Question

Which of the following has the correct order of $2^{nd}$ ionization energies?

• A. Li < Be < B < C < N < O
• B. Li < B < Be < C < O < N
• C. Be < C < B < N < O < Li
• D. Li < Be < C < B < O < N

Answer 1

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 $1s^2$ electronic configuration of $L{i}^{+}$. Hence, the highest eneegy is required in this case . Oxygen is placed second because we need to remove an electron from the stable $2p^3$ orbital. Nitrogen has higer $2^{nd}$ 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 $1s^{2}2s^2$. Thus, less energy is required to remove the second electron of carbon as compared to boron. The lowest $2^{nd}$ ionisation energy is for beryllium because after removing the second electron, it will get a stable electronic configuration $1s^2$ and effective nuclear charge for beryllium is also low.