Grade 12th passInorganic Chemistry
Explain the concept of banana/tau bonding.
Explain the concept of banana/tau bonding.
4 Answers
Harishwar
Bananas are fun things, those of you who might have thought that I have changed into Sean Banancan relax and breathe easy again. I am here to talk about the banana bond. This is the informal name for a bent bond, the banana bond I want to discuss now is a very bent bond which is found in diborane.
Now after we have seen how to draw the Lewis structure of a borohydride anion we can try to make a similar diagram for diboraneusing conventional two atom and two electron bonds. This is the best I can do and it is clear that it is a dismal failure, the boronsonly have six electrons around them and this falls short of the eight electrons which the atoms need to attain a noble gas like full outer shell.[Bad version of diborane, it has only six valence electrons around each boron.]Bad version of diborane, it has only six valence electrons around each boron.
Now what we have in this molecule are two threecentred two electron bonds, these are bonds which involve three atoms and only have two electrons in them. With two sp3and one s orbital we can draw out the three ways the orbitals can be combined to make three molecular orbitals. Here are the three molecular orbitals associated with the three atoms.[The three orbitals in the banana bond]The three orbitals in the banana bond
So after we back the electrons into the two bonding orbitals for the banana bonds, we can now get eight electrons around each boron atom. Thus even for these electron deficient molecules we can still have eight electrons around each boron.[Electrons in a diborane molecule]Electrons in a diborane molecule
We can move ontobigger and better things, for example if we replace the main group boron with a transition metal (such as copper) then to have a noble gas configuration we need 18 electrons around the transition metal. For example we can have a copper complex. How about this complex.[[Cu(BH4)(PPh3)2]][Cu(BH4)(PPh3)2]Now I hold the view that in organometallic chemistry that oxidation state can bea bit of fluid, at times it is not clear. It is possible to assign two different oxidation states to a metal in the same complex. Now if we assume that the copper in our complex has an oxidationstate of zero then the zero valent copper has 11 electrons, ten of these electrons will stay around the copper in its atomic orbitals while one will enter one of the 3 centre 2 electron bonds. Two electrons from the bridging hydrogens will be in these banana bonds while one of the boron electrons will also be in a banana bond. The bonds to the phosphines contain two electrons from the phosphorus atoms (dative bonds) while the terminal hydrogen to boron bonds are simple normal two centre two electron bonds.[The copper borohydride complex with a zero valent copper and a zero valent boron]The copper borohydride complex with a zero valent copper and a zero valent boron
If we make the complex a copper(I) complex which has a boron in the (-I) oxidation state then the boron will have four electrons to start with. The complex will still have four electrons in the banana bonds, four in the dative bonds from the phosphines and four in the “normal” bonds to the terminal hydrides. Again we have 18 valence electrons around the copper and eight around the boron.
[The copper borohydride complex with a copper(I) centre in it.]The copper borohydride complex with a copper(I) centre in it.
Now after we have seen how to draw the Lewis structure of a borohydride anion we can try to make a similar diagram for diboraneusing conventional two atom and two electron bonds. This is the best I can do and it is clear that it is a dismal failure, the boronsonly have six electrons around them and this falls short of the eight electrons which the atoms need to attain a noble gas like full outer shell.[Bad version of diborane, it has only six valence electrons around each boron.]Bad version of diborane, it has only six valence electrons around each boron.
