In the following which of the two are paramagnetic? a.) N2 b.) CO c.) B2 d.) NO2

Correct answer is: a.) a and c b.) b and c c.) c and d d.) b and d

To determine the paramagnetic species, we need to check the presence of unpaired electrons in their molecular orbitals (for diatomic molecules) or electronic configuration (for polyatomic molecules).

### a) \( N_2 \) (Nitrogen molecule):
- Molecular orbital theory helps us analyze whether the molecule is paramagnetic or diamagnetic.
- The electronic configuration of \( N_2 \) is \( 1\sigma^2 1\sigma^*^2 2\sigma^2 2\sigma^*^2 2p\pi^4 2p\pi^*^0 \).
- All the electrons are paired in \( N_2 \), so it is **diamagnetic**.

### b) \( CO \) (Carbon monoxide):
- \( CO \) has a similar molecular orbital configuration to \( N_2 \), as it is isoelectronic with \( N_2 \) (same number of electrons).
- The electronic configuration of \( CO \) is similar to \( N_2 \), so all electrons are paired.
- Thus, \( CO \) is **diamagnetic**.

### c) \( B_2 \) (Boron molecule):
- The electronic configuration of \( B_2 \) is \( 1\sigma^2 1\sigma^*^2 2\sigma^2 2\sigma^*^2 2p\pi^2 2p\pi^*^0 \).
- In this case, the two \( 2p\pi \) electrons are unpaired.
- Thus, \( B_2 \) is **paramagnetic** due to these unpaired electrons.

### d) \( NO_2 \) (Nitrogen dioxide):
- \( NO_2 \) has an odd number of electrons (23 electrons).
- Molecules with an odd number of electrons tend to have at least one unpaired electron, which makes them **paramagnetic**.

### Conclusion:
- \( B_2 \) and \( NO_2 \) are paramagnetic.

Thus, the correct answer is **option c) c and d**.