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Define the term 'bond order' and find bond order of O_(2). |
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Answer» SOLUTION :Bond order is half of the difference of the ELECTRONS present in the bonding and antibonding present in the bonding and antibonding molecular orbitals , i.e., Bond order = `(1)/(2) (N_(b) - N_(a))` E.C. of `O_(2) = [ sigma (1s)]^(2)[sigma ^(**) (1s)]^(2)[sigma (2s)]^(2)[sigma ^(**) (2s)]^(2)[sigma (2p_(Z))]^(2) [ pi(2p_(y))]^(2) [pi^(**)(2p_(x))]^(1) [pi(2p_(y))]^(1)` `therefore ` Bond order = `(1)/(2) (10 - 6) = 2` |
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