CHE141
Chapter 6
Chapter 6
Electronic Structure of Atoms
1. Electromagnetic radiation travels through vacuum at a speed of _________m/s.
(a). 6.626 x 1026
(b). 4186
(c). 3.00 x 108
(d). It depends on wavelength
Explanation: The speed of light (electromagnetic radiation) through vacuum has a
constant value of 3.00 x 108 m/s and is independent of the wavelength in vacuum.
2. Ham radio operators often broadcast on the 6-meter band. The frequency of this
electromagnetic radiation is __________MHz.
(a). 500
(b). 200
(c). 50
(d). 20
Explanation: The frequency and wavelength of electromagnetic radiation are related
by the equation c = λν, where c is the speed of light (=3.00 x 108 m/s ), λ is the
wavelength in m and ν is the frequency is s-1 or Hz. The frequency can be calculated
by rearranging the above formula to get
c 3.00 ! 10 8 m s "1
í = =
= 50 MHz
ë
6m
3. What is the frequency (s-1) of electromagnetic radiation that has a wavelength of 0.53
m __________?
(a). 5.7 x 108
(b). 1.8 x 10-9
(c). 1.6 x 108
(d). 1.3 x 10-33
Explanation: The frequency and wavelength of electromagnetic radiation are related
by the equation c = λν, where c is the speed of light (=3.00 x 108 m/s ), λ is the
wavelength in m and ν is the frequency is s-1 or Hz. The frequency can be calculated
by rearranging the above formula to get
c 3.00 " 10 8 m s !1
í = =
= 5.7 " 10 8 s !1
ë
0.53 m
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CHE141
Chapter 6
4. The energy of a photon of light is __________proportional to its frequency and
__________proportional to its wavelength.
(a). directly, directly
(b). inversely, inversely
(c). inversely, directly
(d).directly, inversely
Explanation: The energy is always directly proportional to the frequency while the
frequency and wavelength of electromagnetic radiation are always inversely related to
each other.
5. Of the following, __________ radiation has the shortest wavelength.
(a). X-ray
(b). radio
(c). microwave
(d). ultraviolet
Explanation: The wavelength and frequency are inversely related to each other, thus
the radiation with the highest frequency (refer to figure 6.4 on page 219 from our
text) will have the shortest wavelength.
6. What is the frequency of light (s-1) that has a wavelength of 1.23 x 10-6 cm
__________?
(a). 3.00 x 108
(b).2.44 x 1016
(c). 4.10 x 10-17
(d). 9.62 x 1012
Explanation: The frequency and wavelength of electromagnetic radiation are related
by the equation c = λν, where c is the speed of light (=3.00 x 108 m/s ), λ is the
wavelength in m and ν is the frequency is s-1 or Hz. Since the wavelength is in cm it
must be converted to meters before rearranging the above formula as follows:
1m
c 3.00 " 10 8 m s !1
1.23 " 10 -6 cm "
= 1.23 " 10 -8 m, í = =
= 2.44 " 1016 s !1
100 cm
ë
1.23 " 10 -8 m
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CHE141
Chapter 6
7. What is the frequency of light (s-1) that has a wavelength of 3.12 x 10-3 cm
__________?
(a). 3.69
(b). 2.44 x 1016
(c). 9.62 x 1012
(d). 4.10 x 10-17
Explanation: The frequency and wavelength of electromagnetic radiation are related
by the equation c = λν, where c is the speed of light (=3.00 x 108 m/s ), λ is the
wavelength in m and ν is the frequency is s-1 or Hz. Since the wavelength is in cm it
must be converted to meters before rearranging the above formula as follows:
1m
c 3.00 " 10 8 m s !1
-3
-5
3.12 " 10 cm "
= 3.12 " 10 m, í = =
= 9.62 " 1012 s !1
-5
100 cm
ë
3.12 " 10 m
8. The wavelength of a photon of energy 5.25 x 10-19 J is __________m.
(a). 2.64 x 106
(b).3.79 x 10-7
(c). 2.38 x 1023
(d). 4.21 x 10-24
Explanation: The wavelength and energy are related by the formula E = hc/λ, where
h is Planck’s constant (= 6.626 x 10-34 Js), c is the speed of light (3.00 x 108 m/s) and
λ is the wavelength in meters. The wavelength can then be calculated by rearranging
the above formula as follows:
hc
hc (6.626 " 10 !34 Js) " (3.00 " 10 8 m/s)
E=
or ë =
=
= 3.79 " 10 !7 m
!19
ë
E
5.25 " 10 J
9. The energy of a photon that has a wavelength of 9.0 m is __________J.
(a). 2.2 x 10-26
(b). 4.5 x 1025
(c). 6.0 x 10-23
(d). 4.5 x 10-25
Explanation: The energy in Joules of any electromagnetic radiation whose
wavelength is known can be calculated by the formula
hc (6.626 " 10 !34 Js) " (3.00 " 10 8 m/s)
E=
=
= 2.2 " 10 ! 26 J
ë
9.0 m
When solving any of these problems pay attention to the units and ensure that they
cancel. Some of these will be compound units (like m/s) that may appear
problematic.
