The energy required to remove an electron from the outermost shell of an isolate gaseous atom is known as `IE_(1)` of that atom. Similarly, the enrgy required for the removal of the electron from the unipositive ion, diapositive ion and tripositive ion are known as `IE_(2),IE_(3)` and `IE_(4)` respectively, and are called successive ionisation energies. The magnitude of the charge depends on the size of the orbital of electron. Electrons in smaller orbitals are on average close with each other and have more repulsion. Thus for `Be(2s^(2))`, the `IE_(1)` and `IE_(2)` are `9.3` and `18.2 eV "atom"^(1)`, whereas for `Ca(4s^(2))`, the values are `6.1` and `11.9 eV`.
Among the following ionisation reactions, which one will have the maximum value of ionisation energy?

Options

A. `Be rarr Be^(o+)`
B. `Be^(o+) rarrBe^(2+)`
C. `Sr rarr Sr^(o+)`
D. `Sr^(o+) rarrSr^(2)`

Answer

B

Explanation

`IE_(2) gt IE_(1)`. Therefore, answer can be (b) or (d).
But both `Be` and `Sr` belong to group `2`, and `IE` decreases down the group `(darr)` so answer is (b).