The internuclear distances in O-O bonds for O | vedclass

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

Question

(D) According to Molecular Orbital Theory,the bond order decreases as we add electrons to antibonding orbitals.The bond orders are:$O_2^+ (2.5), O_2 (

Vedclass · Accepted Answer

$1.12 \ \mathring{A}, 1.21 \ \mathring{A}, 1.30 \ \mathring{A}, 1.49 \ \mathring{A}$

Vedclass · Answer

(D) According to Molecular Orbital Theory,the bond order decreases as we add electrons to antibonding orbitals.The bond orders are:$O_2^+ (2.5), O_2 (2.0), O_2^- (1.5), O_2^{2-} (1.0)$.Since bond length is inversely proportional to bond order,the bond length increases as bond order decreases.The order of bond lengths is: $O_2^+ The corresponding bond lengths are approximately $1.12 \ \mathring{A}, 1.21 \ \mathring{A}, 1.30 \ \mathring{A}, 1.49 \ \mathring{A}$ respectively.

The internuclear distances in $O-O$ bonds for $O_2^+$,$O_2$,$O_2^-$ and $O_2^{2-}$ respectively are

Similar Questions

Molecular orbital theory was developed mainly by

$A$ pair of isoelectronic species having bond order of $1$ is

What is the total number of electrons present in bonding orbitals of $O_2$ molecule according to molecular orbital theory?

With increasing bond order,stability of a bond

$N_2$ and $O_2$ are converted into monocations,$N_2^+$ and $O_2^+$ respectively. Which of the following statements is wrong?

Vedclass Test Series

Exam Paper Generator

Online Exam Module