Bond order in ${\text{He}}_{\text{2}}$ specie is

Atomic number of He is 2. Thus, ${\text{He}}_{\text{2}}$ molecule has total 4 electrons.

The molecular orbital diagram of helium is follows as below.

In  ${\text{He}}_{\text{2}}$, No. of bonding electrons = No. of anti bonding electrons

So, Bond order is

${\text{B.O of He}}_{\text{2}}=\frac{1}{2}[{\text{No.\hspace{0.17em}of bonding e}}^{-}-{\text{No. of antibonding e}}^{-}]$

$=\frac{1}{2}[2-2]=0$