Question

Which of the following configurations represents an atom of the group having the highest second ionization energy?

• A. 1 2 2
• B. 1 2 2
• C. 1 2 2 3
• D. 1 2 2 3

Answer 1

The correct option is C

1 2 2 3

Lets talk about ionization energy. What is ionization energy?

The minimum amount of energy required to remove the most loosely bound electron from an isolated atom in the gaseous state is called ionization energy or enthalpy. Energies required to remove $2^{nd}$ - $3^{rd}$ $e^{-}$ are called successive ionization energy.

The removal of $2^{nd}$ electron from a cation is relatively more difficult because remaining electrons in cation are more effectively pulled towards nucleus and requires more energy.

But in certain cases, removal of one or more electrons from an atom means the removal of the outermost shell completely and thus the configuration of the remaining ion becomes similar to the highly stable noble gas configuration. This makes the next ionization energy abnormally high. For example, $I{E}_2$ $\gg$ $I{E}_1$ for alkali metals. Similarly, $I{E}_3$ $\gg$ $I{E}_2$ for alkaline earth metals.

So, now here if we remove an electron from each configuration

(a) 1$s^2$ 2$s^2$ 2$p^3$ (b) 1$s^2$ 2$s^2$ 2$p^5$ (c) 1$s^2$ 2$s^2$ 2$p^6$ (d) 1$s^2$ 2$s^2$ 2$p^6$ 3$s^2$

Now for second ionization energy, we have to remove one more electron from these configuration . Now, if we look at the configurations.

Option (a) and (c) are very stable configuration but from them [option (a) and (c)] also, option (c) is most stable as it is similar to Neon's electronic configuration.

It will require a lot of energy to remove an electron from that configuration.

That's why option (c) will have highest second ionization energy.