Q.16 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

First ionization energies generally increase across a period in the p-block due to rising effective nuclear charge and decreasing atomic size, with exceptions at half-filled p-orbitals. Both options (A) and (B) correctly capture these trends for the second and third periods.

Option Analysis

Option (A) C < N > O < F
Carbon (1086 kJ/mol) has lower energy than nitrogen (1402 kJ/mol) due to increasing nuclear charge. Nitrogen exceeds oxygen (1314 kJ/mol) because of its stable half-filled 2p³ configuration, despite higher nuclear charge in oxygen. Oxygen’s value is lower than fluorine (1681 kJ/mol) due to electron repulsion in paired p-orbitals. This trend holds correct.

Option (B) Si < P > S < Cl
Silicon (787 kJ/mol) shows lower energy than phosphorus (1012 kJ/mol) from effective nuclear charge increase. Phosphorus surpasses sulfur (1000 kJ/mol) owing to half-filled 3p³ stability. Sulfur rises to chlorine (1251 kJ/mol) as pairing stabilizes less than half-filling. This matches observed values.

Option (C) C < N < O < F
This steady increase ignores nitrogen’s half-filled orbital stability, where N > O occurs. Incorrect due to this anomaly.

Option (D) Si < P < S < Cl
This assumes monotonic rise but overlooks P > S from half-filled stability. Incorrect.

Introduction to First Ionization Energies Trend

First ionization energy represents energy required to remove the outermost electron from gaseous atoms, crucial for CSIR NET periodic trends questions. In p-block elements (C:6, N:7, O:8, F:9; Si:14, P:15, S:16, Cl:17), general increase occurs left-to-right due to nuclear charge rise and size decrease. Half-filled orbitals (N: 2p³, P: 3p³) create exceptions where ionization drops at O and S.

Periodic Trends in p-Block

Across periods, ionization energy rises as protons increase without new shells, pulling electrons tighter. Second period: C (1086) < N (1402) > O (1314) < F (1681) kJ/mol . Third period mirrors: Si (787) < P (1012) > S (1000) < Cl (1251). Half-filled stability resists electron loss more than paired orbitals in O (2p⁴) or S (3p⁴).

Exceptions and Stability

Nitrogen’s 2p³ and phosphorus’s 3p³ configurations lower energy for adjacent elements due to exchange energy and symmetry. Oxygen and sulfur show dips from inter-electronic repulsion in p⁴. These anomalies appear in CSIR NET for trend identification.

CSIR NET Preparation Tips

Practice values: C 1086, N 1402, O 1314, F 1681; Si 787, P 1012, S 1000, Cl 1251 kJ/mol. Recognize both (A) and (B) as correct for multi-correct questions. Use graphs plotting atomic number vs. energy to visualize peaks at group 15.