Ionization Enthalpy

What Is Ionization Enthalpy?

Think of an atom as holding on to its electrons through the nuclear pull. Some of those electrons (the inner, core ones) are gripped very tightly. The outermost electron, on the other hand, is relatively loose — it's the furthest from the nucleus, and it's the one that chemistry "notices" first. Ionization enthalpy is a quantitative measure of how hard it is to rip away that outermost electron.

Definition

The ionization enthalpy ($\Delta_{\text{i}}H$) of an element is the energy required to remove the most loosely bound electron from an isolated, gaseous atom of that element in its ground state.

The process is:

$$\text{X}(g) \;\longrightarrow\; \text{X}^{+}(g) + e^{-} \qquad \Delta_{\text{i}}H_{1} > 0$$

Note the three essential conditions:

1. The atom must be in the gas phase (so we aren't sneaking lattice or bonding energies into the number).
2. It must be isolated (no neighbours influencing it).
3. It must be in its ground-state electronic configuration (not an excited state).

If any of these fails, what you're measuring is not the ionization enthalpy in the strict NCERT sense.

Sign convention and units

Ionization is always endothermic — pulling a negatively charged electron away from a positively charged nucleus costs energy. So $\Delta_{\text{i}}H$ is always positive. In SI units it is reported in kJ mol$^{-1}$ (energy per mole of atoms); older texts often use eV/atom (1 eV/atom ≈ 96.485 kJ/mol).

Two sample values to carry in your head:

Element	$\Delta_{\text{i}}H_{1}$ / kJ mol$^{-1}$
Cs (largest alkali metal)	376
Na	496
H	1312
He	2372

Cs — the easiest stable element on the periodic table to ionize. He — among the hardest.

Enthalpy vs. energy

You'll sometimes see the term "ionization energy" used interchangeably with ionization enthalpy. Strictly, they differ by a small $\Delta(PV)$ term, which for a gas is $RT \approx 2.5$ kJ/mol at 298 K. This difference is almost always negligible compared to the IE values themselves (which run in the thousands of kJ/mol), so NCERT treats the two as equivalent. We will too — just know the distinction exists for formal thermodynamics questions.

[JEE Tip] When a problem says "calculate the energy required to ionize 1 g of H atoms," expect answer in kJ (or J). Multiply the per-mole IE by the number of moles: $E = n \times \Delta_{\text{i}}H$.

Factors That Decide the Value of IE

Five factors, acting together, set the ionization enthalpy of any atom. You have met most of them in earlier sections — here we bring them all together for IE.

Factor 1 — Size of the atom

The farther an electron sits from the nucleus, the weaker the Coulomb pull and the easier it is to remove. So:

Bigger atom → lower IE.

This is the single most important factor. It explains why alkali metals (big atoms with a lone valence electron) have the lowest IEs in any period, and why noble gases (small, tight atoms with fully-filled shells) have the highest.

Factor 2 — Magnitude of the nuclear charge, $Z$

More protons pull all electrons — including the valence one — harder. So:

Higher $Z$ → higher IE (holding other factors equal).

Factor 3 — Shielding by inner electrons, $\sigma$

Inner electrons partially cancel the nuclear charge seen by the valence electron. The net charge felt is the effective nuclear charge, $Z_{\text{eff}} = Z - \sigma$.

More shielding → smaller $Z_{\text{eff}}$ → lower IE.

Factors 1, 2, and 3 are not independent: a bigger atom typically has more inner electrons and hence more shielding, reinforcing the "bigger = lower IE" rule.

Factor 4 — Penetration of the orbital

At the same principal quantum number $n$, orbitals differ in how close they penetrate toward the nucleus:

$$s \;>\; p \;>\; d \;>\; f$$

An $s$-electron spends more time closer to the nucleus (tight inner lobes) than a $p$-electron in the same shell, so it feels higher $Z_{\text{eff}}$. Removing an $s$-electron therefore costs more energy than removing a $p$-electron from the same shell.

