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Question

Find the difference in oxidation numbers of sulphur in the (SO4)2 ion and SO3 compound.


A
2
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B
0.0
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C
-2
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D
1
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Solution

The correct option is B 0

Let the oxidation number of sulphur be x in (SO4)2 and y in SO3.

Now the sulphate ion is given to us has a negative charge of -2 on it. This means that the sum of the oxidation numbers of all the elements in this ion is equal to -2 and not zero. The oxidation number of oxygen is -2. So let's calculate the value of x.

x+4×(2)=2

x8=2

x=+6

Now let's calculate the oxidation number for sulphur in SO3.

y + 3 times (-2) = 0

y - 6 = 0

y = +6

So the oxidation number of sulphur in (SO4)2 ion is +6.

Hence the difference of oxidation number of sulphur in (SO4)2 and SO3 is +6 - (+6) = 0

The answer is 0.


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