In which of the following molecules would you | vedclass

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

Question

(B) Bond length is inversely proportional to bond order. Higher bond order results in a shorter bond length.In $N_2H_4$ $(H_2N-NH_2)$,the $N-N$ bond i

Vedclass · Accepted Answer

$N_2$

Vedclass · Answer

(B) Bond length is inversely proportional to bond order. Higher bond order results in a shorter bond length.In $N_2H_4$ $(H_2N-NH_2)$,the $N-N$ bond is a single bond (bond order = $1$).In $N_2O_4$ $(O_2N-NO_2)$,the $N-N$ bond is a single bond (bond order = $1$).In $N_2O$ $(N \equiv N^+-O^-)$,the $N-N$ bond is a triple bond (bond order = $3$).In $N_2$ $(N \equiv N)$,the $N-N$ bond is a triple bond (bond order = $3$).Comparing $N_2$ and $N_2O$,the $N-N$ bond in $N_2$ is a pure triple bond,whereas in $N_2O$,the resonance structures result in a bond order slightly less than $3$. Therefore,$N_2$ has the shortest $N-N$ bond length.

In which of the following molecules would you expect the nitrogen-to-nitrogen bond length to be the shortest?

Similar Questions

The molecule having one unpaired electron is

With increasing bond order,stability of a bond

Consider the given figure showing the formation of $H^{+}_2$ ion depending on internuclear distance versus potential energy of the system.
Which is the correct statement?

The correct order of energy of $\sigma_{2p_z}$,$\pi_{2p_x}$,$\pi_{2p_y}$,$\pi_{2p_x}^$,$\pi_{2p_y}^$ and $\sigma_{2p_z}^*$ orbitals in $F_2$ is:

The bond order and the magnetic characteristics of $CN^{-}$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module