Hybridization of Atomic Orbitals

• As two Hydrogen atoms approach, their 1s orbitals interact, and the electrons feel the attraction of the other proton.
• Electron density builds between the nuclei.
• A stable H2 molecule forms at an internuclear distance of 74 pm.

Hybridization of Atomic Orbitals

• Atomic orbital overlap applies to polyatomic molecules.
• A bonding scheme must account for molecular geometry.
• VB theory explains bonding in polyatomic molecules.

sp3 Hybridization

• Consider the CH4 molecule and its valence electrons.
• Carbon has two unpaired electrons and can form two bonds with Hydrogen in its ground state.
• CH2 is unstable.
• To account for four C-H bonds in methane, promote an electron from the 2s to the 2p orbital.
• Four unpaired electrons on C will form four C-H bonds.
• Three HCH bond angles would have to be 90°, but HCH angles in methane are actually 109.5°.
• Hybrid orbitals explain methane bonding by combining nonequivalent orbitals.
• Hybridization mixes atomic orbitals in an atom to generate hybrid orbitals.
• Four equivalent hybrid orbitals for carbon can be generated by mixing the 2s orbital and three 2p orbitals.
• The four new orbitals are called sp3 hybrid orbitals because they form from one s and three p orbitals.
• sp3 hybrid orbitals orient towards the corners of a regular tetrahedron.
• Four covalent bonds form between the carbon sp3 hybrid orbitals and the hydrogen 1s orbitals in CH4.
• CH4 has a tetrahedral shape with HCH angles of 109.5°.
• Although hybridization requires energy, the energy released upon C-H bond formation compensates.
• The purple color is made up of the red and blue components of the original solutions, the sp3 hybrid orbitals possess both s and p orbital characteristics.
• For example, ammonia (NH3) uses sp3 hybridization
• The arrangement of four electron pairs is tetrahedral, so the bonding in NH3 can be explained by assuming Nitrogen, like Carbon in CH4 is sp3-hybridized.
• The ground-state electron configuration of Nitrogen is 1s22s22p3, so that the orbital diagram for the sp3 hybrid-ized N atom is
• Three of the four hybrid orbitals form covalent N-H bonds, and the fourth hybrid orbital accommodates the lone pair on nitrogen.
• Repulsion between lone pair electrons and bonding electrons reduces the HNH bond angles from 109.5° to 107.3°.
• Use hybridization to describe bonding only when the arrangement of electron pairs has been predicted using VSEPR.
• The tetrahedral arrangement of electron pairs means we assume that one s and three p orbitals are hybridized to form four sp3 hybrid orbitals.

sp Hybridization

• The beryllium chloride (BeCl2) molecule is linear by VSEPR.
• Beryllium does not form covalent bonds with Chlorine in its ground state because its electrons are paired in the 2s orbital, First, a 2s electron shifts to a 2p orbital to explain Be's bonding behavior.
• Two Beryllium orbitals (2s and 2p) are available for bonding.
• One Chlorine atom shares a 2s electron, and the other Chlorine shares a 2p electron, making two nonequivalent BeCl bonds.
• The 2s and 2p orbitals are hybridized to form two equivalent sp hybrid orbitals because the two BeCl bonds are identical.
• Each of BeCl bonds are formed by the overlap of a Beryllium Sp hybrid orbital, and a Chlorine 3p orbital.

sp2 Hybridization

• Boron trifluoride (BF3) is known to have planar geometry.
• The valence electrons' orbital diagram is considered
• First, a 2s electron promotes to an empty 2p orbital.
• Mixing the 2s orbital with the two 2p orbitals generates three sp2 hybrid orbitals.
• Each BF bond is formed by an overlap of boron sp2 hybrid orbital and fluorine 2p orbital.
• The hybridized structure is planar, and it conforms to experimental findings and VSEPR predictions.

Hybridization and the Octet Rule

• An atom starting with one s and three p orbitals possesses four orbitals and 8 electrons, regardless of the type of hybridization.
• 8 is the maximum number of electrons that an atom can accommodate in the valence shell for elements in the second period.
• Different situation for an atom of a third-period element as it will use 3s and 3p orbitals of the atom to form hybrid orbitals in a molecule, then the octet rule applies.
• In some molecules the same atom may use one or more 3d orbitals, in addition to the 3s and 3p orbitals to form hybrid orbitals.

Hybridization Summary

• Hybridization applies only to covalent bonding and is not applied to isolated atoms.
• Hybridization is mixing at least two nonequivalent atomic orbitals, forming different shapes.
• Generate the same number of hybrid orbitals that participate in the hybridization process.
• Hybridization requires an input of energy, but the system recovers this during bond formation.
• Covalent bonds form from overlapping hybrid orbitals, and hybrid orbitals with unhybridized ones
• A hybridization bonding scheme falls within valence bond theory, and electrons occupy hybrid orbitals of individual atoms.

Hybridization Examples

• To assign a suitable state of hybridization to the central atom in a molecule, you must have some idea about the geometry of the molecule.
• Draw the Lewis structure of the molecule.
• Predict the arrangement of electron pairs using the VSEPR model from table 10.1
• Deduce the hybridization of the central atom by matching the pairs' arrangement from table 10. 4.

Hybridization of s, p, and d Orbitals

• Hybridization explains bonding with s and p orbitals.
• With third period elements can have hybridization involves only s and p orbitals .
• Understanding trigonal bipyramidal and octahedral geometries, one must include the d orbitals in the hybridization concept.
• For the SF6 molecule, the S atom is sp3d2-hybridized where electrons from 3s and 3p orbitals from to 2 of the 3d orbitals
• The six S-F bonds form by overlapping the orbitals of the S atom, and the 2p orbitals of the F atoms.
• The octet rule is violated since there are 12 electrons around the S atom
• Using d orbitals in addition to s and p orbitals to form an expanded octet is valence-shell expansion.
• Second-period elements do not have 2d energy levels.
• Atoms of second-period elements cannot have more than eight electrons in their compounds.

Hybridization in Molecules Containing Double and Triple Bonds

• Double and triple bonds make hybridization useful.
• The C2H4 molecule contains a carbon-carbon double bond and planar geometry.
• The Carbon atom is sp2-hybridized while only the 2px and 2py orbitals combine with the 2s orbital, and that the 2pz orbital remains unchanged.
• Each carbon atom uses the three sp2 hybrid orbitals to form two bonds with the two hydrogen 1s orbitals and one bond with the sp2 hybrid orbital of the adjacent carbon atom, and The two unhybridized 2pz orbitals of the carbon atoms form another bond by overlapping sideways.
• The three bonds from each Carbon atom form sigma bonds, which are covalent bonds extending to end, with the electron density between the nuclei of the bonding atoms.
• The other type is a pi bond where a covalent bond formed by sideways overlapping orbitals with the electron density above and below the plane of the nuclei of the bonding atoms.
• The carbon atoms form a pi bond.

sp-hybridization

• The C atom is sp-hybridized by mixing the 2s with the 2px orbital where the two sp hybrid orbitals of each carbon atom form one sigma bond with a hydrogen 1s orbital.
• There are two pi bonds from the sideways overlap of the hybridized 2py and 2pz orbital bonds that make up one sigma, and two pi bonds

Multiple Bonds Rule

• If the central atom forms a double bond, it is sp2-hybridized, and If it forms two double bonds, or a triple bond, it is sp-hybridized.
• Atoms of third-period elements and beyond that form multiple bonds present a complicated picture.