Multiple homo- and heteronuclear bonding between heavier p-block compounds Flashcards
Why do first row p-block compounds form pi-bonding interactions?
2p orbitals are compact and separation between the two atoms is quite short
Therefore it is energetically favourable
As n increases past 2, pi-interactions become less favourable as orbitals become more diffuse and bonds get longer
How can M=M multiple bonds be formed for heavier p-block elements?
Through appropriate kinetic stabilisation using bulky/sterically demanding ligands
‘Appropriate’ i.e. ligand needs to be sufficiently large to prevent formation of chains of single bonds, but also needs to be sufficiently small and compact that 2 of the ‘carbene’ units can actually come together to form the interaction
Synthesis of compounds containing M=M multiple bonds of heavier p-block elements
Same as carbene analogues (because basically just bringing together 2 carbene analogue units)
- Reduction of higher oxidation state compounds for the lighter elements of groups 13 and 14 and all of group 15
- Metathesis of low valent halides for the heavier elements of group 13 and 14
Structural changes in R2M=MR2 going down group 14
When M=C, the structure is completely planar and all the angles at C sum to 360 degrees (3x120)
Fold angle is 0 degrees because the structure is completely planar (i.e. no trans bending)
C=C bond is 12 % shorter than C-C bond
As you go down the group, the sum of the angles at M decreases because the structures have started to pyramidalise (no longer trigonal planar)
Trans-bending is also introduced and the fold angle increases from Si to Sn
In all cases, M=M is shorter than M-M, but the % shortening decreases down the group from 8 for Si to 4 for Ge to 2 for Sn
Can no longer describe Sn=Sn as a true double bond i.e. bond order decreases from C to Sn
How do Ge, Sn (and Pb) compounds exist in solution?
As monomers (i.e. as the carbene analogues R2M:) R2M=MR2 is only a solid state interaction Therefore undergo the same types of reactions as R2M: i.e. oxidative addition and Lewis base character
Effects of carbon ligands on the dimerisation of R2M:
Carbon donor ligands destabilise the HOMO, which narrows the HOMO/LUMO gap
Therefore R2M: compounds with R=C are more likely to dimerise
When M = Sn, compounds are generally red in solid state and solution (absorb in blue part of visible spectrum)
Effects of nitrogen ligands on the dimerisation of R2M:
Nitrogen = electron-withdrawing so stabilises the HOMO
Nitrogen = pi-donor so destabilises the LUMO
Therefore larger HOMO/LUMO gap and therefore these compounds exist purely as the monomeric species (no tendency to dimerise)
When M=Sn, compounds are generally completely colourless (no longer absorbing in visible part of spectrum)
Simple view of explaining trans-bent structures
Described as a ‘double donor-acceptor interaction’ between the lone pair on R2M: (i.e. the HOMO) and the empty p-orbital (i.e. the LUMO) on another R2M: monomer, providing the p-orbital is of sufficiently close energy to the HOMO
Bonding gets weaker down the group as the orbitals become more diffuse, meaning overlap is less efficient
The lone pair become more ‘lone-pair-like’ down the group, meaning there is less of a tendency for any end-on directionality in the bonding, explaining the increased trans-bending down the group (i.e. formation of ‘banana bonds’)
More sophisticated explanation of trans-bending in R2M=MR2 structures
As you go down the group, orbital overlap becomes less efficient which leads to a narrowing of the HOMO-LUMO gap
In the planar system (D2h), all the MOs are of different symmetry so cannot mix together, even if they are close in energy
When trans-bending is introduced, the symmetries of the MOs change as well as the symmetry of the molecule overall (C2h)
Sigma and pi* orbitals are now the same symmetry (ag) and sigma* and pi orbitals are now the same symmetry (bu) - orbitals of the same symmetry can mix
Pi-bond is weakened and transformed into a non-bonding lone pair through mixing with sigma*
Sigma-bonding orbital also weakened
(this explains why they dissociate into monomers in solution)
Overall, there is an increased tendency to adopt a trans-pyramidalised structure down the group as the orbitals get closer in energy due to 2nd order Jahn-Teller effects
Why does R2B=BR2 not dissociate in solution?
Structure very similar to that of an alkene (B = sp2, trigonal planar structure, 120 degree bond angles)
B = small so there is good orbital overlap and a large HOMO/LUMO splitting, therefore no need to introduce trans-bending to stabilise the bonding electrons overall
B=B is a ‘proper’ 4 electron double bond (like that formed by C) so doesn’t dissociate in solution
Synthesis of M—M triple bonds
Use higher valent Si precursor e.g. tetrabromodisilane
Si-Br bonds are reducible so use a potent reducing agent e.g. KC8
Produces KBr salt as driving force
For heavier group 14, divalent starting materials are available and reduce from M(II) to M(I) using K
Draw
Properties of M—M triple bonds
Do not dissociate in solution
Structural changes in RM—MR going down group 14
Structures are not linear like with alkynes - significant degree of trans-bending introduced as you go down the group
M—M shorter than M-M (and M=M) for Si to Sn but there is less difference in the values as you go down the group - bond order is decreasing
M—M is longer than M-M for Pb!
Fold angle in Pb compound is 94 degrees!
Group 14 M—M bonding description using simple ‘Lappert’ approach
= dimerisation of 2 monoradicaloid fragments
Increasing tendency down the group for the ns electrons to remain paired (the lone pair become more ‘lone-pair-like’ down the group)
This means there is less of a tendency for any end-on directionality in the bonding, explaining the increased trans-bending down the group
When you get to Pb, trans-bending is so great that there is only a weak Pb-Pb single bond and lp-lp repulsion
Sophisticated bonding description of group 14 M—M triple bonds
As you go down the group, orbital overlap becomes less efficient which leads to a narrowing of the HOMO-LUMO gap
In the planar system (Dinfinityh), all the MOs are of different symmetry so cannot mix together, even if they are close in energy
When trans-bending is introduced, the symmetries of the MOs change as well as the symmetry of the molecule overall (C2h)
Sigma-bonding orbital symmetry changes to ag, which is the same as one of the pi* orbitals
In the linear case, both pi-bonding orbitals are degenerate
When trans-bending is introduced, the pi-bond formed from the p(y) orbitals is unchanged in energy, but the pi-bond formed from the p(x) orbitals is stabilised through interaction with the sigma* orbital (both are bu symmetry)
This pi-orbital has increasing ‘lone pair’ (non-bonding) character as the degree of trans-bending increases
The bonding system is stabilised overall, but the introduction of anti-bonding character into the sigma and one of the pi-bonding orbitals has weakened the strength of these interactions
These interactions become increasingly weak down the group
