Which of the following has/have been arranged in order of decreasing oxidation number of sulphur?

H_{2} S_{2} O_{7} > {N a}_{2} S_{4} O_{6} > {N a}_{2} S_{2} O_{3} > S_{8}

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S O^{2 +} > S O_{4}^{2 -} > S O_{3}^{2 -} > H S O_{4}^{⊖}

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H_{2} S O_{5} > H_{2} S O_{3} > S {C l}_{2} > H_{2} S

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H_{2} S O_{4} > S O_{2} > H_{2} S > H_{2} S_{2} O_{8}

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Solution

The correct options are
A $H_{2} S_{2} O_{7} > {N a}_{2} S_{4} O_{6} > {N a}_{2} S_{2} O_{3} > S_{8}$
B $H_{2} S O_{5} > H_{2} S O_{3} > S {C l}_{2} > H_{2} S$
Let $x$ be the oxidation state of S.
a. Oxidation state of $S$ in
$H_{2} S_{2} O_{7} : 2 + 2 x - 14 = 0 \Rightarrow x = 6$
${N a}_{2} S_{4} O_{6} : 2 + 4 x - 12 = 0 \Rightarrow x = 2.5$
${N a}_{2} S_{2} O_{3} : 2 + 2 x - 6 = 0 \Rightarrow x = 2$
$S_{8} : 8 x = 0 \Rightarrow x = 0$
Therefore, the decreasing order is $H_{2} S_{2} O_{7} > {N a}_{2} S_{4} O_{6} > {N a}_{2} S_{2} O_{3} > S_{8}$ .
c. $H_{2} S O_{5}$ (Peroxo linkage)
Oxidation state of $S = + 6$
$H_{2} S O_{3} : 2 + x - 6 = 0$
Oxidation state of $S = + 4$
$S {C l}_{2} : x - 2 = 0$
Oxidation state of $S = + 2$
$H_{2} S : 2 + x = 0$
Oxidation state of $S = - 2$
Therefore, the decreasing order is $H_{2} S O_{5} > H_{2} S O_{3} > S {C l}_{2} > H_{2} S$ .
Similarly, we can check for options b and d, we will find them to be incorrect.

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Similar questions

Out of $H_{2} S O_{4} (I), H_{2} S_{2} O_{7} (I I), H_{2} S O_{5} (I I I)$ and $H_{2} S_{2} O_{3} (I V)$ :

H_{2} S_{2} O_{3}

H_{2} S_{2} O_{4}

H_{2} S O_{5}

H_{2} S_{2} O_{8}

H_{2} S_{2} O_{8}

H_{2} S O_{5}

+ 6

{S F}_{4}

H_{2} S_{2} O_{7}

H_{2} S_{2} O_{8}

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