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There is very little difference in acid strength in the series ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$, ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{3}}}$ and ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{2}}}$ because:
A.Phosphorus in these acids exists in different oxidation states.
B.The hydrogen in these acids are not all bound to the phosphorous
C.Phosphorus is highly electronegative element
D.Phosphorus oxides are less basic

Answer
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Hint: We know that, the higher the oxidation state of the central atom in a compound, the more is the acidity of the compound. This can be connected to increased overall polarity of the molecule.

Complete step by step answer:
Considering the ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ molecule, the phosphorus atom is in $ + 5$ oxidation state, that is, its highest possible oxidation state and thus it tries to pull the ${\text{P}} - {\text{OH}}$ bond electrons towards itself because of the presence of high effective positive charge on it. To compensate for this, the electronegative oxygen atom pulls the electrons of the ${\text{O}} - {\text{H}}$ bond towards itself even more powerfully thus decreasing the bond strength of ${\text{O}} - {\text{H}}$ bond.
So ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ can easily lose a proton and is a stronger acid as compared to ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{3}}}$ in which the ${\text{O}} - {\text{H}}$ bond is a bit stronger and hence, it is a little more difficult to release the ${H^ + }$ ion.
We can make a similar comparison between ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{3}}}$ and ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{2}}}$ .
We can see that difference in the acidic strength depends on the electronegative character of the phosphorus atom. Since the electronegativity of phosphorus is not very high, thus the difference in the acidic strength of the three acids is very less.
Of the given options, the one that corresponds to the low electronegativity of phosphorus atoms is that it forms less basic oxides.

Thus, the correct option is D.

Note:
We can judge the acidities of the oxoacids of phosphorus by another method, that is, the stability of the conjugate bases of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ , ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{3}}}$ and ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{2}}}$ resulting after the loss of a proton.
The conjugate base of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ has three equivalent resonating structures as the negative charge is delocalized over three oxygen atoms. Thus, it has more stability as compared the conjugate base of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{3}}}$ which has two equivalent resonating structures and the conjugate base of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{2}}}$ which has only one resonating structure.