The true statement for the acids of phosphorus, $\mathrm{H}_3 \mathrm{PO}_2, \mathrm{H}_3 \mathrm{PO}_3$ and $\mathrm{H}_3 \mathrm{PO}_4$ is

• A

  the order of their acidity is $\mathrm{H}_3 \mathrm{PO}_4>\mathrm{H}_3 \mathrm{PO}_3>\mathrm{H}_3 \mathrm{PO}_2$
• B

  all of them are reducing in nature
• C

  all of them are tribasic acids
• Correct answer
  D

  the geometry of phosphorus is tetrahedra in all the three.

Solution

$\mathrm{H}_3 \mathrm{PO}_2, \mathrm{H}_3 \mathrm{PO}_3$ and $\mathrm{H}_3 \mathrm{PO}_4$ contain one, two and three ionisable hydrogen atoms respectively. $\begin{aligned} & \mathrm{H}_3 \mathrm{PO}_2 \rightleftharpoons \mathrm{H}^{+}+\mathrm{H}_2 \mathrm{PO}_2^{-} \\ & \mathrm{H}_3 \mathrm{PO}_3 \rightleftharpoons \mathrm{H}^{+}+\mathrm{H}_2 \mathrm{PO}_3^{-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HPO}_3^{2-} \\ & \mathrm{H}_3 \mathrm{PO}_4 \rightleftharpoons \mathrm{H}^{+}+\mathrm{H}_2 \mathrm{PO}_4^{-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HPO}_4^{2-} \\ & \rightleftharpoons \mathrm{H}^{+}+\mathrm{PO}_4^{3-} \\ & \end{aligned}$ But there is very little difference in acidity.
As $\mathrm{P}$ is $s p^3$ hybridised in all therefore all are tetrahedral.

Hence, the correct answer is option: 'D'.