This is why, for instance, Ca's IE$_{1}$ (590 kJ/mol, removing a $4s$ electron) is larger than Ga's IE$_{1}$ (579 kJ/mol, removing a $4p$ electron), even though Ga has a higher $Z$.

Factor 5 — Sub-shell stability (full- and half-filled rule)

This is the factor that produces the famous periodic-table exceptions. Fully-filled and exactly half-filled sub-shells carry a small extra stability from two quantum-mechanical effects:

1. Exchange energy. Electrons with parallel spins in different orbitals of the same sub-shell can "exchange" quantum-mechanically. Each such exchange lowers the energy. Half-filled sub-shells maximise the number of parallel-spin electrons and therefore maximise the exchange energy.
2. Symmetrical distribution. Full (e.g. $p^{6}$) and half-filled (e.g. $p^{3}$) sub-shells produce a spherically symmetric charge distribution, which is energetically favourable.

Half-filled or fully-filled sub-shell → extra-stable → extra-hard to ionize.

This is a small extra energy — enough to reverse a trend, but not enough to reorder elements that differ by two or more positions.

Factor 6 (bonus) — Electronic repulsion in a partially-filled orbital

When a sub-shell is more than half-filled, two electrons share an orbital and repel each other strongly. Removing one of those paired electrons relieves the repulsion, so it is easier than expected. This explains why removing an electron from O (which has $2p^{4}$, with one paired orbital) is easier than from N ($2p^{3}$, all unpaired).

How to use the five factors

In practice, when you face an IE-comparison question, run this checklist:

• Which atom is bigger? → that one has lower IE (Factor 1).
• Which has higher $Z_{\text{eff}}$? → that one has higher IE (Factors 2 + 3).
• Are we removing an $s$, $p$, $d$, or $f$ electron? → $s$ electrons are hardest to remove for the same shell (Factor 4).
• Is either configuration half- or fully-filled? → that one has extra IE (Factor 5).
• Is either sub-shell more-than-half-filled with a paired electron? → that one has less IE than expected (Factor 6).

One or two of these usually dominate and give you the answer in seconds.

[NEET Important] NEET loves to probe Factors 5 and 6 with trick ordering questions like "Order IE: N, O, F, Ne" — the "expected" answer N < O < F < Ne is wrong because of Factors 5/6. The correct order is O < N < F < Ne. Know why before the exam.

Variation of IE Across a Period

The rule is simple: IE increases across a period (left to right).

The numbers

Here are first ionization enthalpies (in kJ/mol) across Periods 2 and 3:

Group	1	2	13	14	15	16	17	18
Period 2	Li 520	Be 899	B 801	C 1086	N 1402	O 1314	F 1681	Ne 2081
Period 3	Na 496	Mg 737	Al 577	Si 786	P 1012	S 1000	Cl 1251	Ar 1521

Across Period 2, IE rises from 520 (Li) to 2081 (Ne) — nearly a factor of 4. The same shape appears in Period 3 (496 → 1521).

Why does IE increase across a period?

Move from Li ($Z = 3$) to Ne ($Z = 10$): the nuclear charge grows by 7 units. The added electrons all enter the same $n = 2$ shell, so they shield each other poorly (each same-shell electron only adds about 0.35 to $\sigma$ per Slater). The net $Z_{\text{eff}}$ seen by the valence electron rises steeply, pulling it closer to the nucleus. Two combined effects:

• Atomic radius shrinks (Section 6 trend).
• The valence electron feels a stronger pull.

Both effects make the electron harder to remove, so IE grows.

But the trend is not perfectly monotonic — the 4 exceptions

Look closely at the values. There are four clear "dips" where IE falls when it should have risen:

1. Be (899) > B (801): boron's IE is lower than beryllium's.
2. N (1402) > O (1314): oxygen's IE is lower than nitrogen's.
3. Mg (737) > Al (577): aluminium's IE is lower than magnesium's.
4. P (1012) > S (1000): sulfur's IE is lower than phosphorus's (a small dip).

