Key Fact
Ionization enthalpy trends
Energy required to remove electron from gaseous atom. Across period: increases. Down group: decreases. Anomalies: B<Be (s vs p removal); O<N (paired vs half-filled p).
-- NCERT Class 11 Chemistry, Ch. 3, p. 89Ionization Enthalpy
8 MCQs3 revision cards9-step worked example
Source: NCERT Classification of Elements and Periodicity in PropertiesPYQ coverage: NEET 2024, 2026Official key: NTA-verifiedLast updated: 24 Sep 2026
Ionization Enthalpy, explained for NEET
Ionization Enthalpy Trends — The Two Anomalies That Cost Marks
Ionization enthalpy (IE₁) is the energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state. The general trend is straightforward: IE₁ increases across a period (rising nuclear charge, same shell) and decreases down a group (new shells, greater shielding). NCERT Class 11 Chemistry Chapter 3, page 89 states this directly.
The trap is in the word "increases." Students read it as monotonic increase — every element higher than the one before it across a period. That is wrong for Period 2, and NEET tests exactly this.
Two anomalies you must memorise:
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Be (1s² 2s²) > B (1s² 2s² 2p¹). Beryllium's fully filled 2s² subshell is more stable than boron's single 2p¹ electron. Removing that lone p-electron from boron is easier, so IE₁(Be) > IE₁(B), breaking the left-to-right increase.
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N (1s² 2s² 2p³) > O (1s² 2s² 2p⁴). Nitrogen's half-filled 2p³ configuration has exchange energy stabilisation. Oxygen has one electron paired in a 2p orbital; the inter-electronic repulsion in that pair makes it easier to remove, so IE₁(N) > IE₁(O).
These are not curiosities — they are the basis for a recurring NEET distractor pattern. When a question asks you to arrange Period 2 elements in order of IE₁, the wrong option almost always presents a smooth monotonic sequence. The correct answer shows the two dips: Li < B < Be and O < N.
For hydrogen-like (single-electron) species, IE is calculated exactly: IE = 13.6 × Z²/n² eV. This formula appears in comparison questions — e.g., "which has higher IE₁: He⁺ or Li²⁺?"
Watch-out: The anomalies are specific to first ionization enthalpy. Second and successive IEs follow different logic (removing electrons from increasingly positive ions). Don't extend the Be > B or N > O pattern to IE₂ without checking the electronic configuration of the resulting ion.
Can you answer these Ionization Enthalpy MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Ionization enthalpy of elements generally __________ across a period and __________ down a group.
Show answer and why every option is right or wrong
Answer: C. Across a period, effective nuclear charge increases with same principal shell → IE increases. Down a group, new shells and greater shielding → IE decreases. (NCERT Class 11 Chemistry Chapter 3, page 89.)
Why A is wrong: A is wrong because IE decreases down a group due to increasing atomic size and shielding, not increases.
Why B is wrong: B is wrong because IE increases across a period (rising Zeff, same shell), not decreases.
Why D is wrong: D is wrong because IE increases across a period — the decrease applies only down a group.
MCQ 2Easy RecallPractice
Ionization enthalpy is defined as the energy required to remove an electron from:
Show answer and why every option is right or wrong
Answer: A. The standard definition specifies an isolated gaseous atom in its ground state. (NCERT Class 11 Chemistry Chapter 3, page 89.)
Why B is wrong: B is wrong because the definition specifies the ground state, not an excited state — an excited electron is already partially removed energetically.
Why C is wrong: C is wrong because IE is defined for gaseous atoms, not solid-state atoms where lattice interactions complicate the measurement.
Why D is wrong: D is wrong because aqueous solution involves solvation energy contributions; IE is a gas-phase property of an isolated atom.
MCQ 3Easy RecallPractice
Which pair of elements demonstrates an anomaly in the expected trend of first ionization enthalpy across Period 2?
Show answer and why every option is right or wrong
Answer: B. Be (2s²) has higher IE₁ than B (2s² 2p¹) because the fully filled s-subshell is more stable. This breaks the expected left-to-right increase. (NCERT Class 11 Chemistry Chapter 3, page 89.)
Why A is wrong: A is wrong because IE₁(Li) < IE₁(Be) follows the expected increasing trend — no anomaly between these two.
Why C is wrong: C is wrong because IE₁(C) < IE₁(N) is the expected increasing trend. The anomaly occurs between N and O, not C and N.
Why D is wrong: D is wrong because IE₁(F) < IE₁(Ne) follows the expected trend — noble gases have the highest IE in their period.
MCQ 4Direct ApplicationPractice
Among the elements B, C, N, O, and F, which has the lowest first ionization enthalpy?
