Which of the following s-block eleent reacts with `NaOH` to give water soluble complex:-

Options

A. `Al`
B. `Ca`
C. `Be`
D. `Li`

Answer

To solve the question "Which of the following s-block elements reacts with NaOH to give a water-soluble complex?", we will follow these steps:

### Step 1: Identify the s-block elements
The s-block elements include Group 1 (alkali metals) and Group 2 (alkaline earth metals) of the periodic table. The common s-block elements are:
- Group 1: Lithium (Li), Sodium (Na), Potassium (K), Rubidium (Rb), Cesium (Cs), Francium (Fr)
- Group 2: Beryllium (Be), Magnesium (Mg), Calcium (Ca), Strontium (Sr), Barium (Ba), Radium (Ra)

### Step 2: Determine the reactivity with NaOH
We need to check which of these s-block elements can react with sodium hydroxide (NaOH) to form a water-soluble complex.

- **Alkali Metals (Group 1)**: Generally, alkali metals do not form stable complexes with NaOH. They react vigorously with water but do not form soluble complexes with NaOH.

- **Alkaline Earth Metals (Group 2)**: Among these, Beryllium (Be) is known to exhibit amphoteric behavior. This means it can react with both acids and bases.

### Step 3: Focus on Beryllium
Beryllium is unique among alkaline earth metals because it can react with NaOH to form a soluble complex. The reaction can be represented as follows:
\[ \text{Be} + 2 \text{NaOH} + 2 \text{H}_2\text{O} \rightarrow \text{Na}_2[\text{Be(OH)}_4] + \text{H}_2 \uparrow \]
This reaction produces a soluble complex called sodium beryllate.

### Step 4: Conclusion
Based on the above analysis, the only s-block element that reacts with NaOH to give a water-soluble complex is **Beryllium (Be)**.

### Final Answer
The answer to the question is **Beryllium (Be)**.

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