52 questions on this topic of Chemical Bonding and Molecular Structure (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.
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Answer: (B) O₂²⁻ > O₂⁻ > O₂ > O₂⁺
Bond order is inversely proportional to bond length. The calculated bond orders are: O₂⁺ = 2.5, O₂ = 2.0, O₂⁻ = 1.5, and O₂²⁻ = 1.0. Therefore, the bond length order is O₂²⁻ > O₂⁻ > O₂ > O₂⁺.
Answer: (D) C₂²⁻ and N₂
Both C₂²⁻ (14 electrons) and N₂ (14 electrons) are isoelectronic. Their molecular orbital configuration has no unpaired electrons (diamagnetic) and both have a bond order of 3.0.
Answer: (A) Statement I is true but Statement II is false.
Statement I is true because sp-mixing pushes σ2p_z above the π2p subshell in B₂, C₂, and N₂. Statement II is false because across a period, increasing nuclear charge stabilizes 2s much more than 2p, making the 2s-2p energy gap large in O and F, which prevents significant sp-mixing.
Answer: (A) O₂⁺ and O₂
Bond dissociation energy increases with increasing bond order. O₂⁺ has the highest bond order (2.5) and thus the highest bond dissociation energy. O₂ has 2 unpaired electrons in degenerate π*2p orbitals, which is more than O₂⁺ (1), O₂⁻ (1), and O₂²⁻ (0).
Answer: (A) N₂
Bond length is inversely proportional to bond order. The bond orders are: N₂ = 3.0, N₂⁺ = 2.5, N₂⁻ = 2.5, and N₂²⁻ = 2.0. Since N₂ has the highest bond order among the given species, it possesses the shortest bond length.
Answer: (D) Be₂ < Li₂ < B₂ < C₂
The calculated bond orders are: Be₂ has bond order 0 (does not exist/least stable); Li₂ has bond order 1 (σ2s bonding); B₂ has bond order 1 (two π bonds, slightly stronger/higher bond energy than Li₂); C₂ has bond order 2. Hence, stability increases as Be₂ < Li₂ < B₂ < C₂.
Answer: (B) 1.73 BM
The superoxide ion O₂⁻ has 17 electrons. Its electronic configuration is KK σ2s² σ*2s² σ2p_z² π2p_x² π2p_y² π*2p_x² π*2p_y¹. It contains n = 1 unpaired electron. The spin-only magnetic moment is μ = √(n(n + 2)) = √(1(3)) = √3 ≈ 1.73 BM.
Answer: (C) N₂ > N₂⁺ = N₂⁻ > N₂²⁻
The number of electrons and bond orders are: N₂ (14 e⁻, BO = 3.0), N₂⁺ (13 e⁻, BO = 2.5), N₂⁻ (15 e⁻, BO = 2.5), and N₂²⁻ (16 e⁻, BO = 2.0). Hence, the order of bond orders is N₂ (3.0) > N₂⁺ (2.5) = N₂⁻ (2.5) > N₂²⁻ (2.0).
Answer: (D) (I) > (II) > (III) > (IV)
Bond energy is directly proportional to bond order. The bond orders are: O₂⁺ = 2.5, O₂ = 2.0, O₂⁻ = 1.5, and O₂²⁻ = 1.0. Therefore, the bond energy decreases in the order: O₂⁺ > O₂ > O₂⁻ > O₂²⁻, which corresponds to (I) > (II) > (III) > (IV).
Answer: (C) s + p_y
When the z-axis is the internuclear axis, an s orbital centered on one atom has zero net overlap with a p_y (or p_x) orbital centered on the other atom because the positive overlap with the positive lobe is exactly canceled by the negative overlap with the negative lobe. Thus, s + p_y results in zero overlap (non-bonding interaction).
Answer: (D) 2s and 2p_x
When the z-axis is the internuclear axis, a 2s orbital overlaps equally with the positive and negative lobes of a 2p_x orbital. Because positive overlap cancels out negative overlap due to differences in symmetry, the net overlap is zero (non-bonding interaction).
Answer: (A) Removed from σ2p_z; bond length increases
For N₂ (14 electrons), the HOMO is σ2p_z. Ionization to form N₂⁺ removes an electron from this bonding σ2p_z orbital, reducing the bond order from 3.0 to 2.5. A decrease in bond order leads to an increase in bond length.