Question

Consider an atom with the electron configuration of $1s^{2}2s^{2}2p^{6}3s^2$.
Which successive ionization energy will be significantly higher than the previous value?

• A. First ionization energy.
• B. Second ionization energy.
• C. Third ionization energy.
• D. Fourth ionization energy.

Answer 1

The correct option is C Third ionization energy.

After removing $2$ electrons, the element achieves configuration of $1s^{2}2s^{2}2p^6$, which is a complete octate and hence is extremely stable. Thus additional energy is required to remove third electron. So third IE is much higher than second, hence answer is C.