South Pasadena • Honors Chemistry
Name
9 • Atomic Structure
Period
STATION
1
Date
– WAVE CALCULATIONS
c =  ν
E=hν
c = 3.0 × 108 m/s
h = 6.626 × 10–34 J·s
The color orange (school colors) has a wavelength of 615 nm.
 Calculate the frequency of orange light.
c 3.0 × 108 m/s 1 nm
ν= =
× –9 = 4.9 × 1014 Hz
612 nm
10 m


1 nm = 109 m
Calculate the energy of a photon of orange light.
E = h· ν = (6.626 × 10–34 J·s)(4.9 × 1014 Hz) = 3.25 × 10–19 J

If this is a wave of ORANGE light, sketch what a wave of
RED light would look like.
Red has lower energy and longer wavelength.

The red light would have a [ higher | lower ] frequency, a
[ longer | shorter ] wavelength, and [ more | less ] energy
compared to orange light.
9  Atomic Structure
STATION
2
Circle the subshells that do NOT exist:
–
SHELLS, SUBSHELLS
4p
__5__ The number of orbitals in a 4d subshell.
__4__ The number of orbitals in the n = 2 shell.
1p
2f
5s
3d
7p
2d
& ORBITALS
3s
Students fill the outer electrons in oxygen in the
following way. Explain what is wrong, if
anything, with each.
__5__ The number of subshells in the n = 5 shell.
p electrons should be unpaired
__18_ The number of electrons in the n = 3 shell.
__7__ The number of orbitals in a 4f subshell.
s orbitals should be filled before p
__3__ The number of subshells in the n = 3 shell.
__6__ The number of electrons in a 6p subshell.
nothing is wrong
9  Atomic Structure
STATION 3


– SHELLS,
ORBITALS & SPECTRA
List the six subshells that make up the n = 6 shell and state how many electrons fit in each subshell:
subshell:
6s
6p
6d
6f
6g
6h
# of electrons
2
6
10
14
18
22
Using your periodic table only (not your notes or an orbital diagram) write the orbitals in order of lowest energy to
highest energy.
_1s_ < _2s_ < _2p_ < _3s_ < _3p_ < _4s_ < _3d_ < _4p_ < _5s_ < _4d_ < _5p_
Consider the emission spectrum of hydrogen:
V

BV
BG
Red
Draw an arrow showing how an electron must change to create the red line in the
spectrum above.
9  Atomic Structure
STATION
4
–
ORBITALS
&
ELECTRON CONFIG.
Write the long form electron configuration for the element selenium, Se (Z = 34).
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p4
_16_ The number of completely filled orbitals.
_2_ The number of half-filled orbitals.
Element
Sulfur, S (Z = 16)
Nickel, Ni (Z = 28)
Total Electrons
16
28
Valence Electrons
6
2
Long Form Electron
Configuration
1s2 2s2 2p6 3s2 3p4
1s2 2s2 2p6 3s2 3p6 4s2 3d8
Short Form Electron
Configuration
[Ne] 3s2 3p4
[Ar] 4s2 3d8
Ion Formed
S2–
Ni2+
Short-Form Electron
Configuration of Ion
[Ne] 3s2 3p6
[Ar] 3d8
9  Atomic Structure
STATION
5
–
USING THE ELECTRON CONFIGURATION
Consider the element vanadium, V (Z=23).
 Fill in the orbital energy diagram for the
electrons in vanadium.
4p
3d
4s
3s
3p

Write the short form electron configuration for V.
1s2 2s2 2p6 3s2 3p6 4s2 3d3

Write the short form electron configuration for V2+.
[Ar] 3d3

Write the equation for the 1st ionization energy of V:
V + IE1 → V+ + e–
2p
2s
The 1st ionization energy removes an electron from the _4s_ orbital.

1s
For vanadium, state:
 Number of half-filled orbitals: _3__
 Number of valence electrons: _2__
 Orbital of highest energy electron:_3d__
 Orbital of electron that is furthest from the
nucleus: _4s_
Write the equation for the 2nd ionization energy of V:
V+ + IE2 → V2+ + e–
The 2nd ionization energy removes an electron from the _4s_ orbital.