Each dip sits at the group 13 or group 16 position — i.e., right after a full $s^{2}$ or a half-filled $p^{3}$. The reason is sub-shell stability (Factor 5 from Note 2), reinforced by Factor 4 (penetration) for the $s \to p$ cases.

The "Be > B" and "Mg > Al" anomaly — in detail

Beryllium has configuration $1s^{2}\,2s^{2}$ — a fully-filled $2s$ sub-shell. Removing an electron from that stable $2s^{2}$ requires extra energy.

Boron, next door, has $1s^{2}\,2s^{2}\,2p^{1}$ — we are now removing the one lonely $2p$ electron. Two things make this easier than removing Be's $2s$:

• The $2p$ orbital is less penetrating than $2s$ (Factor 4), so the $2p$ electron feels a lower effective nuclear charge.
• There's no "break a stable sub-shell" penalty.

Net result: $\Delta_{\text{i}}H_{1}(\text{B}) < \Delta_{\text{i}}H_{1}(\text{Be})$, by about 100 kJ/mol.

The same logic applies one period down — Mg ($3s^{2}$) → Al ($3p^{1}$) — and even shows up more dramatically because the 3s-3p penetration gap is larger.

The "N > O" and "P > S" anomaly — in detail

Nitrogen has configuration $1s^{2}\,2s^{2}\,2p^{3}$ — a half-filled $2p$ sub-shell, with all three $2p$ electrons unpaired and spin-aligned. Exchange energy (Factor 5) gives it extra stability.

Oxygen has $2p^{4}$ — one of the $2p$ orbitals now has two paired electrons. Removing one of those paired electrons has two advantages:

• It relieves the electron-electron repulsion in the doubly-occupied orbital (Factor 6).
• It leaves behind the extra-stable $2p^{3}$ half-filled configuration.

Both effects make it easier to ionize O than N — by about 90 kJ/mol.

The same pattern appears at P ($3p^{3}$) → S ($3p^{4}$).

[JEE Tip] The four anomalies (Be>B, N>O, Mg>Al, P>S) are favourite JEE Main MCQ triggers. Memorise them and their explanations. Note that they don't overturn the overall period trend — F and Ne are still the hardest to ionize in Period 2, even after the dip at O.

Second-period vs third-period: a subtle pattern

Compare: IE$_{1}$(Li) = 520, IE$_{1}$(Na) = 496. Na is slightly lower — expected, because Na has a bigger atom. Now compare IE$_{1}$(Be) = 899, IE$_{1}$(Mg) = 737: Mg is much lower than Be. Then IE$_{1}$(N) = 1402, IE$_{1}$(P) = 1012: again Period 3 lower. So across the board, Period 3 atoms have lower IEs than their Period 2 neighbours in the same group. This is just the "down-a-group" trend, to which we turn now.

Variation of IE Down a Group

The rule is again simple — and the opposite of the period trend:

IE decreases down a group.

The numbers

Group 1 (alkali metals):

Element	Period	$\Delta_{\text{i}}H_{1}$ / kJ mol$^{-1}$
H	1	1312
Li	2	520
Na	3	496
K	4	419
Rb	5	403
Cs	6	376

From Li down to Cs, IE roughly halves. H is an outlier — with its tiny covalent radius and bare proton nucleus, H has an unusually high IE despite being a Group 1 element.

Group 18 (noble gases):

Element	$\Delta_{\text{i}}H_{1}$ / kJ mol$^{-1}$
He	2372
Ne	2081
Ar	1521
Kr	1351
Xe	1170
Rn	1037

Same decreasing pattern.

Why does IE decrease down a group?

Three of our factors conspire:

• Atomic radius increases (new shell each step) — Factor 1, dominant.
• Shielding $\sigma$ grows (more inner electrons) — Factor 3.
• Nuclear charge $Z$ increases too, but is largely offset by the increased shielding — Factor 2 is weaker here.

The net effect: $Z_{\text{eff}}$ seen by the valence electron stays roughly similar (or increases only slightly) down a group, while the distance from the nucleus grows substantially. The Coulomb pull on the valence electron weakens, so IE falls.

