In the following disproportionation reaction $P_{4}(s)+3OH^{-}(aq)+3H_{2}O(l)\rightarrow PH_{3}(g)+3H_{2}PO_{2}^{-}$, the oxidation states of P in $P_{4}$, $PH_{3}$ and $H_{2}PO_{2}^{-}$ are respectively:

Question

Accepted Answer

To determine the oxidation states of phosphorus (P) in the given disproportionation reaction, we need to analyze each compound involved:

1. **$P_{4}(s)$**: In elemental phosphorus, $P_{4}$, each phosphorus atom is in its elemental form. The oxidation state of an element in its standard state is always 0. Therefore, the oxidation state of P in $P_{4}$ is 0.

2. **$PH_{3}(g)$**: In phosphine ($PH_{3}$), hydrogen has an oxidation state of +1. Since there are three hydrogen atoms, the total contribution from hydrogen is +3. To balance this, phosphorus must have an oxidation state of -3, because the sum of oxidation states in a neutral molecule must be zero. Therefore, the oxidation state of P in $PH_{3}$ is -3.

3. **$H_{2}PO_{2}^{-}$**: In the hypophosphite ion ($H_{2}PO_{2}^{-}$), hydrogen has an oxidation state of +1, and there are two hydrogen atoms, contributing a total of +2. Oxygen typically has an oxidation state of -2, and there are two oxygen atoms, contributing a total of -4. The overall charge of the ion is -1. Let the oxidation state of phosphorus be x. The equation for the sum of oxidation states is:

$$
   2(+1) + x + 2(-2) = -1
   $$

Simplifying this gives:

$$
   2 + x - 4 = -1 \
   x - 2 = -1 \
   x = +1
   $$

Therefore, the oxidation state of P in $H_{2}PO_{2}^{-}$ is +1.

Based on this analysis, the oxidation states of P in $P_{4}$, $PH_{3}$, and $H_{2}PO_{2}^{-}$ are 0, -3, and +1, respectively. This matches Option 4: 0, -3, +1.

**Explanation of why other options are incorrect:**

- **Option 1 (4, 3, 2):** This option suggests oxidation states of 4, 3, and 2 for $P_{4}$, $PH_{3}$, and $H_{2}PO_{2}^{-}$, respectively. This is incorrect because the oxidation state of P in $P_{4}$ is 0, not 4. Additionally, the oxidation state of P in $PH_{3}$ is -3, not 3, and in $H_{2}PO_{2}^{-}$, it is +1, not 2.

- **Option 2 (4, -3, -2):** This option incorrectly assigns an oxidation state of 4 to $P_{4}$, which should be 0. It also incorrectly assigns an oxidation state of -2 to $H_{2}PO_{2}^{-}$, which should be +1.

- **Option 3 (0, +3, +1):** This option correctly identifies the oxidation state of P in $P_{4}$ as 0 and in $H_{2}PO_{2}^{-}$ as +1, but it incorrectly assigns an oxidation state of +3 to $PH_{3}$, which should be -3.

Thus, Option 4 is the only correct choice, as it accurately reflects the oxidation states of phosphorus in each compound involved in the reaction.