Which complex is the most stable?

Question

Between $\ce{[Rh(Ph2PCH2PPh2)3]^{3+}}$ and $\ce{[Rh(Ph2PCH2CH2PPh2)3]^{3+}}$, which one is the most stable thermodynamically? I think it's about the size of the chelate ring effecting their stability, but i'm not sure.

Answer 1

It is risky to predict just by looking at the structure, of course there is the ring size effect and one could easily conclude that $\ce{[Rh(Ph2PCH2CH2PPh2)3]3+}$ is the most thermodynamically stable compound since you would have a 5-membered ring instead of a 4-membered as in $\ce{Rh(Ph2PCH2PPh2)3]3+}$. However, there is also the contribution of the electronic effect, which is the increase or decrease in electron density of the metal ion caused by the distribution of ligands around the metal center. The combination of electronic and steric effects was reviewed by C.A Tolman in 1977 for monodentate ligands and he proposed a way of measuring these effects experimentally and quantitatively, they are the so-called Tolman's parameters (Tolman, C. A. (1977). "Steric effects of phosphorus ligands in organometallic chemistry and homogeneous catalysis". Chem. Rev. 77 (3): 313–348.)

For bidentate ligands such as DPPE (2 $\ce{-CH2 -}$ groups between phosphorus atoms) or DPPM (1 $\ce{-CH2 -}$ group between phosphorus atoms) another important parameter is the bite angle "β" (the circular sector angle between the metal and the two phosphorus atoms). For example, in square-planar and octahedral complexes the ideal bite angle is 90º , deviations from this value will electronically destabilize the complex (KUMAR, M. et al. Importance of Long-Range Noncovalent Interactions in the Regioselectivity of Rhodium-Xantphos-Catalyzed Hydroformylation. Organometallics, v. 34, n. 6, p. 1062-1073, Mar 2015.).

Bidentate phosphines with a natural bite angle of 120° preferentially occupy diequatorial sites in a trigonal bipyramidal complex whereas a bidentate phosphine with a natural bite angle of 90° preferentially occupy apical-equatorial positions (Casey, C. P.; Whiteker, G. T.; Melville, M. G.; Petrovich, L. M.; Gavney, J. A.; Powell, D. R. (1992). "Diphosphines with natural bite angles near 120° increase selectivity for n-aldehyde formation in rhodium-catalyzed hydroformylation". J. Am. Chem. Soc. 114 (2): 5535–5543.).

DPPM has a bite angle of 78º while DPPE has a bite angle of 85º (closer to 90º), so considering only the angular aspect, the $\ce{Rh-DPPE}$ complex would be more stable (Ligand Bite Angle Effects in Metal-catalyzed $\ce{C−C}$ Bond Formation Piet W. N. M. van Leeuwen,, Paul C. J. Kamer,Joost N. H. Reek, and, and Peter Dierkes, Chemical Reviews 2000 100 (8), 2741-2770)*. Electronically, I assume that the 2 $\ce{-CH2 -}$ groups of DPPE could donate more electronic density to the P atoms than DPPM, which would somehow strengthen the $\ce{Rh-P}$ bonds.

However, it is the combination of sterical + electronic effects that can unequivocally decide which of the two complexes are more stable, and this would only be possible using computational chemistry or experimentally, though the above can be used when discussing about it on a statement or paper.

Answer 2

As @andselisk stated, the bis-Rh species is expected to be more thermodynamically stable. You also need to consider the metal itself; Rh. Rhodium is famous for often forming $\mathrm{d^8}$ square planar complexes (a lot more than octahedral complexes) which results in a four coordination number complex instead of six - this would strongly suggest that the bis-Rh complex involving bidentate ligands would be thermodynamically more stable based primarily on the chemistry of the metal itself. Whether this is is reflected experimentally I have no idea.