Paper: CSIR-NET December 2015 · Subject: Inorganic Chemistry · Chapter: Chemical Bonding · Topic: Chemical Bonding – General · Marks: 2 · Difficulty: Easy
The correct statement among the following is
(a)$\mathrm{N_2}$ has higher bond order than $\mathrm{N_2^{+}}$ and hence has larger bond length compared to $\mathrm{N_2^{+}}$.
(b)$\mathrm{N_2^{+}}$ has higher bond order than $\mathrm{N_2}$ and hence has larger bond length compared to $\mathrm{N_2}$
(c)$\mathrm{N_2}$ has higher bond order than $\mathrm{N_2^{+}}$ and hence has higher dissociation energy compared to $\mathrm{N_2^{+}}$
(d)$\mathrm{N_2}$ has lower bond order than $\mathrm{N_2^{+}}$ and hence has lower dissociation energy compared to $\mathrm{N_2^{+}}$ energy.
Answer
Answer: C ✓ checked by 4AB · confidence high
Explanation
N₂: (σ2s)²(σ*2s)²(π2p)⁴(σ2p)², bond order 3. N₂⁺ loses one σ2p electron, bond order 2.5. Higher bond order means a shorter bond and a higher dissociation energy, so N₂ has the higher dissociation energy compared with N₂⁺.