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Topic: back bonding  (Read 16560 times)

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Offline geethika

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back bonding
« on: September 06, 2011, 02:33:14 PM »
what is back bonding???

Offline DevaDevil

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Re: back bonding
« Reply #1 on: September 06, 2011, 03:16:26 PM »
you probably mean back-donation?

it is where electrons are entered into a pi-antibonding orbital on a ligand through a chemical bond. A good example is CO ligands bonded to a central metal atom.

Offline geethika

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Re: back bonding
« Reply #2 on: September 07, 2011, 03:18:05 AM »
could you please expline the back bonding in N(SiH3)3

Offline DevaDevil

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Re: back bonding
« Reply #3 on: September 07, 2011, 01:02:31 PM »
I will try to lead you through this, rather than just answer:


which electron(s) is(are) available for back donation in this complex?

Offline geethika

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Re: back bonding
« Reply #4 on: September 07, 2011, 02:14:16 PM »
SiH3 electrons are available


Offline cheese (MSW)

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Re: back bonding
« Reply #5 on: September 07, 2011, 02:32:56 PM »
Hint:  Consider H*OMO-LUMO interactions and unlike NMe3 N(SiH3)3 is trigonal planar.

Offline DevaDevil

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Re: back bonding
« Reply #6 on: September 07, 2011, 04:14:28 PM »
SiH3 electrons are available



are they? What about nitrogen and its 3 bonds?
Consider the lone uncoupled electron pair on nitrogen

Offline cheese (MSW)

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Re: back bonding
« Reply #7 on: September 07, 2011, 06:25:00 PM »
"  ...planar N atoms exist.  Some early ones were in silylamines such as N(SiMe3)3 and ..." (it then gives the bonding away!)

F. A. Cotton, G. Wilkinson, C. A. Murillo, M. Bochmann, Advanced Inorganic Chemistry 6th ed (1999). p 312

Offline geethika

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Re: back bonding
« Reply #8 on: September 09, 2011, 03:02:46 AM »
thanks guys i got the concept

Offline Rucasta

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Re: back bonding
« Reply #9 on: September 09, 2011, 07:32:45 PM »
I think I'm missing a point in this topic...  Isn't back bonding the movement of electron density from a metal (M) into the pi* or sigma* orbitals of the attached ligand?  If this is the case and we are looking at M-NR3 (r=SiH3), I'm pretty sure that back bonding doesn't occur, regardless of the R groups.

Offline geethika

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Re: back bonding
« Reply #10 on: September 11, 2011, 12:48:31 PM »
here Si acts as metal and there exists a d pie - p pie bonding between nitrogen and silicon

Offline DevaDevil

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Re: back bonding
« Reply #11 on: September 11, 2011, 03:31:01 PM »
I think I'm missing a point in this topic...  Isn't back bonding the movement of electron density from a metal (M) into the pi* or sigma* orbitals of the attached ligand?  If this is the case and we are looking at M-NR3 (r=SiH3), I'm pretty sure that back bonding doesn't occur, regardless of the R groups.


back bonding happens through the lone pair of the center nitrogen into the antibonding orbital of the SiH3 ligand

Offline cheese (MSW)

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Re: back bonding
« Reply #12 on: September 12, 2011, 01:01:11 AM »
DevaDevil: the antibonding orbital of the SiH3 ligand
Could you elaborate on that please. cheers.

Offline cheese (MSW)

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Re: back bonding
« Reply #13 on: September 13, 2011, 01:47:03 PM »
(My 2nd –ve rating has been noted)
 π -backbonding
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

Offline shachi

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Re: back bonding
« Reply #14 on: February 05, 2012, 08:32:51 PM »
Hello Geethika

N(Sih3)3  should actually be sp3 hybridisation bt it has sp2 hb as the lone pair of N donates its electrons to the vacant d orbital of Si.
This is a question of Jee. its doesn't take place by actual overlapping of orbitals .
I guess nw u got it!!

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