JEE Chemistry 2025 Question with Solution

Given: The atomic numbers are $87$, $19$, $32$, and $35$.

Find: Which element has the lowest first ionization enthalpy.

Ionization enthalpy decreases down a group because atomic size increases and shielding also increases. Alkali metals have the lowest ionization enthalpy in their respective periods.

The corresponding elements are:

• $87$ = Francium (Fr)
• $19$ = Potassium (K)
• $32$ = Germanium (Ge)
• $35$ = Bromine (Br)

Among these, Francium and Potassium are Group $1$ elements. Within Group $1$, first ionization enthalpy decreases down the group.

Therefore,

$$\text{Fr} < \text{K} < \text{Ge} < \text{Br}$$

in first ionization enthalpy among the given options.

So the element with the lowest first ionization enthalpy is Francium, whose atomic number is $87$.

The correct option is A.

Given: Atomic numbers $87$, $19$, $32$, and $35$.

Find: The element with minimum first ionization enthalpy.

Concept:

• First ionization enthalpy is inversely related to atomic size.
• It decreases down a group.
• It is generally lower on the left side of the periodic table and higher on the right side.

Now compare the options:

1. $87$: Francium, Group $1$, Period $7$
2. $19$: Potassium, Group $1$, Period $4$
3. $32$: Germanium, Group $14$, Period $4$
4. $35$: Bromine, Group $17$, Period $4$

Francium is the largest atom among these and belongs to Group $1$, so it loses its valence electron most easily.

Hence, the ionization enthalpy order is:

$$\text{Fr} (87) < \text{K} (19) < \text{Ge} (32) < \text{Br} (35)$$

Therefore, the required atomic number is $87$ and the correct option is A.