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Question:

Out of $CH_4, CaH_2, FeCl_2$ and $BeH_2$, which compound would produce the most strongly reducing aqueous solution?

Thoughts:

Out of the hydrides, I would say that $CaH_2$ stands a good chance at being the strongest reducing agent, since the Ca-H bond energy is lowest (allowing the extremely reducing hydride to be released more readily into solution). The only rationale I can think up for this is the relative sizes of the other ion in the hydrides (Ca>Be>C) which implies that bond length is greatest in the calcium hydride (thus bond energy is lower). I am unsure as to how I should tackle the chloride compound. Much assistance is needed.

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  • $\begingroup$ You are correct that calcium hydride is good reducing agent. It is also a strong base and reacts with water, rendering its aqueous solutions non-reducing:$$\ce{CaH2 + 2H2O -> Ca(OH)2 + 2H2}$$ $\endgroup$ – Ben Norris Aug 24 '15 at 1:58
  • $\begingroup$ @BenNorris just about any hydride is a good reducing agent. But is it the most? For some reason I think that beryllium hydride might be better but it's not based on anything. $\endgroup$ – Gimelist Aug 24 '15 at 13:41

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