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AP Chemistry
Mr. Ferwerda, Tecumseh HS
06/26/17
Ch. 7 Outline : Atomic Structure and Periodicity
A. Atomic structure and periodicity(need good chart of e-m spectrum)
1 m=109 nm
or 10-9 m 1 nm
1 J = 1 kg x m2/s2
1. Electromagnetic radiation
a. a form of radiant energy
b. three properties of waves
- wavelength ( - Greek lambda) (shorter wavelength = higher energy)
- frequency (v - Greek nu) (Hz - 1/s or s-1)
- speed (c)- equal for all kinds of e-m radiation ( = 2.9979 x 108 m/s)
c. v = c (wavelength and frequency are inversely related)
_____ (T/F) The higher the frequency, the longer the wavelength.
_____ (T/F) Frequency and wavelength are inversely related to each other.
Sample problem : Calculate the wavelength of a 101.5 mHz frequency given off by a radio
station.
Solution :  = c/ v (101.5 mHz = 101.5 x 106 Hz or 101.5 x 106 s-1).
 = 2.9979 x 108 m/s / 101.5 x 106 s-1
= 2.954 m
Example: Calculate the wavelength of a 99.5 MHz frequency given off by a radio station.
2. Nature of matter
a. Planck - energy is quantized E = nhv (n = integer, h = Planck's constant 6.626 x 10-34 J x s)
- energy increases with increased frequency (decreased wavelength)
_____ (T/F) Planck determined that atoms can absorb or emit any amount of energy.
AP Chemistry
Mr. Ferwerda, Tecumseh HS
06/26/17
b. quantum - a small "packet" of energy - a discrete, yet varying amount (not all quanta are the
same)
c. Einstein - photons (packets of light energy) are quantized (Ephoton = hv = hc/) v = c/



d. Einstein : E = mc2
e. dual nature of light/matter
- diffraction (scattering of light from a regular array of lines or points)- property of waves
- photoelectric effect- evidence of particle nature
Sample problems :
1. Calculate the energy of a single photon and a mole of photons emitted from a vapor lamp giving
off light with a wavelength of 412.0 nm.
Solution : Ephoton = hv = hc/
= (6.626 x 10-34 J·s)(2.9979 x 108 m/s) / (412.0 x 10-9 m)
= 4.821 x 10-19 J
E mol = (4.821 x 10-19 J/e-) (6.022 x 1023 e-/mol) = 290.3 kJ
2. The first ionization energy of aluminum is 577 kJ/mol. Is light with a wavelength of 215 nm
capable of ionizing an atom of aluminum in the gas phase?
Solution : kJ/mol are given, first calculate J/atom : 577 kJ/ (6.022 x 1023) = 9.58 x 10-22 kJ/atom
= 9.58 x 10-19 J/atom
Second - calculate wavelength from energy needed and compare to given wavelength :
 = hc / E = (6.626 x 10-34 J·s)(2.9979 x 108 m/s) / 9.58 x 10-19 J/atom
= 2.07 x 10-7 m = 207 nm
Answer : No. The 207 nm wavelength is a higher energy wavelength than the 215 nm
wavelength.
Example: Calculate the energy of a single photon and of a mole of photons emitted from a mercury vapor lamp
with light of a wavelength of 404.7 nm.
AP Chemistry
Mr. Ferwerda, Tecumseh HS
06/26/17
Example 2 : The ionization energy of gold is 890.1 kJ/mol. Is light with a wavelength of 225 nm capable of
ionizing a gold atom in the gas phase ?
f. de Broglie's equation : = h/mv (where m = mass (in kg) and v = velocity in m/s)
Light has a dual nature, what about other forms of matter? (Yes)
e.g. What is the wavelength of an electron moving at 50.% of the speed of light?
3. The atomic spectrum of hydrogen
a. Contrast to a continuous spectrum - all wavelengths
b. emission spectrum of hydrogen -pattern of light emitted from an excited hydrogen atom) - a line
spectrum- indicates quantized nature of hydrogen (only certain wavelengths absorbed or
emitted)
How many electron transitions can take place for a hydrogen atom with an excited electron at n = 6?
AP Chemistry
Mr. Ferwerda, Tecumseh HS
06/26/17
4. The Bohr model - quantum model
a. energy levels available : E = -2.178 x 10-18 J(Z2/n2)
- Z = nuclear charge (number of protons)(=1 for hydrogen atom)
- n = integer- corresponds to energy levels
- neg. sign indicates negative energy (e- is bound to nucleus therefore has lower energy
than free e- where n = )
- J is joules
b. ground state lowest possible energy state
c. calculation of energy of quantization in hydrogen for change in energy : E = Ef - Ei (e.g.
change in E when electron moves from n = 6 to n = 5)
- positive sign indicates E (light) absorbed
- negative sign indicates E (light) emitted
Example 1. : Calculate the wavelength of light emitted from a hydrogen atom when an electron makes a
transition from n = 4 to n = 2.
