I E \approx 13.6 \times \frac{1.7 0 ^{2}}{1 ^{2}} \approx 39.3 eV$ Why this step? Smallest $n$ and high $Z_{eff}$ both push IE up — the extreme "hard to steal" case.
(c) Limit $n \to \infty$ . In $I E = 13.6 Z_{eff}^{2} / n^{2}$ , sending $n \to \infty$ makes $I E \to 0$ . Physically: an electron infinitely far out is already essentially free. Why this step? It shows the machine behaves sensibly at the boundary — a limiting-value sanity check.

Verify: experimental $I E_{1} (He) = 24.6$ eV — again our estimate ( $39$ ) is high but He is the largest neutral-atom IE, and the limit $n \to \infty \Rightarrow I E \to 0$ is exact. ✅

Cell H — real-world word problem

Worked example Ex 8 — Sodium vs potassium street lamps

Sodium-vapour lamps glow orange, potassium is used in some fireworks. A lamp works by exciting/ionizing the metal's outer electron. Which of Na or K needs less energy to lose its outer electron, and does the Coulomb model agree?

Forecast: guess before computing — same group, so which is easier?

Both are Group 1, valence $s^{1}$ . Na $n = 3$ , K $n = 4$ ; both $Z_{eff} \approx 2.2$ (Slater gives similar values for the outer $s$ electron). Why this step? Same group means the group trend (Cell D) applies — larger $n$ wins.
Compare with the machine: $I E \propto 1/ n^{2}$ if $Z_{eff}$ is roughly equal. $\frac{I E ( K )}{I E ( Na )} \approx \frac{1/ 4 ^{2}}{1/ 3 ^{2}} = \frac{9}{16} \approx 0.56$ Why this step? It predicts K should need only about half as much energy — so K ionizes more easily.
Conclusion: potassium loses its electron more readily; both are chosen for lamps/fireworks precisely because their low IE makes them easy to excite.

Verify: real $I E_{1}$ : Na $= 496$ , K $= 419$ kJ/mol → ratio $419/496 \approx 0.845$ . Our $1/ n^{2}$ estimate (0.56) is in the right direction (K lower) though not exact, because $Z_{eff}$ isn't truly equal. ✅

Cell I — the exam twist (two effects fighting)

Worked example Ex 9 — Order

$I E_{1}$ : N, O, F, and explain why it is not strictly increasing Period 2, moving right: N → O → F. The plain trend says "increase all the way," but there's an anomaly hiding.

Forecast: where does the trend break, and does it recover?

Plain trend expectation: $Z$ rises N(7) → O(8) → F(9), so raw pull rises. Why this step? Establishes the default before the twist.
Insert the O anomaly (Cell C): O $= 2 p^{4}$ pays a pairing-repulsion penalty, dropping below N. So the order is not N < O < F. Why this step? Two effects collide: the proton-count push (up) vs pairing penalty (down at O).
Does it recover at F? F $= 2 p^{5}$ has an even higher $Z$ and one more proton than O; the extra nuclear pull outweighs its pairing, so F climbs back above both. Why this step? Shows the anomaly is a local dip, not a permanent reversal.
Final order: $I E_{1} (O) < I E_{1} (N) < I E_{1} (F)$ .

Verify: O $= 1314$ , N $= 1402$ , F $= 1681$ kJ/mol. Indeed $1314 < 1402 < 1681$ — the dip at O then recovery at F. ✅

Active recall

Recall Self-test the matrix (hide answers)

Which cell: "order Li, Na, K"? ::: Cell D — group trend, IE falls down a group. Which cell: "successive IEs jump after 2 removals"? ::: Cell E — jump ⇒ 2 valence ⇒ Group 2. Why does $I E_{2} (H)$ not exist? ::: H has only one electron; nothing left to remove after $I E_{1}$ . In Ex 9, why is the order O < N < F not N < O < F? ::: O's $2 p^{4}$ pairing repulsion drops it below N; F's extra proton lifts it back on top. Why does the Coulomb estimate for B (Ex 6) overshoot? ::: Slater's rules over-count the true pull on a diffuse $2 p$ electron; model gives direction, not precision.

Related: Electron Affinity · Electronegativity · Aufbau, Hund & Pauli — Electron Configuration