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Valency and Oxidation State

Which of the ...

Question

Which of the following have been arrange in order of decreasing oxidation number of sulphur?

A

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

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B

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

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C

H_{2} S O_{5} > H_{2} S O_{3} > S C l_{2} > H_{2} S

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D

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
B $N a_{2} S_{4} O_{6} > H_{2} S_{2} O_{7} > N a_{2} S_{2} O_{3} > S_{8}$
D $H_{2} S O_{5} > H_{2} S O_{3} > S C l_{2} > H_{2} S$
1.Oxidation number of $S$ in $H_{2} S_{2} O_{7}$ is +6,in $N a_{2} S_{4} O_{6}$ +2.5,in $N a_{2} S_{2} O_{3}$ is +2 and in $S_{8}$ is 0.
Oxidation number of sulphur in decreasing order: $H_{2} S_{2} O_{7}$ > $N a_{2} S_{4} O_{6}$ > $N a_{2} S_{2} O_{3}$ > $S_{8}$ .
2.Oxidation number of $S$ in $S O_{2}^{+}$ is +5,in $S O_{4}^{2 -}$ 6,in $S O_{3}^{2 -}$ is +4 and in $H S O_{4}^{-}$ is +6.
Oxidation number of sulphur in decreasing order: $H S O_{4}^{-}$ = $S O_{4}^{2 -}$ > $S O_{2}^{+}$ > $S O_{3}^{2 -}$ .
3.Oxidation number of $S$ in $H_{2} S O_{5}$ is +8,in $H_{2} S O_{3}$ +4,in $S C l_{2}$ is +2 and in $H_{2} S$ is -2.
Oxidation number of sulphur in decreasing order: $H_{2} S O_{5}$ > $H_{2} S O_{3}$ > $S C l_{2}$ > $H_{2} S$ .
4.Oxidation number of $S$ in $H_{2} S O_{4}$ is +6,in $S O_{2}$ +4,in $H_{2} S$ is -2 and in $H_{2} S_{2} O_{8}$ is +7.
Oxidation number of sulphur in decreasing order: $H_{2} S_{2} O_{8}$ > $H_{2} S O_{4}$ > $S O_{2}$ > $H_{2} S$ .
So A and C are correct answer.

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Q. 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:

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0.20

M solution of sulphuric acid?
Given:

H_{2} S O_{4} \to H^{+} + H S O_{4}^{-}

H S O_{4}^{-} ⇌ S O_{4}^{2 -}; K_{2} = 10^{- 2}

Q. Assertion :In the following reaction:

S O_{2} + 2 H_{2} S ⟶ 3 S + 2 H_{2} O

1

S O_{2}

1

H_{2} S

3

moles sulphur. Reason:

1

S O_{2}

2

H_{2} S

3

moles sulphur and

2

H_{2} O

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

In the above reaction:
(a)

H_{2} S

gets oxidized.
(b)

S O_{2}

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(c)

H_{2} S

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(d)

S O_{2}

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