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 both oxidized and reduced. We calculate the oxidation states of P in the reactant and both products.

Step 2: Detailed Explanation:
1. Reactant ($P_4$): Elemental phosphorus has an oxidation state of $0$.
2. Product 1 ($PH_3$): Hydrogen is $+1$. So, $x + 3(+1) = 0 \implies x = -3$.
3. Product 2 ($H_2PO_2^-$): Let the oxidation state of P be $x$.
\[ 2(+1) + x + 2(-2) = -1 \]
\[ 2 + x - 4 = -1 \]
\[ x - 2 = -1 \implies x = +1 \]
Thus, the oxidation state changes from $0$ to $-3$ (reduction) and from $0$ to $+1$ (oxidation).

Step 3: Final Answer:
The changes are 0 to -3 and 0 to +1.