Which of the following will have lowest value of `K_(sp)` at room temperature?

Options

A. `Be(OH)_(2)`
B. `Mg(OH)_(2)`
C. `Ca(OH)_(2)`
D. `Ba(OH)_(2)`

Explanation

To determine which of the given compounds has the lowest value of \( K_{sp} \) at room temperature, we will analyze the solubility products of the hydroxides of the alkaline earth metals: Beryllium hydroxide \((Be(OH)_2)\), Magnesium hydroxide \((Mg(OH)_2)\), Calcium hydroxide \((Ca(OH)_2)\), and Barium hydroxide \((Ba(OH)_2)\).

### Step-by-Step Solution:

1. **Identify the Compounds**: The compounds given are:
- \( Be(OH)_2 \) (Beryllium hydroxide)
- \( Mg(OH)_2 \) (Magnesium hydroxide)
- \( Ca(OH)_2 \) (Calcium hydroxide)
- \( Ba(OH)_2 \) (Barium hydroxide)

2. **Understand Lattice Energy**: The solubility of these hydroxides is influenced by their lattice energy, which is the energy required to separate one mole of a solid ionic compound into its gaseous ions. Higher lattice energy typically means lower solubility.

3. **Analyze Ionic Sizes**: The ionic size increases as we move down the group in the periodic table:
- \( Be^{2+} \) < \( Mg^{2+} \) < \( Ca^{2+} \) < \( Ba^{2+} \)
- This means that \( Be^{2+} \) is the smallest cation and \( Ba^{2+} \) is the largest.

4. **Determine Lattice Energy Trends**:
- Smaller cations (like \( Be^{2+} \)) have higher lattice energy due to stronger electrostatic forces with the anions (in this case, \( OH^{-} \)).
- Larger cations (like \( Ba^{2+} \)) have lower lattice energy, leading to higher solubility.

5. **Evaluate Solubility**:
- \( Be(OH)_2 \): High lattice energy → Low solubility
- \( Mg(OH)_2 \): Moderate lattice energy → Moderate solubility
- \( Ca(OH)_2 \): Lower lattice energy than \( Mg(OH)_2 \) → Higher solubility
- \( Ba(OH)_2 \): Lowest lattice energy → Highest solubility

6. **Conclusion on \( K_{sp} \)**:
- Since \( K_{sp} \) is a measure of solubility, the compound with the lowest solubility will have the lowest \( K_{sp} \).
- Therefore, \( Be(OH)_2 \) will have the lowest \( K_{sp} \) among the given options.

### Final Answer:
The compound with the lowest value of \( K_{sp} \) at room temperature is **Option 1: \( Be(OH)_2 \)**.

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