Question

The atomic number of the element from the following with lowest $ 1^{\text{st}} $ ionization enthalpy is:

• A. 87
• B. 19
• C. 32
• D. 35

Hint:

- Ionization energy \(\propto \frac{1}{\text{atomic size}}\) - Alkali metals have lowest ionization energies in their periods - Within Group 1, ionization energy decreases down the group - Francium has the lowest known ionization energy (380 kJ/mol)

Answer 1

The Correct Option is A

Concept:
The first ionization enthalpy decreases:
- Down a group (atomic size increases)
- From right to left across a period (effective nuclear charge decreases)

Analysis of Options:
Option 1 (87): Francium (Fr) - Group 1, Period 7 element - Largest atomic size in periodic table - Lowest effective nuclear charge on valence electron - Lowest ionization energy among given options.
Option 2 (19): Potassium (K) - Group 1, Period 4 - Higher ionization energy than Fr (smaller size).
Option 3 (32): Germanium (Ge) - Group 14, Period 4 - Much higher ionization energy than alkali metals.
Option 4 (35): Bromine (Br) - Group 17, Period 4 - Highest ionization energy among options (high effective nuclear charge).

Periodic Trend:
Ionization Energy Order: Fr (87) < K (19) < Ge (32) < Br (35)