Ephoton = hv = hc/ or = hc / | ∆E |
∆E = -2.178 x 10-18 J (1/nf2 - 1/ni2)
= -2.178 x 10-18 J (1/22 - 1/42)
= -2.178 x 10-18 J (1/4 - 1/16)
∆E = - 4.084 x 10-19 J
= hc / | ∆E | = (6.626 x 10-34 J x s)(2.998 x 108 m/s) / (4.084 x 10-19 J) = 4.864 x 10-7 m = 486.4 nm
Example 2 : Calculate the maximum wavelength (lowest energy) needed to remove an electron for the
energy state of hydrogen in which n = 3.
ni = 3, nf = ∞, (at n = ∞ , E = 0)
∆E = Ef - EI = 0 - (-2.178 x 10-18 J (1/ni2))
∆E = 0 - (-2.178 x 10-18 J (1/32))
= 2.420 x 10-19 J
= hc / | ∆E | = (6.626 x 10-34 J x s)(2.998 x 108 m/s) / (2.420 x 10-19 J) = 8.209 x 10-7 m = 820.9 nm
d. calculation of change in electron energies E= -2.178 x 10-18 J(1/nfinal2 - 1/ninitial2)
e. spectral transition diagram : ("spectral" because light is usually absorbed or emitted)
e. failure of Bohr model in other elements -the Bohr model did not hold true for polyelectronic
atoms
_____(T/F) The Bohr model worked for all atoms of all elements.
_______________ and __________________ said that electrons bound to nuclei acted as standing waves.
AP Chemistry
Mr. Ferwerda, Tecumseh HS
06/26/17
5. Quantum Mechanical Model of atom
a. Heisenberg, Schrodinger, de Broglie - developed wave , or quantum mechanics
- bound electron is a standing wave
- only certain circumferences will allow whole numbers of half waves
- Schrodinger's equation HΨ = EΨ
- Ψ- wave function, or "orbital" (function of coordinates x, y and z)( does not indicate
pathway of electron)
- H - operator - contains mathematical terms which produce total energy of the atom
- E - total energy of atom (kinetic and potential)
- a specific wave function is an orbital
An _____________________________ is a specific wave function.
The ___________________________________ uncertainty principle states that we can not know both the
position and the momentum of an electron.
b. Heisenberg uncertainty principle-cannot know both momentum and position of electron at a
given time
c. Physical meaning of wave function in light of uncertainty principle : probability of e- location
- probability distribution- represents square of wave function - produces an e- density map
- radial probability distribution - total probability within spherical shells
- orbital - no definite size-defined as area in which 90% of total e- probability
6. Quantum numbers-describe orbitals
a. Principal quantum number (n = 1, 2, 3, ...)
- indicates size and energy of orbital (higher n, higher energy-less bound, less neg. energy)
for n = 1,
n = 2,
n = 3,
n = 4,
AP Chemistry
Mr. Ferwerda, Tecumseh HS
06/26/17
b. Angular momentum number (azimuthal quantum number) (l)
- integral values from 0 to n-1
- indicates shape of orbital (l = 0 = s; l = 1 = p; l = 2 = d; l = 3 = f )
for l = 0, ml =
for l = 0, ml =
for l = 0, ml =
for l = 0, ml =
c. magnetic quantum number (ml)
- integral values from l to - l including 0
- indicates position of orbital in space relative to other orbitals
d. Spin quantum number (ms) - indicates rotation of electron
- -½ or -½
Table 7-1 : Quantum numbers for first four energy levels
Orbital
n
No. of orbitals
ml
l
(n = 1, 2,
designation
(= n2)
(ml = -l …0…+l)
(l = 0 to
3…)
n-1)
1s
2s
2p
3s
3p
3d
4s
4p
4d
4f
1
2
Total number of
electrons
(= 2n2)
0
0
1
2 (2)
0
0
1
2
1
-1, 0, +1
3*
6 (8)
3
0
0
1
2
1
-1, 0, +1
3
6 (18)
2
-2, -1, 0, +1, +2
5**
10
4
0
0
1
2
1
-1, 0, +1
3
6 (32)
2
-2, -1, 0, +1, +2
5
10
3
-3, -2, -1, 0, +1, +2, +3
7
14
* The 3 p orbitals are x, y, and z (e.g. 2px, 2py, 2pz)
** The 5 d orbitals are xz, yz, xy, x2-y2, and the z2 (e.g. 3dxz, 3dyz, 3dxy, 3dx2-y2, and the 3dz2)
AP Chemistry
Mr. Ferwerda, Tecumseh HS
06/26/17
Sample problems :