Copyright © 2006 Dr. Harshavardhan D. Bapat
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CHE141
Chapter 6
10. The energy of a photon that has a wavelength of 13.2 nm is __________J.
(a). 9.55 x 10-25
(b).1.51 x 10-17
(c). 1.99 x 10-25
(d). 4.42 x 10-23
Explanation: The wavelength is given in nm here which needs to be converted to
meters and then the energy can be calculated as follows:
1m
(6.626 " 10 !34 Js) " (3.00 " 10 8 m/s)
13.2nm " 9
= 1.32 " 10 !8 m, E =
= 1.51" 10 !17 J
!8
10 nm
1.32 " 10 m
11. What is the frequency (s-1) of a photon of energy 4.38 x 10-18J?
(a). 438
(b). 1.45 x 10-16
(c). 6.61 x 1015
(d). 2.30 x 107
Explanation: The frequency ν of this photon can be calculated by rearranging the
equation E = h ν where E is the energy, h = Planck’s constant and ν = frequency in
s-1.
E
4.38 " 10 !18 J
E = hí or í = =
= 6.61 " 1015 s !1
h 6.626 " 10 !34 Js
12. What is the wavelength (angstroms) of a photon of energy of 4.38 x 10-18J
__________?
(a). 438
(b). 4.54x 10-8
(c). 454
(d). 1.45 x 108
Explanation: The wavelength calculated here will be in meters which can be easily
converted to angstroms as follows:
hc
hc (6.626 ! 10"34 Js) ! (3.00 ! 108 m/s)
or ë =
=
= 4.538 ! 10"8 m
"18
ë
E
4.38 ! 10 J
1 angstrom
4.538 ! 10"8 m !
= 453.8 = 454 angstrom
10"8 m
E=
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CHE141
Chapter 6
13. A mole of red photons of wavelength 755 nm has __________kJ of energy.
(a). 2.63 x 10-19
(b). 6.022 x 1023
(c). 6.05 x 10-3
(d).158.6
Explanation: The energy of red photons of a wavelength of 755 nm can be easily
calculated using the formula E = hc/λ. This will be the energy of only ONE photon
and will have to be multiplied by Avogadro’s number to convert it to the energy of
one mole of red photons. The wavelength needs to be converted to meters before
performing this calculation:
(6.626 ! 10"34 Js) ! (3.00 ! 108 m/s)
J
photons
E=
= 2.632 ! 10"19
! 6.022 ! 1023
=
1m
photon
mole
755.0 nm ! 9
10 nm
1 kJ
158550.08 J !
= 158.6 kJ
1000 J
14. Of the following, __________ radiation has the longest wavelength and __________
radiation has the greatest energy.
Gamma, ultraviolet or visible
(a). ultraviolet, gamma
(b). visible, ultraviolet
(c). gamma, gamma
(d).visible, gamma
Explanation: From figure 6.4 in our text you can see that the visible radiation will
have the longest wavelength while the gamma will have the highest frequency. Since
energy is directly proportional to frequency, gamma will be the most energetic.
15. What color of visible light has the longest wavelength__________?
(a). blue
(b). violet
(c). red
(d). yellow
16. What color of visible light has the highest energy?
(a). violet
(b). yellow
(c). red
(d). green
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CHE141
Chapter 6
Explanation: From figure 6.4 in our text you can see that violet colored light would
have the highest frequency in the visible region. Since energy is directly proportional
to the frequency, violet light will have the highest energy.
17. An electron is a Bohr hydrogen atom has energy of -1.362 x 10-19J. The value of n
for this electron is __________.
