Unit 6A: Executive Summary
Quantum Mechanics
c = the speed of light = 2.998*108 m/s
λ = wavelength (generally in nm)
h = Planck’s constant = 6.63*10-34 J*s
E = energy (generally in J)
m = mass (mass of an electron = 9.11*10-31 kg)
c=λv
E = hv
λ = h / mv (deBroglie wavelength, takes mass into account)
Colors of light:
 Infrared: 700+ nm
 Red: 630 - 700 nm
 Orange: 590 – 630 nm
 Yellow: 560 – 590 nm
 Green: 490 – 560 nm
 Blue: 450 – 490 nm
 Violet: 400 – 450 nm
 Ultraviolet: < 400 nm
IMFAs
London Dispersion Forces: exist between all atoms and molecules, and are the only forces
between nonpolar atoms and molecules. Are the weakest type of intermolecular forces, caused by
instantaneous dipoles. Dispersion forces tend to increase in strength with increasing molecular
weight.
Dipole-Dipole Forces: exist between polar molecules when the net positive end of one attracts the
net negative end of another. Only exist when molecules are close together.
Ion-Dipole Forces: exist between ions and polar molecules. A cation attracts the negative ends of
polar molecules, or an anion attracts the positive ends.
Hydrogen Bonds: exist only between an H bonded to an O, F, or N in a very polar bond and
usually another O, F, or N, or other very electronegative atom. The strongest type of intermolecular
forces.
 When molecules of 2 substances are relatively the same mass, the different strengths of
attractive forces are due to differences in the length of the dipole moment.
 When molecules of 2 substances are different masses, the different strengths of attractive forces
are due to the strength of dispersion forces (the more massive one generally has stronger
attractive forces).
Types of Bonds
Single bonds: Sharing of 1 electron pair (H-H)
Double bonds: Sharing of 2 electron pairs (O=C=O)
Triple bonds: Sharing of 3 electron pairs (N=N)
Bond length decreases as the number of shared electron pairs increases
Drawing Lewis Structures
1) Pick a central atom. This should be of the element with the lowest electronegativity
(generally from families 3, 4, or 5, or a noble gas)
2) Determine how many bonds are needed by subtracting the valence electrons (number of
family) from the total electrons needed to complete octets (or duets etc) and dividing by two.
3) Connect each atom with a single bond.
4) Add more bonds if needed. Otherwise, complete the octets on the outer atoms.
5) If there are extra electrons, add them to the central atom.
Periodic Trends
Size


Generally, size decreases across a period and increases down a family
As the nuclear charge increases but no new electron orbitals are added, the valence electrons
are held more strongly by the nucleus, decreasing the size
 In transition metals, the electrons in filled d orbitals tend to repel one another strongly
enough to result in a deviation from this trend
Ionization Energy and Electronegativity
 Ionization energy tends to increase across a period and up a family, as does electronegativity
 Across a period, as more protons are added to the nucleus, they hold the valence electrons
more tightly and removing an electron requires more energy
 As more orbitals are added, the valence electrons are farther away from the nucleus,
reducing the pull of the protons on the electrons and allowing them to be removed more
easily.
Central Atom Hybridization / Shape
Shape is determined by the bonding and lone pairs of electrons on the central atom.
Electron Geometry
Linear
Trigonal Planar
Tetrahedral
Trigonal Bipyramidal
Octahedral
Molecular Geometry
Linear
Bent
Trigonal Planar
Bent
Trigonal Pyramidal
Tetrahedral
Linear
T-Shaped
See-Saw
Trigonal Bipyramidal
Square Planar
Square Pyramidal
Octahedral
Bonding Pairs
2
2
3
2
3
4
2
3
4
5
4
5
6
Lone Pairs
0
1
0
2
1
0
3
2
1
0
2
1
0
Hybridization
s
sp
sp
sp2
sp2
sp2
sp2d
sp2d
sp2d
sp2d
sp2d2
sp2d2
sp2d2
IMPORTANT! Only elements with a d orbital may exceed an octet (elements in periods 4 or later,
starting with the transition metals in period 4)