Interpret the phrase

A bond written as "two $\pi$ and half $\sigma$ bonds" means:

• total $\pi$ bond order $=2$
• total $\sigma$ bond order $=\dfrac{1}{2}$

So we need a molecule/ion whose bond order contribution can be split as: $\sigma = \frac{1}{2}, \qquad \pi = 2$

Use molecular orbital (MO) theory

Bond order is given by: $\text{Bond order} = \frac{N_b - N_a}{2}$ where $N_b$ and $N_a$ are the number of electrons in bonding and antibonding orbitals.

We must separately inspect $\sigma$ and $\pi$ contributions.

Option A: $O_2$

MO configuration of $O_2$ (valence shell): $\sigma_{2s}^2\,\sigma_{2s}^{*2}\,\sigma_{2p_z}^2\,(\pi_{2p_x})^2(\pi_{2p_y})^2\,(\pi_{2p_x}^*)^1(\pi_{2p_y}^*)^1$

$\sigma$ contribution

From $2p_z$ orbitals:

• bonding $\sigma_{2p_z}$ has $2$ electrons
• antibonding $\sigma_{2p_z}^*$ has $0$

Hence, $\text{$\sigma$ bond order} = \frac{2-0}{2}=1$

$\pi$ contribution

• bonding $\pi$ electrons = $4$
• antibonding $\pi^*$ electrons = $2$

Hence, $\text{$\pi$ bond order} = \frac{4-2}{2}=1$

So $O_2$ has: $1\,\sigma + 1\,\pi$

Not correct.

Option B: $N_2^+$

First, $N_2$ has valence MO configuration: $\sigma_{2s}^2\,\sigma_{2s}^{*2}\,(\pi_{2p_x})^2(\pi_{2p_y})^2\,\sigma_{2p_z}^2$

Removing one electron for $N_2^+$ gives: $\sigma_{2s}^2\,\sigma_{2s}^{*2}\,(\pi_{2p_x})^2(\pi_{2p_y})^2\,\sigma_{2p_z}^1$

$\sigma$ contribution

From $2p_z$ orbitals:

• bonding $\sigma_{2p_z}$ has $1$ electron
• antibonding $\sigma_{2p_z}^*$ has $0$

Therefore, $\text{$\sigma$ bond order} = \frac{1-0}{2}=\frac{1}{2}$

$\pi$ contribution

• bonding $\pi$ electrons = $4$
• antibonding $\pi^*$ electrons = $0$

Therefore, $\text{$\pi$ bond order} = \frac{4-0}{2}=2$

So $N_2^+$ has: $2\,\pi + \frac{1}{2}\,\sigma$

This matches exactly.

Option C: $O_2^+$

$O_2^+$ is formed by removing one electron from $O_2$: $\sigma_{2s}^2\,\sigma_{2s}^{*2}\,\sigma_{2p_z}^2\,(\pi_{2p_x})^2(\pi_{2p_y})^2\,(\pi_{2p}^*)^1$

$\sigma$ contribution

$\text{$\sigma$ bond order} = \frac{2-0}{2}=1$

$\pi$ contribution

• bonding $\pi$ electrons = $4$
• antibonding $\pi^*$ electrons = $1$

$\text{$\pi$ bond order} = \frac{4-1}{2}=\frac{3}{2}$

So $O_2^+$ has: $1\,\sigma + \frac{3}{2}\,\pi$

Not correct.

Option D: $N_2$

Configuration: $\sigma_{2s}^2\,\sigma_{2s}^{*2}\,(\pi_{2p_x})^2(\pi_{2p_y})^2\,\sigma_{2p_z}^2$

$\sigma$ contribution

$\text{$\sigma$ bond order} = \frac{2-0}{2}=1$

$\pi$ contribution

$\text{$\pi$ bond order} = \frac{4-0}{2}=2$

So $N_2$ has: $1\,\sigma + 2\,\pi$

Not correct, because the question asks for half $\sigma$, not one $\sigma$.

Final conclusion

The species containing two $\pi$ bonds and half a $\sigma$ bond is: $\boxed{N_2^+}$

Therefore, the correct option is B.