Question

$KAl(SO\frac{}{4}) \frac{}{2} .12H\frac{}{2} O$
What is the oxidation number of the underlined element (S)? Please help!
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Answer 1

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Question:

Oxidation state of S in K[Al(SO₄)₂].12H₂O

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Answer and explanation

Leave the water molecules, because their overall charge is zero

Let the charge of S atom be 'x'

As we know that potassium loses one electron, and aluminum loses 3 electrons, and oxygen gains 2 electrons, it will be a negative charge on it

And, there are two sulphate entities in this compound

Also, the total charge of the compound is zero

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So, your answer is:

+1 +3 + 2[x - (4 × 2)] = 0

= +4 +2x - 16 = 0

= 2x = +16 - 4

= 2x = +12

= x = +6

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