The correct order of acidic character of oxides in the third period of the periodic table is:

S i O_{2} < P_{4} O_{10} < S O_{3} < C l_{2} O_{7}

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C l_{2} O_{7} < S O_{3} < P_{4} O_{10} < S i O_{2}

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S O_{3} < C l_{2} O_{7} < P_{4} O_{10} < S i O_{2}

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S i O_{2} < C l_{2} O_{7} < P_{4} O_{10} < S O_{3}

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Solution

The correct option is A $S i O_{2} < P_{4} O_{10} < S O_{3} < C l_{2} O_{7}$
Acidic oxide strength increases as the non-metallic character of the element increases. Therefore it increases from left to right in a period.
$S i O_{2} < P_{4} O_{10} < S O_{3} < C l_{2} O_{7}$

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Similar questions

Among $A l_{2} O_{3}, S i O_{2}, P_{2} O_{3}, a n d S O_{3}$ , the correct order of acid strength is:

P_{4} O_{10}, C O, S O_{2}, C l O_{2}, C l_{2} O, N_{2} O, N O_{2}, C l_{2} O_{6}, C O_{2}, C l_{2} O_{7}, S O_{3}

P_{4} O_{10}, C l_{2} O_{7}, I_{2} O_{5}

N a_{2} O

S i O_{2}

P_{4} O_{10}

N a_{2} O

S i O_{2}

P_{4} O_{10}

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