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) 0_2⁺

To calculate the bond order and determine whether the molecules are paramagnetic or diamagnetic, we will first write the molecular orbital (MO) configuration for each molecule.

(i) NO (Nitric Oxide)

The NO molecule has an odd number of electrons (11 electrons in total). The molecular orbital configuration is:

1. \( \sigma_{1s}^2 \)
2. \( \sigma^*_{1s}^2 \)
3. \( \sigma_{2s}^2 \)
4. \( \sigma^*_{2s}^2 \)
5. \( \sigma_{2p_z}^2 \)
6. \( \pi_{2p_x}^2 \), \( \pi_{2p_y}^2 \)
7. \( \pi^*_{2p_x}^1 \)

The bond order is calculated using the formula:

\[
\text{Bond order} = \frac{1}{2} \left( \text{number of electrons in bonding orbitals} - \text{number of electrons in antibonding orbitals} \right)
\]

In this case:
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