The strength of the oxyacids decreases in the following order

H N O_{3} > H_{3} P O_{4} > H_{3} A s O_{4} > H_{3} S b O_{4}

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H N O_{3} > H_{3} P O_{4} > H_{3} S b O_{4} > H_{3} A s O_{4}

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H N O_{3} > H_{3} P O_{4} = H_{3} A s O_{4} > H_{3} S b O_{4}

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H N O_{3} > H_{3} P O_{4} > H_{3} A s O_{4} = H_{3} S b O_{4}

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Solution

The correct option is A $H N O_{3} > H_{3} P O_{4} > H_{3} A s O_{4} > H_{3} S b O_{4}$
Since all the elements are in the same oxidation state, electronegativity has to be considered. Electronegativity of the elements from N to Sb decreases generally. Thus N with high Electronegativity can donate the proton easily where as for Sb with least Electronegativity becomes difficult to donate proton. Thus (A) is the correct order.

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