Show answer and why every option is right or wrong
Answer: A. Boron has configuration 2s² 2p¹. Its lone p-electron is easier to remove than the fully filled 2s² of Be, and it is the leftmost (lowest Zeff) among the listed elements. The IE₁ order is B (801) < C (1086) < O (1314) < N (1402) < F (1681 kJ/mol), making B the lowest. (Trap: students pick O because of the N > O anomaly, but O is still higher than B.)
Why B is wrong: B is wrong because although O has lower IE₁ than N (the N > O anomaly), O still has higher IE₁ than B. The anomaly makes O lower than expected, not the lowest overall. (Trap: IE anomaly Be > B and N > O.)
Why C is wrong: C is wrong because F, near the right end of Period 2 with high Zeff, has one of the highest IE₁ values — far higher than B.
Why D is wrong: D is wrong because N has half-filled 2p³ stability giving it higher IE₁ than both O and B.
MCQ 5Direct ApplicationPractice
IE₁(N) > IE₁(O) because:
Show answer and why every option is right or wrong
Answer: D. Nitrogen's three unpaired electrons in 2p³ (half-filled) enjoy exchange energy stabilisation. Oxygen's 2p⁴ has one paired electron with inter-electronic repulsion, making that electron easier to remove. Hence IE₁(N) > IE₁(O). (NCERT Class 11 Chemistry Chapter 3, page 89.)
Why A is wrong: A is wrong because nitrogen actually has a lower nuclear charge (Z = 7) than oxygen (Z = 8). The anomaly is not explained by Zeff alone — it is the subshell stability (half-filled 2p³) that overrides the Zeff effect. (Trap: assuming IE correlates only with Z.)
Why B is wrong: B is wrong because having fewer electrons does not inherently make removal harder. The explanation lies in the specific stability of the half-filled 2p³ configuration, not electron count.
Why C is wrong: C is wrong because oxygen has a smaller atomic radius than nitrogen (higher Zeff across period contracts the electron cloud). The anomaly is due to subshell electronic configuration, not atomic size.
MCQ 6Direct ApplicationPractice
Using IE = 13.6 × Z²/n² eV, calculate the ionization energy of He⁺ (Z = 2, n = 1).
Show answer and why every option is right or wrong
Answer: B. IE = 13.6 × (2)²/(1)² = 13.6 × 4 = 54.4 eV. He⁺ is a hydrogen-like species with one electron; the formula applies directly.
Why A is wrong: A is wrong because 13.6 eV is the IE of hydrogen (Z = 1, n = 1). For He⁺, Z = 2, so the Z² factor gives 4 × 13.6 = 54.4 eV. (Trap: forgetting to square Z.)
Why C is wrong: C is wrong because 27.2 = 13.6 × 2, which means Z was not squared. The formula requires Z², not Z. (Trap: using Z instead of Z².)
Why D is wrong: D is wrong because 6.8 = 13.6/2, which would correspond to n = √2 — not a valid quantum number. This likely arises from dividing by Z instead of multiplying by Z².
MCQ 7CalculationPractice
Arrange the following in order of increasing first ionization enthalpy: Li, Be, B, N, O.
Show answer and why every option is right or wrong
Answer: A. The general trend is increasing IE₁ across Period 2, but with two anomalies: Be > B (fully filled 2s² stability) and N > O (half-filled 2p³ stability). The correct order: Li (lowest Zeff) < B (2p¹, lower than Be) < Be (2s² stable) < O (2p⁴, lower than N) < N (2p³ half-filled, highest among these five). Both anomalies must be applied simultaneously.
Why B is wrong: B is wrong because it places O < Be, but IE₁(O) > IE₁(Be). Oxygen, despite the N > O anomaly, still has higher Zeff and higher IE₁ than beryllium. (Trap: over-applying the N > O anomaly to push O too low.)
Why C is wrong: C is wrong because it places Be < B, which is the monotonic-increase error. Be (2s² fully filled) has higher IE₁ than B (2s² 2p¹). This is the most common distractor — it ignores the Be > B anomaly. (Trap: assuming monotonic increase across period.)
Why D is wrong: D is wrong because it places B < Li, but IE₁(B) > IE₁(Li). Boron has higher Zeff than lithium despite having a lone 2p electron.
MCQ 8Concept TrapPractice
A student claims that first ionization enthalpy increases strictly from left to right across Period 2 without exception. Which specific pair of consecutive elements disproves this claim?
Show answer and why every option is right or wrong
Answer: B. IE₁(Be) > IE₁(B), which means moving from Be to B (left to right), IE₁ decreases. This directly disproves a strict monotonic increase. The Be > B anomaly arises from the extra stability of beryllium's fully filled 2s² subshell compared to boron's 2s² 2p¹.
