Question

Arrange Li, Na, K, Rb, Cs in order of increasing ionization energy:
(A) Cs < Rb < K < Na < Li
(B) K < Rb < Cs < Na < Li
(C) Cs < K < Rb < Na < Li
(D) None of these

Answer 1

Hint: Ionisation energy decreases as the nuclear effective charge decreases on the atom. Nuclear effective charge is also called as ${{Z}_{eff}}$. Size of the atom is inversely proportional to the Nuclear effective charge.

Complete step by step answer:
Ionisation energy is the quantity of energy that an isolated, gaseous atom in the group electronic state must absorb to discharge an electron, resulting in a cation. When considering an initial neutral atom, expelling the first electron will require less energy than expelling the second, the second will require less energy than the third, and so on. This is due to the reason that after removal of one electron the atom acquires a positive charge, therefore, the rest of the electrons present on the ion, will experience more nuclear effective force. As we move from top to bottom in the group the ionization energy will be decreased, because the atomic radius will be increased so removal of electrons will be easy.
Li, Na, K, Rb, Cs are IA group elements. Ionisation energy decreases from Li to Rb. So the correct order of ionization energy will be (A) Cs < Rb < K < Na < Li . So, the correct answer is “Option A”.

Note: Ionisation energy is usually expressed in kJ/mol, or the amount of energy it takes for all the atoms in a mole to lose one electron each. The more electrons that are lost, the more positive this ion will be, the harder it is to separate the electrons from the atom, more is the ionization energy.