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प्रश्न
Species having same bond order are:
(i) \[\ce{N2}\]
(ii) \[\ce{N^{-}2}\]
(iii) \[\ce{F^{+}2}\]
(iv) \[\ce{O^{-}2}\]
उत्तर
(iii) \[\ce{F^{+}2}\]
(iv) \[\ce{O^{-}2}\]
Explanation:
The formula of bond order is:
BO = `1/2[N_b - N_a]`
Number of bonding and antibonding electrons can be found from the molecular electronic configuration of the species.
(i) Total number of electrons in \[\ce{N2}\] is 14. So, its molecular electronic configuration will be `σ1s^2, σ^∗1s^2, σ2s^2, σ^∗2s^2, π2px^2, π2py^2, σ2p_z^2`.
The bond order will be:
BO = `1/2(10 - 4)` = 3
(ii) The total number of electrons in \[\ce{N^{-}2}\] is 15. So its molecular electronic configuration will be: `σ1s^2, σ^∗1s^2, σ2s^2, σ^∗2s^2, π2px^2, π2py^2, σ2p_z^2, π^∗2p_x^1`.
The bond order will be:
BO = `1/2(10 - 5)` = 2.5
(iii) The total electrons in \[\ce{F^{+}2}\] is 17. So, its electronic configuration will be: `σ1s^2, σ^∗1s^2, σ2s^2, σ^∗2s^2, σ2pz^2, π2py^2, π2px^2, π^∗2p_y^2, π^∗2p_y^1`.
The bond order will be:
BO = `1/2(10 - 7)` = 1.5
(iv) The total electrons in \[\ce{O^{-}2}\] is 17. So, its electronic configuration will be: `σ1s^2, σ^∗1s^2, σ2s^2, σ^∗2s^2, σ2pz^2, π2py^2, π2px^2, π^∗2p_y^2, π^∗2p_y^1`.
The bond order will be:
BO = `1/2(10 - 7)` = 1.5
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Which of the following options represents the correct bond order?
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(i) \[\ce{CN-}\]
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(iii) \[\ce{O^{-}2}\]
(iv) \[\ce{O^{2-}2}\]
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(i) \[\ce{N2 -> N^{+}2 + e-}\]
(ii) \[\ce{O2 -> O^{+}2 + e-}\]
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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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