Chemistry Hybridization and Bonding

Consider an sp2 hybridized atom

s + p + p orbitals are mixed. Three hybrid orbitals are formed. The ideal angle is 120° Three identical sp2 hybrid orbitals and one unhybridized p orbital

List the four orbitals surrounding an sp hybridized atom.

s + p orbitals are mixed. Two hybrid orbitals are formed. The ideal angle is 180° Two identical sp hybrid orbitals and two unhybridized p orbitals

Why is a C-C (sigma) bond shorter than a C-C (pi) bond?

A (Sigma) bond has better shielding of the atomic nuclei. The bonding electrons in a (Sigma) bond are predominantly located between the two bonding nuclei. This shields the positive charges from one another and allows the bond to be shorter (the nuclei can be closer together).A (Pi) bond has a node (where there is a zero probability of finding an electron) along the bonding axis which reduces the amount of electron density between the bonding atoms. A (Pi) bond has most of its electron density in clouds above and below the plane containing the bonding axis. This reduced shielding results in more nuclear repulsions and a longer bond than the comparable (Sigma) bond.

What is the difference between a bond dipole and a molecular dipole?

A bond dipole results from the uneven distribution of electrons in a single bond. The more electronegative atom of the bonding couple will have a partial negative charge and the less electronegative atom will have a partial positive charge. Molecular dipoles result from the geometric sum of all the individual bond dipoles within the molecule.

Is it possible for a (sigma) bond in a molecule to be longer than a (pi) bond? Explain.

A bond length depends upon the atoms being bonded and the orbitals used as well as the type of bond that is formed. For example, a carbon-iodine (Pi) bond resulting from sp3 hybrid orbital overlap has a bond length of 2.025 angstroms. Most of this is due to the large size of the iodine atom. A carbon-carbon bond has a typical bond length of 1.3 angstroms.

Are all (sigma) bonds identical? Explain your answer

All bonds share common characteristics such as not having a node along, or crossing the bonding axis but not all bonds are identical. Bond lengths will depend upon the sizes of the atoms forming the bond and the orbitals being used. A bond resulting from combining two s atomic orbitals will be shorter than a bond made by the same atoms and using sp3 hybrid orbitals. Additionally, the electron distribution (dipole) within the bond depends upon the electronegativities of the bonding partners.

Forthecarbon-chlorinebond(bond'a')indicatedintheabovemolecule,(1)assignthehybridizationofthecarbonandthe chlorine. (2) List the four orbitals surrounding each of those atoms. (3) Indicate which orbitals are overlapping to form the C-Cl bond. (4) What type of orbital do the non-bonding electrons of the chlorine occupy?

(1) Both the carbon and chlorine are sp3 hybridized.(2) Both atoms have four sp3 hybrid orbitals surrounding them. (3) The C-Cl bond results from overlap between an sp3 hybrid orbital from the carbon with an sp3 hybrid orbital from the chlorine. (4) The three pairs of non-bonding electrons on chlorine occupy sp3 hybrid orbitals.

For the carbon-nitrogen bonds (bonds 'b') in the above molecule, (1) assign the hybridization of the carbon and the nitrogen. (2) List the four orbitals surrounding each of those atoms. (3) Indicate which orbitals are overlapping to form all three C-N bonds. (4) What type of orbital do the non-bonding electrons of the nitrogen occupy?

(1) Both the carbon and nitrogen are sp hybridized. (2) Both atoms have two identical sp hybrid orbitals and two unhybridized p orbitals surrounding them. (3) The C-N (-bond results from overlap between an sp hybrid orbital from the carbon with an sp hybrid orbital from the nitrogen. The two -bonds result from side-to-side overlap between an unhybridized p orbital from carbon with an unhybridized p orbital from nitrogen. (4) The pair of non-bonding electrons on nitrogen occupies an sp hybrid orbital.

For the carbon-oxygen bonds (bonds 'd') in the above molecule, (1) assign the hybridization of the carbon and the oxygen. (2) List the four orbitals surrounding each of those atoms. (3) Indicate which orbitals are overlapping to form all the C-O bonds. (4) What type of orbital do the non-bonding electrons of the oxygen occupy?

