Rank the following oxides in order of increasing acidity in water: ${\mathbf{Sb}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\mathbf{,}{\mathbf{Bi}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\mathbf{,}{\mathbf{P}}_{\mathbf{4}}{\mathbf{O}}_{\mathbf{10}\mathbf{,}}{\mathbf{Sb}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{5}}$

Group 15 elements belonging to the nitrogen family. They include nitrogen, phosphorus, arsenic, antimony, bismuth, and moscovium.

The atomic number of the group 15 elements is as follows;

Nitrogen (N) = 7

Phosphorus (P) = 15

Arsenic (As) = 33

Antimony (Sb) = 51

Bismuth (Bi) = 83

As we move down the group, the atomic number increase. In a group, electronegativity decreases, and metallic character increases as we move down. So that acidic character also decreases down the group. Hence among the four${\mathrm{P}}_4{\mathrm{O}}_{10}$is the most acidic one and as Bi present at the bottom so it is less acidic.

For the different oxides of the same element, acidity increase with an increase in the oxidation state. For ${\mathrm{Sb}}_2{\mathrm{O}}_5$oxidation state of $\mathrm{Sb}$is +5 while for ${\mathrm{Sb}}_2{\mathrm{O}}_3$oxidation state of $\mathrm{Sb}$is +3. Hence ${\mathrm{Sb}}_2{\mathrm{O}}_5$is more acidic than ${\mathrm{Sb}}_2{\mathrm{O}}_3$.