Question

The dibasic oxoacid of phosphorus on disproportionation gives two products A and B. A and B are respectively

• A. HPO$_3$, PH$_3$
• B. H$_3$PO$_2$, H$_2$O
• C. H$_3$PO$_4$, PH$_3$
• D. H$_4$P$_2$O$_6$, H$_3$PO$_2$

Hint:

Oxoacids of phosphorus with P-H bonds are good reducing agents and tend to undergo disproportionation upon heating. Phosphorous acid (H$_3$PO$_3$, oxidation state +3) disproportionates to phosphoric acid (+5) and phosphine (-3). Hypophosphorous acid (H$_3$PO$_2$, +1) also disproportionates. Phosphoric acid (H$_3$PO$_4$, +5) is in the highest oxidation state and does not disproportionate.

Answer 1

The Correct Option is C

First, we must identify the dibasic oxoacid of phosphorus from the common ones.
- Hypophosphorous acid (H$_3$PO$_2$): It has one P-OH bond, so it is monobasic. - Phosphorous acid (H$_3$PO$_3$): It has two P-OH bonds, so it is dibasic. - Orthophosphoric acid (H$_3$PO$_4$): It has three P-OH bonds, so it is tribasic. - Pyrophosphoric acid (H$_4$P$_2$O$_7$): It has four P-OH bonds, so it is tetrabasic.
The dibasic oxoacid of phosphorus is phosphorous acid, H$_3$PO$_3$.
Now, we need to consider the disproportionation reaction of phosphorous acid. Disproportionation is a reaction where an element in an intermediate oxidation state is simultaneously oxidized and reduced.
In H$_3$PO$_3$, the oxidation state of phosphorus is +3. (3(+1) + P + 3(-2) = 0 $\implies$ P = +3).
When phosphorous acid is heated, it disproportionates into orthophosphoric acid and phosphine.
The reaction is: $4\text{H}_3\text{PO}_3 \xrightarrow{\Delta} 3\text{H}_3\text{PO}_4 + \text{PH}_3$.
Let's check the oxidation states in the products:
- In orthophosphoric acid (H$_3$PO$_4$), the oxidation state of P is +5. (Oxidation product) - In phosphine (PH$_3$), the oxidation state of P is -3. (Reduction product)
Since the oxidation state of P (+3) is both increased (+5) and decreased (-3), this is a disproportionation reaction.
The two products A and B are H$_3$PO$_4$ and PH$_3$.