Why A is wrong: A is wrong because IE₁(Li) < IE₁(Be) — this pair follows the expected increasing trend and does not disprove the claim.
Why C is wrong: C is wrong because IE₁(C) < IE₁(N) — this pair follows the expected increasing trend. (The anomaly is between N and O, the next pair.)
Why D is wrong: D is wrong because IE₁(N) < IE₁(F) — this pair follows the expected increasing trend. The student would need the N-O pair or Be-B pair to see an anomaly.
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Ionization Enthalpy: quick recall before you leave
How do you solve a Ionization Enthalpy question? A worked example
- 1
Given
He⁺ (Z = 2) with electron in n = 1. Li²⁺ (Z = 3) with electron in n = 1. Both are hydrogen-like (single-electron) species.
- 2
Required
Which species has higher ionization energy, and by what factor?
- 3
Concept
For hydrogen-like atoms, ionization energy depends on the square of the nuclear charge and inversely on the square of the principal quantum number. Higher Z means a stronger hold on the electron.
- 4
Formula
IE = 13.6 × Z² / n² eV
- 5
Substitution
IE(He⁺) = 13.6 × (2)² / (1)² = 13.6 × 4 / 1
IE(Li²⁺) = 13.6 × (3)² / (1)² = 13.6 × 9 / 1 - 6
Calculation
IE(He⁺) = 54.4 eV
IE(Li²⁺) = 122.4 eV
Ratio: IE(Li²⁺) / IE(He⁺) = 122.4 / 54.4 = 9/4 = 2.25
Note on exact values: 13.6 eV is the standard reference value for hydrogen's ground-state IE. Z and n are exact integers (nuclear charge and quantum number respectively). These exact values do not limit the significant figures of the result. - 7
Final answer
Li²⁺ has higher IE than He⁺ by a factor of 9/4 (= 2.25). IE(Li²⁺) = 122.4 eV; IE(He⁺) = 54.4 eV.
- 8
Common trap
Students sometimes forget to square Z, computing IE(Li²⁺) = 13.6 × 3 = 40.8 eV instead of 13.6 × 9 = 122.4 eV. The Z² dependence is critical — IE scales with the square of nuclear charge.
- 9
Similar NEET-style question
"Among H, He⁺, Li²⁺, and Be³⁺, arrange in order of increasing ionization energy." Answer: H < He⁺ < Li²⁺ < Be³⁺ (follows Z² scaling: 1 < 4 < 9 < 16, all with n = 1).
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What to remember before solving Ionization Enthalpy questions
Which Ionization Enthalpy formulas do you need for NEET?
Ionization energy of hydrogen-like atom
Energy required to ionize an electron from the n-th shell of hydrogen-like atom.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Z | nuclear charge | - |
| n | quantum number | - |
Valid when
- One-electron atom
- Non-relativistic
Where do students lose marks on Ionization Enthalpy?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Inorganic Exception
Student expects monotonic increase in IE across period. Anomalies: Be(s²) > B(s²p¹); N(p³ half-filled) > O(p⁴).
When it triggers
Compare IE values across period 2 (Li, Be, B, C, N, O, F).
How to avoid
Be > B (s² stable; B's p¹ easier to remove). N > O (N has p³ half-filled stability; O loses one to attain p³). Memorise these two anomalies.
Root cause: concept gap
Correction
Be>B (s² stability); N>O (N's p³ half-filled stability). Memorise these two anomalies in period 2.
More in Classification of Elements and Periodicity in Properties: 2 exam traps and mistakes from its other lessons.
Ionization Enthalpy questions from past NEET papers
2 questions from NEET 2024, 2026. Answers verified against NTA official keys.
NEET 2026
1Both A and R are correct and R is the correct explanation of A.
2Both A and R are correct and R is NOT the correct explanation of A.
3A is correct but R is not correct.
4A is not correct but R is correct.
NTA Answer: Option 1(final)
NEET 2024Revised key
1Li < Be < B < C < N
2Li < B < Be < C < N
3Li < Be < C < B < N
4Li < Be < N < B < C
NTA Answer: Option 2(revised_final)
All 10 past-paper questions from Classification of Elements and Periodicity in Properties →
How does NEET ask about Ionization Enthalpy?
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Periodic trends comparison — atomic/ionic radii, ionization energy, electron gain enthalpy, electronegativity.
Direct ApplicationMedium
Common distractors
swapped classes
Tempts surface-level recall.
Sources
NCERT refs: Class 11 Chemistry Chapter 3, p.89
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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