1. Idea: A species is paramagnetic if it has one or more unpaired electrons.

2. We examine each option using molecular orbital (MO) theory / electron count.

Option A: $NO^+$

• Total electrons: $7 + 8 - 1 = 14$
• This is isoelectronic with $N_2$ and CO.
• MO configuration for 14 electrons gives all electrons paired.
• Therefore, $NO^+$ is diamagnetic.

Option B: CO

• Total electrons: $6 + 8 = 14$
• CO is also isoelectronic with $N_2$.
• All electrons are paired in its MO configuration.
• Therefore, CO is diamagnetic.

Option C: $O_2^{2-}$

• Total electrons: $8 + 8 + 2 = 18$
• For dioxygen species:
  • $O_2$ has 2 unpaired electrons,
  • $O_2^-$ has 1 unpaired electron,
  • $O_2^{2-}$ has all electrons paired.
• Thus $O_2^{2-}$ is diamagnetic.

Option D: $B_2$

• Total electrons: $5 + 5 = 10$
• For lighter homonuclear diatomics ($B_2, C_2, N_2$), MO order is: $\sigma(1s),\ \sigma^*(1s),\ \sigma(2s),\ \sigma^*(2s),\ \pi(2p_x)=\pi(2p_y),\ \sigma(2p_z)$
• Filling 10 electrons: $\sigma(1s)^2\,\sigma^*(1s)^2\,\sigma(2s)^2\,\sigma^*(2s)^2\,\pi(2p_x)^1\,\pi(2p_y)^1$
• There are two unpaired electrons in the degenerate $\pi$ orbitals.
• Therefore, $B_2$ is paramagnetic.

3. Conclusion: Only $B_2$ is paramagnetic.

$\boxed{\text{D: } B_2}$