Solution

$$\mu = \sqrt{n(n+2)} = 1.73\ B.M$$

$$\therefore \sqrt{n(n+2)} = 1.73$$

$$\therefore (\sqrt{n(n+2)})^2 = (1.73)^2$$

$$\therefore n(n+2) = 3$$

$$n = 1$$

$$\therefore$$ One unpaired electron

In option B:
$${O_2}^-$$ has $$17$$ electrons.

$$\sigma 1s^2 \sigma^*1s^2\ \ \sigma\ 2s^2\sigma^*2s^2\ \ \sigma 2{p_z}^2\ \ \pi 2{p_x}^2 = \pi 2{p_y}^2\ \ \ \pi^* 2{p_x}^2 = \pi^*2{p_y}^1$$

$$\Rightarrow$$ One unpaired electron

$${O_2}^+$$ has 15 electrons.

$$\sigma 1s^2 \sigma^*1s^2\ \ \sigma\ 2s^2\sigma^*2s^2\ \ \sigma 2{p_z}^2\ \ \pi 2{p_x}^2 = \pi 2{p_y}^2\ \ \ \pi^* 2{p_x}^1 = \pi^*2{p_y}^0$$

$$\Rightarrow $$ One unpaired electron

$$\therefore {O_2}^-$$ & $${O_2}^{+}$$ both of the species contain one unpaired electron. i.e, $$n = 1$$.

Hence, option $$B$$ is correct.