Question

The oxidation number of sulphur in plaster of Paris is?

Answer 1

Your answer bro:

Let oxidation state of S be "x"

Molecular formula of Plaster of Paris = CaSO₄.$\frac{1}{2}$H₂O

Now, the overall charge of the compound is 0, since there is no positive or negative charge over the compound

Calcium loses 2 electrons, so, its oxidation state is +2

Oxygwn gains 2 electrons, and there are 4 oxygen atoms (neglect the water molecule) gain 8 electrons

So, your answer will be:

+2 + x -8 = 0

x= +8-2 = +6 is the oxidation state of sulphur

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