The brilliant red colors seen in fireworks are due to the
emission of light with wavelengths around 650 nm
when strontium salts such as Sr(NO3)2 and SrCO3 are
heated. Calculate the frequency of red light of
wavelength 6.50 × 102 nm.
λ = 650 nm = 6.50 x 10-7 m
ν = c = 2.9979 x 108 m/s = 4.61 x 1014 Hz
λ
6.50 x 10-7 m
The blue color in fireworks is often achieved by heating copper(I)
chloride (CuCl) to about 1200°C. Then the compound emits blue
light having a wavelength of 450 nm. What is the increment of
energy (the quantum) that is emitted at 4.50 × 102 nm by CuCl?
λ = 450 nm = 4.50 x 10-7 m
Ephoton = hc = (6.626 x 10-34 Js)(2.9979 x 108 m/s)
λ
4.50 x 10-7 m
= 4.417 x 10-19 J
Calculate the energy required to excite the hydrogen electron
from level n = 1 to level n = 2. Also calculate the wavelength of
light that must be absorbed by a hydrogen atom in its ground
state to reach this excited state.
E = -2.178 x 10-18 J(Z2/n2)
for H, Z = 1 so E = -2.178 x 10-18 J(1/n2)
From n = 1 to n = 2
1
1
-18
ΔE = -2.178 x 10 J( 2 - 2)
= -2.178 x
λ=
𝑛2
1
-18
10 J( 2
2
-
𝑛1
1
)
2
1
= 1.634 x 10-18 J
hc
= (6.626 x 10-34 Js)(2.9979 x 108 m/s)
Ephoton
1.634 x 10-18 J
= 1.217 x 10-7 m
Calculate the energy required to remove the electron from a
hydrogen atom in its ground state.
E = -2.178 x 10-18 J(Z2/n2)
for H, Z = 1 so E = -2.178 x 10-18 J(1/n2)
From n = 1 to n = ∞ since the electron is being removed
ΔE = -2.178 x
= -2.178 x
10-18
10-18
1
J( 2
𝑛2
1
J( 2 ∞
= 2.178 x 10-18 J
-
1
)
2
1
1
)
𝑛12
= -2.178 x 10-18 J(0 - 1)
EXERCISE 5 ELECTRON SUBSHELLS
 For principal quantum level n = 5, determine the
number of allowed subshells (different values of ℓ),
and give the designation of each.
n = 5 so l = 0, 1, 2, 3, 4 or s, p, d, f, and then g
l = 0  5s
l = 1  5p
l = 2  5d
l = 3  5f
l = 4  5g (theoretical, not actually observed)
EXERCISE 6
Give the full electron configuration and orbital
diagram for each of the following elements.
Sulfur
Iron
1s22s22p63s23p4
Lithium
1s22s1
Magnesium
1s22s22p63s2
1s22s22p63s23p64s23d6
Aluminum
1s22s22p63s23p1
Silicon
1s22s22p63s23p2
EXERCISE 6 CONTINUED
Orbital Diagram—Sulfur
1s
2s
2p
3s
3p
EXERCISE 6 CONTINUED
Orbital Diagram—Lithium
1s
2s
EXERCISE 6 CONTINUED
Orbital Diagram—Magnesium
1s
2s
2p
3s
EXERCISE 6 CONTINUED
Orbital Diagram—Iron
1s
2s
4s
3s
2p
3d
3p
EXERCISE 6 CONTINUED
Orbital Diagram—Aluminum
1s
2s
2p
3s
3p
EXERCISE 6 CONTINUED
Orbital Diagram—Silicon
1s
2s
2p
3s
3p
EXERCISE 6 CONTINUED—OTHER SIDE OF
PAGE
Write the noble gas configuration for the
following ions.
S 2- A sulfide ion has two more electrons than a sulfur atom.
S2-  [Ne]3s23p6
isoelectronic with argon, Cl-, P3-, K+, Ca2+
 Al 3+
 Mg 2+
[He]2s22p6
isoelectronic with neon, Mg2+, Na+, F-, O2-, N3-
[He]2s22p6
isoelectronic with neon, Al3+, Na+, F-, O2-, N3-
EXERCISE 6 CONTINUED—OTHER SIDE OF
PAGE
Write the noble gas configuration for the
following ions.
 Fe 3+
[Ar]3d5
Electrons are lost from 4s first
 Br -
[Ar]4s24p6
Isoelectronic with Kr, Se2-, As3-, Rb+, Sr2+
EXERCISE 7
Determine the number of unpaired
electrons in each element from
Exercise 6.
EXERCISE 7
Orbital Diagram—Sulfur
1s
2s
2p
3s
3p
EXERCISE 7
Orbital Diagram—Lithium
1s
2s
EXERCISE 7
Orbital Diagram—Magnesium
1s
2s
2p
3s
No unpaired
electrons
EXERCISE 7
Orbital Diagram—Iron
1s
2s
4s
3s
2p
3d
3p
EXERCISE 7
Orbital Diagram—Aluminum
1s
2s
2p
3s
3p
EXERCISE 7
Orbital Diagram—Silicon
1s
2s
2p
3s
3p
EXERCISE 8 TRENDS IN IONIZATION
ENERGIES
The first ionization energy for phosphorus is 1060 kJ/ mol, and
the first ionization energy for sulfur is 1005 kJ/mol. Why ?
P
S
First IE = removing “last” electron
For phosphorous, this means removing the electron that makes the 3p
subshell half-full. A filled OR half-filled subshell is stable; the atom does not
“want” to lose this stability.
For sulfur, removing the last electron results in a half-filled 3p subshell.
Since stability is gained, the first IE is lower for sulfur than for phosphorous.
EXERCISE 9 IONIZATION ENERGIES
Consider atoms with the following electron
configurations:
a. 1s 2 2s 2 2p 6 Neon Largest first IE (noble gas, stable)
b. 1s 2 2s 2 2p 6 3s 1 Sodium
c. 1s 2 2s 2 2p 6 3s 2 Magnesium Smallest second IE because it
gains a noble gas configuration
when it loses 2 electrons
Identify each atom. Which atom has the largest
first ionization energy, and which one has the
smallest second ionization energy? Explain your
choices.
EXERCISE 10 TRENDS IN RADII
Predict the trend in radius for the following
ions: Be 2+ , Mg 2+ , Ca 2+ , and Sr 2+ . Explain the
trend as well.
Be2+ < Mg2+ < Ca2+ < Sr2+
Beryllium has the smallest number of principal
energy levels, so it is the smallest. As the number of
principal energy levels (n) increases, the size of the
atom increases.