Question

Dimagnetic species are those which contain no unpaired electrons. Which among the following are dimagnetic?

(i) \[\ce{N2}\]

(ii) \[\ce{N^{2-}2}\]

(iii) \[\ce{O2}\] 

(iv) \[\ce{O^{2-}2}\]

Answer 1

(i) \[\ce{N2}\]

(iv) \[\ce{O^{2-}2}\]

Explanation:

(i) Electronic configuration of \[\ce{N2}\] = `σ1s^2, σ^∗1s^2, σ2s^2, σ^∗2s^2, π2p_x^2 ≈ π2p_y^2, σ2p_z^2`.

 It has no unpaired electron. This indicates it is a diamagnetic species.

(ii) Electronic configuration of \[\ce{N^{2-}2}\] ion = `σ1s^2, σ^∗1s^2, σ2s^2, σ^∗2s^2, π2p_x^2  ≈ π2p_y^2, σ2p_z^2,  π2p_x^1 ≈ π^∗2p_y^1`.

It has two unpaired electrons, so it is paramagnetic in nature.

(iii) Electronic configuration of \[\ce{O2}\] = `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`.

(iv) Electronic configuration of \[\ce{O^{2-}2}\] ion = `σ1s^2, σ^∗1s^2, σ2s^2, σ^∗2s^2, σ2p_z^2, π2p_x^2 ≈ σ^∗2p_y^2, π^∗2p_x^2 ≈ π^∗2p_y^2`.

It contains no unpaired electron, therefore, it is diamagnetic in nautre.