Molecular Geometry and Bonding Theories. Brown, LeMay Ch 9 AP Chemistry Monta Vista High School. Rationale for VSEPR Theory. Lewis structure is a flat drawing showing the relative placement of atoms, bonds etc. in a molecule, but does not tell anything about the shape of the molecule.
Brown, LeMay Ch 9
Monta Vista High School
Even though the VSEPR model is useful in predicting the shapes of molecules, it does not differentiate between single, double and triple bonds and does not account for bond strengths.
• In complex molecules that contain polar covalent bonds, the three-dimensional geometry and the compound’s symmetry determine if there is a net dipole moment
• Mathematically, dipole moments are vectors; they possess both a magnitude and a direction
• Dipole moment of a molecule is the vector sum of the dipole moments of the individual bonds in the molecule
• If the individual bond dipole moments cancel one another, there is no net dipole moment
• Molecular structures that are highly symmetrical (tetrahedral and square planar AB4, trigonal bipyramidal AB5, and octahedral AB6) have no net dipole moment; individual bond dipole moments completely cancel out
• In molecules and ions that have V-shaped, trigonal pyramidal, seesaw, T-shaped, and square pyramidal geometries, the bond dipole moments cannot cancel one another and they have a nonzero dipole moment
:Molecular Dipole Moments Contd.
Two factors must be considered in the polarity of a molecule:
m = Q r
2. If individual bonds are polar, then do individual dipole moments cancel out or not. A molecule is polar if its centers of (+) and (-) charge do not cancel out- generally happens in a distorted molecule (molecules with lone pairs of e on the central atom.)
How to determine if a molecule is polar?
The sum of the bond dipole moments in a molecule determines the overall polarity of the molecule.
Draw molecular geometires, bond dipole moments, and overall dipole
moments. Also, name the e- domain geometry and the molecular
CO2 BF3 H2O
CCl4 NH3 PH3
The basic principle of VB theory is that a covalent bond forms when the orbitals of two atoms overlap. Three central themes of VB theory derive from this principle:
1. Opposing spins of e pairs: In accordance with Pauli’s exclusion principle, an orbital can have max of two e with opposite spins.
2. Maximum overlap of bonding orbitals: The bond strength depends upon the attraction of nuclei for the shared e, so the greater the overlap, the stronger the bond.
3. End to end overlap of the atomic orbitals form a sigma bondand allows the free rotation of the parts of the molecule. Side-to-side overlap forms a pi bond, which restricts rotation. A multiple bond consists of one sigma bond and rest pi bonds.
Sigma (s) bond:
Pi (p) bond:
Ex: O2 N2
Figure : Formation of bond in H2
1. The number of hybrid orbitals obtained equals the number of atomic orbitals mixed.
2. The type of hybrid orbitals obtained varies with the types of atomic orbitals mixed.
Hybridization of s and p Orbitals
Combination of an ns and two nporbitals produces three equivalent sp2 hybrid orbitals.
• Atomic orbitals other than ns orbitals can interact to form molecular orbitals
• p orbitals are not spherically symmetrical — need to define a coordinate system to know which lobes are interacting in
• For each np subshell, there are npx, npy, and npz orbitals
– All have the same energy and are degenerate but have different spatial orientations
• Just as with ns orbitals, molecular orbitals can be formed from np orbitals by taking their mathematical sum and difference
• With this approach, the electronic structures of homonuclear diatomic molecules (molecules with two identical atoms), can be understood.
• Most substances contain only paired electrons like F2.
• F2 has a total of 14 valence electrons; starting at the lowest energy level, the electrons are placed in the orbitals according to the Pauli’s principle and Hund’s rule.
– Ttwo electrons each fill the 2sand *2s orbitals, two fill the 2pz orbital, four fill two degenerate orbitals, and four fill two degenerate * orbitals.
– There are eight bonding and six antibonding electrons, giving a bond order of 1.
• The O2 molecule contains two unpaired electrons and is attracted into a magnetic field.
• A similar procedure can be applied to molecules with two dissimilar atoms, called heteronuclear diatomic molecules.
• When two nonidentical atoms interact to form a chemical bond, the interacting atomic orbitals do not have the same energy.
• Use a molecular orbital energy-level diagram that is skewed or tilted toward the more electronegative element.
• An odd number of valence electrons: NO
– Nitric oxide (NO) is an example of a heteronuclear diatomic molecule.
– Molecular orbital theory is able to describe the bonding in molecules with an odd number of electrons, such as NO, whereas Lewis electron structures cannot.
– The molecular orbital energy-level diagram for NO is similar to that for O2.
– Molecular orbital theory can also describe the chemistry of molecules, such as NO.
• Nonbonding Molecular Orbitals
– Molecular orbital theory can explain the presence of lone pairs of electrons by determining the presence of nonbonding molecular orbitals (nb).
– A nonbonding molecular orbital occupied by a pair of electrons is the molecular orbital equivalent of a lone pair of electrons.
Draw Lewis structures. For C’s: label hybridization, molecular geometry, and unique bond angles
Ex: ethene; C-C s-bonds and C-H s-bonds result from axial overlap of H s-orbitals and C sp2-orbitals
p orbital bonds side-by-side = p bond
sp2 hybrids bond axially = s bonds
Ex: ethyne (a.k.a. acetylene) C-C s-bond and C-H s-bonds result from axial overlap of H s-orbitals and C sp-orbital
p orbital bonds side-by-side = p bonds
sp hybrids bond axially = s bonds
Ex: benzene; C-C s-bonds and C-H s-bonds result from axial overlap of H s-orbitals and C sp2-orbitals
(delocalized – MINIMUM OF 4 c’S)
Hybrid orbital theory assumes that all bonds are formed with localized electrons, which is not true. MO (Molecular Orbital) theory explains bonding in terms of delocalized orbitals as well.
½ (# bonding e- - # antibonding e-)
B.O. (N2) = ½ (10 – 4) =6 / 2 = 3 (triple bond)
½ (# bonding e- - # antibonding e-)
B.O. (O2) = ½ (10 – 6) == 2 (double bond)
In an element or compound:
Ex: N O