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H\(_2\)S\(_{(g)}\)  +  Cl\(_2\)\(_{(g)}\) → 2HCl\(_{(g)}\) + S\(_{(s)}\) What is the change in oxidation state of sulphur...

Chemistry
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H\(_2\)S\(_{(g)}\)  +  Cl\(_2\)\(_{(g)}\) → 2HCl\(_{(g)}\) + S\(_{(s)}\)

What is the change in oxidation state of sulphur from reactant to product?

  • A. -3 to -1
  • B. -3 to 1
  • C. -2 to 0
  • D. -1 to 0
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Correct Answer: Option C
Explanation

Oxidation state of S in  H\(_2\)S = 0

H = +1,                         (+1 X 2) + S = 0

                                       + 2 + S = 0

                                         S = - 2 ( from the reactant)

Already, from the equation, S is in solid form,as an element, therefore the oxidation state of elements in their free state is 0

The change in oxidation state of sulphur is from - 2 to 0 

There is an explanation video available below.


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Explanation Video

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