The bond order and the magnetic characteristic of CN- are

🧠 $\ce{CN-}$ has 14 electrons (C: 6, N: 7, +1 from charge). This makes it isoelectronic with $\ce{N2}$.

MO filling (same as $\ce{N2}$): \sigma_{1s}^2\sigma^*_{1s}^2\sigma_{2s}^2\sigma^*_{2s}^2\pi_{2p}^4\sigma_{2p}^2

BO = (10−4)/2 = 3 (triple bond)

All electrons are paired → diamagnetic (no unpaired electrons).

⚡ The 14-electron isoelectronic series: $\ce{N2}$, CO, $\ce{NO+}$, $\ce{CN-}$, $\ce{O2^{2+}}$ — all have BO = 3 and are diamagnetic.

$\boxed{\text{Answer: (b) 3, diamagnetic}}$