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:

• A. A is not correct but R is correct
• B. Both A and R are correct and R is the correct explanation of A
• C. Both A and R are correct and R is NOT the correct explanation of A
• D. A is correct but R is not correct

Hint:

Exception in ionization energy trend: Nitrogen = stable half-filled configuration Oxygen = paired electron repulsion lowers ionization energy Always remember: \[ IE_O \lt IE_N \] despite moving left to right in periodic table.

Answer 1

The Correct Option is B

Step 1: Write the configurations.
N is \(1s^2 2s^2 2p^3\), O is \(1s^2 2s^2 2p^4\), and F is \(1s^2 2s^2 2p^5\).
Step 2: See why nitrogen resists ionization.
Nitrogen has an exactly half-filled \(2p^3\) set, with one electron in each p orbital and parallel spins. This is an extra-stable arrangement, so removing an electron from N needs relatively high energy.
Step 3: See why oxygen ionizes more easily.
Oxygen's \(2p^4\) has one orbital holding a pair, and the extra electron-electron repulsion in that pair makes one electron easier to remove. Losing it gives \(O^+ = 2p^3\), a stable half-filled state.
Step 4: Bring fluorine into the comparison.
Fluorine has a higher nuclear charge than oxygen, pulling its electrons in more tightly, so its ionization enthalpy is higher.
Step 5: Order the energies.
Thus \(IE_O < IE_N < IE_F\), so oxygen's first ionization enthalpy is indeed the lowest of the three.
Step 6: Judge assertion and reason.
Assertion A (O lower than N and F) is true; Reason R (removing an electron from O gives a stable half-filled \(p^3\)) is true and correctly explains A.
\[ \boxed{\text{Both A and R are correct and R is the correct explanation of A.}} \]