The correct option is C Be < C < B < N < O < Li
The correct answer is option (c).
The second ionization involves the removal of an electron from the 1s2 electronic configuration of Li+. Hence, the highest eneegy is required in this case . Oxygen is placed second because we need to remove an electron from the stable 2p3 orbital. Nitrogen has higer 2nd ionization energy than boron because of the effective nuclear charge. In carbon, after the removal of the second electron, it will have an electronic configuration 1s22s2. Thus, less energy is required to remove the second electron of carbon as compared to boron. The lowest 2nd ionisation energy is for beryllium because after removing the second electron, it will get a stable electronic configuration 1s2 and effective nuclear charge for beryllium is also low.