Question

Among the following the number of compounds that can react with $PC{{l}_{5}}$ to give $POC{{l}_{3}}$ are:
${{O}_{2}},\text{ }C{{O}_{2}},\text{ }S{{O}_{2}},\text{ }{{H}_{2}}O,\text{ }{{H}_{2}}S{{O}_{4}},\text{ }{{P}_{4}}{{O}_{10}}$
(A) 4
(B) 5
(C) 3
(D) 6

Answer 1

Hint: The compounds that are most stable in nature will not react with $PC{{l}_{5}}$ to form $POC{{l}_{3}}$. The reaction of sulfuric acid with $PC{{l}_{5}}$ forms very unusual compounds.

Complete step by step solution:
Let us see all the reactions of the given compound with $PC{{l}_{5}}$.
(i)- ${{O}_{2}}$
This is oxygen. It is a very stable compound. The oxidation state is 0. So, when it combines with $PC{{l}_{5}}$ it doesn’t form $POC{{l}_{3}}$. Because if it forms $POC{{l}_{3}}$, the oxidation state of oxygen is -2, which means that it is gaining electrons but there is no donor atom. So, this reaction doesn’t occur.
(ii)- $C{{O}_{2}}$
This is carbon dioxide. It is also a very stable compound and it also does not react with $PC{{l}_{5}}$ to form $POC{{l}_{3}}$.
(iii)- $S{{O}_{2}}$
This compound is sulfur dioxide. It combines with $PC{{l}_{5}}$ to form products. The products formed are $POC{{l}_{3}}$ and $SOC{{l}_{2}}$. The reaction is given below:
$PC{{l}_{5}}+S{{O}_{2}}\to POC{{l}_{3}}+SOC{{l}_{2}}$
(iv)- ${{H}_{2}}O$
This compound is water. It combines with $PC{{l}_{5}}$ to form products. The products formed are $POC{{l}_{3}}$ and $HCl$ . The reaction is given below:
\[PC{{l}_{5}}+{{H}_{2}}O\to POC{{l}_{3}}+2HCl\]
(v)- ${{H}_{2}}S{{O}_{4}}$
This compound is sulfuric acid. It combines with $PC{{l}_{5}}$ to form products. They give very unusual products. The products formed are $S{{O}_{2}}C{{l}_{2}}$, $POC{{l}_{3}}$ and $HCl$. The reaction is given below:
$PC{{l}_{5}}+{{H}_{2}}S{{O}_{4}}\to S{{O}_{2}}C{{l}_{2}}+2POC{{l}_{3}}+2HCl$
(vi)- ${{P}_{4}}{{O}_{10}}$
This is known as phosphorus pentoxide. It combines with $PC{{l}_{5}}$ to form products. The product formed is $POC{{l}_{3}}$. The reaction is given below:
$6PC{{l}_{5}}+{{P}_{4}}{{O}_{10}}\to 10POC{{l}_{3}}$
So from the above discussion 4 compounds react with $PC{{l}_{5}}$ to form $POC{{l}_{3}}$.

Therefore, the correct answer is an option (A) 4.

Note: The compound formed $S{{O}_{2}}C{{l}_{2}}$ when sulfuric acid reacts with $PC{{l}_{5}}$ is known as Sulfuryl chloride. $PC{{l}_{5}}$ also reacts with carboxylic acids and alcohols to give $POC{{l}_{3}}$ as a by-product.