The Student Room Group

Multiple choice Q

Hi, can someone explain why C is not the correct answer for this question. I thought both C or D could happen unless the acid base reaction requires concentrated sulfuric acid? A96A021B-EB67-4C89-8C86-F1643C942A78.jpg.jpeg
Original post by subbhy
Hi, can someone explain why C is not the correct answer for this question. I thought both C or D could happen unless the acid base reaction requires concentrated sulfuric acid? A96A021B-EB67-4C89-8C86-F1643C942A78.jpg.jpeg

The reducing powers of the halides increase down group 7, so KAt would be a very strong reducing agent indeed - more so than KI, which reduces H2SO4 (l) to H2S with I2 as the oxidation product. Since KAt is an exceptionally strong reducing agent, oxidation to At (s) would be expected (even with dilute H2SO4, rather than hot concentrated H2SO4) and HAt would likely not form.

I suppose another explanation you could use (maybe a little above A level) is that HAt would be a much stronger acid than H2SO4, so protonation of the astatide ion to HAt would be extremely difficult to do.
(edited 6 months ago)
Reply 2
Original post by UtterlyUseless69
The reducing powers of the halides increase down group 7, so KAt would be a very strong reducing agent indeed - more so than KI, which reduces H2SO4 (l) to H2S with I2 as the oxidation product. Since KAt is an exceptionally strong reducing agent, oxidation to At (s) would be expected (even with dilute H2SO4, rather than hot concentrated H2SO4) and HAt would likely not form.

I suppose another explanation you could use (maybe a little above A level) is that HAt would be a much stronger acid than H2SO4, so protonation of the astatide ion to HAt would be extremely difficult to do.


So would KAt not react with the sulfuric acid in an acid base reaction?

I undertand the second explanation! Thanks
Original post by subbhy
So would KAt not react with the sulfuric acid in an acid base reaction?
I undertand the second explanation! Thanks

It’s unlikely KAt would be able to react with H2SO4 to form KHSO4 and HAt. The hydrogen halides become more and more acidic down group 7 and so it becomes harder to protonate some halide ions to hydrogen halides down the group, even with strong acids like sulphuric acid.

Essentially the reducing power argument is that astatine is lower down group 7 than iodine is and so the astatide ion is going to be larger than the iodide ion and therefore going to have weaker nuclear attraction to the electron to be removed. As such, it would be expected to reduce sulphuric acid as far as possible (H2S is as far as you can reduce it, as sulphur’s maximum oxidation state is -2 - convince yourself of this by looking at its electronic structure). Because the reducing power of astatide is greater, it is likely that it will need milder conditions to reduce sulphuric acid fully than the iodide ion (which requires the sulphuric acid to be hot and concentrated), so this is why the complete reduction is likely to take place in a more dilute solution.
(edited 6 months ago)
Reply 4
Original post by UtterlyUseless69
It’s unlikely KAt would be able to react with H2SO4 to form KHSO4 and HAt. The hydrogen halides become more and more acidic down group 7 and so it becomes harder to protonate some halide ions to hydrogen halides down the group, even with strong acids like sulphuric acid.

Essentially the reducing power argument is that astatine is lower down group 7 than iodine is and so the astatide ion is going to be larger than the iodide ion and therefore going to have weaker nuclear attraction to the electron to be removed. As such, it would be expected to reduce sulphuric acid as far as possible (H2S is as far as you can reduce it, as sulphur’s maximum oxidation state is -2 - convince yourself of this by looking at its electronic structure). Because the reducing power of astatide is greater, it is likely that it will need milder conditions to reduce sulphuric acid fully than the iodide ion (which requires the sulphuric acid to be hot and concentrated), so this is why the complete reduction is likely to take place in a more dilute solution.


Makes sense. Thank you so much

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