The correct order of acidic strength is :

C O_{2} > N_{2} O_{5} > S O_{3}

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

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K_{2} O > C a O > M g O

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N a_{2} O > M g O > A l_{2} O_{3}

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Solution

The correct option is B $C l_{2} O_{7} > S O_{3} > P_{4} O_{10}$
Acidic strength is directly propertional to the oxidation state
(A)
$C l_{2} O_{7} : 2 x + 7 (- 2) = 0 \Rightarrow x = \frac{14}{2} = + 7$
$S O_{3} : x + 3 (- 2) = 0$
$\Rightarrow x = + 6$
$P_{4} O_{10} : 4 x + 10 (- 2) = 0$
$\Rightarrow 4 x = 20$
$\Rightarrow x = + 5$
Acidic strength order will be $C l_{2} O_{7} > S O_{3} > P_{4} O_{10}$

(B)
$C O_{2} : x + 2 (- 2) = 0$
$\Rightarrow x = + 4$
$N_{2} O_{5} : 2 x + 5 (- 2) = 0$
$\Rightarrow x = \frac{10}{2} = + 5$
$S O_{3} : x - 6 = 0 \Rightarrow x = + 6$

(C)
$N a_{2} O : 2 (x) + (- 2) = 0$
$\Rightarrow x = + 1$
$M g O : x - 2 = 0 \Rightarrow x = + 2$
$A l_{2} O_{3} : 2 x + 3 (- 2) = 0$
$\Rightarrow x = \frac{6}{2} = + 3$
$Acidic strength order A l_{2} O_{3} > M g O > N a_{2} O$

(D)
$K_{2} O : 2 (x) + (- 2) = 0$
$\Rightarrow x = + 1$
$C a O : Oxidation state of C a = + 2$
$M g O : Oxidation state of M g = + 2$
$Acidic strength : C a O \approx M g O > K_{2} O$
Hence, Correct option is (A)

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