A subtle anomaly: Ga > Al (almost)

For the p-block, the down-group trend is not always perfectly smooth. Compare IE$_{1}$ of Group 13:

Element	$\Delta_{\text{i}}H_{1}$ / kJ mol$^{-1}$
B	801
Al	577
Ga	579
In	558
Tl	589

Ga's IE is slightly higher than Al's — against the naive trend. The reason: between Al (Period 3) and Ga (Period 4) we cross the 3d-block (Sc through Zn). The $3d^{10}$ filling shields Ga's $4p^{1}$ electron poorly, so Ga's $Z_{\text{eff}}$ is higher than simple shell-counting would predict. This "d-block contraction" partly cancels the normal shell-growth effect, leaving Ga with an IE close to (and slightly above) Al's.

A similar effect from lanthanoid contraction makes Tl's IE slightly higher than In's.

Combining with periodic trends

Putting the two rules together:

• Up and to the right of the periodic table → highest IE (→ F, Ne, He corners).
• Down and to the left → lowest IE (→ Cs, Fr, Rb).

[Board Level Important] A favourite Board question is "Among Li, Be, B, C, N, O, F, Ne, which has the highest / lowest IE$_{1}$?" — Highest: Ne. Lowest: Li. Memorise the endpoints.

Successive Ionization Enthalpies

So far we have only talked about removing the first electron. But an atom can be ionized more than once: after ripping off the first electron, we can rip off the second, the third, and so on. Each step has its own energy cost:

$$\begin{aligned} \text{X}(g) &\longrightarrow \text{X}^{+}(g) + e^{-} \quad &\Delta_{\text{i}}H_{1}\\ \text{X}^{+}(g) &\longrightarrow \text{X}^{2+}(g) + e^{-} \quad &\Delta_{\text{i}}H_{2}\\ \text{X}^{2+}(g) &\longrightarrow \text{X}^{3+}(g) + e^{-} \quad &\Delta_{\text{i}}H_{3} \end{aligned}$$

The second, third, fourth… values are called successive (or higher-order) ionization enthalpies.

The golden rule

Successive ionization enthalpies always increase: $\Delta_{\text{i}}H_{1} < \Delta_{\text{i}}H_{2} < \Delta_{\text{i}}H_{3} < \ldots$

Why? After each electron is removed, the ion becomes smaller and has a higher $Z/e^{-}$ ratio. The remaining electrons are held more tightly, so the next one is harder to remove than the one before.

The diagnostic "big jump" — reading the group number

The beauty of successive IEs is that the pattern of increases tells you which group the element belongs to. Take sodium ($Z = 11$, configuration $1s^{2}\,2s^{2}\,2p^{6}\,3s^{1}$):

Step	kJ/mol	Orbital being emptied
IE$_{1}$	496	$3s^{1}$ (valence)
IE$_{2}$	4560	$2p^{6}$ (core)
IE$_{3}$	6910	$2p^{5}$
IE$_{4}$	9540	$2p^{4}$
…	…	…
IE$_{10}$	141360	$1s^{2}$ (innermost)

Notice the huge jump from IE$_{1}$ = 496 to IE$_{2}$ = 4560 — a factor of 9.2×. Between these two, sodium has lost its last valence electron and is now dipping into the core $2p^{6}$ shell. Removing a core electron is orders of magnitude harder than removing a valence electron, because the core is much closer to the nucleus and much more tightly held.

The position of the "big jump" tells you the group:

• Group 1 (e.g. Na): jump at IE$_{1}$ → IE$_{2}$.
• Group 2 (e.g. Mg): jump at IE$_{2}$ → IE$_{3}$.
• Group 13 (e.g. Al): jump at IE$_{3}$ → IE$_{4}$.
• Group 14 (e.g. Si): jump at IE$_{4}$ → IE$_{5}$.

In general, the first "big jump" appears at IE$_{n+1}$ where $n$ is the number of valence electrons.

