Question

the oxidation number of sulphur in S8, S2F2 ,H2S ARE

Answer 1

Oxidation number of sulfur in

1. $S_8$

• The rule of oxidation number states that, free element has zero as its oxidation number.
• As $S_8$ is a free element, the oxidation number of sulfur in it is 0.

2. $S_2F_2$

• As fluorine belongs to group 17, and it is present as binary compound, its oxidation number is -1.
• To find the oxidation number of S in $S_2F_2$ the following equation must be written.

2 (S) + 2(F) = 0

2 (X) + 2 (-1) = 0

2X - 2 = 0

2X = 2

X = $\frac{2}{2} = 1$

• Thus the oxidation state of S in $S_2F_2$ is +1.

3. $H_2S$

• As hydrogen is bonded with non-metal sulfur, its oxidation number is 1.
• Let the oxidation number of S be X.

• To find the oxidation number of S in $H_2S$ the following equation must be written

2 (H) + S = 0

2 (1) + X = 0

2 + X = 0

X = -2

• Thus the oxidation number of S in $H_2S$ is identified as -2.

Learn more about oxidation number

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Answer 2

Answer:

The answer is 0,+1, -2 .

Explanation:

Solution:

Solution:In S, oxidation number of S is 0, elemental state.

Solution:In S, oxidation number of S is 0, elemental state.In S, F is in - 1 oxidation state, hence S is in + 1 oxidation

Solution:In S, oxidation number of S is 0, elemental state.In S, F is in - 1 oxidation state, hence S is in + 1 oxidationstate.In HS, H is in +1 oxidation state, hence S is in - 2 oxidation

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