The correct sequence of bond order is

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(B) According to molecular orbital theory,the bond order is calculated as: $Bond \ order = \frac{(	ext{Number of } e^{-} 	ext{ in bonding MO}) - (

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

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(B) According to molecular orbital theory,the bond order is calculated as: $Bond \ order = \frac{(	ext{Number of } e^{-} 	ext{ in bonding MO}) - (	ext{Number of } e^{-} 	ext{ in antibonding MO})}{2}$Calculating bond orders for the given species:$O_2^{+} (15 \ e^{-}): \sigma 1s^2, \sigma^{*} 1s^2, \sigma 2s^2, \sigma^{*} 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^{*} 2p_x^1$. $B.O. = \frac{10-5}{2} = 2.5$$O_2 (16 \ e^{-}): \sigma 1s^2, \sigma^{*} 1s^2, \sigma 2s^2, \sigma^{*} 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^{*} 2p_x^1 = \pi^{*} 2p_y^1$. $B.O. = \frac{10-6}{2} = 2.0$$O_2^{-} (17 \ e^{-}): \sigma 1s^2, \sigma^{*} 1s^2, \sigma 2s^2, \sigma^{*} 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^{*} 2p_x^2 = \pi^{*} 2p_y^1$. $B.O. = \frac{10-7}{2} = 1.5$$O_2^{2-} (18 \ e^{-}): \sigma 1s^2, \sigma^{*} 1s^2, \sigma 2s^2, \sigma^{*} 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^{*} 2p_x^2 = \pi^{*} 2p_y^2$. $B.O. = \frac{10-8}{2} = 1.0$Comparing the values: $2.5 > 2.0 > 1.5 > 1.0$.Thus,the correct sequence is $O_2^{+} > O_2 > O_2^{-} > O_2^{2-}$.

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