The "second jump" — into a deeper shell

After the core electrons of shell $n-1$ are gone, the next electron comes from shell $n-2$ — an even deeper shell — and produces a second big jump. For sodium, this second jump comes at IE$_{9}$ → IE$_{10}$ (from $2s$ into $1s$): it rises from 28,930 to 141,360 kJ/mol, a 5× jump.

Why this matters for chemistry

The size of $\Delta_{\text{i}}H_{1}$ tells you how easily the element gives up its outermost electron — i.e., how metallic or reducing it is. The step from $\Delta_{\text{i}}H_{1}$ to $\Delta_{\text{i}}H_{2}$ tells you how willing the atom is to form multiply-charged cations:

• If the step is small (as for alkaline earths like Ca: $\Delta_{\text{i}}H_{1} = 590$, $\Delta_{\text{i}}H_{2} = 1145$), forming Ca$^{2+}$ is thermodynamically feasible.
• If the step is huge (as for alkali metals like Na: $\Delta_{\text{i}}H_{1} = 496$, $\Delta_{\text{i}}H_{2} = 4560$), forming Na$^{2+}$ is prohibitive — Na exists only as Na$^{+}$.

This single observation explains almost all the common ion charges you learned in earlier classes.

[JEE Tip] A classic JEE-style question: "Four successive IEs of an element are 578, 1817, 2745, and 11578 kJ/mol. Identify the group." — The big jump sits between IE$_{3}$ and IE$_{4}$, so the element is in Group 13 (aluminium, in this case). Learn to spot these.

Trend Map, Exam Traps, and Strategy

The master arrow map

Combine everything into one picture:

Every IE-ordering question you will ever see is solved with these two arrows + the five-factor checklist + the four known exceptions. Let's organize them as a one-page cheat-sheet.

Cheat-sheet table

Situation	Rule	Reason
Across a period (L → R)	IE increases	$Z_{\text{eff}}$ rises; radius shrinks
Down a group	IE decreases	new shell each step; $Z_{\text{eff}}$ roughly constant
Group 13 vs Group 2 (same period)	IE(Group 13) lower than IE(Group 2)	$p^{1}$ removal vs breaking stable $s^{2}$
Group 16 vs Group 15 (same period)	IE(Group 16) lower than IE(Group 15)	paired-$p$ repulsion + loss of $p^{3}$ stability
Noble gas of any period	highest IE in that period	fully-filled $p^{6}$; tight radius
Alkali metal of any period	lowest IE in that period	$s^{1}$ electron; big radius
IE$_{1}$ vs IE$_{2}$ for same atom	IE$_{2}$ always larger	smaller cation; higher $Z/e^{-}$
Across d-block (Sc → Zn)	rises only slightly	imperfect shielding by added d-electrons

The six most-tested exceptions

Put these at the top of your exam-prep flashcard stack:

1. Be > B (reason: $2s^{2}$ stable; $2p^{1}$ less penetrating)
2. N > O (reason: $2p^{3}$ half-filled stable; $2p^{4}$ has paired repulsion)
3. Mg > Al (reason: $3s^{2}$ stable; $3p^{1}$ less penetrating)
4. P > S (reason: $3p^{3}$ half-filled stable; $3p^{4}$ has paired repulsion)
5. H is an outlier of Group 1 (IE = 1312 — way above Li = 520, because H has tiny size and no shielding)
6. Ga ≥ Al (reason: d-block contraction; $3d^{10}$ shields $4p^{1}$ poorly)

Six classic exam traps

Trap 1. The transition-metal comparison trap. "Which has higher IE: K or Cu?" K and Cu are both in Period 4, but Cu experiences much larger effective nuclear attraction because of its much higher nuclear charge and the imperfect shielding by the filled $3d^{10}$ subshell. So Cu has higher IE than K.

Trap 2. "Lowest IE in the periodic table." The answer is Cs among stable elements. If the question allows francium, Fr is usually quoted as the lowest, but its value is less commonly used because of radioactivity and limited data.