(1) The carbon is sp2 hybridized and the oxygen is sp2 hybridized. (2) Carbon has three identical sp2 hybrid orbitals and one unhybridized p orbital surrounding it and the oxygen has three identical sp2 hybrid orbitals and one unhybridized p orbital surrounding it. (3) The C-O (Sigma)-bond results from overlap between an sp2 hybrid orbital from the carbon with an sp2 hybrid orbital from the oxygen. The C-O (Pi)-bond results from side-to-side overlap between an unhybridized p orbital from carbon with an unhybridized p orbital from oxygen. (4) The two pairs of non-bonding electrons on oxygen occupy sp2 hybrid orbitals.

For the carbon-nitrogen bond (bond 'c') in the above molecule, (1) assign the hybridization of the carbon and the nitrogen. (2) List the four orbitals surrounding each of those atoms. (3) Indicate which orbitals are overlapping to form the C-N bond. (4) What type of orbital do the non-bonding electrons of the nitrogen occupy?

(1) The carbon is sp3 hybridized and the nitrogen is sp2 hybridized. (2) Carbon has four identical sp3 hybrid orbitals surrounding it and the nitrogen has three identical sp2 hybrid orbitals and one unhybridized p orbital surrounding it. (3) The C-N (Sigma)-bond results from overlap between an sp3 hybrid orbital from the carbon with an sp2 hybrid orbital from the nitrogen (4) The pair of non-bonding electrons on nitrogen occupy an sp2 hybrid orbital.

Describe the requirements for making a (Pi) bond.

(Pi) Bonds result from side-to-side overlap between unhybridized atomic p orbitals. The molecular orbitals created will contain the characteristics of the atomic orbitals that were combined. The side- to-side overlap uses orbital lobes on either side of a node at the atomic nucleus. This node will therefore be present in the resulting molecular orbital that is formed. This is why the (Pi) bonding molecular orbital contains a nodal plane running along the bond axis. This is also the reason why (Pi)bonds do not have free rotation.

Describe the requirements for making a (Sigma) bond

(Sigma) Bonds result from end-to-end overlap between atomic orbitals (either hybrid or pure). Adding atomic orbitals in this fashion creates a molecular orbital with no node between the bonding nuclei. This results in a bond with free rotation.

Give an example of a molecule which has a bond dipole but no molecular dipole.

Carbon tetrachloride (CCl4) has four polar bonds (Cl is more electronegative than C) but no significant net molecular dipole because the tetrahedral shape of the molecule results in a "cancellation" of the dipoles.

What is the difference between a covalent and an ionic bond?

Covalent bonding results from the sharing of electrons between two atoms. This sharing occurs by overlapping (adding) atomic orbitals together to produce molecular orbitals which encompass both of the atoms participating in the bond. Ionic bonds are the result of electrostatic attractions between ions of opposite charge. There is no sharing of any electrons between the ions.

Fully describe the bonding in ethene (C2H4). Identify the types of bonds present and indicate which atomic or hybid orbitals are mixing together to form each molecular bond.

Each of the four C-H (Sigma) bonds in ethylene are identical. They result from overlap between the s atomic orbital of hydrogen with an sp2 hybridized orbital on a carbon. There are two different types of C-C bonds. A C-C (Sigma) bond that results from end-to-end overlap between sp2 hybridized orbitals on each carbon and a C-C (Pi) bond that is formed from side-to-side overlap of the unhybridized p orbitals on each of the sp2 hybridized carbons.

Fully describe the bonding in methane (CH4). Identify the types of bonds present and indicate which atomic or hybid orbitals are mixing together to form each molecular bond.

Methane contains four identical (sigma) bonds which result from the s atomic orbitals of the hydrogen overlapping with the four sp3 hybridized orbitals of the carbon.

Why do we need to invoke (or use) atomic hybridization when we describe bonding in molecules?

The experimentally observed shapes of molecules do not correlate with the use of only pure atomic orbitals in molecular bonding. Hybridization easily explains the observed shapes of molecules.

What is atomic hybridization?

The mathematical mixing of pure atomic orbitals (s, p, d) to generate new "hybrid" orbitals which have different shapes and orientations than the starting atomic orbitals. This mathematical model gives us the ability to predict non-ideal bond angles so we have a more accurate understanding of the shape of molecules.

Consider an sp3 hybridized atom.

s + p + p + p orbitals are mixed. Four hybrid orbitals are formed. The ideal angle is 109.5° Four identical sp3 hybridized orbitals