The set of species having only fractional bon | vedclass

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Question

(B) Bond order $(BO) = \frac{1}{2} 	imes (	ext{Number of electrons in bonding } MOs - 	ext{Number of electrons in anti-bonding } MOs)$.For species

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$O_2^{+}, O_2^{-}, N_2^{+}$

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(B) Bond order $(BO) = \frac{1}{2} 	imes (	ext{Number of electrons in bonding } MOs - 	ext{Number of electrons in anti-bonding } MOs)$.For species to have fractional bond order,the total number of electrons must be odd.$(A) C_2^{2-} (14e^-), N_2 (14e^-), O_2^{2-} (18e^-)$: All have even electrons,so bond orders are integers $(3, 3, 1)$.$(B) O_2^{+} (15e^-), O_2^{-} (17e^-), N_2^{+} (13e^-)$: All have odd electrons,so bond orders are fractional $(2.5, 1.5, 2.5)$.$(C) O_2^{2+} (14e^-), O_2 (16e^-), C_2^{2-} (14e^-)$: $O_2$ has $16e^-$ (even),bond order is $2$.$(D) Li_2 (6e^-), H_2^{+} (1e^-), C_2 (12e^-)$: $Li_2$ and $C_2$ have even electrons,bond orders are integers $(1, 2)$.Thus,only option $(B)$ contains species with fractional bond orders.

Pair of species	Identical property
$A$. $B_2 \& O_2$	$P$. Bond order $= 2.5$
$B$. $Be_2 \& H_2^{2-}$	$Q$. Paramagnetic nature
$C$. $N_2^{+} \& N_2^{-}$	$R$. Diamagnetic nature
$D$. $O_2^{+} \& O_2^{-}$	$S$. Doesn't exist

The set of species having only fractional bond order values is

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Match the following:

Pair of species Identical property

$A$. $B_2 \& O_2$ $P$. Bond order $= 2.5$

$B$. $Be_2 \& H_2^{2-}$ $Q$. Paramagnetic nature

$C$. $N_2^{+} \& N_2^{-}$ $R$. Diamagnetic nature

$D$. $O_2^{+} \& O_2^{-}$ $S$. Doesn't exist

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