Write the equation for the 3rd ionization energy of V:
V2+ + IE3 → V3+ + e–
The 3rd ionization energy removes an electron from the _3d_ orbital.
9  Atomic Structure
STATION 6
Element
Total
e–
Valence
–
ELEMENTS & ELECTRON CONFIG.
e–
Long Form
Short Form
Cl
17
7
1s2 2s2 2p6 3s2 3p5
[Ne] 3s2 3p5
Sr
38
2
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2
[Kr] 5s2
Sb
51
5
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p3
[Kr] 5s2 4d10 5d3
B
5
3
1s2 2s2 2p1
[He] 2s2 2p1
Ge4+
32-4=28
0
1s2 2s2 2p6 3s2 3p6 3d10
[Ar] 3d10
For the examples of the 8 families of the representative elements, state the # of valence electrons and what ion they form.
Element:
Li
Be
B
C
N
O
F
Ne
# valence electrons:
ion formed:
1
2
3
4
5
6
7
8
Li+
Be2+
B3+
–
N3–
O2–
F–
–
9  Atomic Structure
STATION
7
–
Z AND
Z
e f f
Consider the element aluminum, Al. The nuclear charge (Z) for Al is _+13_. Al has _3_
valence electrons and _10_ core electrons. An outer electron of Al feels _+3_ (Zeff) because the
outer electron is attracted by _13_ protons and repelled by _10_ core electrons.
When one electron is removed, an outer electron of Al+ has a Zeff = _+3_.
When a second electron is removed, an outer electron of Al2+ has a Zeff = _+3_.
When a third electron is removed, an outer electron of Al3+ has a Zeff = _+11_.
Fill in the following chart:
Element:
Li
Be
B
C
N
O
F
Ne
Nuclear charge (Z):
+3
+4
+5
+6
+7
+8
+9
+10
# of core electrons:
2
2
2
2
2
2
2
2
+1
+2
+3
+4
+5
+6
+7
+8
Zeff:
Moving across a period (e.g. from Li to Ne), the atomic radius [ increases | decreases ] and the first ionization energy
[ increases | decreases ]. Use the Zeff above, explain briefly.
Going across a period, the Zeff increases as the number of protons increases but the shielding by the core
electrons remains constant, so outer electrons are more attracted and closer to the nucleus.
Moving down a family (e.g. from B to Al), the atomic radius [ increases | decreases ] and the first ionization energy
[ increases | decreases ]. Explain briefly.
Going down a family, the outer electron is in a higher electron shell, so it is further from and less attracted to
the nucleus, resulting in larger atomic radii and lower IE.
9  Atomic Structure
STATION
8
–
TRENDS IN SIZE
For each pair of elements, circle the element with the larger atomic radius. Explain briefly.
 Mg
Ca
Outer electrons of Ca are on a higher electron shell than those of Mg.

N
O
N has fewer protons (and lower Zeff) than O.

Sn
As
Outer electrons of Sn are on a higher electron shell than those of As.

O2–
Mg2+
Both have same # of electrons, but O2– has fewer protons than Mg2+.

K
K+

I
I–
Both have same # of protons, but K has more electrons  more e–-e– repulsions.
(Outer e– of K is in 4s and outer e– of K+ is in 3p, so outer e– of K is in higher shell)
Both have same # of protons, but I– has more electrons  more e–-e– repulsions.
Put these five elements in order from smallest atomic radius to largest atomic radius. F Br Ca K Cs
Smallest
F
Br
Ca
K
Cs
Largest
9  Atomic Structure
STATION
9
–
TRENDS IN IONIZATION
For each pair of elements, circle the element with the larger
ionization energy. Explain briefly.
 F
Cl
Outer e– of F is in a lower shell
than that of Cl.

Na
Be
Outer e of Be is in a lower shell
than that of Na.

Al
Si
Si has more protons than Al.
(or Si has greater Zeff than Al.)


C
Ca+
N
Ca2+
–
ENERGY
The Period 3 Elements are:
Na
Mg
Al
Si
P
S
Cl
Ar
Consider the successive ionization energy data for
unknown elements X and Y in Period 3 (in kJ/mol):
IE1
IE2
IE3
IE4
IE5
Element X: 736
1445
7730 10,600 13,600
Element Y: 787
1575
3220
4350 16,100
Which Period 3 element is X?
Mg
N has more protons than C.
(or N has greater Zeff than C.)
Which Period 3 element is Y?
Si
Outer e– of Ca2+ is in lower shell
than that of Ca+.
Which Period 3 element has the largest 3rd ionization
energy?
Mg