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Standard XII
Chemistry
Question
The oxidation state of sulphur in the anions
S
O
2
−
3
,
S
2
O
2
−
4
and
S
2
O
2
−
6
follows the order :
S
O
2
−
3
<
S
2
O
2
−
4
<
S
2
O
2
−
6
S
2
O
2
−
6
<
S
2
O
2
−
4
<
S
O
2
−
3
S
2
O
2
−
4
<
S
O
2
−
3
<
S
2
O
2
−
6
S
2
O
2
−
4
<
S
2
O
2
−
6
<
S
O
2
−
3
A
S
2
O
2
−
6
<
S
2
O
2
−
4
<
S
O
2
−
3
B
S
O
2
−
3
<
S
2
O
2
−
4
<
S
2
O
2
−
6
C
S
2
O
2
−
4
<
S
2
O
2
−
6
<
S
O
2
−
3
D
S
2
O
2
−
4
<
S
O
2
−
3
<
S
2
O
2
−
6
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Solution
Verified by Toppr
O.S. of S in
S
O
2
−
3
,
x
−
6
=
−
2
x
=
+
4
O.S. of S in
S
2
O
2
−
4
,
2
x
−
8
=
−
2
x
=
3
O.S. of S in
S
2
O
2
−
6
,
2
x
−
12
=
2
x
=
+
5
The correct order of increasing O.S. is;
S
2
O
2
−
4
<
S
O
2
−
3
<
S
2
O
2
−
6
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The oxidation states of sulphur in the anions
S
O
2
−
3
,
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−
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a
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−
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and follows the order:
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