CHEM 110, Chapter 10 summary
♥: challenging contents
Electron domains
One pair of non-bonding electron pair = 1 domain
One bond (no matter single, double or triple) = 1 domain
X
Example: for the structure of
, the atom X has 5 domains.
♥VSEPR
Valence Shell Electron Pair Repulsion
♥Electron domains try to stay away from each other as much as possible.
To minimize the repulsions (because they are all negative charged)
3D structure of molecules
Caused by VSEPR
Determined by number of domains
2 domains → linear
3 domains → trigonal planar
4 domains → tetrahedral (Td)
5 domains → Trigonal bipyramidal
6 domains → Octahedral (Oh)
♥May vary from ideal structure, depends on the nature of electron domains
Fatness of electron domains: lone pair > triple bond > double bond > single bond
Polarity of a molecule
♥Dipole moment ()
One arrow (bond moment) for each bond
Arrow points from less electronegative atom to more electronegative atom
♥Polarity of the molecule determined by the 3D structure
Opposite bond moments of same magnitudes cancel
Practice on 3D structure: Is IF4‾ a polar ion or non-polar ion?
(1) Draw the Lewis structure;
(2) Find its 3D structure;
(3) Draw the bond moment for each bond;
(4) Cancel the opposite pairs;
(5) Polar or non-polar?
♥Valence bond (VB) theory
♥Orbital hybridization
To prepare for bonding (form electron clouds overlap with another atom)
Type of hybridization is determined by number of electron domains for that atom:
2 domains → sp hybridization (total 2 hybridized orbitals)
3 domains → sp2 hybridization (total 3 hybridized orbitals)
4 domains → sp3 hybridization (total 4 hybridized orbitals)
5 domains → sp3d hybridization (total 5 hybridized orbitals)
6 domains → sp3d2 hybridization (total 6 hybridized orbitals)
Type of bonds
 bond
mostly formed with hybridized orbitals
end-to-end (overlap along the bond axis)
good overlap  strong bond
♥ bond
mostly formed with un-hybridized orbitals
sideway (overlap above and below the bond axis, no overlap along the axis)
low overlap  weak bond
Single bond =  bond
Double bond =  bond +  bond
Triple bond =  bond + 2 ( bond)
Flowchart for a typical VB question:
(1) Draw the Lewis structure;
(2) Count the number of domains for the atom;
(3) Determine the 3D structure of the molecule;
(4) Determine the type of hybridization of the center atom.
Practice on VB theory: What is the kind of orbital hybridization of I atom in IF4‾ ?
♥Molecular orbital (MO) theory
♥Draw the molecular orbitals (MO) diagram from atomic orbitals (AO)
Bonding orbital (such as 1s) and anti-bonding orbital 1s
s + s (2 AO)   +  (2 MO)
p + p (6 AO)  + 2 2  (6 MO)
Count the total electrons in the molecule, and fill them in the MO diagram
Electron configuration of a molecule (homo-nuclear diatomic molecule of 2nd period)
Bond order of a molecule = ½ (number of bonding e‾ – number of anti-bonding e‾)
Higher bond order  higher stability
Magnetic properties (diamagnetic or paramagnetic) of a molecule
Practice on MO theory: Which is more stable, N2, N2+, or N2‾?
(1) Draw the MO diagram for N2;
(2) Count the total number of electrons N2 has;
(3) Insert all electrons in the diagram;
(4) Count the bonding electrons and non-bonding electrons;
(5) Calculate the bond order for N2;
(6) Repeat step1-6 for N2+;
(7) Repeat step1-6 for N2‾;
(8) Compare the bond orders for all three species, the higher bond order, the
more stable.