Question

In the following redox reaction,  \(P_{4(s)} + 3OH^-_{(aq)} + 3H_2O_{(l)} \)\(\rightarrow PH_{3(g)} + \)\(3H_2PO_2^-\)  the oxidation state of phosphorus changes from:

• A. 0 to -3 and 0 to -1
• B. 0 to +1 and 0 to +3
• C. 0 to -1 and 0 to +5
• D. 0 to -3 and 0 to +1

Hint:

In a disproportionation reaction involving an element in its standard state ($0$), look for one negative oxidation state (reduction) and one positive oxidation state (oxidation).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This is a disproportionation reaction where the same element (Phosphorus) is simultaneously oxidized and reduced.
Step 2: Key Formula or Approach:
Calculate the oxidation state of Phosphorus (P) in reactants and products.
Elemental form (P\textsubscript{4}) = 0.
In PH\textsubscript{3}: \( x + 3(+1) = 0 \implies x = -3 \).
Step 3: Detailed Explanation:
1. In P\textsubscript{4}, the oxidation state of P is 0.
2. In PH\textsubscript{3} (Phosphine), P is -3. (Reduction)
3. In H\textsubscript{2}PO\textsubscript{2}{-} (Hypophosphite ion):
Let \( x \) be the state of P: \( 2(+1) + x + 2(-2) = -1 \)
\( 2 + x - 4 = -1 \implies x - 2 = -1 \implies x = +1 \). (Oxidation)
The state changes from 0 to -3 and from 0 to +1.
Step 4: Final Answer:
The changes are 0 to -3 and 0 to +1.