How Amorphous boron of 95% to 98% purity is obtained ?

Answer

To obtain amorphous boron with a purity of 95% to 98%, the following steps can be followed:

### Step-by-Step Solution:

1. **Identify the Starting Material**:
- The process begins with boric oxide (B2O3), which is the oxide form of boron.

**Hint**: Remember that boron does not exist in its pure form as a stable element; it often exists as an oxide.

2. **Reduction Process**:
- The boric oxide (B2O3) is subjected to a reduction reaction. This is typically done using magnesium (Mg) as the reducing agent.

**Hint**: Reduction is a chemical reaction that involves the gain of electrons or a decrease in oxidation state, often involving a reducing agent.

3. **Chemical Reaction**:
- The chemical reaction can be represented as follows:
\[ \text{B}_2\text{O}_3 + 3\text{Mg} \rightarrow 2\text{B} + 3\text{MgO} \]
- In this reaction, boric oxide reacts with magnesium to produce amorphous boron (B) and magnesium oxide (MgO).

**Hint**: Pay attention to the stoichiometry of the reaction; it tells you the ratio of reactants to products.

4. **Formation of Amorphous Boron**:
- The product of this reaction is amorphous boron, which is not crystalline and has a purity level between 95% to 98%.

**Hint**: Amorphous substances lack a defined structure, which differentiates them from crystalline forms.

5. **Byproduct**:
- The byproduct of this reaction is magnesium oxide (MgO), which can be separated from the amorphous boron.

**Hint**: Always consider byproducts in chemical reactions; they can often be useful or need to be disposed of properly.

### Summary:
Amorphous boron of 95% to 98% purity is obtained by the reduction of boric oxide (B2O3) with magnesium (Mg), resulting in the formation of amorphous boron and magnesium oxide.