The set of species having only fractional bon | vedclass

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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

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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.

The set of species having only fractional bond order values is

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