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$\ce{CrCl3(bpy)3}$ (A) when heated forms (B) which has the formula $\ce{CrCl3(bpy)2}$ and has one free $\ce{Cl-}$ per mole of (B). B can react with ethylene diamine ($\ce{en}$) to form (C) which has the formula $\ce{CrCl3(bpy)2(en)}$.

(B) can also be converted into (D) which also has the formula $\ce{CrCl3(bpy)2}$ but cannot react with $\ce{en}$.

Identify compounds A–D and draw their structures

I know that B is cis-$\ce{[CrCl2(bpy)2]Cl}$
C is $\ce{[Cr(bpy)2(en)]Cl3}$. The $\ce{en\bond{->}metal}$ bonds are where the $\ce{Cl-}$ ions in B are.
D is the trans isomer of B.

What I dont understand is how you're meant to know what A is. I assume it's $\ce{[Cr(bpy)3]Cl3}$ as a later part of the question asks about relative molar conductivity of A and B.

Why would heating that cause a $\ce{bpy}$ to leave the complex and two chlorides to replace it? Or is it $\ce{[CrCl2(bpy)2]Cl(bpy)}$ in which case, why does heating it not form the more stable complex I've stated above?

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It is Cr(III) not Co(III). Cr(III) has reasonable substitution kinetics. Your structure for A is good. Heating drives off a mole of volatile bipy, then two chlorides inner sphere coordinate to take its place. The third chloride remains outer sphere and readily exchangeable. Bidentate en comes in to displace two inner sphere chlorides outer sphere again.

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  • $\begingroup$ This does not answer the question. $\endgroup$ – Jan Nov 5 '16 at 19:29
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If you are given only the options $\ce{Cr(bpy)3Cl3}$, then it is mandatory for the bipyridine ligand to actually be attached to the chromium centre. If it were not, as in your suggestion $\ce{[Cr(bpy)2Cl2]Cl . bpy}$, this additional bipyridine can dissociate away as there is nothing keeping it close to the complex. On the other hand, isolateable substances are always neutral, thus if your complex cation is $\ce{[Cr(bpy)3]^3+}$, then the three chloride anions will remain in the same crystal structure and you can isolate a complex salt of the type $\ce{[Cr(bpy)3]Cl3}$. This molecular structure can be proven by being able to precipitate three moles of chloride per mole of complex, e.g. with $\ce{AgNO3}$.

Why would heating that cause a $\mathrm{bpy}$ to leave the complex and two chlorides to replace it? Or is it $\ce{[CrCl2(bpy)2]Cl(bpy)}$ in which case, why does heating it not form the more stable complex I've stated above?

This somehow assumes that $\ce{[Cr(bpy)3]Cl3}$ is the most stable of these complexes. However, the chemical literature thinks different. Burstall and Nyholm first reported the synthesis of $\ce{[Cr(bpy)2Cl2]Cl}$ from anhydrous chromium(III) chloride using three (!) equivalents of $\ce{bpy}$ and heating to reflux.[1]

On the other hand, traditional methods to generate $\ce{[Cr(bpy)3]^3+}$ salts — most commonly perchlorates $\ce{[Cr(bpy)3](ClO4)3}$ — typically started from a corresponding $\ce{[Cr(bpy)3]^2+}$ and employed oxidation.[1] (This was apparantly first published in 1932 by Barbieri and Tettamanzi in an obscure journal that I can’t find.)[2] Relatively recently, a new method to synthesise the $\ce{[Cr(bpy)3](ClO4)3}$ complexes from $\ce{CrCl3 . 6 H2O}$ was published — which, however, builds on completely removing any chloride from the solution by using silver(I) salts as in equation (1).[3]

$$\begin{multline}\ce{CrCl3 . 6 H2O + 3 [Ag(bpy)2](ClO4) ->\\ [Cr(bpy)3](ClO4)3 + 3 AgCl v + 3 bpy}\tag{1}\end{multline}$$

Seemingly, having chloride ions in solution around the $\ce{[Cr(bpy)3]^3+}$ ion can’t be trusted.

Partially due to SciFinder being down for maintenance as of now, I was unable to find any references for the synthesis or properties of $\ce{[Cr(bpy)3]Cl3}$, but given the Burstall-Nyholm method I would not assume the cation to be more stable than $\ce{[Cr(bpy)2Cl2]+}$.

A likely explanation for the lower stability could be the charge balance. The dichloridobis(bipyridine)chromium(III) has an overall charge of $1-$, which is generally okay. In tris(bipyridine)chromium(III), there is a very small $\ce{Cr^3+}$ ion in the centre of a complex surrounded by otherwise neutral ligands. Any method to relieve the charge could well be favourable; amoung them reduction of the metal or ligand exchange with anionic ligands. Remember that charge balance is a thing: $\ce{[CoCl4]-}$ forms when cobalt(III) reacts with chloride ions mainly due to charge balance reasons.


References:

[1]: F. H. Burstall, R. S. Nyholm, J. Chem. Soc. 1952, 3570. DOI: 10.1039/JR9520003570.

[2]: Barbieri, Tettamanzi, Atti R. Accad. Lincei. 1932, 15 877. Reference as cited in [1].

[3]: W. Kharmawphlang, S. Choudhury, A. K. Deb, S. Goswami, Inorg. Chem. 1995, 34, 3826. DOI: 10.1021/ic00118a035.

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Nicholas, Cr(III) is generally 6-coordinate and both bipy and en are bidentate. So 3 bipy in (A) connect to the metal to form the cation, leaving 3 $\ce{Cl-}$ in the outer sphere. Then in (B) one bipy leaves and its two positions are occupied by $\ce{Cl}$, and as you say the configuration must be cis (both because a bidentate ligand departed and because another can easily come in). Isomerization of (B) to (D), the trans isomer, probably occurs slowly, and (D) cannot accept en because the $\ce{Cl}$ sites are opposite instead of adjacent.

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  • $\begingroup$ This does not answer the question. $\endgroup$ – Jan Nov 5 '16 at 19:29

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