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

Step 1 — Two charges, one pull

WHAT: We draw the simplest possible atom — one positive nucleus, one negative electron a distance $r$ apart — and mark the force pulling them together.

WHY: Before we talk about shells, shielding, or anomalies, we need the raw ingredient. Everything else is a correction to this one picture. The tool that measures "how hard two charges pull" is Coulomb's law — we pick it because it is the only law that connects charge and distance to force, which is precisely the two things we suspect matter.

$F = k \frac{q _{1} q _{2}}{r ^{2}}$

$F$ ::: the strength of the pull (bigger $F$ = harder to separate).
$q_{1}$ ::: charge of the nucleus. For a nucleus with == $Z$ protons== (its true nuclear charge), $q_{1} = + Z e$ , where $e$ is the charge on one proton.
$q_{2}$ ::: charge of the electron, $q_{2} = - e$ (negative).
$r$ ::: the gap between them; it is squared, so doubling the distance quarters the pull.
$k$ ::: Coulomb's constant — a fixed number ( $k = \frac{1}{4 π ε _{0}}$ ) that turns "charges over distance-squared" into an actual force. It carries all the units and it will not vanish; we track it right through to the $13.6$ eV.

PICTURE (s01): A lavender nucleus marked $+ Z e$ and a coral electron marked $- e$ , separated by a labelled gap $r$ , with a red arrow showing the pull $F$ and the formula $F = k q_{1} q_{2} / r^{2}$ floating above. The figure is the raw force we spend the rest of the page converting into energy.

Look at the red arrow in the figure: it is the pull. The whole rest of this page is the question "how much energy does it take to drag the electron away against that arrow, all the way to $r = \infty$ ?"

Step 2 — Adding up the pull: from force to a potential-energy curve

WHAT: We add up (integrate) the attractive force from the electron's position $r$ all the way out to infinity. That total work is the potential energy $U (r)$ — how much energy the electron has stored by sitting at distance $r$ .

WHY: Work done against the pull, summed over the whole trip out, is precisely the energy you must supply to free the electron. Doing the sum (integral) of $k q_{1} q_{2} / r^{2}$ gives a clean $1/ r$ shape:

$U (r) = - \int_{\infty}^{r} F d r^{'} = k \frac{q _{1} q _{2}}{r} = - k \frac{Z e ^{2}}{r}$

$\int_{\infty}^{r}$ ::: "sum the force over every tiny step from far away in to $r$ " — this is the tool that turns force into stored energy.
$k \frac{Z e ^{2}}{r}$ ::: notice the power of $r$ dropped from 2 to 1 — integrating $1/ r^{2}$ gives $1/ r$ . This is why energy falls off more gently than force.
The minus sign ::: because $q_{1} q_{2} = (+ Z e) (- e)$ is negative, the electron sits below zero — it is trapped in a dip. Free electron = zero energy at the top.

PICTURE (s02): The lavender $- 1/ r$ curve — a well that plunges near the nucleus and flattens to zero far away. The electron rests at the bottom; the green arrow shows the climb back up to the zero line. That climb height is the energy we must pay.

So far this is pure Coulomb — no quantum physics yet. The curve says the electron could sit at any depth. Nature says otherwise, which is Step 3.

Step 3 — Quantization: why only certain depths are allowed, and where $13.6$ eV is born

WHAT: We label the allowed rungs by an integer $n = 1, 2, 3, \dots$ — the principal quantum number. It counts how many wave-humps fit, which is the same as which shell the electron lives in.

WHY: Bohr's condition (only whole wavelengths allowed) forces the allowed radii and energies to depend on $n^{2}$ . Feeding Coulomb's $U (r) = - k Z e^{2} / r$ through that condition and solving gives the allowed energies:

$E_{n} = - \frac{k ^{2} e ^{4} m _{e}}{2 ℏ ^{2}} \cdot \frac{Z ^{2}}{n ^{2}}$

$\frac{k ^{2} e ^{4} m _{e}}{2 ℏ ^{2}}$ ::: this bundle of fundamental constants — Coulomb's $k$ , the electron charge $e$ , its mass $m_{e}$ , and Planck's constant $ℏ$ — is a fixed number. Evaluate it and it equals $13.6$ eV. This is where the mystery constant comes from: it is not arbitrary, it is $k$ , $e$ , $m_{e}$ , $ℏ$ multiplied together. Notice $k$ is still here, alive inside the $13.6$ .
$Z^{2}$ ::: the charge appears squared — once from the pull, once from how tightly the orbit shrinks. Stronger nucleus ⇒ dramatically deeper rungs.
$\frac{1}{n ^{2}}$ ::: bigger shell (larger $n$ ) ⇒ shallower rung. The quantization put this $1/ n^{2}$ here; Coulomb alone could not.

