Classification NEET 2026: Oxygen Ionization Enthalpy

What does the Classification NEET 2026 assertion–reason question ask?

Oxygen requires less energy to lose its first electron than either nitrogen or fluorine. This classification NEET 2026 question asks whether oxygen’s configuration after electron removal explains that comparison. The chapter is Classification of Elements and Periodicity in Properties; the topic is ionization enthalpy and electronic configuration.

Assertion: Removing the first electron requires less energy for oxygen than for either nitrogen or fluorine.
Reason: Removing an electron from oxygen produces a half-filled p subshell.

Choose the correct relationship:

The question bank classifies this as medium and gives an expected solving time of 75 seconds. That is its practice target, not an official exam time limit.

Which electronic configurations should you write first?

Start with the neutral configurations of nitrogen, oxygen and fluorine. In all three, the first electron removed comes from the 2p subshell. First ionization enthalpy is the energy required to remove one electron from an isolated gaseous atom:

X (g) \to X^{+} (g) + e^{-}

The full configurations are:

N : & 1 s^{2} 2 s^{2} 2 p^{3} O : & 1 s^{2} 2 s^{2} 2 p^{4} F : & 1 s^{2} 2 s^{2} 2 p^{5}

By Hund’s rule, nitrogen has one electron in each of its three 2p orbitals. Oxygen has one pair and two singly occupied 2p orbitals.

Three p orbitals can hold two electrons each. Their full capacity and half-filled electron count are: $Full capacity = 3 \times 2 = 6$

Half-filled electron count = \frac{6}{2} = 3

A half-filled 2p subshell therefore contains three singly occupied orbitals. The reason refers to this arrangement, not to one half-filled orbital.

Why is oxygen easier to ionize than nitrogen?

Removing an electron from oxygen’s paired 2p orbital reduces electron–electron repulsion and leaves a half-filled subshell. Nitrogen already has that stable half-filled arrangement, which electron removal disrupts. These effects explain why oxygen is easier to ionize despite the general increase across a period.

For gaseous nitrogen, the removal process is:

\underset{1 s^{2} 2 s^{2} 2 p^{3}}{\underset{⏟}{N (g)}} ⟶ \underset{1 s^{2} 2 s^{2} 2 p^{2}}{\underset{⏟}{N^{+} (g)}} + e^{-}

Nitrogen begins with three singly occupied 2p orbitals. Ionization breaks this stable half-filled configuration, leaving only two electrons in the subshell.

For gaseous oxygen, the process is:

\underset{1 s^{2} 2 s^{2} 2 p^{4}}{\underset{⏟}{O (g)}} ⟶ \underset{1 s^{2} 2 s^{2} 2 p^{3}}{\underset{⏟}{O^{+} (g)}} + e^{-}

The two electrons sharing oxygen’s paired orbital repel each other. Removing one relieves that repulsion and leaves three singly occupied orbitals, giving the oxygen ion a half-filled 2p subshell.

The electron count after removal is: $4 - 1 = 3$

Nitrogen loses a half-filled arrangement; oxygen gains one. Paired-electron repulsion and half-filled stability support the same conclusion:

{I E}_{1} (O) < {I E}_{1} (N)

Why is fluorine harder to ionize, and why is option A correct?

Fluorine has a higher effective nuclear charge than oxygen, so it holds its outer electrons more strongly. This explains the oxygen–fluorine comparison. Oxygen’s half-filled product alone is not a complete explanation for both comparisons.

{I E}_{1} (O) < {I E}_{1} (F)

Now check the statements and their link:

Assertion: true. Oxygen has lower first ionization enthalpy than both nitrogen and fluorine.
Reason: true. Oxygen’s first ion has a half-filled 2p subshell.
Explanatory link: The reason supplies the stability explanation for the oxygen–nitrogen exception, while effective nuclear charge supports the fluorine comparison.

The oxygen ion’s configuration confirms the reason:

O^{+} : 1 s^{2} 2 s^{2} 2 p^{3}

Final answer: option A. Both statements are correct, and the reason explains the assertion, consistent with the supplied official solution.

Why does choosing option B miss the explanatory link?

Option B follows from an incomplete method: verify both statements, then dismiss the half-filled oxygen ion as an unrelated fact about the product. Both truth checks pass, but the explanatory link gets rejected. The mistake is separating the product configuration from the electron-removal event that creates it.

Trace that event instead:

Oxygen contains a paired 2p electron.
Repulsion within that pair makes removal easier than in nitrogen.
Removing that electron leaves a stable, half-filled 2p subshell.

The product is part of the explanation, not a disconnected observation. Use configuration and pairing for oxygen versus nitrogen; use effective nuclear charge for oxygen versus fluorine. The two comparisons need different supporting arguments.

Can you apply the same method to Be versus B and P versus S?

Beryllium and phosphorus have the higher first ionization enthalpies, respectively. These are original related practice questions, not additional verified NEET PYQs. The first tests subshell energy; the second tests the same pairing effect used for oxygen.

Which has the higher first ionization enthalpy, Be or B?

Beryllium has the higher value. Boron loses a higher-energy 2p electron, whereas beryllium must lose a 2s electron. Their configurations are:

B e : & 1 s^{2} 2 s^{2} B : & 1 s^{2} 2 s^{2} 2 p^{1}

Boron’s higher-energy electron is easier to remove. Therefore:

{I E}_{1} (B e) > {I E}_{1} (B)

This is a subshell-energy exception, not a pairing exception. Boron has no paired 2p electron whose removal could relieve pairing repulsion.

Which has the higher first ionization enthalpy, P or S?

Phosphorus has the higher value. Removing an electron disrupts its stable half-filled 3p subshell. Sulfur instead loses an electron from a paired 3p orbital, relieving electron–electron repulsion.

P : & [N e] 3 s^{2} 3 p^{3} S : & [N e] 3 s^{2} 3 p^{4}

Removing that paired electron from sulfur produces:

S^{+} : [N e] 3 s^{2} 3 p^{3}

This directly parallels oxygen becoming its first positive ion: paired-electron removal leaves a half-filled p subshell. Hence:

{I E}_{1} (P) > {I E}_{1} (S)

Practise identifying the electron removed before choosing a trend: Periodic Table Practice Questions NEET: 6 Explained MCQs.

Frequently asked questions

Why is oxygen's first ionization enthalpy lower than nitrogen's?

Oxygen loses an electron from a paired 2p orbital, reducing electron–electron repulsion and leaving a half-filled 2p subshell. Nitrogen already has a stable half-filled 2p subshell, which electron removal disrupts. These effects make oxygen easier to ionize than nitrogen.

Why is fluorine harder to ionize than oxygen?

Fluorine has a higher effective nuclear charge than oxygen and holds its outer electrons more strongly. Oxygen therefore has a lower first ionization enthalpy. The half-filled configuration formed by oxygen after electron removal is not, by itself, a complete explanation for this comparison.

What is the answer to the oxygen ionization assertion–reason question?

The supplied solution selects option A: both statements are true, and the reason explains the assertion. Formation of a half-filled 2p subshell supports the oxygen–nitrogen comparison. Fluorine's higher effective nuclear charge is also needed to explain the oxygen–fluorine comparison.

Why does beryllium have a higher first ionization enthalpy than boron?

Boron loses a higher-energy 2p electron, while beryllium must lose a 2s electron. Boron's electron is easier to remove, so beryllium has the higher first ionization enthalpy. This is a subshell-energy exception, not a paired-electron repulsion effect.