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What is the Oxidation state of sulphur in thionic acid(h2s2o3)?Please explain it with the help of fundamental concept and with it's structure.
Dear AshishĀ I am trying to give goosd explanation.Na2S2O3Oxidation state of 'S' you do not know, okOxidation state for sodium = 1, So, Two sodium gets (1+1)= 2Your question is oxidation state for sulphur, So unknown, There are two sulphur atoms are there, we will write as 2xOxidation state for oxygen = -2.There are there are three oxygen atoms, SoĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā (-2*3)= -6Just apply in the avove molecular fomula,Na2S2O3(2+2x-6) = 0(2-6)+2x= 0-4+2x= 0x " means sulphur atom = 2, SoOsidation state of sulphur is "2"
Oxidation state of sulpher is Ā +4 . Because S attached to other S with =bond lonely. and oxigen (with =bond),two OH gp attached with one of the S.so that one S has zero oxidn state and other has +4Ā
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