By Lauren and Joe
Electromagnetic Spectrum
From www.lcse.umn.edu/specs/labs/images/spectrum.gif
Electromagnetic Spectrum
 Frequency and wavelength are inversely related, as
demonstrated by the equation
c=λν
 c-speed of light constant-2.9979x108m/s
 λ-wavelength(m)
 ν-frequency(Hz or s-1)
Quantum Theory
 Einstein—Light behaves as if it consists of quantized
energy packets, meaning that energy can have only
certain allowed values given by the equation
Ephoton=hν
 Ephoton-(J)
 h-Planck’s constant-6.626x10-34 J-sec
 ν-frequency (Hz or s-1)
Another Equation to Remember
Ephoton=Eremove electron/threshold + Ekinetic
Quantum Numbers
 n-principal quantum number (shell)
 i.e. 3s, n=3
 l-azimuthal quantum number (sub-shell)
 The value of l corresponds to the sub-shell of the orbital
 s=0, p=1, d=2, f=3
 i.e. if n=3, can have 3s, 3p, 3d. Accordingly, you can have
l values 0, 1, and 2.
More Quantum Numbers!
 ml-magnetic quantum number
 Each orbital has number from –L to +L
 i.e. 3p4 , occupies 1st orbital, ml= -1
 ms-magnetic spin quantum number
 Value is ±1/2
 If electron points up, +1/2
 If electron points down, -1/2
 i.e. 3p4, points down, ms= -1/2
DeBroglie Wavelength
 Matter has a characteristic wavelength that depends
on its momentum, mv
λ=h/mv
 λ-wavelength(m)
 h-Planck’s constant-6.626x10-34 J-s
 m-mass of particle(kg) (e-=9.11x10-31kg)
 v-velocity (m/s)
Bohr’s Model
 A model of the hydrogen atom that explains its line
spectrum
 Light emitted when the electron drops from a higher
energy state to a lower energy state
 Light must be absorbed to excite the electron from a
lower energy state to a higher energy state
Electron Configuration
 Use the periodic table to write electron configurations
 Core electron configuration—use largest noble gas that
is smaller than atom/ion, then write additional electrons
 Remember that each orbital can hold 2 electrons each
 Place 1 electron in each orbital before putting a second
one
 Electron configurations are most stable when the
orbitals are full or half-full
Electron Configuration
 D Block (Transition metals)
 -5 orbitals @ 2 electrons each=10 electrons
 D block 1 behind s/p block
 F Block
 7 orbitals @2 electrons each=14 electrons
 F block 1 behind D block, 2 behind s/p block
Electron Configuration of Ions
 When determining configurations for cations, remove
electrons first from the orbital with the largest
quantum number n
 For example, Sn=[Ar] 4s23d104p2
Sn3+= =[Ar]4s13d10
Hybridization
 Mixing of s, p, and d orbitals to form hybrid orbitals
 A particular mode of hybridization corresponds with
each of the five common electron-domain geometries
 note: electron domain geometry is arrangement of
electron domains around a central atom. Each bond,
whether it is single, double, or triple, and each lone pair
is one electron domain.
Hybridization
 Linear-2 electron domains-sp hybridization
 Trigonal planar-3 electron domains- sp2
 Tetrahedral-4 domains, sp3
 Trigonal bipyramidal- 5 domains, sp3d
 Octahedral- 6 domains, sp3d2
Valence Bond Theory
 Bonds form when atomic orbitals overlap between two
atoms
 The greater the overlap between two orbitals, the stronger
the bond
 Sigma Bond

Covalent bonds formed from end to end overlap of s orbitals
 Pi Bond

Bond formed from the sideways overlap of p orbitals
Molecular Orbital Theory
 Electrons exist in allowed energy states called
molecular orbitals (MOs)
 Like an atomic orbital, an MO can hold two electrons
of opposite spin
 Occupation of bonding MOs favors bond formation
 Occupation of antibonding MOs (denoted with an *)
is unfavorable
Molecular Orbital Theory
 Bond Order
 Bond Order = ½(# of electrons bonding - # of electrons
anti-bonding)
 The principle of anti-bonding sets molecular orbital
theory apart from valence bond theory
Paramagnetism and Diamagnetism
 paramagnetism—an attraction of a molecule by a
magnetic field due to unpaired electrons
 diamagnetism—a weak repulsion from a magnetic
field by paired electrons
Question 1
What is the core electron configuration of
Pb3+?
Answer
1
14
10
[Xe]6s 4f 5d
Question 2
What type of hybridization does the central atom in the
following compounds assume?
a. NH3
b. SF6
c. ClF3
Answers
 a. Sp3
 b. Sp3d2
 c. Sp3d
Final Question!
 What score will you get on the AP Chem Exam after
seeing this presentation?
ANSWER
5