Question

The ionisation energy will be maximum for the process :

• A. $Ba⟶{Ba}^{2+}$
• B. $Be⟶{Be}^{2+}$
• C. $Cs⟶{Cs}^{+}$
• D. $Li⟶{Li}^{+}$

Answer 1

Answer: (B)

$Be⟶{Be}^{2+}$

The ionisation energy will be maximum for the process, $Be⟶{Be}^{2+}$. Smaller is the atom, more is the energy needed to remove an electron, i.e., ionisation energy. Also, removal of two electron needs more energy.
Atomic number of $Be$ is 4. To convert into $B{e}^{2+}$, two electrons are released from $2s$, which are tightly attached to the nucleus, thus, it requires a large amount of ionisation energy.