The correct option is
B Graphite
As might be expected based on the periodic trend for the effective nuclear charge, ionization energy generally increases to the right across a period and decreases down a given group. The IE again increases progressing across period two from
Li to Be, but then decreases as we move from
Be to B. Then, as predicted, the IE increases as we move to
C and N, but decreases when we move to
O. The ionization of
Be requires removal of an electron from the
2s subshell. Ionization of
B, however, requires removal of an electron from a
2p subshell. The
2p subshell is higher in energy than the
2s subshell resulting in a lower than expected first ionization energy for boron. On comparing
N and O, we see that oxygen has a fourth electron in its
2p subshell. Since this subshell contains only three orbitals, this fourth electron must occupy an orbital that already contains an electron. Electrons, being negatively charged particles, repel each other and it is this increased electron-electron repulsion that causes oxygen to have a lower than expected first ionization energy.
Therefore the order of IE is B<Be<C<O<N.