Find the total number of molecules in which total number of antibonding $e^{-}$ are more than the total number of bonding $e^{-}$.

${\mathrm{O}}_2,{\mathrm{C}}_2,{\mathrm{B}}_2,{\mathrm{F}}_2,{\mathrm{N}}_2$

Nitrogen has greater molecules in which total number of antibonding e- are more than the total number of bonding e- because :

The total number of molecules of F₂ is 12g. Thus are 10 electrons in bonding orbitals and 8 electrons in antibonding

The total number of molecules of N₂ is 14g. Thus are 9 electrons in bonding orbitals and 4 electrons in antibonding

The total number of molecules of O₂ is 8g. Thus are 10 electrons in bonding orbitals and 6 electrons in antibonding

The total number of molecules of B₂ is 10g. Thus are 4 electrons in bonding orbitals and 6 electrons in antibonding

The total number of molecules of C₂ is 12g. Thus are 4 electrons in bonding orbitals and 6 electrons in antibonding

In all the molecules antibonding electron is different from the bonding electron. Thus, the correct answer is 0.