White phosphorus reacts with caustic soda. The products are $P{{H}_{3}}$ and $Na{{H}_{2}}P{{O}_{2}}$. This reaction is an example of ----
(A) oxidation
(B) reduction
(C) neutralization
(D) disproportionation
Answer
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Hint: To get the answer, first try to write a balanced chemical reaction of phosphorus when it reacts with caustic soda. Then find the oxidation number of phosphorus in all the reactants and products containing it. If the oxidation number increases, the reaction is said to be oxidised while when it decreases, the reaction is reduced and when both occurs, the reaction is said to be disproportionate.
Complete step by step solution:
Given that.
The reactants are phosphorous with formula ${{P}_{4}}$ and caustic soda with chemical formula $NaOH$. Caustic soda is also called sodium hydroxide.
So, when phosphorus reacts with caustic soda, it produces phosphine ($P{{H}_{3}}$) and sodium hypophosphite ($Na{{H}_{2}}P{{O}_{2}}$). This is shown in the following balanced chemical reaction,
${{P}_{4}}+3NaOH+3{{H}_{2}}O\to P{{H}_{3}}+Na{{H}_{2}}P{{O}_{2}}$
Now, let us determine the oxidation numbers:
The oxidation number of phosphorus in white phosphorus (${{P}_{4}}$) is $0$.
Let us consider the oxidation number of phosphorus in phosphine ($P{{H}_{3}}$) be “X” while in sodium hypophosphite ($Na{{H}_{2}}P{{O}_{2}}$) be “Y”.
So, the oxidation number of hydrogens in phosphine is $+1$ and we can see that there are three hydrogen atoms in phosphine. Therefore, the oxidation number of phosphorus in phosphine will be $X+1\times 3=0$.
So, $X+3=0$
Thus, $X=-3$
Similarly, the oxidation number of sodium and hydrogen in sodium hypophosphite is $+1$ while that of oxygen is $-2$ and we can see that there is only one sodium atom, two hydrogen atoms and two oxygen atoms. So, the oxidation number of phosphorus will be $Y+1\times 1+1\times 2+(-2)\times 2=0$.
So, $Y+1+2-4=0$
Then, $Y-1=0$
And, $Y=+1$.
We know, when the oxidation number increases, the element is said to be oxidised while when the oxidation number decreases, the element is said to be reduced. And here, in the same reaction, the phosphorus is getting both oxidised and reduced and this type of reaction, where the same species is both oxidised and reduced is called a disproportionation reaction.
Hence, the correct option is (D).
Note: You need to be clear with the concepts to find the oxidation numbers and should have knowledge about the different reactions. Remember, when there is an increase in the oxidation number of a particular species in a reaction, the reaction is said to be oxidised while when the oxidation number decreases, the reaction is said to be reduced. When there is an occurrence of both oxidation and reduction, that reaction is said to be a redox or disproportionation reaction.
Complete step by step solution:
Given that.
The reactants are phosphorous with formula ${{P}_{4}}$ and caustic soda with chemical formula $NaOH$. Caustic soda is also called sodium hydroxide.
So, when phosphorus reacts with caustic soda, it produces phosphine ($P{{H}_{3}}$) and sodium hypophosphite ($Na{{H}_{2}}P{{O}_{2}}$). This is shown in the following balanced chemical reaction,
${{P}_{4}}+3NaOH+3{{H}_{2}}O\to P{{H}_{3}}+Na{{H}_{2}}P{{O}_{2}}$
Now, let us determine the oxidation numbers:
The oxidation number of phosphorus in white phosphorus (${{P}_{4}}$) is $0$.
Let us consider the oxidation number of phosphorus in phosphine ($P{{H}_{3}}$) be “X” while in sodium hypophosphite ($Na{{H}_{2}}P{{O}_{2}}$) be “Y”.
So, the oxidation number of hydrogens in phosphine is $+1$ and we can see that there are three hydrogen atoms in phosphine. Therefore, the oxidation number of phosphorus in phosphine will be $X+1\times 3=0$.
So, $X+3=0$
Thus, $X=-3$
Similarly, the oxidation number of sodium and hydrogen in sodium hypophosphite is $+1$ while that of oxygen is $-2$ and we can see that there is only one sodium atom, two hydrogen atoms and two oxygen atoms. So, the oxidation number of phosphorus will be $Y+1\times 1+1\times 2+(-2)\times 2=0$.
So, $Y+1+2-4=0$
Then, $Y-1=0$
And, $Y=+1$.
We know, when the oxidation number increases, the element is said to be oxidised while when the oxidation number decreases, the element is said to be reduced. And here, in the same reaction, the phosphorus is getting both oxidised and reduced and this type of reaction, where the same species is both oxidised and reduced is called a disproportionation reaction.
Hence, the correct option is (D).
Note: You need to be clear with the concepts to find the oxidation numbers and should have knowledge about the different reactions. Remember, when there is an increase in the oxidation number of a particular species in a reaction, the reaction is said to be oxidised while when the oxidation number decreases, the reaction is said to be reduced. When there is an occurrence of both oxidation and reduction, that reaction is said to be a redox or disproportionation reaction.
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