Trap 3. Transition-metal ionization order. The 4s electrons are removed before the 3d electrons when a transition metal is ionized. So Fe's IE$_{1}$ and IE$_{2}$ remove the two $4s$ electrons; IE$_{3}$ starts removing a $3d$ electron.

Trap 4. "Highest IE in the Periodic Table." Many students answer F. Wrong: He has the highest first IE (2372 kJ/mol). Among Period-2 atoms, Ne is higher than F.

Trap 5. "IE$_{2}$ of Na vs IE$_{1}$ of Ne — which is larger?" Na$^{+}$ and Ne are isoelectronic (both [Ne], 10 electrons). IE$_{2}$ of Na removes an electron from Na$^{+}$ (which has Z = 11), and IE$_{1}$ of Ne removes an electron from Ne (which has Z = 10). Na$^{+}$ holds its electrons more tightly, so IE$_{2}$(Na) > IE$_{1}$(Ne).

Trap 6. "Ionization energy of the 3d-block increases gradually." It increases only slowly across Sc-to-Zn because the added 3d electrons do not change the effective attraction on the outer electron as dramatically as same-shell p-electrons do. The trend is upward overall, but much flatter than in the p-block.

Strategy summary

If you can (a) recite the master arrow map, (b) apply the five factors, (c) remember the four dips and the common traps — you will solve almost all Board- and JEE-Mains-level IE questions quickly.

Solved Examples

Example 1: Unit conversion

The first ionization enthalpy of hydrogen is 13.6 eV per atom. Express this in kJ per mole.

Solution:

Use the conversion 1 eV/atom = 96.485 kJ/mol.

$$\Delta_{\text{i}}H_{1}(\text{H}) = 13.6 \times 96.485 = \mathbf{1312.2\ \text{kJ/mol}}$$

This matches the standard NCERT value of 1312 kJ/mol.

Takeaway: when numerical problems give IEs in different units, always convert to a single system before comparing.

Example 2: Explain — Be > B

Explain why the first ionization enthalpy of boron (801 kJ/mol) is lower than that of beryllium (899 kJ/mol), despite B having a higher nuclear charge.

Solution:

Electronic configurations:

• Be: $1s^{2}\,2s^{2}$ — a completely filled $2s$ sub-shell.
• B: $1s^{2}\,2s^{2}\,2p^{1}$ — one lonely $2p$ electron outside a filled $2s^{2}$ core.

Two reasons make B's ionization easier:

1. Sub-shell stability: Be's filled $2s^{2}$ is an extra-stable configuration, so removing an electron from it requires extra energy beyond what simple $Z_{\text{eff}}$ would predict.
2. Lower orbital penetration: a $2p$ orbital penetrates toward the nucleus less effectively than a $2s$ orbital at the same $n$. So the $2p^{1}$ electron in B feels a lower effective nuclear charge than the $2s$ electron in Be.

Both effects make B's outermost electron easier to remove, lowering its IE despite the higher $Z$.

Takeaway: this is the first of the two classical "group 13 dips" — and is the direct cause of the similar Mg > Al anomaly in Period 3.

Example 3: Explain — N > O

The first ionization enthalpy of oxygen (1314 kJ/mol) is lower than that of nitrogen (1402 kJ/mol). Explain.

Solution:

Electronic configurations:

• N: $1s^{2}\,2s^{2}\,2p^{3}$ — half-filled $2p$ sub-shell, all three electrons unpaired and spin-parallel.
• O: $1s^{2}\,2s^{2}\,2p^{4}$ — one of the $2p$ orbitals now has two paired electrons.

Two reasons oxygen ionizes more easily:

1. Loss of half-filled stability: nitrogen has the exceptionally stable half-filled $2p^{3}$ configuration, which maximises exchange energy. Removing N's electron destroys this extra-stable state, making N's IE higher than expected.
2. Paired-electron repulsion: oxygen's fourth $2p$ electron shares an orbital with one of the others, and the pair experience strong mutual repulsion. Removing one of the paired electrons relieves this repulsion — and leaves behind the stable half-filled $2p^{3}$ configuration. Both effects push O's IE lower.

