Tin(II) chloride is often used as a reducing agent in multiple reactions. I gather it may be due to greater stability of the stannic ion over the stannous ion. This seems to be against the inert pair effect, where lower oxidation states are more stable for higher elements of groups 13, 14 and 15. So why is this so?

  • $\begingroup$ That is what makes the existence of Sn(II) possible in the first place. You’re comparing stability of Sn(II) against its lighter congeners, like Si or Ge. Have you ever seen anybody using SiCl2 as a reducing agent? No? That’s because it’s too unstable to even exist. And you can get Ge(II) into aqueous solution, but that’s a terribly strong reducing agent, much stronger than Sn(II). So there is no contradiction. Nobody said that Sn(II) must be more stable than other things like Cu(II) or Mg(II). $\endgroup$ – orthocresol Jan 4 at 13:09

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