According to MO theory,the number of species/ | vedclass

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(B) The bond order for each species is calculated as follows:$CN^{-}$ ($14$ electrons): Bond order $= \frac{10-4}{2} = 3$$NO^{+}$ ($14$ electrons): Bo

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$3$

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(B) The bond order for each species is calculated as follows:$CN^{-}$ ($14$ electrons): Bond order $= \frac{10-4}{2} = 3$$NO^{+}$ ($14$ electrons): Bond order $= \frac{10-4}{2} = 3$$O_{2}$ ($16$ electrons): Bond order $= \frac{10-6}{2} = 2$$O_{2}^{+}$ ($15$ electrons): Bond order $= \frac{10-5}{2} = 2.5$$O_{2}^{2+}$ ($14$ electrons): Bond order $= \frac{10-4}{2} = 3$Thus,$CN^{-}, NO^{+},$ and $O_{2}^{2+}$ have an identical bond order of $3$.The total number of such species is $3$.

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

According to $MO$ theory,the number of species/ions from the following having identical bond order is $.........$
$CN^{-}, NO^{+}, O_{2}, O_{2}^{+}, O_{2}^{2+}$

Similar Questions

Match the molecules in List-$I$ with their respective bond orders in List-$II$.
List-$I$ List-$II$
$A. Li_2$ $i. 3$
$B. N_2$ $ii. 1.5$
$C. Be_2$ $iii. 1.0$
$D. O_2$ $iv. 0$
$v. 2$

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According to $MO$ theory,the number of species/ions from the following having identical bond order is $.........$$CN^{-}, NO^{+}, O_{2}, O_{2}^{+}, O_{2}^{2+}$