Question

The third ionization energy amongst $Li,Be,B$ and $C$ is lowest for:

• A. $Li$
• B. $Be$
• C. $B$
• D. $C$

Answer 1

The correct option is C $B$

It is the energy needed to remove the second electron from each ion in 1 mole of gaseous 1+ ions to give gaseous 2+ ions. You can then have as many successive ionisation energies as there are electrons in the original atom. That's a lot of energy.

The third I.E. amongst $Li,Be,B$ and $C$ is lowest for B.

Because after removing the two electron its electronic configuration Have become $1s^2,2s^1$.

After removing this electron from 2s orbital it will attain the stable electronic configuration of helium hence the third ionization energy of boron is lowest.