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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. 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 Ashish 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, So (-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 +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 +4
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