(a). 1
(b). 2
(c). 3
(d).4
Explanation: The energy of an electron in a particular energy state in the hydrogen
atom can be calculated by using the formula E = (-2.18 x 10-18 J)/n2, where n is the
principal quantum number for the energy state. The value of n can be found by
rearranging the above formula as follows:
n=
" 2.18 ! 10"18 J
=4
" 1.362 ! 10"19 J
18. A spectrum containing only specific wavelengths is called a __________ spectrum.
(a). continuous
(b).line
(c). visible
(d). Bohr
19. The n = 2 to n = 6 transition in the Bohr hydrogen atom corresponds to the
_________ of a photon with a wavelength of __________ nm.
(a). emission, 411
(b).absorption, 411
(c). absorption, 657
(d). emission, 389
Explanation: There are 2 parts to this question. Since the electron is moving from a
smaller value of n (ni) to a larger value of n (nf), it must be absorbing energy. The
wavelength responsible for this transition can be calculated by using the formula:
' 1
1
1 $
1$
'1
'1 1 $
= R H %% 2 ( 2 "" = 1.097 ! 107 m (1 % 2 ( 2 " = 1.097 ! 107 m (1 % ( "
ë
6 #
&2
& 4 36 #
& ni nf #
1
1
= 1.097 ! 107 m (1 (0.222) = 2435340m (1 and ë =
= 4.11 ! 10( 7 m
(1
ë
2435340 m
= 411 nm
Copyright © 2006 Dr. Harshavardhan D. Bapat
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CHE141
Chapter 6
20. The n =5 to n = 3 transition in the Bohr hydrogen atom corresponds to the
__________ of a photon with a wavelength of __________ nm.
(a). absorption, 657
(b). absorption, 1280
(c). emission, 657
(d).emission, 1282
Explanation: There are 2 parts to this question. Since the electron is moving from a
larger value of n (ni) to a smaller value of n (nf), it must be emitting energy. The
wavelength responsible for this emission can be calculated by using the formula:
( 1
1
1 %
(1 1%
= R H && 2 ! 2 ## = 1.097 " 107 m !1 & 2 ! 2 # = 1.097 " 107 m !1 (!0.0711)
ë
'5 3 $
' ni nf $
1
= !780088.88 m, ignore the negative sign here and then ë = 1282 nm
ë
21. The deBroglie wavelength of a particle is given by __________.
(a). h + mv
(b). hmv
(c). h/mv
(d). mv/c
22. What is the de Broglie wavelength (m) of a 2.0 kg object moving at a speed of 50 m/s
__________?
(a). 6.6 x 10-36
(b). 1.5 x 1035
(c). 5.3 x 10-33
(d). 2.6 x 10-35
Explanation: The de Broglie wavelength of an object can be calculated by
remembering that 1 J = 1 kg•m2•s-2, which changes h = 6.626 x 10-34 kg •m2•s-1. Now
using the formula that relates the mass and velocity of an object to its wavelength, the
wavelength can be calculated as follows:
h
6.626 " 10!34 kg # m 2s !1
ë=
=
= 6.626 " 10! 36 m
mv
2.0 kg " 50 m/s
23. At what speed (m/s) must a 3.0 mg object be moving in order to have a de Broglie
wavelength of 5.4 x 10-29 m?
(a). 1.6 x 10-28
(b). 3.9 x 10-4
(c). 2.0 x 1012
(d).4.1
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CHE141
Chapter 6
Explanation: The speed of this object can be calculated by remembering that 1 J = 1
kg•m2•s-2, which changes h = 6.626 x 10-34 kg •m2•s-1. Another change that must be
made in this question is converting the mass from mg to kg.
1 kg
3.0 mg !
= 3 ! 10" 6 kg
6
1 ! 10 mg
v=
6.626 ! 10" 34 kg m 2 s "1
= 4.1 m/s
(3 ! 10" 6 kg) ! (5.4 ! 10" 29 m/s)
24. The de Broglie wavelength of an electron is 8.7 x 10-11 m. The mass of an electron is
9.1 x 10-31 kg. The velocity of this electron is __________ m/s.
(a). 8.4 x 103
(b). 1.2 x 10-7
(c). 6.9 x 10-5
(d).8.4 x 106
Explanation: The speed of this electron can be calculated by remembering that 1 J =
1 kg•m2•s-2, which changes h = 6.626 x 10-34 kg •m2•s-1.
6.626 ! 10"34 kgm 2s "1
v=
= 8.4 ! 106 m/s
(9.1 ! 10" 31 kg) ! (8.7 ! 10"11 m)
25. The __________ quantum number defines the shape of an orbital.
(a). spin
(b). magnetic
(c). principal
(d).azimuthal
26. There are _________ orbitals in the third shell
(a). 25
(b). 4
(c). 9
(d). 16
Explanation: The number of orbitals in a shell is easily calculated by the formula #
of orbitals = n2 where n = principal quantum number, which is 3 in this case.
