Question

A balanced equation for the conversion of phosphorus to phosphoric acid is given by:
A. $P + 5HN{O_3} \to {H_2}O + {H_3}P{O_2} + 5N{O_2}$
B. $P + 5HN{O_3} \to {H_2}O + {H_2}P{O_5} + 5N{O_2}$
C. $P + 3HN{O_3} \to {H_2}O + {H_3}P{O_4} + 3N{O_2}$
D. $P + 5HN{O_3} \to {H_2}O + {H_3}P{O_4} + 5N{O_2}$

Answer 1

Hint: Phosphorus belongs to group $15$ of the periodic table and is placed just below Nitrogen.
-Its compounds with oxygen and hydrogen are known as oxoacids of phosphorus.
-One of its allotropes reacts with Nitric Acid to form an oxoacid of phosphorus which is tribasic.

Complete step by step solution:
In the given question in all the four reactions an allotrope of phosphorus is reacting with Nitric Acid $\left( {HN{O_3}} \right)$ to give an oxoacid of phosphorus, water and Nitrogen dioxide
As we Know that only one allotrope of phosphorus reacts with nitric acid to give an oxoacid of phosphorus which is “Red Phosphorus”
This is done by heating red phosphorus with $50\% \,\,HN{O_3}$ as follow:
$P + 5HN{O_3} \to {H_2}O + {H_3}P{O_4} + 5N{O_2}$

Hence, option “D” is the correct answer.

Additional Information:
- Red phosphorus is odorless and has a specific gravity of $2.1$. It is non-poisonous in nature, in contrast to the white phosphorus allotrope. Upon heating to temperatures above $673K$ red phosphorus undergoes crystallization.
-It does not glow in the dark and is insoluble in water.

Note:
-Remember “C” cannot be the answer even though the reactants and products are same as in option “D”
This is because the reaction is not balanced as shown below
$P + 3HN{O_3} \to {H_2}O + {H_3}P{O_4} + 3N{O_2}$
In this reaction we can see that the number of hydrogen atoms in the reactant side is three while on the product side it is five making the reaction unbalanced.
Red Phosphorus has a polymeric structure. It consists chains of ${P_4}$ tetrahedral linked together
-This polymeric links make red Phosphorus more stable and less reactive
This is the reason that the reaction
 $P + 5HN{O_3} \to {H_2}O + {H_3}P{O_4} + 5N{O_2}$
is done in the presence of a crystal of iodine which acts as a catalyst.