Net result: $\Delta_{\text{i}}H_{1}(\text{O}) < \Delta_{\text{i}}H_{1}(\text{N})$ by about 90 kJ/mol.

Takeaway: the same pattern repeats at P > S. Anywhere a half-filled $p^{3}$ is followed by a $p^{4}$, the second element has a lower IE than the first.

Example 4: Identify the group from successive IEs

The first four ionization enthalpies of an element X are 577, 1817, 2745, and 11578 kJ/mol. Identify the group to which X belongs.

Solution:

Look for the big jump in successive IEs — this tells us when we have moved from valence to core electrons.

Step	kJ/mol	Jump ratio from previous
IE$_{1}$	577	—
IE$_{2}$	1817	3.1×
IE$_{3}$	2745	1.5×
IE$_{4}$	11578	4.2× ← big jump

The big jump sits between IE$_{3}$ and IE$_{4}$: after 3 successive removals, the next electron is much harder to take. This means X has 3 valence electrons; the 4th removal dips into a core shell.

Three valence electrons → Group 13.

The actual element is Al ($Z = 13$, configuration $3s^{2}\,3p^{1}$). Check: losing all three valence electrons gives [Ne] (noble gas core), and the next step (IE$_{4}$) would start stripping the $2p^{6}$ core — hence the huge jump.

Takeaway: the position of the first big jump in successive IEs tells you the number of valence electrons.

Example 5: Compare — isoelectronic pair

Compare $\Delta_{\text{i}}H_{2}$ of Na and $\Delta_{\text{i}}H_{1}$ of Ne. Which is larger, and why?

Solution:

Note first that Na$^{+}$ and Ne are isoelectronic — both have 10 electrons with [Ne] = $1s^{2}\,2s^{2}\,2p^{6}$.

• $\Delta_{\text{i}}H_{2}(\text{Na})$ removes an electron from Na$^{+}$ (10 electrons, $Z = 11$).
• $\Delta_{\text{i}}H_{1}(\text{Ne})$ removes an electron from Ne (10 electrons, $Z = 10$).

Same electron configuration; Na$^{+}$ has one more proton. Within an isoelectronic set, the species with higher $Z$ binds its electrons more tightly:

$$\Delta_{\text{i}}H_{2}(\text{Na}) > \Delta_{\text{i}}H_{1}(\text{Ne})$$

Actual values: 4560 kJ/mol (Na IE$_{2}$) vs 2081 kJ/mol (Ne IE$_{1}$). Na's IE$_{2}$ is much larger than Ne's IE$_{1}$.

Takeaway: isoelectronic logic applies to ionization too. Larger $Z$ → harder to ionize.

Example 6: Ordering a mixed set

Arrange in increasing order of first ionization enthalpy: O, N, F, Ne, Cl, Ar.

Solution:

Group the elements:

• Period 2: N (1402), O (1314), F (1681), Ne (2081)
• Period 3: Cl (1251), Ar (1521)

Within each period:

• Period 2: O < N < F < Ne
• Period 3: Cl < Ar

Now compare across periods using the values. Final increasing order:

$$\text{Cl} < \text{O} < \text{N} < \text{Ar} < \text{F} < \text{Ne}$$

Takeaway: when elements come from different periods and groups, the safest way is to combine periodic logic with the known anomaly values.

Example 7: Energy to ionize 1 g of atoms

How much energy, in kJ, is required to ionize 1.00 g of gaseous hydrogen atoms? (IE$_{1}$(H) = 1312 kJ/mol; atomic mass of H = 1.008 g/mol.)

Solution:

Use:

$$E = n \times \Delta_{\text{i}}H_{1}$$

Moles of H atoms in 1.00 g:

$$n = \frac{1.00\ \text{g}}{1.008\ \text{g/mol}} = 0.992\ \text{mol}$$

Energy needed:

$$E = 0.992 \times 1312 = \mathbf{1302\ \text{kJ}}$$

Takeaway: large ionization enthalpies still need to be scaled by the number of moles present.

