Question

Upon heating ${\rm{KCl}}{{\rm{O}}_{\rm{3}}}$ in the presence of catalysis amount of ${\rm{Mn}}{{\rm{O}}_{\rm{2}}}$, a gas is formed ${\rm{W}}$ is formed . Excess amount of ${\rm{W}}$ reacts with white phosphorus to give ${\rm{X}}$. The reaction of ${\rm{X}}$ with ${\rm{HN}}{{\rm{O}}_{\rm{3}}}$ gives ${\rm{Y}}$and ${\rm{Z}}$. ${\rm{W}}$ and ${\rm{X}}$ are, respectively
(A) ${{\rm{O}}_2}$and ${{\rm{P}}_4}{{\rm{O}}_{10}}$
(B) ${{\rm{O}}_2}$and ${{\rm{P}}_4}{{\rm{O}}_6}$
(C) ${{\rm{O}}_3}$and ${{\rm{P}}_4}{{\rm{O}}_6}$
(D) ${{\rm{O}}_3}$and ${{\rm{P}}_4}{{\rm{O}}_{10}}$

Answer 1

As we know that ${\rm{KCl}}{{\rm{O}}_{\rm{3}}}$when heated it release a gas which is diatomic. Phosphorus reacts very fast with this gas and forms a compound which is very shiny in the dark. It is also found where the human body burns.

Complete step by step answer
When ${\rm{KCl}}{{\rm{O}}_{\rm{3}}}$ is heated in the presence of catalyst it releases oxygen gas as
${\rm{2}}\,{\rm{KCl}}{{\rm{O}}_{\rm{3}}} \to \,{\rm{2}}\,{\rm{KCl + }}\,{\rm{3}}\,{{\rm{O}}_{\rm{2}}}{\rm{(g)}}$
So, ${\rm{W}}$ is oxygen gas.
When an excess amount of oxygen gas reacts with white phosphorus, it gives a compound whose name is phosphorus penta-oxide. The reaction occurs as-
${{\rm{P}}_{\rm{4}}}{\rm{(white}}\,{\rm{phosphorous) + }}\,{\rm{5}}{{\rm{O}}_{\rm{2}}} \to \,{{\rm{P}}_{\rm{4}}}{{\rm{O}}_{{\rm{10}}}}{\rm{(Phosphorous}}\,{\rm{pentaoxide)}}$
phosphorus penta-oxide is a compound which is the oxidized form of phosphorus, so when the human body burns phosphorus it reacts with oxygen, so due to oxidation of phosphorus, phosphorus penta-oxide shines in the dark.
Therefore, ${\rm{X}}$ is ${{\rm{P}}_{\rm{4}}}{{\rm{O}}_{{\rm{10}}}}$.
Now, ${\rm{X}}$ reacts with ${\rm{HN}}{{\rm{O}}_{\rm{3}}}$, it gives as
${{\rm{P}}_{\rm{4}}}{{\rm{O}}_{{\rm{10}}}}\,{\rm{ + }}\,{\rm{HN}}{{\rm{O}}_{\rm{3}}} \to {{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}{\rm{ + }}\,{\rm{HP}}{{\rm{O}}_{\rm{3}}}$
Where ${{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{5}}}$ is ${\rm{Y}}$ and ${\rm{HP}}{{\rm{O}}_{\rm{3}}}$ is ${\rm{Z}}$. Both ${\rm{Y}}$ and ${\rm{Z}}$ are known as nitrogen-pentoxide and metaphosphoric acid respectively. ${\rm{Y}}$ is also known as oxide of nitrogen and ${\rm{Z}}$ is known as oxoacid of phosphorus.

**Therefore, our correct option is option(A).

Note: **
When any atom reacts with oxygen then it is known as oxidation, in the oxidation process atom loses its electrons because oxygen is a second most electronegative atom in periodic table after fluorine. We can calculate that an atom loses how many electrons from its outermost shell by calculating oxidation number.