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 $O_2^{2-} < O_2^- < O_2 < O_2^+$.

List-$I$ (Molecule / ion)	List-$II$ (Bond order)
$1$. $NO$	$i$. $1.5$
$2$. $CO$	$ii$. $2.0$
$3$. $O_2^{-}$	$iii$. $2.5$
$4$. $O_2$	$iv$. $3.0$

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

Similar Questions

Match List-$I$ with List-$II$ and select the correct option:
List-$I$ (Molecule / ion) List-$II$ (Bond order)
$1$. $NO$ $i$. $1.5$
$2$. $CO$ $ii$. $2.0$
$3$. $O_2^{-}$ $iii$. $2.5$
$4$. $O_2$ $iv$. $3.0$

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

Explain in short: $MO$ occupancy and molecular properties for $B_2, C_2, N_2, O_2, F_2, Ne_2$.

The bond order of $O_2^{+}$ is $x$. The bond orders of $O_2^{-}$ and $O_2^{2+}$ are respectively

What is the magnetic nature of dioxygen $(O_2)$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module