Answer
Please see the work below.
Work Step by Step
We know that total electrons
$2\times 8+1=18$
The electronic configuration is $(\sigma2s)^2(\sigma 2s^\star)^2(\pi_{2p})^4(\sigma_{2p})^2(\pi_{2p}^{\star})^4$
Bond order
$\frac{1}{2}(n_b-n_a)$
We plug in the known values to obtain:
$\frac{1}{2}(10-8)=1$
