A) \[{{H}_{3}}P{{O}_{2}}\]
B) \[{{H}_{3}}P{{O}_{3}}\]
C) \[{{H}_{3}}P{{O}_{4}}\]
D) \[{{H}_{4}}{{P}_{2}}{{O}_{7}}\]
Correct Answer: C
Solution :
\[{{K}_{2}}S{{O}_{4}}+2N{{a}_{2}}S{{O}_{4}}+\underset{\text{Chromyl chloride}\,\,\,\,\,\,\,\,\,\,\,}{\mathop{2Cr{{O}_{2}}C{{l}_{2}}+3{{H}_{2}}O}}\,\] it is ionizes in three steps because three -OH group are present.
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