The correct order of bond dissociation energy | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The bond order is directly proportional to the bond dissociation energy. Calculating the bond order using Molecular Orbital Theory: For $N_2$ ($14

Vedclass · Accepted Answer

$N_2 > O_2 > O_2^-$

Vedclass · Answer

(C) The bond order is directly proportional to the bond dissociation energy. Calculating the bond order using Molecular Orbital Theory: For $N_2$ ($14$ electrons): Bond order = $(10-4)/2 = 3$. For $O_2$ ($16$ electrons): Bond order = $(10-6)/2 = 2$. For $O_2^-$ ($17$ electrons): Bond order = $(10-7)/2 = 1.5$. Since the bond order follows the order $N_2 (3) > O_2 (2) > O_2^- (1.5)$,the bond dissociation energy follows the same order: $N_2 > O_2 > O_2^-$.

The correct order of bond dissociation energy among $N_2, O_2, O_2^-$ is shown in which of the following arrangements?

Similar Questions

Among the following,the paramagnetic compound is

The molecule having one unpaired electron is

Of the species,$NO$,$NO^{+}$,$NO^{2+}$,$NO^{-}$,the one with minimum bond strength is:

$N_2$ and $O_2$ are converted into $N_2^+$ and $O_2^+$ respectively. Which of the following is not correct?

Identify the set containing isoelectronic species.

Vedclass Test Series

Exam Paper Generator

Online Exam Module