Chemistry - PERIODIC TABLE Question with Solution | TestHub

${\mathrm{N}}_2$ has more I.E than N as removal of electron is from bonding molecular orbital while O has more I.E than ${\mathrm{O}}_2$ as the electron is removed from anti bonding molecular orbital in ${\mathrm{O}}_2$ Though ${\mathrm{B}}_2$ has pi bond and ${\mathrm{Li}}_2$ has sigma bond still bond energy of ${\mathrm{B}}_2$ is more than ${\mathrm{Li}}_2$ due to smaller size of it.