* Ionization Energy: The energy required to remove the most loosely bound electron from a gaseous atom is called ionization energy.

* Periodic Trends: Ionization energy generally increases as you move:

* Across a period (left to right): Atoms have a stronger hold on their electrons due to increasing nuclear charge.

* Up a group (top to bottom): Electrons are further from the nucleus, making them easier to remove.

Let's analyze the options:

* Sn (Tin): Group 14, Period 5

* Sr (Strontium): Group 2, Period 5

* Be (Beryllium): Group 2, Period 2

* Br (Bromine): Group 17, Period 4

Reasoning:

* Strontium (Sr) and Beryllium (Be): Both are in Group 2 (alkaline earth metals). Sr is further down the group than Be, so its outermost electron is further from the nucleus and easier to remove.

* Tin (Sn) and Bromine (Br): Sn is in Group 14, Br is in Group 17. Bromine will have a higher ionization energy due to its greater nuclear charge and stronger attraction to its electrons.

Conclusion:

Based on periodic trends, Strontium (Sr) requires the least energy to remove its most loosely bound electron.

Therefore, the answer is (b).