Which of the following statement is correct for s-block elements ?

Question

Accepted Answer

Option 1: Be has the smallest atomic size in group-2.

Explanation:
1. **Justification for Option 1:**
   - In the periodic table, atomic size generally increases as you move down a group due to the addition of electron shells. In group-2, which consists of alkaline earth metals, beryllium (Be) is at the top of the group. Therefore, it has the smallest atomic size compared to other elements in the same group, such as magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra). This is because Be has fewer electron shells and a smaller atomic radius.

2. **Explanation for why other options are incorrect:**

- **Option 2: Li is most metallic.**
     - Lithium (Li) is in group-1, not group-2. While Li is metallic, it is not the most metallic element even within its own group. Metallic character increases down the group, so elements like cesium (Cs) and francium (Fr) are more metallic than lithium.

- **Option 3: Mg imparts red color to the flame.**
     - Magnesium (Mg) does not impart a red color to the flame. Instead, it typically burns with a bright white flame. The red color in flame tests is usually associated with elements like strontium (Sr) or lithium (Li).

- **Option 4: Cs is most reducing in water.**
     - While cesium (Cs) is indeed a very reactive metal and can act as a strong reducing agent, this statement is not relevant to s-block elements in the context of group-2. Additionally, the focus here is on group-2 elements, and Cs is a group-1 element. The statement does not pertain to the properties of group-2 elements.

In conclusion, Option 1 is the correct statement regarding s-block elements, specifically for group-2, as beryllium (Be) indeed has the smallest atomic size in its group.