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← IIT-JEE 2011 · Paper 1
ChemistryInteger+4Question 17 of 69

IIT-JEE 2011 · Paper 1, Chemistry question 17

The difference in the oxidation numbers of the two types of sulphur atoms in NaX2SX4OX6\ce{Na2S4O6} is

Answer and worked solution

Official answer: 5

Tetrathionate X−X22−OX3S−S−S−SOX3X−\ce{^-O3S-S-S-SO3^-}: the two terminal S are +5 and the two bridging S are 0 (total 2(5)+2(0)=102(5)+2(0)=10, balancing 6(−2)+26(-2)+2).

Difference 5.

Answer: 5

Topics:Oxidation number
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