Q3 · CSIR NET Chemistry, December 2015
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)$N_2$ has higher bond order than $N_2^+$ and hence has larger bond length compared to $N_2^+$.
(b)$N_2^+$ has higher bond order than $N_2$ and hence has larger bond length compared to $N_2$
(c)$N_2$ has higher bond order than $N_2^+$ and hence has higher dissociation energy compared to $N_2^+$
(d)$N_2$ has lower bond order than $N_2^+$ and hence has lower dissociation energy compared to $N_2^+$ energy.
Answer
Answer (as printed): C
Explanation
+ The Mo configuration of N2and N2 are 2 2 2 2 2 N2 = σ2s σ2s* π2px π2py σ2pz 𝟖#𝟐 B.O.= =3 𝟐 + 2 2 2 2 1 N2 = σ2s σ2s* π2px π2py σ2pz 𝟕#𝟐 B.O.= =2.5 𝟐 N₂→ N₂ (Bond order decreases) - -
Explanation as extracted from the printed page; notation may be imperfect.
Study loop for Chemical Bonding