Arrange the following in the increasing order | vedclass

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(A) Bond dissociation energy is directly proportional to the bond order.Calculate the bond order for each species:$O_2$ ($16$ electrons): Bond order =

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Bond dissociation energy is directly proportional to the bond order.Calculate the bond order for each species:$O_2$ ($16$ electrons): Bond order = $(10-6)/2 = 2.0$$O_2^+$ ($15$ electrons): Bond order = $(10-5)/2 = 2.5$$O_2^-$ ($17$ electrons): Bond order = $(10-7)/2 = 1.5$$O_2^{2-}$ ($18$ electrons): Bond order = $(10-8)/2 = 1.0$Since bond order follows the order $O_2^{2-} (1.0) Therefore,the correct order is $O_2^{2-} < O_2^- < O_2 < O_2^+$.

List-$I$	List-$II$
$A. Li_2$	$i. 3$
$B. N_2$	$ii. 1.5$
$C. Be_2$	$iii. 1.0$
$D. O_2$	$iv. 0$
	$v. 2$

Arrange the following in the increasing order of their bond dissociation energy: $O_2$,$O_2^+$,$O_2^-$,and $O_2^{2-}$.

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