The most basic hydroxide from the following is

Options

A. `Pr(OH)_3(Z=59)`
B. `Sm(OH)_3(Z=62)`
C. `Ho(OH)_3(Z=67)`
D. `La (OH)_3(Z=57)`

Answer

To determine the most basic hydroxide among the given options, we can follow these steps:

### Step 1: Identify the Elements
The question provides us with four elements: PR (Praseodymium), SM (Samarium), HO (Holmium), and LA (Lanthanum). We need to identify their positions in the periodic table and their respective atomic numbers.

### Step 2: Understand Basicity of Hydroxides
Basicity of hydroxides generally increases with increasing ionic character. Ionic character is influenced by the size of the cation; smaller cations tend to have a higher covalent character, which corresponds to lower basicity. Conversely, larger cations tend to exhibit more ionic character, leading to higher basicity.

### Step 3: Compare the Sizes of Cations
The order of the elements based on their atomic size (from largest to smallest) is:
1. La (Lanthanum)
2. Pr (Praseodymium)
3. Sm (Samarium)
4. Ho (Holmium)

Since all these elements form +3 cations, we can conclude that:
- La³⁺ is the largest cation among the four.
- Ho³⁺ is the smallest cation among the four.

### Step 4: Analyze Covalent and Ionic Character
According to Fajans' rule:
- Smaller cations (like Ho³⁺) have higher polarizing power, leading to more covalent character in their hydroxides, which decreases basicity.
- Larger cations (like La³⁺) have lower polarizing power, leading to more ionic character in their hydroxides, which increases basicity.

### Step 5: Determine the Most Basic Hydroxide
Since La³⁺ is the largest cation, it will have the least covalent character and the most ionic character, making La(OH)₃ the most basic hydroxide among the options provided.

### Conclusion
The most basic hydroxide from the given options is **La(OH)₃**.

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