27. The azimuthal quantum number is 3 in __________ orbitals.
(a). s
(b). p
(c). d
(d). f
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CHE141
Chapter 6
28. The n = 1 shell contains __________p orbitals. All the other shells contain
__________p orbitals.
(a). 3, 6
(b).0, 3
(c). 6, 2
(d). 3, 3
Explanation: If n = 1, then the only possible value of l is 0 which means that n = 1
can contain only s orbitals. When n > 1, the value of l = 1 is possible making the
existence of 3 p orbitals possible.
29. The principal quantum number of the first d subshell is __________.
(a). 1
(b). 2
(c). 3
(d). 4
Explanation: The value of l for a d subshell is 2 which means that the first shell in
which one would come across this value will be n = 3.
30. The total number of orbitals in a shell is given by __________.
(a). L2
(b).n2
(c). 2n
(d). 2n + l
31. Each p-subshell can accommodate a maximum of __________electrons.
(a). 6
(b). 2
(c). 10
(d). 3
Explanation: There are 3 different p orbitals: px, py and pz. Each of these can contain
2 electrons leading to the maximum number of electrons as 6.
32. How many quantum numbers are necessary to designate a particular electron in an
atom __________?
(a). 3
(b).4
(c). 2
(d). 1
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CHE141
Chapter 6
33. The 3p subshell in the ground state of atomic xenon contains __________ electrons.
(a). 2
(b).6
(c). 36
(d). 10
Explanation: Since Xe is a noble gas, its subshells will be completely filled
regardless of their principal quantum number. Thus the 3p subshell will contain 6
electrons.
34. [Ar]4s23d104p3 is the electron configuration of a(n) __________ atom.
(a). As
(b). V
(c). P
(d). Sb
Explanation: The easiest way to answer this question is to count the total number of
electrons and find which element that number corresponds to. The total number of
electrons is = 18 (for the Ar) + 2 +10 + 3 = 33 which corresponds to As.
35. There are __________ unpaired electrons in a ground state phosphorus atom.
(a). 0
(b). 1
(c). 2
(d).3
Explanation: You will need the electronic configuration of the phosphorus atom to
answer this question which is 1s22s22p63s23p3. According to Hund’s rule the 3
electrons in the 3p subshell will have the same spin and be unpaired.
36. In a ground-state manganese atom, the __________ subshell is partially filled.
(a). 3s
(b). 4s
(c). 4p
(d).3d
Explanation: The electronic configuration of manganese will help you answer this
question which is [Ar]4s23d5, this is an example of an anomalous electron
configuration where the 3d subshell is half-filled. The 3d subshell can accommodate
a total of 10 electrons but here has only 5.
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CHE141
Chapter 6
37. The principal quantum number for the outermost electrons in a Br atom in the ground
state is __________.
(a). 2
(b). 3
(c). 4
(d). 5
Explanation: The electronic configuration of bromine is [Ar]3d104s24p5 shows that
the outermost electrons are in the s and p orbitals in the 4th energy level making the
principal quantum number = 4.
38. The azimuthal quantum number for the outermost electrons in a nitrogen atom in the
ground state is __________.
(a). 0
(b).1
(c). 2
(d). 3
Explanation: The electronic configuration of nitrogen is [He]2s22p3, showing that
the outermost electrons are in the 2p orbital. The azimuthal quantum number (l) for
any p orbital is 1.
39. All of the __________ have a valence shell electron configuration ns1.
(a). noble gases
(b). halogens
(c). chalcogens
(d).alkali metals
40. The elements in the __________ period of the periodic table have a core-electron
configuration that is the same as the electron configuration of neon.
(a). sixth
(b). second
(c). third
(d). fifth
Explanation: The core electrons are electrons that are not in the valence shells. The
electron configuration of elements in the 3rd period can be written with neon’s
configuration as their core electrons.
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CHE141
Chapter 6
41. Elements in group __________ have a np6 electron configuration in the outer shell.
(a). 4A
(b). 6A
(c). 7A
(d).8A
42. Which group in the periodic table contains elements with the valence electron
configuration of ns2np1 __________?
(a). 1A
(b). 2A
(c). 3A
(d). 4A
Explanation: The electron configurations of elements in group 3A are written with
the valence electron configuration of ns2np1.
43. Which one of the following is correct?
(a). ν + λ = c
(b). ν/λ = c
(c). λ = cν
(d).νλ = c
44. In the Bohr model of the atom, __________.
(a). electrons travel in circular paths called orbitals
(b). electrons can have any energy
(c). electron energies are quantized
(d). electron paths are controlled by probability
45. According to the Heisenberg Uncertainty Principle, it is impossible to know precisely
both the position and the __________ of an electron.
