(i) `SiO_(2) + NaOH to ?`
(ii) `SiO_(2) + HF to ? `
The products of (ii) & (i) respectively are

Options

A. `H_(2) SiF_(6) , SiO_(4)^(4-)`
B. `SiF_(4) , Na_(2) SiO_(3)`
C. `Na_(2) SiO_(3) , SiF_(4)`
D. `Na_(2) SiO_(4) , H_(2) SiF_(6)`

Answer

To solve the given reactions step by step, we will analyze each reaction separately.

### Step 1: Reaction of SiO2 with NaOH
1. **Write the reactants**: Silicon dioxide (SiO2) and sodium hydroxide (NaOH).
2. **Identify the reaction**: When SiO2 reacts with NaOH, it forms sodium silicate (Na2SiO3) and water (H2O).
3. **Write the balanced equation**:
\[ SiO_2 + 2NaOH \rightarrow Na_2SiO_3 + H_2O \]
- Here, 1 mole of SiO2 reacts with 2 moles of NaOH to produce 1 mole of Na2SiO3 and 1 mole of water.

### Step 2: Reaction of SiO2 with HF
1. **Write the reactants**: Silicon dioxide (SiO2) and hydrofluoric acid (HF).
2. **Identify the reaction**: When SiO2 reacts with HF, it forms silicon tetrafluoride (SiF4) and water (H2O).
3. **Write the balanced equation**:
\[ SiO_2 + 4HF \rightarrow SiF_4 + 2H_2O \]
- Here, 1 mole of SiO2 reacts with 4 moles of HF to produce 1 mole of SiF4 and 2 moles of water.

### Final Products
- From the first reaction (SiO2 + NaOH), the product is **Na2SiO3** (sodium silicate).
- From the second reaction (SiO2 + HF), the product is **SiF4** (silicon tetrafluoride).

### Conclusion
The products of the reactions are:
1. From (ii) SiO2 + HF: **SiF4**
2. From (i) SiO2 + NaOH: **Na2SiO3**

Thus, the final answer is:
- The products of (ii) and (i) respectively are **SiF4** and **Na2SiO3**.

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