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. 16Ionization Enthalpy
8 MCQs1 revision card9-step worked example
Source: NCERT Classification of Elements and Periodicity in PropertiesPYQ coverage: NEET 2021, 2024Official key: NTA-verifiedLast reviewed: May 2026
Lesson
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 16 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:
-
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.
-
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.
Practice 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.
MCQ 2Easy RecallPractice
Ionization enthalpy is defined as the energy required to remove an electron from:
MCQ 3Easy RecallPractice
Which pair of elements demonstrates an anomaly in the expected trend of first ionization enthalpy across Period 2?
MCQ 4Direct ApplicationPractice
Among the elements B, C, N, O, and F, which has the lowest first ionization enthalpy?
MCQ 5Direct ApplicationPractice
IE₁(N) > IE₁(O) because:
MCQ 6Direct ApplicationPractice
Using IE = 13.6 × Z²/n² eV, calculate the ionization energy of He⁺ (Z = 2, n = 1).
MCQ 7CalculationPractice
Arrange the following in order of increasing first ionization enthalpy: Li, Be, B, N, O.
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?
Quick recall before you leave
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). ---
Before solving, remember these
Formulas
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
Exam Traps & Common Mistakes
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 includes inert-gas radius in atomic-radius trends. But inert gases use van der Waals radius (much larger than covalent), making 'monotonic decrease across period' look broken.
When it triggers
Atomic radius comparison includes a noble gas or trends across period 2/3.
How to avoid
Compare like with like: covalent radii for non-noble gases. Noble gas radii are van der Waals (no covalent bond). Don't compare noble-gas radius directly to halogen.
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.
Root cause: concept gap
Correction
Don't compare different radius types. Noble gases use vdW radius (much larger); halogens use covalent radius. Compare like-with-like.
Past Year Questions
3 questions from NEET 2021, 2024. Answers verified against NTA official keys.
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)
NEET 2024Revised key
1Si < C < N < O < F
2Si < C < O < N < F
3O < F < N < C < Si
4F < O < N < C < Si
NTA Answer: Option 1(revised_final)
NEET 2021
1CO < SiO : Increasing 2 2 < SnO < PbO oxidizing power 2 2
2HF < HCl : Increasing acidic < HBr < HI strength
3H O < H S : Increasing pK 2 2 a < H Se < H Te values 2 2
4NH < PH : Increasing 3 3 < AsH < SbH acidic character 3 3
NTA Answer: Option 3(final)
How NEET usually asks this
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.16
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