Question:
In the reaction of phosphorus with conc. \(HNO_3\), the oxidized and reduced products respectively are
In the reaction of phosphorus with conc. \(HNO_3\), the oxidized and reduced products respectively are
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Concentrated \(HNO_3\) usually gets reduced to \(NO_2\), while dilute \(HNO_3\) commonly forms \(NO\).
Updated On: Jun 22, 2026
- \(H_3PO_4,\; NO_2\)
- \(H_3PO_2,\; NO\)
- \(H_3PO_3,\; N_2O\)
- \(HPO_3,\; NO\)
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The Correct Option is A
Solution and Explanation
Step 1: Understand the nature of concentrated nitric acid.
Concentrated nitric acid \((HNO_3)\) is a strong oxidizing agent.
When phosphorus reacts with concentrated nitric acid, phosphorus gets oxidized while nitric acid gets reduced.
Step 2: Determine the oxidized product of phosphorus.
Elemental phosphorus has oxidation state:
\[ 0 \] During the reaction, phosphorus is oxidized to phosphoric acid:
\[ H_3PO_4 \] Now calculate the oxidation state of phosphorus in \(H_3PO_4\):
Let oxidation state of phosphorus be \(x\).
\[ 3(+1)+x+4(-2)=0 \] \[ 3+x-8=0 \] \[ x=+5 \] Thus, phosphorus changes from:
\[ 0 \longrightarrow +5 \] Hence, phosphorus undergoes oxidation and forms:
\[ H_3PO_4 \]
Step 3: Determine the reduced product of nitric acid.
In concentrated nitric acid, nitrogen has oxidation state:
\[ +5 \] Concentrated \(HNO_3\) generally gets reduced to nitrogen dioxide \((NO_2)\).
Now calculate oxidation state of nitrogen in \(NO_2\):
\[ x+2(-2)=0 \] \[ x-4=0 \] \[ x=+4 \] Thus, nitrogen changes from:
\[ +5 \longrightarrow +4 \] Hence, \(HNO_3\) is reduced to:
\[ NO_2 \]
Step 4: Write the reaction.
The reaction can be represented as:
\[ P + HNO_3 \longrightarrow H_3PO_4 + NO_2 + H_2O \] Thus, the oxidized product is \(H_3PO_4\) and the reduced product is \(NO_2\).
Step 5: Match with the given options.
The correct option is:
\[ (1)\; H_3PO_4,\; NO_2 \]
Step 6: Final conclusion.
Hence, the correct answer is:
\[ \boxed{H_3PO_4,\; NO_2} \]
Concentrated nitric acid \((HNO_3)\) is a strong oxidizing agent.
When phosphorus reacts with concentrated nitric acid, phosphorus gets oxidized while nitric acid gets reduced.
Step 2: Determine the oxidized product of phosphorus.
Elemental phosphorus has oxidation state:
\[ 0 \] During the reaction, phosphorus is oxidized to phosphoric acid:
\[ H_3PO_4 \] Now calculate the oxidation state of phosphorus in \(H_3PO_4\):
Let oxidation state of phosphorus be \(x\).
\[ 3(+1)+x+4(-2)=0 \] \[ 3+x-8=0 \] \[ x=+5 \] Thus, phosphorus changes from:
\[ 0 \longrightarrow +5 \] Hence, phosphorus undergoes oxidation and forms:
\[ H_3PO_4 \]
Step 3: Determine the reduced product of nitric acid.
In concentrated nitric acid, nitrogen has oxidation state:
\[ +5 \] Concentrated \(HNO_3\) generally gets reduced to nitrogen dioxide \((NO_2)\).
Now calculate oxidation state of nitrogen in \(NO_2\):
\[ x+2(-2)=0 \] \[ x-4=0 \] \[ x=+4 \] Thus, nitrogen changes from:
\[ +5 \longrightarrow +4 \] Hence, \(HNO_3\) is reduced to:
\[ NO_2 \]
Step 4: Write the reaction.
The reaction can be represented as:
\[ P + HNO_3 \longrightarrow H_3PO_4 + NO_2 + H_2O \] Thus, the oxidized product is \(H_3PO_4\) and the reduced product is \(NO_2\).
Step 5: Match with the given options.
The correct option is:
\[ (1)\; H_3PO_4,\; NO_2 \]
Step 6: Final conclusion.
Hence, the correct answer is:
\[ \boxed{H_3PO_4,\; NO_2} \]
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