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प्रश्न
Which of the following pair is expected to have the same bond order?
विकल्प
\[\ce{O2, N2}\]
\[\ce{O^{+}2 , N^{-}2}\]
\[\ce{O^{-}2 , N^{+}2}\]
\[\ce{O^{-}2 , N^{-}2}\]
उत्तर
\[\ce{O^{+}2 , N^{-}2}\]
Explanation:
(i) The electronic configuration of \[\ce{O2}\] is:
`σ1s^2, σ^∗1s^2, σ2s^2, σ^∗2s^2, σ2p_z^2, π2p_x^2 ,π2p_y^2, π^∗2p_x^1, π^∗2p_y^1`
The bond order of \[\ce{O2}\] will be:
BO = `1/2[N_b - N_a]`
BO = `1/2[10 - 6]` = 2.0
The electronic configuration of \[\ce{N2}\]:
`σ1s^2, σ^∗1s^2, σ2s^2, σ^∗2s^2, π2p_x^2, π2p_y^2, σ2p_z^2`
The bond order of \[\ce{N2}\] will be:
BO = `1/2[N_b - N_a]`
BO = `1/2[10 - 4]` = 3.0
So, the bond order of \[\ce{O2, N2}\] will not be equal.
(ii) The electronic configuration of \[\ce{O^{+}2}\] is:
`σ1s^2, σ^∗1s^2, σ2s^2, σ^∗2s^2, σ2p_z^2, π2p_x^2 ,π2p_y^2, π^∗2p_x^1`
The bond order of \[\ce{O^{+}2}\] will be:
BO = `1/2[N_b - N_a]`
BO = `1/2[10 - 5]` = 2.5
The electronic configuration of \[\ce{N^{-}2}\] is:
`σ1s^2, σ^∗1s^2, σ2s^2, σ^∗2s^2, π2p_x^2, π2p_y^2, σ2p_z^2, π^∗2p_x^1`
The bond order of \[\ce{N^{-}2}\] will be:
BO = `1/2[N_b - N_a]`
BO = `1/2[10 - 5]` = 2.5
Hence, the bond order of \[\ce{O^{+}2, N^{-}2}\] will be equal.
(iii) The electronic configuration of \[\ce{O^{-}2}\] is:
`σ1s^2, σ^∗1s^2, σ2s^2, σ^∗2s^2, σ2p_z^2, π2p_x^2, π2p_y^2, π^∗2p_x^2, π^∗2p_y^1`
The bond order of `\[\ce{O^{-}2}\] will be:
BO = `1/2[N_b - N_a]`
BO = `1/2[10 - 7]` = 1.5
The electronic configuration of \[\ce{N^{+}2}\] is:
`σ1s^2, σ^∗1s^2, σ2s^2, σ^∗2s^2, π2p_x^2, π2p_y^2, σ2p_z^1`
The bond order of \[\ce{N^{+}2}\] will be:
BO = `1/2[N_b - N_a]`
BO = `1/2[9 - 4]` = 2.5
Hence, the bond order of \[\ce{O^{-}2, N^{+}2}\] will not be equal.
(iv)The electronic configuration of \[\ce{O^{-}2}\] is:
`σ1s^2, σ^∗1s^2, σ2s^2, σ^∗2s^2, σ2p_z^2, π2p_x^2, π2p_y^2, π^∗2p_x^2, π^∗2p_y^1`
The bond order of \[\ce{O^{-}2}\] will be:
BO = `1/2[N_b - N_a]`
BO = `1/2[10 - 7]` = 1.5
The electronic configuration of \[\ce{N^{-}2}\] will be:
`σ1s^2, σ^∗1s^2, σ2s^2, σ^∗2s^2, π2p_x^2, π2p_y^2, σ2p_z^2, π^∗2p_x^1`
The bond order of \[\ce{N^{-}2}\] will be:
BO = `1/2[N_b - N_a]`
BO = `1/2[10 - 5]` = 2.5
Hence, the bond order of \[\ce{O^{-}2, N^{-}2}\] will not be equal.
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