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the oxidation number of sulphur in S8,S2F2,H2S respectively are
  • a)
    0,+1 and -2
  • b)
    +2, +1 and -2 
  • c)
    0, +1 and +2
  • d)
    -2, +1 and -2
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
the oxidation number of sulphur in S8,S2F2,H2S respectively area)0,+1 ...
(i) Oxidation state of element in its free state is zero.
(ii) Sum of oxidation states of all atoms in compound is zero.
O.N of S in S8=0;    O.N of S in S2F2=+1
O.N of S in H2S=-2
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Most Upvoted Answer
the oxidation number of sulphur in S8,S2F2,H2S respectively area)0,+1 ...
Understanding Oxidation Numbers of Sulphur
To determine the oxidation number of sulphur in S8, S2F2, and H2S, we assess each compound individually.
1. Oxidation Number in S8
- S8 is a molecular form of elemental sulfur.
- In elemental forms, the oxidation number is always 0.
- Therefore, for S8, the oxidation number of sulfur is 0.
2. Oxidation Number in S2F2
- In S2F2, we have sulfur and fluorine.
- Fluorine is the most electronegative element and has an oxidation number of -1.
- With two fluorine atoms, the total contribution from fluorine is -2.
- To balance this and achieve a zero charge for the molecule, the two sulfur atoms must collectively contribute +2.
- Hence, each sulfur atom has an oxidation number of +1.
3. Oxidation Number in H2S
- In H2S, hydrogen has an oxidation number of +1.
- With two hydrogen atoms, the total contribution from hydrogen is +2.
- To balance the total charge of the molecule (which is neutral), sulfur must have an oxidation number of -2.
- Thus, in H2S, the oxidation number of sulfur is -2.
Conclusion
- The oxidation numbers for sulfur in each compound are:
- S8: 0
- S2F2: +1
- H2S: -2
Thus, the correct answer is option A: 0, +1, and -2 respectively.
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