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Question
Which of the following have identical bond order?
(i) \[\ce{CN-}\]
(ii) \[\ce{NO+}\]
(iii) \[\ce{O^{-}2}\]
(iv) \[\ce{O^{2-}2}\]
Solution
(i) \[\ce{CN-}\]
(ii) \[\ce{NO+}\]
Explanation:
Bond order depends on the electronic configuration of the molecular species, and the electronic configuration depends on the number of electrons a particular atom is contributing in its molecular structure.
The \[\ce{CN-}\] species has a total of 14 electrons; 6 electrons from carbon atom, 7 electrons from nitrogen atom and 1 electron for the negative charge.
The \[\ce{NO+}\] species has a total of 14 electrons; 8 from oxygen atoms, 7 from nitrogen atom and 1 electron will be subtracted due to the positive charge on the species.
The \[\ce{O^{-}2}\] species has a total of 17 electrons; 8 from each oxygen atom and 1 for the negative charge on the species.
The \[\ce{O^{2-}2}\] species has a total of 18 electrons; 8 from each oxygen atom and 2 for the negative charge on the species.
Hence, the molecular species/ion having the same bond order will be \[\ce{CN-}\] and \[\ce{NO+}\].
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| Column I | Column II |
| (i) \[\ce{NO}\] | (a) 1.5 |
| (ii) \[\ce{CO}\] | (b) 2.0 |
| (iii) \[\ce{O^{-}2}\] | (c) 2.5 |
| (iv) \[\ce{O2}\] | (d) 3.0 |
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