
White P reacts with caustic soda, the products are $\text{P}{{\text{H}}_{3}}\text{ and Na}{{\text{H}}_{2}}\text{P}{{\text{O}}_{2}}$. This reaction involves:
The question has multiple correct questions.
A. Oxidation
B. Disproportionation
C. Reduction
D. Neutralisation
Answer
486.3k+ views
Hint: Disproportionation reactions are those reactions in which a single reaction can undergo the oxidation as well as reduction. Reduction reactions are those reactions in which the loss of oxygen or gain of hydrogen take place and vice versa.
Complete step by step answer:
- It is given in the question that white phosphorus reacts with caustic soda, here the molecular formula of white phosphorous and caustic soda are
$\begin{align}& {{\text{P}}_{4}}\text{ and NaOH} \\ & \\ \end{align}$.
- White phosphorus is one of the allotropes of the phosphorus.
- When white phosphorus reacts with caustic soda it yields phosphine, a toxic gas and sodium phosphite.
- The balanced chemical reaction is:
${{\text{P}}_{4}}\text{ + NaOH }\to \text{ P}{{\text{H}}_{3}}\,\text{+ Na}{{\text{H}}_{3}}\text{P}{{\text{O}}_{2}}$
- Here, the oxidation state of the phosphorus on the reactant side is zero because single elements always have zero oxidation state.
- But in the product side, the oxidation of phosphorus in phosphine and sodium phosphite is: -3 and +1 i.e.
$\begin{align}
& \text{Phosphine Sodium Phosphite} \\
& \text{x + 3 = 0 1 + 2 + x - 4 = 0} \\
& \text{x = -3 x = +1} \\
\end{align}$
- So, we can see that there are only molecules of phosphate on the reactant side which undergoes reduction as well as an oxidation reaction.
-Because the oxidation state of phosphorus changes from 0 to -3 and +1 from reactant to the product side.
-Moreover, we know that those compounds which undergo both oxidation and reduction reaction, then the reaction is known as disproportionation reaction.
Therefore, option A, B and C are the correct answer.
Note: White phosphorus is the allotrope of phosphorus other than yellow and red phosphorus. White phosphorus is useful to work as a weapon in the munitions. On burning it gives red phosphorus.
Complete step by step answer:
- It is given in the question that white phosphorus reacts with caustic soda, here the molecular formula of white phosphorous and caustic soda are
$\begin{align}& {{\text{P}}_{4}}\text{ and NaOH} \\ & \\ \end{align}$.
- White phosphorus is one of the allotropes of the phosphorus.
- When white phosphorus reacts with caustic soda it yields phosphine, a toxic gas and sodium phosphite.
- The balanced chemical reaction is:
${{\text{P}}_{4}}\text{ + NaOH }\to \text{ P}{{\text{H}}_{3}}\,\text{+ Na}{{\text{H}}_{3}}\text{P}{{\text{O}}_{2}}$
- Here, the oxidation state of the phosphorus on the reactant side is zero because single elements always have zero oxidation state.
- But in the product side, the oxidation of phosphorus in phosphine and sodium phosphite is: -3 and +1 i.e.
$\begin{align}
& \text{Phosphine Sodium Phosphite} \\
& \text{x + 3 = 0 1 + 2 + x - 4 = 0} \\
& \text{x = -3 x = +1} \\
\end{align}$
- So, we can see that there are only molecules of phosphate on the reactant side which undergoes reduction as well as an oxidation reaction.
-Because the oxidation state of phosphorus changes from 0 to -3 and +1 from reactant to the product side.
-Moreover, we know that those compounds which undergo both oxidation and reduction reaction, then the reaction is known as disproportionation reaction.
Therefore, option A, B and C are the correct answer.
Note: White phosphorus is the allotrope of phosphorus other than yellow and red phosphorus. White phosphorus is useful to work as a weapon in the munitions. On burning it gives red phosphorus.
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