A) \[HP{{O}_{3}}\]
B) \[{{H}_{3}}P{{O}_{3}}\]
C) \[{{H}_{3}}P{{O}_{4}}\]
D) \[P{{H}_{3}}\]
Correct Answer: C
Solution :
Cone. \[HN{{O}_{3}}\] oxidizes non-metals (C, P, I etc.) into their oxyacids. Hence, phosphorus is oxidised to orthophosphoric acid, when treated with cone. \[HN{{O}_{3}}\]. \[P+5HN{{O}_{3}}\,\,\,\,\xrightarrow{{}}\underset{\begin{smallmatrix} \text{Orthophoshoric} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{acid} \end{smallmatrix}}{\mathop{{{H}_{3}}P{{O}_{4}}}}\,+5N{{O}_{2}}+{{H}_{2}}O\]
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