Example 8: IE order — alkali metals

Arrange in decreasing order of first ionization enthalpy: Li, Na, K, Rb, Cs.

Solution:

All five are Group 1. Going down a group, IE decreases because atomic radius grows and shielding increases.

Element	IE$_{1}$ / kJ mol$^{-1}$
Li	520
Na	496
K	419
Rb	403
Cs	376

Decreasing order:

$$\text{Li} > \text{Na} > \text{K} > \text{Rb} > \text{Cs}$$

Example 9: Apparent anomaly — IE(Ga) ≥ IE(Al)

Explain why the first ionization enthalpy of gallium (579 kJ/mol) is approximately equal to that of aluminium (577 kJ/mol), even though Ga is in Period 4 and Al is in Period 3.

Solution:

Naively, moving from Al (Period 3) to Ga (Period 4) should lower the IE, because a whole new shell is added. The actual values are almost identical.

The reason is the d-block contraction. Between Al and Ga lies the entire 3d-series (Sc through Zn). The 10 added $3d^{10}$ electrons shield Ga's outer $4p^{1}$ electron poorly, so $Z_{\text{eff}}$ for Ga's valence electron is much higher than naive shell-counting would predict.

The higher $Z_{\text{eff}}$ cancels out the expansion of the valence shell, leaving Ga's IE roughly equal to Al's.

Takeaway: whenever a group crosses the transition metals (e.g. Al→Ga), expect a flat or slightly upward IE anomaly.

Example 10: Successive IEs of magnesium

The first four successive IEs of magnesium (kJ/mol) are 738, 1451, 7733, 10540. Comment on the pattern and explain the "big jump."

Solution:

Step	kJ/mol	Ratio to previous
IE$_{1}$	738	—
IE$_{2}$	1451	2.0×
IE$_{3}$	7733	5.3× ← big jump
IE$_{4}$	10540	1.4×

Magnesium has $Z = 12$, configuration $1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}$. Its valence shell has two electrons.

• IE$_{1}$ and IE$_{2}$ remove the two $3s$ electrons.
• IE$_{3}$ must now remove a core $2p$ electron, which is much more tightly held.

Hence the large jump between IE$_{2}$ and IE$_{3}$.

Takeaway: a big jump after the second IE is the fingerprint of a Group-2 element.

Example 11: IE within isoelectronic species

Arrange in increasing order of the energy required to remove one electron from: Na$^{+}$, Mg$^{2+}$, Al$^{3+}$, Ne, F$^{-}$, O$^{2-}$.

Solution:

All six species are isoelectronic (10 electrons, [Ne]). The energy required to remove one electron increases with nuclear charge $Z$.

Species	$Z$
O$^{2-}$	8
F$^{-}$	9
Ne	10
Na$^{+}$	11
Mg$^{2+}$	12
Al$^{3+}$	13

So the increasing order is:

$$\text{O}^{2-} < \text{F}^{-} < \text{Ne} < \text{Na}^{+} < \text{Mg}^{2+} < \text{Al}^{3+}$$

Takeaway: same configuration, larger nuclear charge → more tightly bound electrons.

Example 12: IE reasoning — "H is Group 1 but doesn't fit"

Hydrogen is placed in Group 1 on most periodic tables, yet its first ionization enthalpy (1312 kJ/mol) is far higher than any other alkali metal. Explain.

Solution:

The alkali metals (Li, Na, K, Rb, Cs) all have low IEs because they have large atomic radii and substantial inner-shell shielding. Hydrogen is the opposite:

• Tiny size
• No inner electrons, so zero shielding
• A very tightly held $1s$ electron

All three factors make hydrogen's electron much harder to remove than the valence electron of any alkali metal.

Takeaway: H is placed above Group 1 by configuration, but its ionization enthalpy reflects its unique atomic structure.