Red P+Alkali `toNa_(4)P_(2)O_(6)+P_(2)H_(4)`

Options

A. For disproportionation reaction.
B. For comproportionation reaction.
C. For either intermolecular redox reaction or displacement reaction
D. For either thermal combination redox reaction or thermal decomposition redox reaction.

Answer

To solve the question regarding the reaction of red phosphorus with alkali to form sodium metaphosphate and phosphine, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
- The reactant is red phosphorus, which can be represented as \( P_4 \).
- The alkali is sodium hydroxide (\( NaOH \)).
- The products are sodium metaphosphate (\( Na_4P_2O_6 \)) and phosphine (\( P_2H_4 \)).

2. **Write the Reaction**:
- The reaction can be written as:
\[ P_4 + 4 NaOH \rightarrow Na_4P_2O_6 + P_2H_4 \]

3. **Determine the Oxidation States**:
- For \( P_4 \): The oxidation state of phosphorus in its elemental form is 0.
- For \( Na_4P_2O_6 \):
- Sodium (\( Na \)) has an oxidation state of +1.
- Oxygen (\( O \)) has an oxidation state of -2.
- Let the oxidation state of phosphorus be \( x \).
- The equation becomes:
\[ 4(+1) + 2x + 6(-2) = 0 \implies 4 + 2x - 12 = 0 \implies 2x = 8 \implies x = +4 \]
- Thus, the oxidation state of phosphorus in \( Na_4P_2O_6 \) is +4.

- For \( P_2H_4 \):
- Hydrogen (\( H \)) has an oxidation state of +1.
- Let the oxidation state of phosphorus be \( y \).
- The equation becomes:
\[ 2y + 4(+1) = 0 \implies 2y + 4 = 0 \implies 2y = -4 \implies y = -2 \]
- Thus, the oxidation state of phosphorus in \( P_2H_4 \) is -2.

4. **Analyze the Changes in Oxidation States**:
- The phosphorus in \( P_4 \) changes from 0 to +4 (oxidation).
- The phosphorus in \( P_4 \) also changes to -2 (reduction).
- This indicates that one part of the phosphorus is oxidized while another part is reduced.

5. **Classify the Reaction**:
- Since a single species (phosphorus) is undergoing both oxidation and reduction, this reaction is classified as a **disproportionation reaction**.

### Final Answer:
The reaction of red phosphorus with alkali to produce sodium metaphosphate and phosphine is a **disproportionation reaction**.

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