The increasing order of bond order of O_2, O_ | vedclass

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

Question

(D) According to Molecular Orbital Theory $(MOT)$,the bond order is calculated as $BO = \frac{1}{2}(N_b - N_a)$.For $O_2$ $(16 \ e^-)$: $BO = \frac{10

Vedclass · Accepted Answer

$O_2^{2-}, O_2^-, O_2, O_2^+$

Vedclass · Answer

(D) According to Molecular Orbital Theory $(MOT)$,the bond order is calculated as $BO = \frac{1}{2}(N_b - N_a)$.For $O_2$ $(16 \ e^-)$: $BO = \frac{10-6}{2} = 2$.For $O_2^+$ $(15 \ e^-)$: $BO = \frac{10-5}{2} = 2.5$.For $O_2^-$ $(17 \ e^-)$: $BO = \frac{10-7}{2} = 1.5$.For $O_2^{2-}$ $(18 \ e^-)$: $BO = \frac{10-8}{2} = 1$.Thus,the increasing order of bond order is $O_2^{2-} < O_2^- < O_2 < O_2^+$.

The increasing order of bond order of $O_2, O_2^+, O_2^-$,and $O_2^{2-}$ is:

Similar Questions

Although $CN^{-}$ ion and $N_2$ molecule are isoelectronic,yet $N_2$ molecule is chemically inert because of

Which of the following is planar due to back bonding?

Which of the following species,$O_2^-$ or $O_2^{2-}$,has a higher bond order?

What is the correct sequence of bond order for the given species?

According to the Molecular Orbital Theory,which of the following properties does the $O_2^+$ species possess?

Vedclass Test Series

Exam Paper Generator

Online Exam Module