(a). mass
(b). color
(c). momentum
(d). speed
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CHE141
Chapter 6
46. All of the orbitals in a given electron shell have the same value of the __________
quantum number.
(a). principal
(b). azimuthal
(c). magnetic
(d). spin
47. Which one of the following is not a valid value for the magnetic quantum number of
an electron in a 5d subshell?
(a). 2
(b).3
(c). 0
(d). 1
Explanation: For an electron in the 5d subshell the value of l = 2 and the magnetic
quantum number ml can have values from –l,…0,…+l, meaning ml could not have a
value = 3.
48. Which of the subshells below do not exist due to the constraints upon the azimuthal
quantum number?
(a). 2s
(b).2d
(c). 2p
(d). none of the above
Explanation: The values of the azimuthal quantum number “l” are decided by the
values of the principal quantum number “n”. The values of l will only be from 0…n
– 1. Thus for n = 2, only the values of 0 and 1 will be possible for l, which means
that only the 2s and 2p orbitals will be possible.
49. An electron cannot have the quantum numbers n = _________, l = __________, ml =
__________.
(a). 2, 0, 0
(b). 2, 1, -1
(c). 3, 1, -1
(d).1, 1, 1
Explanation: The values of 1, 1, 1 would be impossible since if n = 1, the only value
of l would be = 0.
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CHE141
Chapter 6
50. An electron cannot have the quantum numbers n = __________, 1 = __________, ml
= __________.
(a). 6, 1, 0
(b).3, 2, 3
(c). 3, 2, -2
(d). 1, 0, 0
Explanation: The values of 3, 2, 3 will be impossible since n and ml cannot have the
same values.
51. Which quantum number determines the energy of an electron in a hydrogen atom?
(a). n
(b). n and l
(c). ml
(d). l
52. In a px orbital, the subscript x denotes the __________ of the electron.
(a). energy
(b). spin of the electrons
(c). axis along which the orbital is aligned
(d). size of the orbital
Explanation: There are in all 3 degenrate “p” orbitals possible that are typically
distinguished by the axis along which they are oriented. This is achieved by
including the letter denoting the axis as a subscript.
53. Which electron configuration represents a violation of the Pauli exclusion principle?
(a).
1s
2s
2p
1s
2s
2p
1s
2s
2p
1s
2s
2p
(b).
(c).
(d).
Explanation: According to the Pauli exclusion principle no two electrons in an atom
cannot have the same 4 quantum numbers. The2 electrons in the 2s orbital have the
same value for their ms which is not allowed.
Copyright © 2006 Dr. Harshavardhan D. Bapat
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CHE141
Chapter 6
54. Which electron configuration represents a violation of the Pauli exclusion principle?
(a).
1s
2s
2p
1s
2s
2p
1s
2s
2p
1s
2s
2p
(b).
(c).
(d).
Explanation: According to the Pauli exclusion principle no two electrons in an atom
cannot have the same 4 quantum numbers. The2 electrons in the 2p orbital have the
same value for their ms which is not allowed.
55. Which of the following is a valid set of four quantum numbers? (n, l, ml, ms)
(a). 2, 0, 0, + ½
(b). 2, 2, 1, - ½
(c). 1, 0, 1, + ½
(d). 2, 1, +2, + ½
Explanation: Here is why only option (a) is the correct answer: In option (b), n =l
which is not allowed, in (c) ml ≠ 1 since l = 0 and in (d) ml > l which are all not
allowed.
56. Which of the following is not a valid set of four quantum numbers? (n, l, ml, ms)
(a). 2, 0, 0, + ½
(b). 2, 1, 0, - ½
(c). 1, 1, 0, + ½
(d). 1, 0, 0, + ½
Explanation: Since n can never be equal to l, option c is the only set that is not valid.
57. Which one of the following configurations depicts an excited oxygen atom?
(a). 1s22s22p2
(b).1s22s22p23s2
(c). 1s22s22p1
(d). 1s22s22p4
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Chapter 6
Explanation: The ground state electron configuration of the oxygen atom is shown in
option (d), while option (b) shows that two electrons from the 2p orbital have been
promoted to the 3s orbital.
58. Which one of the following configurations depicts an excited carbon atom?
(a). 1s22s22p13s1
(b). 1s22s22p3
(c). 1s22s22p2
(d). 1s22s23s1
Explanation: The ground state electron configuration is shown in (c), while option
(a) shows that one of the electrons has been excited from the 2p to the 3s orbital.
Copyright © 2006 Dr. Harshavardhan D. Bapat
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