Relativistic effect: d-electrons in metallorganic complexes

Question

With higher period the d-electrons of the metal are less strong bonded and therefore oxidative addition is easier for $\ce{Ir(I)}$ than for $\ce{Rh(I)}$ and much easier than for $\ce{Co(I)}$.

For metal carbonyl hydrides, electrons with higher period are more stronger bound to the nucleus, so $\mathrm{p}K_\mathrm{a}$ increases in this order: $\ce{HCrCp(CO)3 < HMoCp(CO)3 < HWCp(CO)3}$ This explanation sounds logical!

I also wrote that for $\ce{CO}$-complexes because of the higher effective atomic number for higher period atoms (higher ionization energy) the $\pi$-backbonding is weak and therefore less stable as e.g. for $\ce{Cr(3d)}$. This explanation sounds logical as well!

But is the written tendency of the oxidative addition not a contradiction to the other tendencies?

I don't think that d-shell electrons will have any relativistic effect. "Relativistic effect" doesn't mean relative to other elements in the same column of the periodic table, but to Einstein's theory of relativity. In simple terms 1S electrons of heavy elements are bound so tightly (energetically) that it is if the electron increases in mass. So in ab initio calculations of the binding energies of 1S electrons versus atomic number you could throw in a correction for the effect. — MaxW, Oct 25 '15 at 01:38
I have some decent guesses.
(1) The square planar complexes with Ir(I), Rh(I), and Co(I) decrease in reactivity to oxidative addition. Suppose we add H2; here, this may be due to how iridium has the largest d orbitals and thus the least compact electron density. Thus, it is easiest to polarize its d electron density towards the added hydrogens and Ir(I), for instance, can become Ir(III) more easily than Co(I) can.

(2) For your m-C hydrides, since the hydrogen only has a bonding AO, maybe the larger radius of the metal increases the bond length and thus makes the W-C hydride most acidic. — timaeus222, May 05 '16 at 01:58
@MaxW: Relativistics affects chemistry. The main 'mechanism' is that due to the scalar (non-four component) effects in deed the innermost electrons orbitals get contracted ($e^-$ gets fast => heavy => radius smaller) thus screen much more effectively the nuclear charge, thus the outer shells expand/ get less strongly bound. — Raphael J.F. Berger, Aug 12 '16 at 17:06
I'm pretty sure the answer to this question lies not in the general bonding trends (and EN trends) caused by relativistic effects but in the specific electronic configuration effects. At least this is the reason reductive elimination is easier for Pd (d10) than Pt (s1d9); however, the Co, Rh, Ir series has different electronic configuration than the Ni, Pd, Pt series. I'll read some papers and try to expand on this... — gannex, Feb 10 '17 at 18:06
I don't think your trends are correct. iridium is a stronger π-donor than rhodium, for instance, despite it being in a "higher" period. Is this what you mean by higher? — gannex, Feb 10 '17 at 18:43
Echoing the previous commenters, I also don't see why relativity should play any role in this. I'm not sure how accurate trend #3 is, either. I was at least taught that backbonding increases from 3d < 4d < 5d because of more diffuse d-orbitals (hence stronger overlap with CO $\pi^*$, as @timaeus222 already noted quite a while ago. The C-O stretching frequency is not very diagnostic, however: Elschenbroich 3rd ed. gives them as 2000, 2004, and 1998 wavenumber units for Cr(CO)6, Mo, and W respectively. — orthocresol, Apr 05 '17 at 23:33
I was wrong about the relativistic effect. For example see paper Relativistic Effects and the Chemistry of Gold on high Z atoms. I was thing only of direct effect. Rather the effect is indirect. Because of the tighter binding of the inner shells the outer shells are more shielded and hence less bound. — MaxW, Jul 11 '18 at 23:13

score 2 · Answer 1 · edited Jul 25 '18 at 10:36

First off, be aware that oxidative addition occurs by several mechanisms which are governed by different factors.^[1]

As you move down a period, the outer electrons become more easily accessed, as the outer electrons are increasingly screened from the nuclear charge by the inner electrons.^[2] This is why both ionization energy and electron affinity decreases down a period.

Classical organometallic chemistry uses metal carbonyl stretching frequencies as a direct measure of π back-bonding and electron density.^[3] For a recent discussion, see: Chem. Sci. 2016, 7, 1174. Compare the stretching frequency of the group VI metal hexacarbonyls ^[4]: $\ce{Cr(CO)6}$: $\pu{2000 cm^-1}$ and $\ce{W(CO)6}$: $\pu{1987 cm^-1}$. Tungsten donates more electron density back into the carbonyl than does $\ce{Cr}$.

The importance of relativistic effects in structure calculations is frequently the subject of reviews.^[5]

References:

Crabtree, R. H. The Organometallic Chemistry of the Transition Metals, Ch. 6
Waldron, K. A.; Fehringer, E. M.; Streeb, A. E.; Trosky, J. E.; Pearson, J. J. Screening Percentages Based on Slater Effective Nuclear Charge as a Versatile Tool for Teaching Periodic Trends. J. Chem. Educ. 2001, 78 (5), 635. DOI: 10.1021/ed078p635.
Tolman, C. A. Steric effects of phosphorus ligands in organometallic chemistry and homogeneous catalysis. Chem. Rev. 1977, 77 (3), 313–348. DOI: 10.1021/cr60307a002.
Clark, R.; Crociani, B. Solvent effects on the infrared spectra of chromium, Molybdenum and tungsten hexacarbonyls. Inorganica Chimica Acta 1967, *1undefined 12–16. DOI: 10.1016/S0020-1693(00)93131-1.
Pyykko, P. Relativistic effects in structural chemistry. Chem. Rev. 1988, 88 (3), 563–594. DOI: 10.1021/cr00085a006.

Relativistic effect: d-electrons in metallorganic complexes

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