Now what we have in this molecule are two threecentred two electron bonds, these are bonds which involve three atoms and only have two electrons in them. With two sp3and one s orbital we can draw out the three ways the orbitals can be combined to make three molecular orbitals. Here are the three molecular orbitals associated with the three atoms.[The three orbitals in the banana bond]The three orbitals in the banana bond
So after we back the electrons into the two bonding orbitals for the banana bonds, we can now get eight electrons around each boron atom. Thus even for these electron deficient molecules we can still have eight electrons around each boron.[Electrons in a diborane molecule]Electrons in a diborane molecule
We can move ontobigger and better things, for example if we replace the main group boron with a transition metal (such as copper) then to have a noble gas configuration we need 18 electrons around the transition metal. For example we can have a copper complex. How about this complex.[[Cu(BH4)(PPh3)2]][Cu(BH4)(PPh3)2]Now I hold the view that in organometallic chemistry that oxidation state can bea bit of fluid, at times it is not clear. It is possible to assign two different oxidation states to a metal in the same complex. Now if we assume that the copper in our complex has an oxidationstate of zero then the zero valent copper has 11 electrons, ten of these electrons will stay around the copper in its atomic orbitals while one will enter one of the 3 centre 2 electron bonds. Two electrons from the bridging hydrogens will be in these banana bonds while one of the boron electrons will also be in a banana bond. The bonds to the phosphines contain two electrons from the phosphorus atoms (dative bonds) while the terminal hydrogen to boron bonds are simple normal two centre two electron bonds.[The copper borohydride complex with a zero valent copper and a zero valent boron]The copper borohydride complex with a zero valent copper and a zero valent boron
If we make the complex a copper(I) complex which has a boron in the (-I) oxidation state then the boron will have four electrons to start with. The complex will still have four electrons in the banana bonds, four in the dative bonds from the phosphines and four in the “normal” bonds to the terminal hydrides. Again we have 18 valence electrons around the copper and eight around the boron.
[The copper borohydride complex with a copper(I) centre in it.]The copper borohydride complex with a copper(I) centre in it.
ANMOL SETH
It was a bit complicated.. However.. I have read that Banana bond is a result of partial sharing of two hydrogen atoms by two boron atoms..
And the bent that appears, is it because of the repulsive force that acts between the hydrogen atoms?(Like the attractive force between H-B overriding the repulsive force of H-H ?)
Manjot Kaur
Tau bond is a banana bond too.
"Banana bond" is not necessarily only for 3c2e bonds. Any bent bond is called a "banana" bond or tau bond. Tau-bonds also provide an alternative valid description of electron density in alkenes and alkynes.
jyoti bhatia
A chemical bond is the phenomenon of chemical species like atoms held together by electrostatic or electronic forces. Of several types of chemical bonds, a special type exists in the molecule of boron hydride BH3 which exists as a dimer B2H6 molecule. It contains two types of hydrogen atoms. Four hydrogen atoms are of one type, which are used in making four normal covalent bonds with two boron atoms. Remaining two H atoms form bridges between the two boron atoms through three-center electron-pair bonds. This type of bond involves three atoms but only two electrons. Since shape of electron cloud resembles a banana, it is called a banana bond.
In organic chemistry, a bent bond, also known as a banana bond, is a type of covalent chemical bond with a geometry reminiscent of a banana. The term itself is a general representation of electron density or configuration resembling a similar ‘bent’ structure within small ring molecules like cyclopropane or as a representation of double or triple bonds within a compound which is an alternative to sigma and pi bond model.
Bent bonds are a special type of chemical bonding in which ordinary hybridization state of two atoms making up a chemical bond are modified with increased or decreased s-orbital character in order to accommodate a particular molecular geometry. Bent bonds are found in strained organic compounds like cyclo-propane, oxirane and aziridine.
In these compounds, it is not possible for carbon atoms to assume the 109.5 degrees bond angles with standard sp3 hybridization. Increasing the p-character to sp5 (i.e. 1/6 s-density and 5/6 p-density) makes it possible to reduce bond angles to 60°. At the same time, carbon-to hydrogen bonds gain more s-character, which shortens them. In cyclo-propane, maximum electron density between two carbon atoms does not correspond to inter-nuclear axis, hence the name ‘bent bond’. In cyclo-propane, inter-orbital angle is 104°. This bending can be observed experimentally by X-ray diffraction of certain cyclopropane derivatives: deformation density is outside the line of centers between carbons. The carbon-carbon bond lengths are shorter than in a regular alkane bond: 151 pm versus 153 pm.
Cyclobutane is a larger ring but still has bent bonds. In this molecule, the carbon bond angles are 90° for planar conformation and 88° for puckered one. Unlike in cyclopropane, the C-C bond lengths increase rather than decrease; this is due to 1,3-nonbonded steric repulsion. In terms of reactivity, cyclobutane is relatively inert and behaves like ordinary alkanes.
Walsh orbital model:
An alternative model utilizes semi-localized Walsh orbitals in which cyclopropane is described as a carbon sp2 sigma bonding and in-plane pi bonding system. But, this model does not represent the ground state of cyclopropane as it cannot be transformed into localized or fully delocalized descriptions via a unitary transformation.
