Question

When phosphorus reacts with caustic soda, the products are $P{H_3}$ and $Na{H_2}P{O_2}$. This is reaction is an example of
(A) Oxidation
(B) Reduction
(C) Oxidation and reduction (Redox)
(D) Neutralization

Answer 1

Hint: The simplest way to recognize whether a reaction is an oxidation reaction or a reduction reaction is to find its oxidation number.
We can rule out the possibility of option (D) Neutralization as the neutralization reaction involves the mixture of acid and base to form salt and water.
Here, ${P_4}$ i.e. Phosphorus is basic in nature as well as $NaOH$ i.e. caustic soda is also basic.

Complete step by step answer:
The reaction between phosphorus and caustic soda is given as follows –
${P_4} + 3NaOH + 3{H_2}O \to 3Na{H_2}P{O_2} + 2P{H_3}$
We first need to find the oxidation number of reactants and products
The oxidation number of Phosphorus in${P_4}$
For elements in a free state, the oxidation number is 0
Here ${P_4}$ is in a free state
Hence, the Oxidation number of ${P_4}$ $ = $0
The oxidation number of Phosphorus in $P{H_3}$
$P + 3\left( H \right) = 0$
$P + 3\left( { + 1} \right) = 0$
$\Rightarrow P + 3 = 0$
$\Rightarrow P = - 3$
Hence, the oxidation number of phosphorus in $P{H_3}$ is $ - $3
The oxidation number of phosphorus in $Na{H_2}P{O_2}$
$Na + 2\left( H \right) + P + 2\left( O \right) = 0$
$\Rightarrow \left( { + 1} \right) + \left( { + 2} \right) + P + 2\left( { - 2} \right) = 0$
$\Rightarrow 1 + 2 + P - 4 = 0$
$\Rightarrow P - 1 = 0$
$\Rightarrow P = + 1$
Hence, the oxidation number of phosphorus in $Na{H_2}P{O_2}$ is $ + $1
Now, we know that the increase in oxidation number indicates an oxidation reaction.
And similarly, the decrease in oxidation number indicates a reduction reaction.
So, the formation of $P{H_3}$ from ${P_4}$ (change in oxidation number from 0 to $ - $ 3) is a reduction reaction. (decrease in oxidation number)
And the formation of $Na{H_2}P{O_2}$ from ${P_4}$ (change in oxidation number from 0 to $ + $1) is an oxidation reaction. (Increase in oxidation number)
So clearly, this reaction involves both oxidation and reduction reactions simultaneously.
Such reactions are called a redox reaction.

So, the correct answer is Option C .

Note: To be more precise, this reaction is a disproportionation reaction, a type of redox reaction.
Redox reaction involves simultaneous oxidation and reduction of two different reactants in the same reaction.
For example, $CuO + {H_2} \to Cu + {H_2}O$
Here, $CuO$is reduced to $Cu$and ${H_2}$ is oxidized to${H_2}O$. $CuO$ and ${H_2}$ are two different molecules.
Disproportionation reaction involves simultaneous oxidation and reduction of the same reactant in the same reaction.
For example, $2{H_2}{O_2} \to {H_2}O + {O_2}$
Here, ${H_2}{O_2}$ is undergoing both reduction and oxidation reactions.