What is the difference in the oxidation numbers of the two different bonding types of sulphur atoms — Redox Reactions and Volumetric Analysis Chemistry Question
Question
What is the difference in the oxidation numbers of the two different bonding types of sulphur atoms found within the tetrathionate ion ($Na_2S_4O_6$)?
Answer: A
💡 Solution & Explanation
In the tetrathionate ion ($S_4O_6^{2-}$), the structure is $^-O_3S-S-S-SO_3^-$. The terminal sulphur atoms are bonded to oxygens and have an actual oxidation state of +5. The central sulphur atoms are bonded only to other sulphurs, having an actual oxidation state of 0. The difference is $ | 5 - 0 | = 5$.
💬Ask on WhatsApp →
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes