Q.7 The decreasing order of bond lengths for 𝐎𝟐, 𝐁𝟐, 𝐍𝟐 and 𝐂𝟐 is

(A) B2 > C2 > N2 > O2

(B) B2 > C2 > O2 > N2

(C) N2 > C2 > O2 > B2

(D) B2 > O2 > N2 > C2

The correct answer is (C) N₂ > C₂ > O₂ > B₂.

This order reflects the experimentally observed bond lengths: N₂ (1.09 Å), C₂ (1.24 Å), O₂ (1.21 Å), and B₂ (1.59 Å).​

Bond Orders

Bond order determines bond strength and inversely correlates with bond length in these homonuclear diatomic molecules. Higher bond order means more shared electrons pulling atoms closer.

• B₂ (10 electrons): Bond order = 1 (σ2s² σ*2s² π2p⁴).​

• C₂ (12 electrons): Bond order = 2 (adds σ2p²).​

• N₂ (14 electrons): Bond order = 3 (KK (σ2s)² (σ*2s)² π2p⁴ σ2p²).​

• O₂ (16 electrons): Bond order = 2 (π*2p² antibonding electrons weaken it).​

Why This Order?

Pure bond order predicts N₂ (shortest) < C₂ ≈ O₂ < B₂ (longest), but atomic size refines it. Oxygen’s smaller radius shortens O₂ below C₂ despite same bond order.

Bond lengths confirm: B₂ longest (lowest order, largest atoms), then O₂ (double bond, small atoms), C₂, N₂ shortest (triple bond).​

Option Analysis

• (A) B₂ > C₂ > N₂ > O₂: Wrong; N₂ shorter than C₂, O₂ not longest.

• (B) B₂ > C₂ > O₂ > N₂: Wrong; violates N₂ shortest and O₂ < C₂.

• (C) N₂ > C₂ > O₂ > B₂: Correct; matches experiment (1.09 < 1.24 > 1.21 < 1.59 Å).

• (D) B₂ > O₂ > N₂ > C₂: Wrong; N₂ not between O₂ and C₂.

The decreasing order of bond lengths for O2, B2, N2, and C2 is a core CSIR NET Life Sciences topic in molecular orbital theory. Candidates often confuse it due to bond order similarities in C2 and O2. This guide explains the correct sequence—N2 > C2 > O2 > B2—using valence electrons, MO diagrams, and real bond lengths.

Bond Order Calculation

Molecular orbital (MO) theory fills orbitals as σ2s, σ2s, π2p, σ2p (for B2-C2), then π2p, σ*2p (O2 onward).​

Higher bond order shortens bonds; O2’s antibonding π* electrons and smaller atomic radius make it shorter than C2.​

Why Not Bond Order Alone?

Atomic radius decreases B→O, overriding same-order comparisons. N2’s triple bond dominates for shortest length.​

Exam Tips for CSIR NET

• Remember: For second-period diatomics, N2 < O2 < C2 for doubles; B2 longest (single).​

• Practice MO diagrams: B2 paramagnetic, O2 too; N2/C2 diamagnetic.

• Common trap: Ignoring size effect on C2 vs O2.

This order (option C) appears in CSIR NET; verify with NIST CCCBDB for values.​