The oxidation state can be calculated as :

H₄P₂O₅ : +4 + 2x + 5(– 2) = 0 $$ \Rightarrow $$ 2x – 6 = 0 $$ \Rightarrow $$ x = +3

H₄P₂O₆ : +4 + 2x + 6(– 2) = 0 $$ \Rightarrow $$ 2x – 8 = 0 $$ \Rightarrow $$ x = +4

H₄P₂O₇ : +4 + 2x + 7(– 2) = 0 $$ \Rightarrow $$ 2x – 10 = 0 $$ \Rightarrow $$ 2x = 10 $$ \Rightarrow $$ x = +5

The correct order of increasing bond angles is

ClO$$_2^ - $$ < Cl₂O < ClO₂

In BF₃, p-p overlap between B and F is maximum due to identical size and energy of p-orbitals, so electron deficiency in boron of BF₃ is neutralized partially to the maximum extent by back donation. Also, the tendency to back donate decreases from F to I. So the order will be:

BBr₃ > BCl₃ > BF₃

Among the given molecules, only diborane(B₂H₆) is electron deficient i.e., it does not complete octet.

Thus, it acts as a Lewis acid.

NH₃ and H₂O being electron rich species behave as Lewis base.