ChemistryCBSE Class 11 · Classification of Elements and Periodicity in Properties

Ionisation Enthalpy and Electron Gain Enthalpy Class 11

Updated 17 October 2026 · Fact-checked

The ionisation enthalpy is the energy to remove an electron from a gaseous atom, and it generally increases across a period and decreases down a group, with dips at group 13 (after 2) and group 16 (after 15). A large jump between successive ionisation enthalpies shows that the next electron comes from an inner shell, which fixes the number of valence electrons. Electron gain enthalpy is the enthalpy change on adding an electron to a gaseous atom; halogens have the most negative values.

Ionisation enthalpy rises across period 2, but B dips below Be and O dips below N because of orbital type and p⁴ repulsion.

Illustration drawn for QuestionBuddy.

Understand Ionisation Enthalpy and Electron Gain Enthalpy

**Ionisation enthalpy ( $Δ_{i} H$ ).** The energy needed to remove the outermost electron from an isolated gaseous atom, $X (g) \to X^{+} (g) + e^{-}$ (kJ mol⁻¹). It is always endothermic. Factors: nuclear charge (higher means higher $Δ_{i} H$ ), atomic size (larger means lower), shielding by inner electrons (greater means lower), and the type of orbital ( $s > p > d > f$ for the same shell, because s electrons penetrate more).

Across a period it generally increases; down a group it decreases.
*Anomalies in period 2/3:* Be > B (the 2p electron is easier to remove than a filled 2s electron), N > O (half-filled $p^{3}$ is stable; $p^{4}$ has electron-electron repulsion); likewise Mg > Al and P > S.

Successive ionisation enthalpies. $Δ_{i} H_{2} > Δ_{i} H_{1}$ because the electron comes from a more positive ion. A very large jump between $Δ_{i} H_{n}$ and $Δ_{i} H_{n + 1}$ shows that the $(n + 1)$ th electron comes from a stable inner shell, so the atom has $n$ valence electrons (group number $= n$ for groups 1–2, $n + 10$ for groups 13–18).

**Electron gain enthalpy ( $Δ_{e g} H$ ).** The enthalpy change when a gaseous atom accepts an electron, $X (g) + e^{-} \to X^{-} (g)$ . A negative value means energy is released. It becomes more negative across a period (up to group 17) and generally less negative down a group; but in group 17 the second period element F has a less negative value than Cl because of its small size and strong electron-electron repulsion in the compact 2p subshell. Noble gases have large positive values (the electron must enter a new shell), and N, Be, Mg have small or positive values (stable half-filled or filled subshell). Second and later electron gain (e.g. O⁻ → O²⁻) is always endothermic.

Using the values. Ionising $n$ moles of atoms needs $n \times Δ_{i} H$ . The minimum photon energy per atom $= Δ_{i} H / N_{A}$ , so $λ_{ma x} = h c N_{A} / Δ_{i} H$ .

Ionisation Enthalpy and Electron Gain Enthalpy: formulas and key facts

Ionisation: $X (g) \to X^{+} (g) + e^{-}$; $Δ_{i} H > 0$ .
Electron gain: $X (g) + e^{-} \to X^{-} (g)$; $Δ_{e g} H$ usually negative.
Energy per atom: $E = \frac{Δ _{i} H}{N _{A}}$; kJ/mol to J/atom.
Maximum wavelength: $λ_{ma x} = \frac{h c N _{A}}{Δ _{i} H}$; Photon able to ionise.
Total energy for n moles: $Q = n \sum Δ_{i} H_{k}$; For removal of several electrons.
Valence electrons: $big jump after Δ_{i} H_{n} \Rightarrow n valence electrons$; Successive data.

How to solve Ionisation Enthalpy and Electron Gain Enthalpy questions

Compare the data with the electronic configuration: a jump signals a new inner shell, and a dip signals a filled or half-filled subshell.

1Write the electronic configuration of each element.
2For trends, compare nuclear charge, size and shielding across or down.
3Look for the exceptions: filled s (Be, Mg) and half-filled p (N, P) are harder to ionise than their right-hand neighbours.
4For successive ionisation enthalpies, find the biggest ratio between neighbours.
5For energy numericals, convert kJ/mol to J/atom using N_A.
6Quote signs correctly for electron gain enthalpy (energy released is negative).

Ionisation Enthalpy and Electron Gain Enthalpy in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Street lamp metals

Suppose a sodium lamp works by exciting sodium atoms, and a student wonders how easily Na loses an electron: IE₁ = 496 kJ/mol.

Find the energy to ionise 0.020 mol of gaseous sodium atoms.

Try it first, then show the solutionSolution

Q = n ΔiH = 0.020 × 496 = 9.92 kJ.