Bundling the constants into the shorthand $13.6 eV$ :

$E_{n} = - \frac{Z ^{2}}{n ^{2}} \times 13.6 eV$

PICTURE (s03): The same Coulomb well from s02, now with horizontal rungs drawn at $E_{1}, E_{2}, E_{3}$ — the only depths the electron is allowed to occupy. The $n = 1$ rung is deepest ( $- 13.6$ eV for hydrogen); rungs crowd toward zero as $n$ grows because of the $1/ n^{2}$ .

For hydrogen ( $Z = 1$ , $n = 1$ ): $E_{1} = - 13.6$ eV. That single number is the depth of the deepest well nature builds — every other atom's IE is a rescaling of it.

Step 4 — Flipping the sign: the ionization energy itself

WHAT: We flip the minus sign of $E_{n}$ into a plus — the energy we pay in.

WHY: $E_{n}$ is where the electron is (negative, in the well). $I E$ is what we add to lift it to zero. So $I E = - E_{n} = ∣ E_{n} ∣$ . This is why IE is always positive: you can only ever pay to climb out of a dip.

The formula has two knobs: $Z$ on top (squared → very sensitive) and $n^{2}$ underneath.
Turn $Z$ up ⇒ deeper rung ⇒ bigger climb ⇒ $I E ↑$ .
Turn $n$ up ⇒ higher, shallower rung ⇒ smaller climb ⇒ $I E ↓$ .

PICTURE (s04): Two wells side by side — a deep, narrow one (high $Z$ , big green climb arrow) and a shallow, wide one (high $n$ , small climb arrow). This single contrast previews both periodic trends (Steps 6–7).

Step 5 — The one honest fix for real atoms: $Z \to Z_{eff}$

$Z$ ::: the true proton count (from Step 1) — the full raw pull.
$S$ ::: the shielding, how much the other electrons cancel out; estimated by Slater's Rules.
$Z_{eff}$ ::: what's left — the real grip the outer electron feels.

WHAT: We keep the whole hydrogen result but replace $Z$ with $Z_{eff}$ , giving the formula the parent note used:

$I E \approx 13.6 eV \times \frac{Z _{eff}^{2}}{n ^{2}}$

WHY: This one swap lets a one-electron formula stand in for a messy many-electron atom — we pretend the outer electron orbits a nucleus of charge $Z_{eff}$ . See Effective Nuclear Charge & Shielding.

PICTURE (s05): The nucleus $+ Z e$ , a lavender inner cloud casting a mint "shadow" that blocks part of the pull, and the coral outer electron feeling only the shortened arrow $Z_{eff} = Z - S$ .

Step 6 — Reading the trend ACROSS a period

WHAT: Fix $n$ ; slide $Z_{eff}$ up step by step and watch the well deepen.

WHY: Same-shell electrons are poor shielders of each other, so nearly all the extra proton's pull survives ⇒ $Z_{eff}$ rises steadily. This also explains why the atomic radius shrinks across a period — the deeper well pulls the electron closer.

PICTURE (s06): A bar chart of $I E_{1}$ for Li→Ne marching upward, with two coral bars (B and O) sticking down out of line — the anomalies we tackle in Steps 7–8.

The bars march upward left→right. Each rightward step deepens the well by a notch — that notch is the extra $Z_{eff}$ .

Step 7 — Reading the trend DOWN a group

WHAT: Fix $Z_{eff}$ (roughly); raise $n$ and watch the well flatten out.

WHY: Distance wins. In our formula $n$ is squared underneath, so going from $n = 2$ to $n = 3$ shrinks $I E$ by a factor near $\frac{4}{9}$ before shielding even acts. The electron is simply too far from the nucleus.

PICTURE (s07): Three wells stacked by shell — $n = 2$ deep, $n = 3$ shallower, $n = 4$ shallowest — with the green climb arrow shrinking each time.

Step 8 — The B < Be anomaly: a different orbital breaks the march

WHAT: We compare which orbital the escaping electron leaves, not just how many protons pull.

