Question

Phosphorus acid ($\mathrm{H_3PO_3}$) on heating undergoes disproportionation to give:

• A. $\mathrm{H_3PO_4}$ and $\mathrm{PH_3}$
• B. $\mathrm{HPO_3}$ and $\mathrm{P_2O_5}$
• C. $\mathrm{H_4P_2O_7}$ and $\mathrm{PH_3}$
• D. $\mathrm{H_3PO_2}$ and $\mathrm{P_4}$

Hint:

Remember the important disproportionation reactions: \[ 4\mathrm{H_3PO_3} \rightarrow 3\mathrm{H_3PO_4} +\mathrm{PH_3} \] \[ 2\mathrm{H_3PO_2} \rightarrow \mathrm{H_3PO_4} +\mathrm{PH_3} \] These are frequently asked in competitive examinations.

Answer 1

The Correct Option is A

Concept: Disproportionation is a redox reaction in which the same element present in one oxidation state simultaneously undergoes oxidation as well as reduction to form products containing higher and lower oxidation states. Phosphorous acid, $\mathrm{H_3PO_3}$, contains phosphorus in the oxidation state +3. On heating, a part of phosphorus is oxidized to +5 oxidation state while another part is reduced to $-3$ oxidation state.

Step 1: Determine oxidation state of phosphorus in $\mathrm{H_3PO_3}$. Let oxidation state of phosphorus be $x$. \[ x+3(+1)+3(-2)=0 \] \[ x+3-6=0 \] \[ x=+3 \] Thus phosphorus exists in the +3 oxidation state.

Step 2: Write the disproportionation reaction. Upon heating, \[ 4\mathrm{H_3PO_3} \longrightarrow 3\mathrm{H_3PO_4} +\mathrm{PH_3} \] Here, \[ \mathrm{P}(+3)\rightarrow \mathrm{P}(+5) \] and \[ \mathrm{P}(+3)\rightarrow \mathrm{P}(-3) \] showing simultaneous oxidation and reduction.

Step 3: Identify the products. The products formed are: \[ \mathrm{H_3PO_4} \] and \[ \mathrm{PH_3}. \] Hence, \[ \boxed{\text{Option (1)}} \]