Two different explanations for the nature of double & triple covalent bonds in organic molecules were proposed in the 1930s. Linus Pauling proposed that double bond results from two equivalent tetrahedral orbitals from each atom, which came to be called ‘banana bonds’ or ‘tau bonds’. Erich Huckel proposed a representation of the double bond as a combination of a sigma bond plus a pi bond. Although a conclusive statement cannot be made on basis of currently available information, one can continue to consider the σ/π and bent-bond descriptions of ethylene to be equivalent. Overall distribution of electrons is exactly the same in the two models.
The bent theory can also explain other phenomena in organic molecules. In fluoromethane (CH3F), experimental F-C-H bond angle is 109 degrees, which is greater than calculated value. This is because according to Bent’s rule, C-F bond gains p-orbital character leading to high s-character in C-H bonds, less for the F-C-H bond angle. Difference is again explained in terms of bent bonds.
Bent bonds also come into play in the gauche effect, explaining preference for gauche conformations in substituted alkanes and the alkene cis effect associated with unusually stable alkene cis isomers.
In Diborane B-H-B bridge, which is formed by sharing of two electrons, is called banana bond or tau bond. Diborane contains two banana bonds.
In organic chemistry, a bent bond, also known as a banana bond, is a type of covalent chemical bond with a geometry reminiscent of a banana. The term itself is a general representation of electron density or configuration resembling a similar ‘bent’ structure within small ring molecules like cyclopropane or as a representation of double or triple bonds within a compound which is an alternative to sigma and pi bond model.
Bent bonds are a special type of chemical bonding in which ordinary hybridization state of two atoms making up a chemical bond are modified with increased or decreased s-orbital character in order to accommodate a particular molecular geometry. Bent bonds are found in strained organic compounds like cyclo-propane, oxirane and aziridine.
In these compounds, it is not possible for carbon atoms to assume the 109.5 degrees bond angles with standard sp3 hybridization. Increasing the p-character to sp5 (i.e. 1/6 s-density and 5/6 p-density) makes it possible to reduce bond angles to 60°. At the same time, carbon-to hydrogen bonds gain more s-character, which shortens them. In cyclo-propane, maximum electron density between two carbon atoms does not correspond to inter-nuclear axis, hence the name ‘bent bond’. In cyclo-propane, inter-orbital angle is 104°. This bending can be observed experimentally by X-ray diffraction of certain cyclopropane derivatives: deformation density is outside the line of centers between carbons. The carbon-carbon bond lengths are shorter than in a regular alkane bond: 151 pm versus 153 pm.
Cyclobutane is a larger ring but still has bent bonds. In this molecule, the carbon bond angles are 90° for planar conformation and 88° for puckered one. Unlike in cyclopropane, the C-C bond lengths increase rather than decrease; this is due to 1,3-nonbonded steric repulsion. In terms of reactivity, cyclobutane is relatively inert and behaves like ordinary alkanes.
Walsh orbital model:
An alternative model utilizes semi-localized Walsh orbitals in which cyclopropane is described as a carbon sp2 sigma bonding and in-plane pi bonding system. But, this model does not represent the ground state of cyclopropane as it cannot be transformed into localized or fully delocalized descriptions via a unitary transformation.
Two different explanations for the nature of double & triple covalent bonds in organic molecules were proposed in the 1930s. Linus Pauling proposed that double bond results from two equivalent tetrahedral orbitals from each atom, which came to be called ‘banana bonds’ or ‘tau bonds’. Erich Huckel proposed a representation of the double bond as a combination of a sigma bond plus a pi bond. Although a conclusive statement cannot be made on basis of currently available information, one can continue to consider the σ/π and bent-bond descriptions of ethylene to be equivalent. Overall distribution of electrons is exactly the same in the two models.
The bent theory can also explain other phenomena in organic molecules. In fluoromethane (CH3F), experimental F-C-H bond angle is 109 degrees, which is greater than calculated value. This is because according to Bent’s rule, C-F bond gains p-orbital character leading to high s-character in C-H bonds, less for the F-C-H bond angle. Difference is again explained in terms of bent bonds.
Bent bonds also come into play in the gauche effect, explaining preference for gauche conformations in substituted alkanes and the alkene cis effect associated with unusually stable alkene cis isomers.
In Diborane B-H-B bridge, which is formed by sharing of two electrons, is called banana bond or tau bond. Diborane contains two banana bonds.
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