Q.63 Chemistry · Section A ⭐ 360° Option Elimination Active

NEET 2024 Chemistry Question 63 Solution

Topic: Classification of Elements and Periodicity in Properties · Official NTA Booklet Code T1

Question

Arrange the following elements in increasing order of first ionization enthalpy: Li, Be, B, C, N Choose the correct answer from the options given below:

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Verified Against NTA Final Answer Key

Option (4) : Li < B < Be < C < N

📚 Classification of Elements and Periodicity in Properties Easy

📋 Given Parameters & Problem Setup

Elements Li, Be, B, C, N (Period 2)

Required order Increasing order of first ionization enthalpy (Δ_i H_1)

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The 30-Second Exam Hall Shortcut: ⚡ Period 2 Ionization Enthalpy Anomaly Rule

⏱️ Save ~30 seconds

Across Period 2, IE generally increases with Z_eff, BUT Be (2s² full subshell) > B (2s² 2p¹ single outer p-electron)! Order: Li < B < Be < C < N!

1 General periodic trend across Period 2: IE increases from left to right.

2 Check subshell stability anomaly between Be (Z=4) and B (Z=5):

3 - Be: 1s² 2s² (fully filled stable 2s subshell, high penetration).

4 - B: 1s² 2s² 2p¹ (electron removed from higher energy 2p orbital, easily shielded by 2s).

5 - Hence, IE of Be (899 kJ/mol) > IE of B (801 kJ/mol)!

6 For C (Z=6, 1086 kJ/mol) and N (Z=7, 1402 kJ/mol half-filled 2p³), IE continues increasing.

7 Ascending sequence: Li (520) < B (801) < Be (899) < C (1086) < N (1402).

8 Option (4) in 5 seconds!

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Why You Can 100% Trust This Shortcut: NCERT Chemistry · Classification of Elements and Periodicity in Properties (Ionization enthalpy anomalies: Be vs B and N vs O).

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Electronic Configurations of Period 2 Elements

Writing ground-state electronic configurations: - Li (Z = 3): 1s² 2s¹ - Be (Z = 4): 1s² 2s² (completely filled 2s orbital) - B (Z = 5): 1s² 2s² 2p¹ - C (Z = 6): 1s² 2s² 2p² - N (Z = 7): 1s² 2s² 2p³ (half-filled, stable 2p subshell)

\text{Be: } [\text{He}]\,2s^2; \quad \text{B: } [\text{He}]\,2s^2\,2p^1

Step 2

Explanation of the Beryllium versus Boron Anomaly

Although nuclear charge increases from Be (+4) to B (+5), the 2p electron in boron is shielded from the nucleus by the inner 2s electrons and occupies a higher energy level than 2s electrons. Furthermore, removing an electron from Be disrupts a completely filled 2s² subshell with greater penetration power. As a consequence, it requires less energy to remove the 2p¹ electron of boron than the 2s² electron of beryllium:

\Delta_i H_1(\text{Be}) = 899\;\text{kJ/mol} > \Delta_i H_1(\text{B}) = 801\;\text{kJ/mol}

Step 3

Comprehensive Ordering Across the Period

Comparing all values: - Li: 520 kJ/mol - B: 801 kJ/mol - Be: 899 kJ/mol - C: 1086 kJ/mol - N: 1402 kJ/mol Arranging in strictly increasing order:

\text{Li} < \text{B} < \text{Be} < \text{C} < \text{N}

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360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 (1)

✗ Incorrect

Verdict & Scientific Reason:

Places B after C, which completely violates effective nuclear charge trends.

⚠️

Student Fallacy (Why candidates pick this trap): Random placement of boron.

2 (2)

✗ Incorrect

Verdict & Scientific Reason:

Places N before B and C.

⚠️

Student Fallacy (Why candidates pick this trap): Misunderstanding nitrogen’s half-filled extra stability.

3 (3)

✗ Incorrect

Verdict & Scientific Reason:

Classic straight-line periodic trend trap ignoring the 2s² penetration anomaly between Be and B.

⚠️

Student Fallacy (Why candidates pick this trap): Assuming ionization enthalpy increases monotonically across a period without subshell exceptions.

4 (4)

✓ 100% Correct

Verdict & Scientific Reason:

Correctly accounts for the Be > B subshell anomaly.

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Why this is airtight: None.

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NTA Trap Alert: Option (3) Monotonic Trend Trap

Option (3) arranges elements strictly by atomic number (Li < Be < B < C < N). This is the most common trap in periodic properties questions!

🛡️ How to avoid the -1 mark: Always flag the two famous Period 2 anomalies: Be > B (full 2s² vs 2p¹) and N > O (half-filled 2p³ vs 2p⁴).

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Core NEET Syllabus Standard

Chemistry Core Inorganic Chemistry · Classification of Elements and Periodicity in Properties Standard Benchmark

Core Scientific Principle Tested:

"First ionization enthalpy generally increases across a period, but Be is higher than B because the 2s electron has higher penetration and belongs to a completely filled subshell."

Syllabus Focus: Periodic Trends in Chemical Properties: Ionization Enthalpy

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

Boron dips below Beryllium’s full shell: Li < B < Be < C < N rings the bell!

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NEET 2024 Chemistry Question 63 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Period 2 Ionization Enthalpy Anomaly Rule

Step-by-Step Rigorous Workflow & Mathematical Proof

Electronic Configurations of Period 2 Elements

Explanation of the Beryllium versus Boron Anomaly

Comprehensive Ordering Across the Period

360° Option Elimination Engine Key Feature

NTA Trap Alert: Option (3) Monotonic Trend Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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