Which one of the following molecules is paramagnetic?

To determine which molecule is paramagnetic, we need to check for unpaired electrons in their molecular orbital configurations. Paramagnetic substances have at least one unpaired electron.

Let's analyze each option step by step:

Option A: N₂
Nitrogen atom has atomic number 7. Each N atom has 7 electrons, so N₂ has 14 electrons. The molecular orbital configuration for N₂ (considering valence electrons: 2s and 2p orbitals) is:
$$\sigma(2s)^2 \sigma^*(2s)^2 \pi(2p_x)^2 \pi(2p_y)^2 \sigma(2p_z)^2$$
All electrons are paired (no unpaired electrons). Thus, N₂ is diamagnetic.

Option B: O₃
Ozone (O₃) has three oxygen atoms. Each oxygen atom has 8 electrons, so total electrons = 3 × 8 = 24. O₃ has a bent structure with resonance. The central oxygen is sp² hybridized. The molecule has 18 valence electrons (6 per oxygen atom). In ozone, all electrons are paired in the molecular orbitals due to resonance stabilization. O₃ is diamagnetic.

Option C: CO
Carbon monoxide has carbon (atomic number 6) and oxygen (atomic number 8). Total electrons = 6 + 8 = 14. CO is isoelectronic with N₂ (14 electrons). Its molecular orbital configuration is:
$$\sigma(2s)^2 \sigma^*(2s)^2 \pi(2p_x)^2 \pi(2p_y)^2 \sigma(2p_z)^2$$
All electrons are paired. Thus, CO is diamagnetic.

Option D: NO
Nitric oxide has nitrogen (atomic number 7) and oxygen (atomic number 8). Total electrons = 7 + 8 = 15. The molecular orbital configuration for NO is:
$$\sigma(2s)^2 \sigma^*(2s)^2 \sigma(2p_z)^2 \pi(2p_x)^2 \pi(2p_y)^2 \pi^*(2p_x)^1$$
Here, the last electron occupies the π* orbital alone, resulting in one unpaired electron. Therefore, NO is paramagnetic.

Hence, the correct answer is Option D.