Question

Choose the CORRECT trend(s) of the first ionization energies among the following. (Given: Atomic numbers C: 6; N: 7; O: 8; F: 9; Si: 14; P: 15; S: 16; Cl: 17)

• A. C < N > O < F
• B. Si < P > S < Cl
• C. C < N < O < F
• D. Si < P < S < Cl

Hint:

Across a period IE increases, but remember the classic exceptions from half-filled \(p\) subshells: \( \mathrm{O}<\mathrm{N} \) and \( \mathrm{S}<\mathrm{P} \). When in doubt, check for \(p^3\) stability.

Answer 1

The Correct Option is A, B

Step 1: Periodic trend. Across a period, first ionization energy generally increases (effective nuclear charge rises), and down a group it generally decreases (greater size & shielding).
Step 2: Key anomaly (half-filled stability). Elements with half-filled \(p\) subshells are relatively more stable, so removing an electron requires more energy. Hence:
\(\bullet\) In Period~2: \( \mathrm{N}(2p^3) \) has higher IE than \( \mathrm{O}(2p^4) \) because removing an electron from O relieves inter-electronic repulsion in the paired \(p\) orbital \(⇒\) \( \mathrm{O}<\mathrm{N} \).
\(\bullet\) In Period~3: \( \mathrm{P}(3p^3) \) has higher IE than \( \mathrm{S}(3p^4) \) for the same reason \(⇒\) \( \mathrm{S}<\mathrm{P} \).
Step 3: Check each option.
(A) \( \mathrm{C}<\mathrm{N}>\mathrm{O}<\mathrm{F} \): Correct—captures the anomaly \( \mathrm{O}<\mathrm{N} \) and the overall rise towards F. \(\checkmark\)
(B) \( \mathrm{Si}<\mathrm{P}>\mathrm{S}<\mathrm{Cl} \): Correct—captures the anomaly \( \mathrm{S}<\mathrm{P} \) and rise towards Cl. \(\checkmark\)
(C) \( \mathrm{C}<\mathrm{N}<\mathrm{O}<\mathrm{F} \): Incorrect—ignores \( \mathrm{O}<\mathrm{N} \). \(\times\)
(D) \( \mathrm{Si}<\mathrm{P}<\mathrm{S}<\mathrm{Cl} \): Incorrect—ignores \( \mathrm{S}<\mathrm{P} \). \(\times\)
\(\boxed{\text{Correct options: (A) and (B)}}\)