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This question has been troubling me since my professor taught this concept. Points that she taught:

  1. pi back bonding can only happen amongst elements of 2nd and 3rd period. At least one of them needs to be in 2nd period.
  2. Fluorine forms the best back bonding inspite of its electronegativity. (when asked why, didn't get any convincing answer)
  3. Back bonding increases bond angle and decreases bond length.
  4. When there's a clash between decrease of a bond angle according to Bent's rule and increase of the same angle by considering back bonding, one should consider back bonding to dominate.
  5. In some cases, possibility of back bonding drives the hybridisation to change.
  6. According to requirements for pi-backbonding, there needs to be a sigma bond between both the atoms and requirement of one empty orbital (p or d) for allowing the electrons (in p orbital) of the side atom to delocalize into it .

Trisilylamine should have been tetrahedral in shape but due to back bonding the hybridisation is $\mathrm{sp}^2$ as Nitrogen lone pair is delocalised into the empty d orbitals of silicon. Similarly, disilyl ether must have bond angle lesser than or equal to $120^\circ$. Less, because there will be a lone pair of oxygen that is not delocalized which should lead to a repulsion between the bond pair and lone pair. But disilyl ether has a bond angle of $144^\circ$.

In conclusion, the main question is as follows:

When do we consider back bonding and when will it change hybridisation?

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$\mathrm p$π-$\mathrm d$π bonding is very weak and normally not considered to be very effective in bonding. Rather, the bonding can be explained via ionic bonding instead due to the large electronegativity differences. This can be seen in siloxanes and trisilylamine.

According to J. Am. Chem. Soc. 1980, 102 (24), 7241–7244

Since our ab initio calculations reproduce the geometry and dipole moments of the three molecules very well, conclusions concerning the electronic structure of the siloxane bond derived from these calculations should be at least qualitatively correct. (1) (p-d)π bonding is of minor importance for explaining the unusual properties of the siloxane bond. [...] A (p-d)π bond order of 0.4 has been deduced from experimental bond lengths

Now referring to Inorg. Chem. 1997, 36 (14), 3031–3039, the results section state the following.

The expected number and arrangement of charge concentrations is not, therefore, observed for ligands less electronegative than carbon, that is, for the molecules that are predominately ionic, and it just for these molecules that the bond angle is substantially larger than the tetrahedral angle.

Further more, it also talks about the assumptions taken in VSEPR and why it doesn't work here.

However, with decreasing ligand electronegativity, each of the four electron pair domains becomes larger in extentsmore spread out or diffusesand so overlaps its neighbors to an increasing extent until, in the limiting case of an oxide ion, there is no electron localization, and each domain may be considered to occupy the whole of the valence shell

This is also the case in disiloxane as explained later in the paper,

That the bond angle in $\ce{H3SiOSiH3}$ ($148.3$°) is somewhat larger than that in $\ce{H2BOBH2}$ ($135.4$°), which in turn is slightly larger than that in $\ce{H2POPH2}$ ($129.8$°), is consistent with the electronegativity values of Si (1.7), B(2.0), and P(2.1). As in $\ce{H2BOBH2}$, there is only one nonbonding maximum in the oxygen valence shell in $\ce{H2POPH2}$ and in $\ce{H3SiOSiH3}$.

Basically the last three paragraphs state that the VSEPR model only works if the attached atoms are sufficiently electronegative enough so that covalent bonds can form. In case of disiloxane, the electronegativity difference leads to it being a more ionic bond than it should be. This is seen also in $\ce{Li2O}$ which has a $180^\circ$ bond angle because of the high electronegativity difference.

For trimethylsilane, see Jan's answer here

Instead, I think we are dealing with something you may call ‘inverse hyperconjugation’. Remember that $\chi(\ce{Si}) = 1.9$ which is less than hydrogen, meaning that the $\ce{Si-H}$ bonds are polarised towards hydrogen. This in turn means that $\sigma^*_{\ce{Si-H}}$ is a silicon-centred orbital with its primary lobe pointing towards nitrogen. Therefore, nitrogen’s $\mathrm p$ orbital can favourably interact with the antibonding $\sigma_{\ce{Si-H}}^*$ orbital, increasing the $\ce{Si-N}$ bond order and decreasing the $\ce{Si-H}$ bond order. The effects are the same as with hyperconjugation stabilisation of secondary or tertiary carbocations but the electronic demand is reversed. We could attempt to draw the following resonance structures in Lewis formalism to explain this:

$$\ce{H-SiH2-N(SiH3)2 <-> \overset{-}{H}\bond{...}SiH2=\overset{+}{N}(SiH3)2}\tag{1}$$

In this Lewis formalism, the double bond would be generated from a $\mathrm p$ orbital on both silicon and nitrogen.

Using backbonding to explain such differences is very non-trivial since it is like trying to build a modern-era supercomputer with vaccum tubes (may be exaggerating here). Chemistry is a science where we observe and then hypothesise.

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