Chemistry - Why is chromate stable in basic medium and dichromate stable in acidic medium?

Solution 1:

I will be using an approach which has been enlisted in the following book for answering this question: Arrow Pushing in Inorganic Chemistry ;A Logical Approach to the Chemistry of the Main-Group Elements

The preface of the book says:

The approach: These reactions represent important facets of the elements involved but are typically presented as no more than facts. (Why does boiling white phosphorus in alkali lead to hypophosphite and not phosphate?—Current texts make no attempt to address such questions.) Arrow pushing demystifies them and places them on a larger logical scaffolding. The transformative impact of this approach cannot be overstated. Almost to a person, students who have gone through our introductory course say that they cannot imagine how someone today could remain satisfied with a purely descriptive, nonmechanistic exposition of inorganic main-group chemistry.

Potential concerns: Given the plethora of advantages of a mechanistic approach, it’s worth reflecting why it has never been adopted for introductory inorganic chemistry. A plausible reason is that, in contrast to common organic functional groups, simple p-block compounds such as hydrides, oxides, halides, and so forth, tend to be much more reactive and their vigorous and even violent reactions have been much less thoroughly studied. As good scientists, inorganic chemists may have felt a certain inhibition about emphasizing an approach that has little grounding in experimental fact. This is a legitimate objection, but hardly a dealbreaker, in our opinion, for the following reasons.

1. Our ideas on main-group element reactivity are not taken out of the blue but are based on parallels with well-studied processes in organic and organoelement chemistry.
2. Second, it no longer necessarily takes a prohibitive amount of resources to test amechanistic proposal, at least in a preliminary way. Quantum chemical calculations, particularly based on density functional theory (DFT), very often provide an efficient and economical way of evaluating reaction mechanisms.
3. Third, and perhaps most important, it’s vastly better to be able to formulate a hypothesis on how a reaction might happen than to have no inkling whatsoever about the mechanism

What all of this basically means is that in this approach, you use the classical organic chemistry concepts of electrophiles and nucleophiles to write an arrow-pushing mechanism for the reaction. It will only be a probable mechanism, but often, it's not much far away from the reality, and this methodology can serve to be a pretty great teaching tool for inorganic chemistry, so long as you accept the fact that the mechanism you have made can only be proved by a research paper. But having an idea is also good.

Having got the introduction and disclaimer of this method out of the way, we can now start to explain your phenomenon by arrow-pushing diagrams. The dichromate anion $\ce{Cr2O7^2-}$ has got a structure consisting of a $\ce{Cr-O-Cr}$ bridge, four $\ce{Cr=O}$ bonds, and two $\ce{Cr-O^-}$ bonds.

Here, the $\ce{Cr=O}$ bonds can be said to be structurally similar to the carbonyl group.

Hence, like the carbonyl group, it can be said that there will be high partial positive charge on the $\ce{Cr}$ atoms due to the electron-withdrawing effect of two $\ce{Cr=O}$ groups per $\ce{Cr}$. Hence, they can serve as the ideal electrophilic site for the attack of a nucleophile like $\ce{OH-}$ in basic medium, and lead to bond cleavage as follows:

The $\ce{HCrO4^-}$ formed can again be deprotonated in the basic medium to form $\ce{CrO4^2-}$

In acidic medium, the $\ce{CrO4^2-}$ anion can be protonated to form $\ce{HCrO4-}$ , which can react with another $\ce{CrO4^2-}$ anion in the following manner to give $\ce{Cr2O7^2-}$ and expelling water as a leaving group:

Any alternate protonation pathways, for example, protonating $\ce{CrO4^2-}$ twice to form $\ce{H2CrO4}$ and then reacting it with $\ce{CrO4^2-}$ might also give the same product with a few extra proton transfers here and there, but the way I have presented thing would be the most probable way, as the reaction is progressing stepwise with each protonation.

Note: Remember that these are just probable pathways we have drawn to reach our product, but there's a good chance they resemble the reality( which can only be determined by kinetic experiments) and as the authors have stated, it is a good learning tool, and not too far off the mark.

Solution 2:

The chromate and dichromate ions exist in equilibrium as follows: $$\ce{2CrO4^2- + 2H+ <=> Cr2O7^2- + H2O}$$ In a basic medium, $\ce{[H+]}$ is less, which favours the formation of reactant $\ce{CrO4^2-}$ whereas in an acidic medium, $\ce{[H+]}$ is high, which favours the formation of product $\ce{Cr2O7^2-}$

Sources: Wikipedia