WHY: Be's outer electron leaves a snug, low $2 s$ orbital. B's outer electron sits in a higher, more spread-out $2 p$ orbital that is shielded by the filled $2 s^{2}$ pair beneath it. That extra shielding drops $Z_{eff}$ for that specific electron enough to beat the single extra proton. In the well picture: B's $2 p$ electron rests higher up the wall than Be's $2 s$ electron, so its climb is shorter. This is the same effect that makes filled subshells extra stable — and it is exactly the "orbital shape" limitation flagged in Step 5.

PICTURE (s08): Two wells — Be's $2 s$ electron sitting deep (big climb arrow) versus B's $2 p$ electron sitting on a ledge higher up the wall (short climb arrow), so it leaves more easily despite the extra proton.

Step 9 — The O < N anomaly: repulsion pops an electron out

WHAT: We add the last piece our Coulomb well cannot contain — electron–electron repulsion inside the same orbital (the very limitation named in Step 5).

WHY: Two electrons crammed into one $2 p$ orbital shove each other (both negative). Removing one relieves that shove, so it leaves more easily. Meanwhile N's half-filled $2 p^{3}$ is symmetric and extra stable (see Hund's rule and Half-filled and Fully-filled Stability). Both facts push $I E (O)$ below $I E (N)$ .

PICTURE (s09): Orbital boxes — N's three calm single arrows (spread out, stable) versus O's one doubled-up box where two arrows repel (coral spark), with an arrow showing that crowded electron is "easy to pull out".

The one-picture summary

PICTURE (s10): One Coulomb well carrying the whole story — its depth set by $Z_{eff}$ (rises across a period), its width set by $n$ (grows down a group), plus two coral marks: the ledge where B's shielded $2 p$ electron sits higher than expected, and the shove where O's paired electrons repel. The green climb arrow's height is, always, the ionization energy.

Recall Feynman: tell the whole walkthrough to a friend

Picture a marble sitting at the bottom of a bowl — that's the electron, and the bowl is carved by the nucleus pulling on it (Coulomb's law). If you add up that pull all the way out to nowhere, you get the shape of the bowl (that's the integral, giving a $- 1/ r$ dip). Now the strange part: the marble can't rest at just any depth — it can only sit on certain shelves inside the bowl, numbered $1, 2, 3 \dots$ , because the electron is secretly a wave and only whole waves fit. Those shelves sit at depths $- 13.6/ n^{2}$ times the charge squared, and that $13.6$ isn't magic — it's just the fundamental constants (Coulomb's $k$ , the electron's charge and mass, Planck's constant) all multiplied together. Ionization energy is how far you lift the marble off its shelf to get it out of the bowl. A stronger nucleus makes a deeper bowl → harder lift → high IE (that's moving right across a period). A bigger, farther shelf makes a shallow bowl → easy lift → low IE (that's going down a group). Two tricks break the pattern: sometimes the marble already sits on a little ledge higher up (boron's shielded $2 p$ ), and sometimes two marbles are jammed in one dimple shoving each other (oxygen's pair), so one nearly jumps out on its own. And the honest footnote: for the heavy d- and f-block atoms the shelves get so crowded and the shoving so tangled that this simple bowl picture stops predicting the exact order — there you need the full quantum accounting.

Recall Rebuild the formula from scratch

Coulomb force $F = k q_{1} q_{2} / r^{2}$ (with $q_{1} = + Z e$ ) → integrate over distance to get potential energy $U (r) = - k Z e^{2} / r$ → impose quantization (whole waves fit, Bohr) to get allowed rungs $E_{n} = - (k^{2} e^{4} m_{e} /2 ℏ^{2}) (Z^{2} / n^{2}) = - 13.6 eV Z^{2} / n^{2}$ → flip sign for $I E = + ∣ E_{n} ∣$ → patch $Z \to Z_{eff} = Z - S$ for real atoms. Depth $\to$ period trend, width $\to$ group trend, ledge $\to$ B<Be, shove $\to$ O<N, and it breaks for d/f blocks.

Related vault topics: Coulomb's Law · Effective Nuclear Charge & Shielding · Slater's Rules · Atomic Radius Trends · Half-filled and Fully-filled Stability · Aufbau, Hund & Pauli — Electron Configuration · Electron Affinity · Electronegativity