Answer: 9.92 kJ.

Memory hook: Energy scales with moles.

Chlorine and the first row

Suppose a student compares the electron gain enthalpy of Cl (−349 kJ/mol) and F (−328 kJ/mol).

Which releases more energy per mole, and why?

Try it first, then show the solutionSolution

Cl releases more energy (−349 vs −328 kJ/mol).
F is very small; the added electron suffers strong repulsion from the compact 2p electrons, while Cl's larger 3p subshell accommodates it better.

Answer: Chlorine; F is too compact.

Memory hook: Size can reduce electron gain enthalpy.

Tricks to remember Ionisation Enthalpy and Electron Gain Enthalpy

Short memory aids. Say each one aloud once, then use it on a question the same day.

Jump means inner shell
Look for the big ratio in successive IE values.
Use it when: Group identification.
Be > B, N > O
Filled and half-filled subshells resist ionisation.
Use it when: Period 2 anomalies.
Cl beats F
Electron gain enthalpy: Cl is the most negative.
Use it when: Halogen question.
Positive EGE
Noble gases, Be, Mg, N have positive or small values.
Use it when: Exceptions list.

Ionisation Enthalpy and Electron Gain Enthalpy: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Period 3 ordering · 3 marks · Short answerBoard level

Arrange Na, Mg, Al, Si, P and S in increasing order of first ionisation enthalpy, and explain the two exceptions to the general trend.

Try it first, then show the solutionSolution

Order: Na < Al < Mg < Si < S < P. (1)
Mg > Al: Mg's 3s² is a filled subshell; Al loses the 3p¹ electron, which is higher in energy and shielded by 3s. (1)
P > S: P has a half-filled 3p³; S (3p⁴) loses a paired electron, relieving electron repulsion. (1)

Answer: Na < Al < Mg < Si < S < P.

Example 2: Valence electrons from successive data · 3 marks · Short answerHarder board level

Suppose an element E shows successive ionisation enthalpies (kJ/mol): 1090, 2350, 4620, 6220, 37800, 47300. Find the number of valence electrons, the group and the formulas of its oxide and chloride.

Try it first, then show the solutionSolution

The largest jump (6220 to 37800) comes after the 4th electron: 4 valence electrons. (1)
Group 14 (valence 4). (1)
Oxide EO₂ and chloride ECl₄. (1)

Answer: 4 valence electrons; group 14; EO₂ and ECl₄.

Example 3: Energy for double ionisation · 3 marks · Short answerHarder board level

The first and second ionisation enthalpies of magnesium are 738 and 1451 kJ/mol. Find the energy needed to convert 2.4 g of gaseous magnesium atoms (24 g/mol) into Mg²⁺ ions, and the energy per ion in joules.

Try it first, then show the solutionSolution

n = 2.4/24 = 0.10 mol. (1)
Q = 0.10 × (738 + 1451) = 218.9 kJ. (1)
Per ion: 2189 × 10³/6.022 × 10²³ = 3.64 × 10⁻¹⁸ J. (1)

Answer: 218.9 kJ; 3.64 × 10⁻¹⁸ J per ion.

Example 4: Electron gain enthalpy of halogens · 4 marks · Long answerHarder board level

The electron gain enthalpy of chlorine is −349 kJ/mol. (a) Find the energy released when 3.55 g of chlorine atoms (Cl = 35.5) each gain an electron. (b) Explain why chlorine has a more negative value than fluorine. (c) Explain why nitrogen has a near-zero value despite its high nuclear charge.

Try it first, then show the solutionSolution

(a) n = 3.55/35.5 = 0.100 mol; energy released = 0.100 × 349 = 34.9 kJ. (1)
(b) Fluorine's small 2p subshell causes strong electron-electron repulsion for the added electron; Cl's larger 3p subshell is less crowded. (1)
(c) Nitrogen has a stable half-filled 2p³ configuration. (1)
An added electron must pair up in a p orbital, which costs energy, so the value is close to zero. (1)

Answer: (a) 34.9 kJ; (b) the small 2p subshell of F; (c) the half-filled 2p³ of N.

Example 5: Identifying consecutive elements · 4 marks · Long answerHOTS / competency

Four elements W, X, Y and Z have consecutive atomic numbers, and their first ionisation enthalpies are 1681, 2081, 496 and 738 kJ/mol. (a) Identify the elements. (b) Explain the large drop from X to Y. (c) Predict how the second ionisation enthalpy of Y compares with its first. (d) Write the formula of the compound between Y and W.

