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Transcript
Hybridization – mixing of two or more atomic
orbitals to form a new set of hybrid orbitals.
1. Mix at least 2 nonequivalent atomic orbitals (e.g. s and
p). Hybrid orbitals have very different shape from
original atomic orbitals.
2. Number of hybrid orbitals is equal to number of pure
atomic orbitals used in the hybridization process.
3. Covalent bonds are formed by:
a. Overlap of hybrid orbitals with atomic orbitals
b. Overlap of hybrid orbitals with other hybrid orbitals
•Methane: known to have four identical C-H bonds.
•How is this possible if the valence orbitals of a carbon atom
are 2s and 2p orbitals?
•Carbon adopts a set of atomic orbitals other than its
“native” 2s and 2p orbitals to bond to the hydrogen atoms
in methane.
•Combine the carbon 2s and 2p orbitals to form “hybrid”
orbitals.
•The four new orbitals are sp3 orbitals because they are
formed from one 2s and three 2p orbitals (s1p3).
•The hybridization of the carbon 2s and 2p orbitals can be
represented by an orbital energy-level diagram.
sp3 Hybridization
• Combination of one s and three p orbitals.
• Whenever a set of equivalent tetrahedral atomic
orbitals is required by an atom, the localized electron
model assumes that the atom adopts a set of sp3
orbitals; the atom becomes sp3 hybridized.
Tetrahedral set of Four
sp3 Orbitals
sp2 Hybridization
• Combination of one s and two p orbitals.
• Gives a trigonal planar arrangement of atomic
orbitals.
• One p orbital is not used.
 Oriented perpendicular to the plane of the sp2
orbitals.
7
•The hybridization of the s, px, and py atomic orbitals results
in the formation of three sp2 orbitals centered in the xy
plane.
•The large lobes of the orbitals lie in the plane at angles of
120 degrees and point toward the corners of a triangle.
•An orbital energy-level diagram for sp2 hybridization.
•Note that one p orbital remains unchanged.
•Ethylene (C2H4) is an important starting material in the
manufacture of plastics.
•The C2 H4 molecule has 12 valence electrons and the
following Lewis structure.
•The double bond acts as one effective pair, so in the
ethylene molecule each carbon is surrounded by three
effective pairs.
•An “effective pair” of electrons is either a lone pair, a single
bond, a double bond, or a triple bond.
•In other words, an effective pair is an electron group used
for the purposes of determining molecular geometry.
•sp2 hybridization can be used to account for the bonds in
ethylene.
•The three sp2 orbitals on each carbon can be used to share
electrons.
•In each of these bonds, the electron pair is shared in an
area centered on a line running between the atoms.
•This type of covalent bond is called a sigma (σ) bond.
•In ethylene, the σ bonds are formed using sp2 orbitals on
each carbon and the 1s orbital on each hydrogen atom.
•How can we explain the double bond between the carbon
atoms?
•In the σ bond the electron pair occupies the space between
the carbon atoms.
•The second bond must therefore result from sharing an
electron pair in the space above and below the σ bond.
•This type of bond can be formed using the 2p orbital
perpendicular to the sp2 hybrid orbitals on each carbon
atom.
•The parallel p orbitals can share an electron pair, which
occupies the space above and below a line joining the atoms,
to form a pi (π) bond.
•Note that σ bonds are formed from orbitals whose lobes
point toward each other, but π bonds result from parallel
orbitals.
•Figure (a): The orbitals used to form the bonds in
ethylene.
•Figure (b): The Lewis structure of ethylene.
•A double bond always consists of one sigma (σ)
bond, where the electron pair is located directly
between the atoms, and one pi (π) bond, where the
shared pair occupies the space above and below the
σ bond.
•Important!!
•Whenever an atom is surrounded by three effective
pairs, a set of sp2 hybrid orbitals is required.
sp Hybridization
•Combination of one s and one p orbital.
•Gives a linear arrangement of atomic orbitals.
•Two p orbitals are not used.
Needed to form the  bonds.
•When one s orbital and one p orbital are
hybridized, a set of two sp orbitals oriented at 180
degrees results.
•The orbital energy-level diagram for the formation
of sp hybrid orbitals on carbon.
•Consider carbon dioxide, which has the following Lewis
structure:
•In the CO2 molecule the carbon atom has two effective
pairs that will be arranged at an angle of 180 degrees.
•We need a pair of atomic orbitals oriented in opposite
directions (sp orbitals).
•Two effective pairs around an atom will always require sp
hybridization of that atom.
•The sp hybrid orbitals are used to form the σ bonds
between the carbon and oxygen atoms in CO2.
•In CO2 each oxygen atom has three effective pairs around
it, requiring a trigonal planar arrangement of the pairs.
•A trigonal set of hybrid orbitals requires sp2 hybridization
so each oxygen atom is sp2 hybridized.
•One p orbital on the oxygen is unchanged and is used for
the π bond with the carbon atom.
•The sp orbitals on carbon form σ bonds with the sp2 orbitals
on the two oxygen atoms.
•The remaining sp2 orbitals on oxygen hold lone pairs.
•The π bonds between the carbon atom and each oxygen
atom are formed by the overlap of parallel 2p orbitals.
dsp3 Hybridization
•Combination of one d, one s, and three p orbitals.
•Gives a trigonal bipyramidal arrangement of five
equivalent hybrid orbitals.
•Consider phosphorus pentachloride (PCl5) whose Lewis
structure is shown below.
•Five electron pairs requires a trigonal bipyramidal
arrangement, therefore, we need a trigonal bypyramidal set
of orbitals on phosphorus.
•A set of five effective pairs around a given atom always
requires a trigonal bypyramidal arrangement, which in turn
requires dsp3 hybridization of that atom.
•The five P-Cl σ bonds are formed by sharing electrons
between a dsp3 orbital on the phosphorus and an sp3 orbital
on each chlorine.
•The other sp3 orbitals on chlorine hold lone pairs.
d2sp3 Hybridization
•Combination of two d, one s, and three p orbitals.
•Gives an octahedral arrangement of six equivalent hybrid
orbitals.
•Consider sulfur hexafluoride (SF6) whose Lewis structure is
shown below.
•Sulfur has six pairs of electrons and requires an octahedral
arrangement of pairs and an octahedral set of six hybrid
orbitals, or d2sp3 hybridization.
•Fluorine has four pairs and is assumed to be sp3 hybridized.
•Six electron pairs around an atom are always arranged
octahedrally and require d2sp3 hybridization of the atom.
•Localized Electron Model
1. Draw the Lewis structure(s).
2. Determine the arrangement of electron pairs
using the VSEPR model.
3. Specify the hybrid orbitals needed to
accommodate the electron pairs.
The relationship of
the number of
effective pairs, their
spatial arrangement,
and the hybrid
orbital set required.