1. Which of the following sets are incorrect?
a. n = 2, l = 2, ml = -1
b. n = 3, l = 2, ml = -3
c. n = 4, l = 2, ml = -1
2. How many electrons can have the quantum numbers n = 5, ml = +1?
3. How many orbitals can have the designation n = 3?
4. How many electrons can have the designation n = 3?
7. Orbital shapes and energies
a. nodal surface (nodes)- regions of zero probability of finding an eb. hydrogen atom - orbitals in energy levels (same n value) are degenerate - have the same energy
8. Electron Spin and the Pauli Principle
a. Pauli exclusion principle - in a given atom, no two electrons can have the same set of four
quantum numbers (i.e. -only two electrons per orbital)
9. Polyelectronic atoms
a. three forces affecting distribution of electrons
- kinetic energy of electrons (outward)
- potential energy of attraction of nucleus (inward)
- potential energy of repulsion of electrons
b. electron correlation problem-since paths of electrons cannot be known exactly, repulsions
cannot be calculated exactly- must make approximations
c. shielding - inner electrons shield outer electrons from nuclear charge (held less tightly)
AP Chemistry
Mr. Ferwerda, Tecumseh HS
06/26/17
d. orbitals in polyelectronic atoms are not degenerate (s < p < d < f)
e. penetration effect -ex. In general, 2p electrons are closer to the nucleus and therefore have a
lower energy than 2s, however, 2s electrons spend a small, but significant amount of time
closer to the nucleus and are said to "penetrate" and thus are considered to have a lower energy
state and are filled first
d. Electron configuration problems
Rules governing electron configurations :
- Aufbau rule
- Hund's rule
- Pauli exclusion principle
AP Chemistry
Mr. Ferwerda, Tecumseh HS
06/26/17
Write the orbital electron configuration of calcium, and then give its summary electron configuration.
Summary :
What is the summary electron configuration of yttrium? (at. no. 39)?
Sample problems :
1. What is the electron configuration of neon?
2. What is the electron configuration of chromium ?
3. What is the identity of the element with the electron configuration 1s22s22p63s13p1 in an excited state ?
4. Write the electron configuration for the following orbital notation for an atom with 8 protons, identify the
species (atom, ion etc) and write its formula .
5. How many unpaired electrons are found in nitrogen in its ground state?
AP Chemistry
Mr. Ferwerda, Tecumseh HS
06/26/17
6. Give a set of values for the four quantum numbers of the valence electrons of phosphorus in its ground state.
Electrons
n
l
Write the summary electron configuration for the element zinc.
Electron configuration of tin (at. no. 50)?
Electron configuration of osmium (at. no. 76)?
Write the summary electron configuration for chromium :
_____ (T/F) Phosphorus is paramagnetic.
_____ (T/F) Phosphorus can form the compound PF5.
ml
ms
AP Chemistry
Mr. Ferwerda, Tecumseh HS
06/26/17
The following shows an element in its ground state. Is this a metal or nonmetal?
Characteristics of this element?
have luster or nonlustrous?
conductor or insulator?
relatively high or low melting point?
electropositive or electronegative?
state?
10. History of periodic table
a. Mendeleev given most of the credit because of his emphasis on using it as a tool to predict yet
unknown elements.
