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Question

(b) $(1)$

Bond order $= \frac{10-6}{2} = 2.0$

(Two unpaired electrons in antibonding molecular orbital)

$(2)$

Bond order $= \frac{10-5}{2}

Vedclass · Accepted Answer

Paramagnetic and bond order >${O_2}$

Vedclass · Answer

(b) $(1)$

Bond order $= \frac{10-6}{2} = 2.0$

(Two unpaired electrons in antibonding molecular orbital)

$(2)$

Bond order $= \frac{10-5}{2} = 2.5$

(One unpaired electron in antibonding molecular orbital so it is paramagnetic)

${O_2} : {(\sigma 1s)^2}{({\sigma ^*}1s)^2}{(\sigma 2s)^2}{({\sigma ^*}2s)^2}{({\sigma ^*}2p_z)^2}$

$(\pi 2p_x^2 \equiv \pi 2p_y^2)\;({\pi ^*}2p_x^1 \equiv {\pi ^*}2p_y^1)\dots{(1)} $

${O_2^+} : {(\sigma 1s)^2}{({\sigma ^*}1s)^2}{(\sigma 2s)^2}{({\sigma ^*}2s)^2}{({\sigma ^*}2p_z)^2}$

$(\pi 2p_x^2 \equiv \pi 2p_y^2)\;({\pi ^*}2p_x^1 \equiv {\pi ^*}2p_y^0)\dots{(2)} $

According to molecular orbital theory which of the following statement about the magnetic character and bond order is correct regarding $O_2^ + $

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