Bond-order of ${\mathrm{Be}}_2$ and ${\mathrm{C}}_2$ respectively are:

${\mathrm{Be}}_2\left(8\right)\to {\sigma }_{1{\mathrm{s}}^{\ast }}{\sigma }_{1{\mathrm{s}}^{\ast }}^{\ast }{\sigma }_{2\mathrm{s}}{\sigma }_{2\mathrm{s}}^{\ast }$

Bond order (B.O.) of ${\mathrm{Be}}_2=\frac{1}{2}[4-4]=0$

Carbon (C) has an electronic arrangement of $1s^{2}2s^{2}2p^2$ There are 6 + 6 = 12 total electrons in a ${\mathrm{C}}_2$ molecule. The electronic arrangement of  ${\mathrm{C}}_2$'s molecular orbitals is

$\left[\sigma (1s{)}^2\right]\left[{\sigma }^{\ast }(1s{)}^2\right]\left[\sigma (2s{)}^2\right]\left[{\sigma }^{\ast }(2s{)}^2\right]\left[\pi (2px{)}^2\right]\left[\pi (2py{)}^2\right]$

8 electrons make up the bonding orbital. 4 electrons make up the anti-bonding orbital. Consequently, bond order is $5=\frac{1}{2}(8-4)=2$ As a result, the ${\mathrm{C}}_2$ molecule has a double bond and has a bond order of 2.

Hence correct option is A.