Question

Which of the following elements will have highest second ionization enthalpy?

• A. $1s^{2}2s^{2}2p^{6}3s^2$
• B. $1s^{2}2s^{2}2p^{6}3s^1$
• C. $1s^{2}2s^{2}2p^{6}3s^{2}3p^2$
• D. $1s^{2}2s^{2}2p^{6}3s^{2}3p^3$

Answer 1

The correct option is B $1s^{2}2s^{2}2p^{6}3s^1$

The element with electronic configuration $1s^{2}2s^{2}2p^{6}3s^1$ will have highest second ionization enthalpy. $1s^{2}2s^{2}2p^{6}3s^1$ loses one electron to form $1s^{2}2s^{2}2p^6$. It is stable electronic configuration of noble gas $Ne$. Removal of second electron will require large amount of energy as it will break stable noble gas electronic configuration.