Question

Match the species in Column I with the bond order in Column II.

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

Answer 1

Column I	Column II
(i) \[\ce{NO}\]	(c) 2.5
(ii) \[\ce{CO}\]	(d) 3.0
(iii) \[\ce{O^{-}2}\]	(a) 1.5
(iv) \[\ce{O2}\]	(b) 2.0

Explanation:

(i) The electronic configuration of \[\ce{NO}\] 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 \[\ce{NO}\] will be:

BO = `1/2[N_b - N_a]`

BO = `1/2[10 - 5]` = 2.5

(ii) The electronic configuration of  \[\ce{CO}\] is:

`σ1s^2, σ^∗1s^2, σ2s^2, σ^∗2s^2, π2p_x^2, π2p_y^2, σ2p_z^2`

The bond order of \[\ce{CO}\] will be:

BO = `1/2[N_b - N_a]`

BO = `1/2[10 - 4]` = 3.0

(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`

BO = `1/2[N_b - N_a]`

BO = `1/2[10 - 7]` = 1.5

(iv) 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`

BO = `1/2[N_b - N_a]`

BO = `1/2[10 - 6]` = 2.0