(My 2nd –ve rating has been noted)
A conventional π bond (e.g., C=C) is formed by the overlap of two p AOs that each contain one e⁻:
Apx(↑) + (↓)pxB → Apπ(↑↓)pπB. The e⁻ pair is approximately equally shared by atoms A and B.
There are however numerous (and important) interactions that are thought to involve donation of an e⁻ pair on one atom to a vacant AO on a second atom:
A(↑↓) + (0)B → Aπ(↑↓)πBThe classic case(mentioned previously) is the M-C≡O bonding in metal carbonyls such as Cr(CO)6 (stronger Cr-L bonds than in [Cr(NH3)6]^3+). There is believed to be donation of the CO σ3 e⁻ pair to an empty σ AO on the metal (not an d^2sp^3 hybrid AO), but there is backbonding from the filled 3d AOs to the π* MOs on CO (two interactions: xz, yz planes) hence why these cmplxs are stable in the 0 oxdn state. This is called synergic bonding since the back donation prevents the build up of e⁻ density on the metal. Note that this interaction can be small or significant giving fractional bond orders for the M-CO bond (partial double-bond character). The ν(CO) of free CO is 2143 cm^-1; in [Mn(CO)6^]+ it is 2090 cm^-1, in Cr(CO)6 ~2000 cm^-1, and [Ti(CO)6]^2- cm^-1 (earlier question on this forum) for Me2C=O ν(CO) = ~1720 cm^-1.
Also included in backbonding however is where the donation is in the same direction as the dative bonds but opposite to the e⁻ flow to the electronegative atom. Examples Fpπ→ 3d(t2g)M (LFT) explains why F^- is a weak field ligand when CFT would indicates it should be a strong field ligand; the stable [MnO4]^- ion: formally Mn(VII), that is, 3d^0 but back donation from Opπ→3dMn; π→π* in visible and strongly allowed, unlike d→d transitions; would you like to try OsO4? (formally Os(VIII)). In main group chem. BF3 Fpπ→Bpπ explains why BF3 is a weaker Lewis acid than BI3 (contrary to electronegativity arguments). And why the B-F is the strongest single bond known (F does not form E=F bonds does it?); Opπ→3d explains why the ν(PO) in F3P=O is some 200 cm^-1 higher than that in Me3P=O which would not be expected if the bond were entirely R3P:→O; the S-O bond lengths in tetrahedral [SO4]^- are too short for S-O single bonds, but S has used all it available 3pAOs to form the σ bonds.
PF3 has bonding properties almost identical to CO and most chemists would agree that this is because the electronegative F atoms places δ+ on the central P atom that lowers the energy of the P 3d AOs so that they more closely match the energy of the filled 3d AOs (PMO theory) enhancing backbonding.
I think it was ~1985 when it was proposed that for ligands such as PMe3 the 3d AOs are too high in energy to participate in such π bonding and that using a proper combination of the P-C σ* MOs could give a combination that has suitable symmetry and energy for π overlap with the filled 3d AOs on the TM. And it is not a simple matter of σ* lobes pointing out the back of each P-C bond (hybridization is out). I have been concerned that these arguments may ignore that the initial donation of the lone pair on PMe3 to give hence P δ+. Certainly the Cr-P bond in Cr(CO)5(PMe3) is a much stronger than the Cr-NMe3 bond in Cr(CO)5NMe3 (HSAB) (the opposite would be expected from σ-donation arguments).
Some years ago my group and I published a paper regarding our observations on cmpds prepared by us and others with M-SiR3 bonds. We found no evidence for the participation of Si-R σ* AOs in the bonding, but strong evidence in support that some SiR3 ligands have strong π-acceptor properties.
Turning to N(SiH3)3. There is rapid inversion in pyramidal NR3 cmps that indicates that the form with planar N is at only at a slightly higher energy than the pyramidal ground state. That the N atom in N(SiH3)3 is planar could be simply due to steric reasons. Note that electronegativity of Si is less than H and hence there is more e⁻ density on the Hs in the silyl deriv than the Me deriv hence more repulsion. But considering the H*OMO-LUMO interactions (a variation of Fukui’s frontier orbital concept): the H*OMO on N is the lone pair that occupies the pz AO for maximum π overlap. There is Npπ→πSi overlap (“backbonding”). The nature of the acceptor π MOs (AOs) on Si is a matter of controversy. The most accepted explanation is that they are the Si 3d AOs that are involved, but it could also involve Si-R σ* MOs (or a mixture of both!).
"I believe the chemical bond is not so simple as some people seem to think." Robert S. Mulliken