Question

Assertion (A): \(16^{th}\) group elements have higher ionization enthalpy values than \(15^{th}\) group elements in the corresponding periods.
Reason (R): \(15^{th}\) group elements have half-filled stable electronic configurations.

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

Hint:

Half-filled and completely filled electronic configurations are extra stable. Hence, group \(15\) elements generally have slightly higher ionization enthalpy than group \(16\) elements.

Answer 1

The Correct Option is D

Step 1: State the assertion and reason.
Assertion (A): Elements of group 16 have higher ionisation energy than group 15 elements in the corresponding periods. Reason (R): Group 15 elements have half-filled, stable electronic configurations.
Step 2: Evaluate the Reason (R).
Group 15 elements (e.g., N, P) have the outer configuration $ns^2 np^3$. A half-filled $p$ subshell is extra stable due to exchange energy. This stability makes it harder to remove an electron. So R is CORRECT.
Step 3: Evaluate the Assertion (A).
Because group 15 elements have extra-stable half-filled $p$ subshells ($np^3$), their ionisation energies are anomalously HIGH. Group 16 elements ($ns^2 np^4$) have one electron beyond the half-filled shell, and that extra electron experiences increased electron-electron repulsion, making it easier to remove.
Step 4: Compare IE values for confirmation.
For example, in period 2: IE of N (group 15) $\approx 1402$ kJ/mol, while IE of O (group 16) $\approx 1314$ kJ/mol. So group 15 has HIGHER IE than group 16, not the other way around.
Step 5: Conclude on Assertion A.
Assertion A claims group 16 has higher IE than group 15. This is INCORRECT. The truth is group 15 has higher IE than group 16 due to the half-filled stability.
Step 6: Select the correct option.
A is incorrect, R is correct. This corresponds to option 4.
\[ \boxed{\text{A is incorrect, R is correct (Option 4)}} \]