Q.70 Chemistry ⭐ 360° Option Elimination Active

NEET 2026 Chemistry Question 70 Solution

Topic: Classification of Elements and Periodicity in Properties · Official NTA Booklet Code 50

Question

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: The first ionization enthalpy of O is lower than that of N and F. Reason R: The loss of an electron from O leads to stable half-filled p orbital. In light of the above statements, choose the most appropriate answer from the options given below:

(1)Both A and R are correct and R is the correct explanation of A.
(2)Both A and R are correct and R is NOT the correct explanation of A.
(3)A is correct but R is not correct.
(4)A is not correct but R is correct.

✓

Verified Against NTA Final Answer Key

Option (1) : Both A and R are correct and R is the correct explanation of A.

📚 Classification of Elements and Periodicity in Properties Easy

📋 Given Parameters & Problem Setup

Assertion (A) The first ionization enthalpy of O is lower than that of N and F.

Reason (R) The loss of an electron from O leads to stable half-filled p orbital.

⚡

The 30-Second Exam Hall Shortcut: ⚡ Half-Filled Stability & Electron Pairing Penalty

⏱️ Save ~40 seconds

Oxygen is 2p⁴: losing 1 electron relieves pairing repulsion and yields a rock-solid half-filled 2p³ subshell!

1 Electronic configurations: N is 2s² 2p³ (stable half-filled); O is 2s² 2p⁴; F is 2s² 2p⁵.

2 Ionization values: N = 1402 kJ/mol, O = 1314 kJ/mol, F = 1681 kJ/mol.

3 Assertion A: O is indeed lower than BOTH N and F -> TRUE.

4 Reason R: In oxygen (2p⁴), two electrons share the same 2p_x orbital. Repulsion between these paired electrons makes it easier to remove one electron, resulting in a stable half-filled 2p³ configuration -> TRUE.

5 Does R explain A? Yes, the extra stability of 2p³ and release of pairing energy directly explains the dip in IE! Option (1) is locked!

🛡️

Why You Can 100% Trust This Shortcut: NCERT Periodicity textbook core concept: Oxygen has a lower first ionization enthalpy than nitrogen because removing an electron from 2p⁴ gives a stable half-filled 2p³ subshell.

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Analyze Assertion A (Ionization Enthalpy Comparison)

Across the second period from left to right, first ionization enthalpy generally increases with increasing nuclear charge. However, oxygen has an anomalous dip: - Nitrogen (Z = 7): Δ_i H₁ = 1402 kJ mol⁻¹ - Oxygen (Z = 8): Δ_i H₁ = 1314 kJ mol⁻¹ - Fluorine (Z = 9): Δ_i H₁ = 1681 kJ mol⁻¹ Thus, Δ_i H₁ of O is lower than that of both N and F.

\Delta_i H_1(\text{N}) > \Delta_i H_1(\text{O}) < \Delta_i H_1(\text{F})

• Assertion A is factually correct.

Step 2

Analyze Reason R (Electronic Origin of the Anomaly)

Examine the valence electron configurations: - Nitrogen: [He] 2s² 2p_x¹ 2p_y¹ 2p_z¹ (half-filled p subshell, highly stable, high exchange energy). - Oxygen: [He] 2s² 2p_x² 2p_y¹ 2p_z¹ (contains one pair of electrons in the 2p_x orbital). In oxygen, the two electrons in the 2p_x orbital repel each other strongly. Removing one electron relieves this interelectronic repulsion and produces the symmetric, exchange-stabilized half-filled 2p³ configuration of O⁺.

\text{O } (2s^2 2p^4) \xrightarrow{-e^-} \text{O}^+ (2s^2 2p^3) \quad (\text{Stable half-filled subshell})

• Reason R is factually correct.

Step 3

Evaluate Reason as the Proper Scientific Explanation

Because the product ion O⁺ achieves a stable half-filled subshell and interelectronic repulsion in 2p⁴ is relieved, less energy is required to ionize oxygen than nitrogen. Therefore, Reason R directly explains Assertion A.

\text{Both A and R are correct and R is the correct explanation of A (Option 1)}

• Matches Option (1).

⭐

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 A

✓ 100% Correct

Verdict & Scientific Reason:

Exact scientific justification matching NCERT Periodicity curriculum.

💡

Why this is airtight: None; textbook mastery of atomic structure anomalies.

2 B

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. R is the precise cause of the anomalous ionization dip.

⚠️

Student Fallacy (Why candidates pick this trap): Failing to link electronic configuration stability to ionization energy magnitude.

3 C

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Reason R correctly states that O⁺ has a stable half-filled 2p³ subshell.

⚠️

Student Fallacy (Why candidates pick this trap): Doubting the stability of half-filled subshells.

4 D

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Assertion A is a well-established experimental fact.

⚠️

Student Fallacy (Why candidates pick this trap): Assuming ionization energy strictly increases across a period without exceptions.

⚠️

NTA Trap Alert: Assuming Ionization Energy Increases Strictly Across a Period

The periodic trend of increasing ionization energy has two hallmark exceptions in the second period: B < Be (due to 2s² penetration) and O < N (due to 2p³ half-filled stability).

🛡️ How to avoid the -1 mark: Remember: Nitrogen (2p³) is harder to ionize than Oxygen (2p⁴)!

📖

Core NEET Syllabus Standard

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

Core Scientific Principle Tested:

"Half-filled and completely filled subshells possess extra stability; electron pairing in 2p increases repulsion."

Syllabus Focus: Periodic trends in ionization enthalpy and electronic configuration stability

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

Nitrogen holds its three tight; Oxygen drops its fourth without a fight!

Ready to attempt under real exam pressure?

Test your speed and accuracy in the authentic 180-minute timed NTA simulator.

Launch NEET 2026 CBT Test →

← Previous (Q.69) All Chemistry Questions Grid Next (Q.71) →

🔗 Related Questions — Classification of Elements and Periodicity in Properties

All 10 Classification of Elements and Periodicity in Properties PYQs →

NEET 2025 Q47

Truth Value of Statements on Melting Point, Electronegativity, Isoelectronic Ions & IE Trends

View Solution →

NEET 2025 Q53

Identification of Main Group (Representative) Elements from Electronic Configurations

View Solution →

NEET 2024 Q63

First Ionization Enthalpy Trends in Second Period Elements

View Solution →

NEET 2024 Q64

Electronegativity Trends Across Periods and Groups

View Solution →

Spotted a mistake in this solution? Report an error ([email protected])

⚖️ Legal & Fair Dealing Notice: Sourced from publicly released NEET 2026 question papers and NEET answer keys for private study, research, and non-commercial educational practice under Fair Dealing (Section 52(1)(a) of the Indian Copyright Act, 1957). Not affiliated with or endorsed by NTA or NMC.

NEET 2026 Chemistry Question 70 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Half-Filled Stability & Electron Pairing Penalty

Step-by-Step Rigorous Workflow & Mathematical Proof

Analyze Assertion A (Ionization Enthalpy Comparison)

Analyze Reason R (Electronic Origin of the Anomaly)

Evaluate Reason as the Proper Scientific Explanation

360° Option Elimination Engine Key Feature

NTA Trap Alert: Assuming Ionization Energy Increases Strictly Across a Period

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

🔗 Related Questions — Classification of Elements and Periodicity in Properties

Truth Value of Statements on Melting Point, Electronegativity, Isoelectronic Ions & IE Trends

Identification of Main Group (Representative) Elements from Electronic Configurations

First Ionization Enthalpy Trends in Second Period Elements

Electronegativity Trends Across Periods and Groups

Solve 4 NEET Questions a Day — One per Subject