JEE Chemistry 2026 Question with Solution

Given: Two statements about periodic trends in first ionization enthalpy and electron gain enthalpy.

Find: Which option correctly evaluates Statement I and Statement II.

Step 1: Analyze Statement I. Across a period, first ionization enthalpy generally increases due to increasing nuclear charge. However, there is an exception between nitrogen and oxygen. Nitrogen has a half-filled $2p^3$ configuration, which is more stable than oxygen's $2p^4$ configuration.

Therefore,

is the correct trend, not $\mathrm{C < O < N < F}$. Hence, Statement I is false.

Step 2: Analyze Statement II. Electron gain enthalpy generally becomes less negative down a group due to increasing atomic size. However, oxygen has a much lower magnitude of electron gain enthalpy than sulfur because of strong inter-electronic repulsion in compact $2p$ orbitals.

Thus, the correct order of magnitude is:

Hence, Statement II is true.

Conclusion: Statement I is false but Statement II is true.

Therefore, the correct option is A.

Given: Periodic trend statements involving C, O, N, F and S, Se, Te, Po, O.

Find: Whether each statement is true or false.

The usual trend for first ionization enthalpy across a period is increasing value from left to right. But nitrogen is an important exception because its half-filled $2p^3$ arrangement is especially stable. Removing one electron from oxygen with configuration $2p^4$ is easier than from nitrogen.

So the order implied by Statement I is incorrect.

For electron gain enthalpy, the common down-group trend is decreasing magnitude. But oxygen is anomalous because its very small size causes greater electron-electron repulsion when an incoming electron enters the compact $2p$ orbital. Therefore sulfur has greater magnitude of electron gain enthalpy than oxygen.

So the order in Statement II is correct.

Therefore, Statement I is false and Statement II is true, so the answer is A.