Find the oxidation states of S atoms in the following compounds and arrange them in decreasing order of maximum O-S- shown by S in them-A- H 2 S 2 O 7- Na 2 S 4 O 6- SO 2- S 8- H 2 SB- H 2 S 2 O 7- Na 2 S 4 O 6- SO 2- H 2 S - S 8- Na 2 S 4 O 6- H 2 S 2 O 7- SO 2- H 2 S - S 8D- Na 2 S 4 O 6- H 2 S 2 O 7- SO 2- S 8- H 2 S

The correct option is A H_2S_2O_7 gt;Na_2S_4O_6 gt;SO_2 gt;S_8 gt;H_2Snbsp;Molecule amp;H_2S_2O_7 amp; Na_2S_4O_6 amp; SO_2 amp;S_8 amp;H_2 ...

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Covalent Bond

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Question

Find the oxidation states of S atoms in the following compounds and arrange them in decreasing order of maximum O.S. shown by S in them:

H_{2} S_{2} O_{7} > N a_{2} S_{4} O_{6} > S O_{2} > S_{8} > H_{2} S

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H_{2} S_{2} O_{7} > N a_{2} S_{4} O_{6} > S O_{2} > H_{2} S > S_{8}

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N a_{2} S_{4} O_{6} > H_{2} S_{2} O_{7} > S O_{2} > H_{2} S > S_{8}

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N a_{2} S_{4} O_{6} > H_{2} S_{2} O_{7} > S O_{2} > S_{8} > H_{2} S

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S -

s p^{3}

S O_{3}

H_{2} S

S O_{2}

S_{8}

$H_{2} S$ acts only as a reducing agent while $S O_{2}$ can act both as a reducing and oxidizing agent because

2 H_{2} S + S O_{2} \to 3 S + 2 H_{2} O

H_{2} S

S O_{2}

H_{2} S

S O_{2}

S O_{2}

S O_{2} + H_{2} S \to S + H_{2} O

H_{2} S

S O_{2}

2 H_{2} S + S O_{2} \to 2 H_{2} O + 3 S

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