Question

\(P_4\) reacts with \(X\) selectively to give \(P_4O_6\). The \(X\) is

• A. A mixture of \(O_2\) and \(N_2\)
• B. \(KMnO_4\)
• C. Jones reagent
• D. \(K_2Cr_2O_7\)

Hint:

Phosphorus forms \(P_4O_6\) in limited oxygen and \(P_4O_{10}\) in excess oxygen.

Answer 1

The Correct Option is A

Concept:
Phosphorus forms two important oxides: \[ P_4O_6 \] and \[ P_4O_{10} \] The oxide formed depends on the amount of oxygen available. In limited oxygen, phosphorus gives: \[ P_4O_6 \] In excess oxygen, phosphorus gives: \[ P_4O_{10} \]

Step 1: Write the formation of \(P_4O_6\).
When phosphorus burns in limited oxygen: \[ P_4+3O_2 \rightarrow P_4O_6 \] So controlled oxygen is required.

Step 2: Why mixture of \(O_2\) and \(N_2\) is used.
A mixture of oxygen and nitrogen provides diluted oxygen. Nitrogen acts as an inert diluent. This controls the oxidation and selectively gives: \[ P_4O_6 \]

Step 3: Why other reagents are not suitable.
\(KMnO_4\), Jones reagent, and \(K_2Cr_2O_7\) are strong oxidizing agents. They are not used for selective preparation of \(P_4O_6\). Strong oxidation may lead to higher oxidation products.

Step 4: Check the options.
Option (A) is correct because controlled oxygen gives \(P_4O_6\).
Option (B), (C), and (D) are not suitable for this selective oxidation. Hence, the correct answer is: \[ \boxed{(A)\ \text{A mixture of }O_2\text{ and }N_2} \]