Chemistry, asked by gellisurabhi, 1 year ago

what is the oxidation state of sulphur in Na2S4O6

Answers

Answered by harini2002
2
the answer is 2.5.....hope it helps

harini2002: oh sry...wait
gellisurabhi: in above compound sulphur os is
gellisurabhi: ok
musaddiqbijle2: but its a bit complicated tooo
musaddiqbijle2: don't forget that
harini2002: its 2...i guess
musaddiqbijle2: avarage is +2.5
gellisurabhi: ya it's 2
gellisurabhi: tq
harini2002: u r most welcome!...
Answered by musaddiqbijle2
2

As a consequence, this compound must have sulfur atoms with mixed oxidation states. Since it is normal for sulfur to have oxidation states of -2, 0, +2, +4, and +6, it is most likely that there are three sulfurs with a +2 oxidation state and one sulfur that is +4. The average of these is +2.5.


Since everyone here seems to agree that the oxidation state of sulfur would be 2.5, I won’t redo the math, but let me say that this oxidation state is impossible. No atom can have a fractional oxidation state - they must be whole numbers (you can’t lose half of an electron). As a consequence, this compound must have sulfur atoms with mixed oxidation states. Since it is normal for sulfur to have oxidation states of -2, 0, +2, +4, and +6, it is most likely that there are three sulfurs with a +2 oxidation state and one sulfur that is +4. The average of these is +2.5.

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