Match List-I with List-II and select the corr | vedclass

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(A) The bond order is calculated using the formula: $	ext{Bond Order} = \frac{1}{2} [	ext{Number of bonding electrons} - 	ext{Number of antibonding

Vedclass · Accepted Answer

$1-iii, 2-iv, 3-i, 4-ii$

Vedclass · Answer

(A) The bond order is calculated using the formula: $	ext{Bond Order} = \frac{1}{2} [	ext{Number of bonding electrons} - 	ext{Number of antibonding electrons}]$.$1$. For $NO$ ($15$ electrons): $	ext{Bond Order} = \frac{1}{2} [10 - 5] = 2.5$ (Matches $iii$).$2$. For $CO$ ($14$ electrons): $	ext{Bond Order} = \frac{1}{2} [10 - 4] = 3.0$ (Matches $iv$).$3$. For $O_2^{-}$ ($17$ electrons): $	ext{Bond Order} = \frac{1}{2} [10 - 7] = 1.5$ (Matches $i$).$4$. For $O_2$ ($16$ electrons): $	ext{Bond Order} = \frac{1}{2} [10 - 6] = 2.0$ (Matches $ii$).Thus,the correct matching is $1-iii, 2-iv, 3-i, 4-ii$.

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

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

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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$. $3.0$