11. Aufbau ("building up") principle and the periodic table
a. Aufbau principle - as protons are added to successive atoms of elements, so are electrons
b. Hund's rule - the lowest energy configuration for an atom is the one having the most unpaired
electrons allowed by the Pauli principle in degenerate orbitals
c. Valence electrons - electrons in the outermost principal quantum energy level of an atom
d. core electrons - "inner" electrons
e. configurations of representative elements vs. transition metals and inner transition metals
(lanthanides and actinides)
f. exceptions to expected configurations - chromium [Ar] 4s13d5 and copper [Ar] 4s13d10
g. arrangement of periodic table- representative elements, transition metals, lanthanides, actinides,
s block, p block, d block and f block
h. IUPAC periodic table- numbers Groups 1-18 instead of 1A-8A and "B" designation for
transition metals
i. determination of electron configurations
12. Periodic trends in atomic properties
a. major factors affecting trends
- nuclear charge - increases across a period causing a greater attraction for electrons
- shielding - additional energy levels down a group causes a decrease in the attraction of the
nucleus for the outermost electrons
- electron configuration - most important - groups have similar electron configurations giving
them similar chemical and physical properties
AP Chemistry
Mr. Ferwerda, Tecumseh HS
06/26/17
a. ionization energy- the energy needed to remove an electron from a gaseous atom (X(g) -->X+(g)
+ e-) in its ground state
- first ionization energy (I1)- energy needed to remove the first electron
- second ionization energy (I2) - energy needed to remove second electron
- note pattern in aluminum [Ne]3s23p1
I1- removes p1electron = 580 kJ/mol
I2 removes one 3s electron = 1815 kJ/mol - higher due to positive charge on Al+ ion
I3 removes remaining 3s electron = 2740 kJ/mol (Al2+ )
I4 removes core 2p electron = 11,600kJ/mol - core electrons come from stable noble gas
configuration
- periodic trend is for ionization energy to increase across a period - increased nuclear charge
with no increase in shielding
-exceptions-due to electron repulsion i.e. nitrogen and oxygen
N:
O:
- easier to remove outer 2p
electron from oxygen than nitrogen because of doubly occupied 2porbital
-group trend is to decrease down a group - increased shielding, outer electrons are farther from
nucleus
b. electron affinity - the energy change associated with the addition of an electron to a gaseous
atom : X(g) + e- ---> X (g)
- sign of energy change- negative if exothermic, positive if endothermic
- periodic trend - negative electron affinities increase across a period (more energy released)
- exceptions : carbon has a negative electron affinity (becomes more stable) and nitrogen has a
positive electron affinity (becomes unstable- higher energy)
C:
N:
- an electron added to carbon goes
into an empty orbital whereas an electron added to nitrogen goes into an occupied
orbital resulting in electron repulsion
- Group trend - electron affinities become more positive (less negative) because less energy is
released due to shielding and increased distance from nucleus (less attraction for e-)
Note relationship between ionization energy ( energy change for X(g) -->X+(g) + e-) and electron
affinity ( energy change for X(g) + e- ---> X (g)). Since the changes are opposite their energies are
opposite.
Sample problem : What is the electron affinity for the ionization of the Al2+ ion (The ionization energy
for Al+(g) + e-  Al2+(g) = 1850 kJ/mol)?
Answer : -1850 kJ/mol : Electron affinity is the energy change associated with the addition of an
electron to a gaseous atom making the reaction : Al2+(g) + e-  Al+(g). The electron affinity is
therefore opposite (EA = -IE) and equals -1850 kJ/mol.
c. Atomic radii - defined as half the distance between the nuclei of diatomic molecules of the
same element (e.g. O2)
- periodic trend - atomic radii decrease across a period - due to increased nuclear charge
- Group trend - atomic radii increase down a group - large n value, increased shielding
d. ion size
- metallic ions lose electrons and become smaller due to loss of volume occupied by lost
electrons and by the increased ratio of positive to negative charge
- nonmetallic ions gain electrons and increase in size for reasons opposite of metallic ions
AP Chemistry
Mr. Ferwerda, Tecumseh HS
06/26/17
- trend for metallic ions is to decrease in size (consider 3rd period metallic ions Na+, Mg2+,
Al3+)
- trend for nonmetallic ions is to also decrease in size (consider the 3rd period nonmetallic ions
P3-, S2-, Cl-)
13. Information contained in the Periodic Table
a. number of valence electrons - main determiner of chemical properties - know well
b. electron configuration - know representative elements and Cr and Cu
c. know : metals, nonmetals, alkali metals, alkaline earth elements, halogens, noble gases,
transition metals, inner transition metals ( lanthanides and actinides) and metalloids
(semimetals)
d. characteristics of metals and nonmetals
- metals - low ionization energies, low (less negative) electron affinities, tend to form cations
- nonmetals - high ionization energies, higher (more negative) electron affinities, tend to form
anions
14. The Alkali metals - Group 1A H, Li, Na, K, Rb, Cs, and Fr
a. hydrogen is a nonmetal mainly because of its small size - holds electron tightly
b. density increases down group - typical of all groups - mass increases at a faster rate than size
c. melting points and boiling points generally decrease
d. tend to lose outer e- making them reducing agents - lower ionization energy makes them
stronger reducing agents thus order of reducing strength is Cs > Rb > K > Na > Li for solid
metals only
- in aqueous solution the reducing strength is Li > K > Na Why change in position for lithium
in aqueous solution?- Lithium's small size gives it a high charge density which makes it more
able to bond to water (hydration energy - energy released when a substance bonds to water in
solution) and form the Li+ ion (and thus lose its valence electron) more readily than K or Na.
-If lithium is a stronger reducing agent and bonds to water more strongly than K or Na, then
why do experiments show that K and Na react more violently with water? - Na and K have
lower melting points causing them to react with water in the liquid state giving more surface
area resulting in a faster reaction. Main point of this - thermodynamics (energy change) tells
us if a reaction will be spontaneous or not, kinetics (study of reaction mechanisms/rates) tells
us how fast a reaction will occur.