Question

5. a) By writing molecular orbital configuration for each of following molecules calculate the
bond order and also determine whether it is paramagnetic or diamagnetic.
(i) NO (ii) CO (iii) O2
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Answer 1

5. a) By writing molecular orbital configuration for each of following molecules calculate the bond order and also determine whether it is paramagnetic or diamagnetic.

(i) NO (ii) CO (iii) O2

+

Solution

(i) NO

AO (N)

MO (NO)

AO (O)

Energy

Bond order $= \frac{1}{2} (2 - 2 + 2 + 4 - 1) = 2.5$

Paramagnetic (has unpaired electron on MO)

(ii) CO

Electron configuration: $(\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\pi_{2p})^4 (\sigma_{2p})^2$

Bond order $= \frac{1}{2} (2 - 2 + 2 + 4) = 3$

Diamagnetic (hasn't unpaired electron on MO)

(iii) Diagrams for $\mathsf{O}_2^+$ and $\mathsf{O}_2$ are shown as it is not clear from the task which one is required to describe

Electron configuration: $(\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi_{2p}^*)^1$

Bond order $= \frac{1}{2} (2 - 2 + 2 + 4 - 1) = 2.5$

Paramagnetic (has unpaired electron on MO)

Energy

Electron configuration: $(\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi_{2p}^*)^2$

Bond order $= \frac{1}{2} (2 - 2 + 2 + 4 - 2) = 2$

Paramagnetic (has unpaired electrons on MO)

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