Try it first, then show the solutionSolution

(a) F, Ne, Na, Mg (Z = 9 to 12). (1)
(b) X is a noble gas with a complete shell; Y begins a new shell, and its 3s electron is far from the nucleus and shielded. (1)
(c) IE₂ is much larger than IE₁ for Na (about nine times) because the electron now comes from the stable Ne-like core. (1)
(d) Na and F form NaF. (1)

Answer: (a) F, Ne, Na, Mg; (d) NaF.

Where students lose marks in Ionisation Enthalpy and Electron Gain Enthalpy

The slips examiners see most often, and what to write instead.

Mistake: Calling ionisation enthalpy exothermic.
Mixing with electron gain enthalpy.
Do this: Removing an electron always needs energy.
Usually costs: The sign mark.
Mistake: Saying N has a more negative EGE than O.
Using the general trend.
Do this: N has a stable half-filled p³; its value is near zero.
Usually costs: Exception mark.
Mistake: Misreading the big jump in successive IE values.
Taking the first big value.
Do this: The valence number is the count of electrons before the largest jump.
Usually costs: The group answer.
Mistake: Forgetting to divide by N_A for a single atom.
Unit slip.
Do this: Energy per atom = kJ/mol × 10³ / N_A.
Usually costs: The numerical.

Exam tips

Write the configuration of each element before explaining the trend.
Name the factor (nuclear charge, size, shielding, stable subshell) in each explanation.
Give the exceptions explicitly with the elements involved.
Convert kJ/mol to J/atom with N_A.
Use signs consistently for electron gain enthalpy.

Practice questions on Ionisation Enthalpy and Electron Gain Enthalpy

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Among F, Cl, Br and I, the element with the most negative electron gain enthalpy is
(a) F (b) Cl (c) Br (d) I
Hint
Small size lowers F's value.
Answer
(b) Cl.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The second ionisation enthalpy of sodium is much greater than its first.
Reason (R): Removal of the second electron breaks into the stable noble-gas core of Na⁺.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Na⁺ is Ne-like.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
Explain why the first ionisation enthalpy of beryllium (900 kJ/mol) is greater than that of boron (801 kJ/mol).
Hint
Compare 2s² and 2p¹.
Answer
Be has the filled 2s² configuration, which is stable; the 2p¹ electron of B is higher in energy and shielded by the 2s electrons, so it is removed more easily.
Q4 · 3 marks · Short answerHarder board level
The first ionisation enthalpy of lithium is 520 kJ/mol. Find the maximum wavelength of light that can ionise a gaseous lithium atom (h = 6.626 × 10⁻³⁴ J s, c = 3.0 × 10⁸ m/s, N_A = 6.022 × 10²³). In which region does it lie?
Hint
E per atom = ΔiH/N_A.
Answer
E = 520 × 10³/6.022 × 10²³ = 8.64 × 10⁻¹⁹ J; λ = hc/E = 2.30 × 10⁻⁷ m = 230 nm, ultraviolet.
Q5 · 4 marks · Long answerHOTS / competency
(a) Explain the dips in first ionisation enthalpy at boron and at oxygen in the second period. (b) The second ionisation enthalpies of Li and Be are 7298 and 1757 kJ/mol. Find the ratio and explain it. (c) Which of Li and Be forms the more stable 2+ ion?
Hint
Think of the configuration of the ion.
Answer
(a) B (2s²2p¹): the 2p electron is easier to remove than a 2s electron; O (2p⁴): the paired electron is removed more easily than from the half-filled N. (b) 7298/1757 = 4.2; Li⁺ is 1s², a noble-gas core, so removal is very hard; Be⁺ (2s¹) loses an outer electron. (c) Be, since only two electrons are removed and the 2+ ion has the stable 1s² core.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Trend of IE across a period?Show answer
Generally increases.
2.Exceptions in IE?Show answer
Be > B, N > O (and Mg > Al, P > S).
3.Most negative EGE?Show answer
Chlorine.
4.Why does IE₂ exceed IE₁?Show answer
The electron is removed from a positive ion.
5.What does a big jump in IE show?Show answer
The next electron is from an inner shell.
6.Units of ΔiH?Show answer
kJ mol⁻¹.

Ionisation Enthalpy and Electron Gain Enthalpy: frequently asked questions

Why is ionisation enthalpy always positive?

Energy must be supplied to overcome the attraction of the nucleus.

Why does IE decrease down a group?

The outer electron is farther from the nucleus and more shielded.

What is the difference between electron gain enthalpy and electron affinity?

Electron gain enthalpy is the enthalpy change on adding an electron; electron affinity is the magnitude of the energy released (opposite sign convention).

Why do noble gases have positive electron gain enthalpy?

The added electron must enter a new, higher shell.

Why is the second electron gain of oxygen endothermic?

The O⁻ ion repels the incoming electron.