A) \[{{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{3}}}\]
B) \[{{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{2}}\]
C) \[{{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{4}}\]
D) \[{{H}_{4}}{{P}_{2}}{{O}_{7}}\]
Correct Answer: C
Solution :
With excess of water, both \[\frac{{{\text{V}}_{\text{A}}}}{{{\text{V}}_{\text{N}}}}\text{=}\frac{\frac{\text{4}}{\text{3}}\text{ }\!\!\pi\!\!\text{ r}_{\text{A}}^{\text{3}}}{\frac{\text{4}}{\text{3}}\text{ }\!\!\pi\!\!\text{ }_{\text{N}}^{\text{3}}}\] and \[={{\left( \frac{{{r}_{A}}}{{{r}_{N}}} \right)}^{3}}={{\left( \frac{{{10}^{-10}}}{{{10}^{-15}}} \right)}^{3}}\] give \[\Rightarrow \] \[{{V}_{A}}:{{V}_{N}}={{10}^{15}}:1\] \[{{T}_{1/2}}=4.47\times {{10}^{8}}yr\]
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