Question

A student proposes that the expected Keq is less than 1.0 for this reaction? Is the student right or wrong? Explain

Answer 1

\[[Cr(D_2O)_6]^{2+}+[Ru(ND_3)]^{3+} \rightarrow [Cr(D_2O)_6]^{3+}+[Ru(ND_3)]^{2+}\]

Answer 2

My answer was that no, the Keq. I argued that since electron is being taken from the eg set of the orbital on Ru, since its more energetically costly removing electrons from the antibonding orbital that the Keq would be expected to be lower? I'm confused actually on this question. @aaronq

Answer 3

hm im not sure either. Keq < 1 means that most of the substance will lie on the side of reactant but I'm not sure what criteria you're supposed to base your answer on.

I think it's actually less energy-costly to remove an electron from an anti bonding orbital since it's of higher energy (unstable) and more reactive by that same coin.


so Cr is d4 and Ru d5 -> Cr d3 + Ru d6

so i would say that Keq IS less than one, simply because of the stability of the d5 Ru3+ and water being a hard base would stabilize the higher oxidation state on Ruthenium

Answer 4

That actually makes a lot of sense. Wow. Thanks man.

Answer 5

no problem dude

Answer 6

I finally got the answer. Actually, the Keq > 1
I think removal from the eg set is easier (I think), hence it stabilizes the complex more. And since its just one electron being removed forming an inert complex (d3) and (d6) complex after electrons are trnasfered, the reaction seems to be product favored, making the students guess wrong and Keq >1

Answer 7

so the reaction favours the products, Cr (+3) d3 and Ru (+2) d6 ?

i guess that would make sense if Ru was low spin and all the t2g orbitals were filled

Answer 8

could explain further though, i'd like to know