Worked Solutions
Chapter 4
Question 20
For the hydrogen atom, the energies of the allowed levels are given by the formula E = -21.76
x 10-19 / n2 joules.
(a) Calculate the energies for the n = 1, n = 2, n = 3, n = 4 and n = 5 energy levels as
multiples of 10-19.
(c) Fill in the missing words or numbers in the following:
The difference in energy between the n = 1 and n = 2 energy levels in the hydrogen
atom is ---------- x 10-19 joules. If an electron in the n = 1 energy level gained this
amount of energy, the value of n for its new energy level would be ----. If another
hydrogen atom, with its electron in the n = 2 energy level, emits 16.32 x 10-19 joules,
the new value of n will be ---.
Answer:
(a) E = 21.76 x 10-19 / n2 joules
n = 1 ⇒ E = -21.76 x 10-19 / 12 = -21.76 x 10-19 joules
n = 2 ⇒ E = -21.76 x 10-19 / 22 = -21.76 x 10-19 / 4
= -5.44 x 10-19 joules
n = 3 ⇒ E = -21.76 x 10-19 / 32 = -21.76 x 10-19 / 9
= -2.42 x 10-19 joules
n = 4 ⇒ E = -21.76 x 10-19 / 42 = -21.76 x 10-19 / 16
= -1.36 x 10-19 joules
n = 5 ⇒ E = -21.76 x 10-19 / 52 = -21.76 x 10-19 / 25
= -0.87 x 10-19 joules
(c) The difference in energy between the n = 1 and n = 2 energy levels in the hydrogen
atom is 16.32 x 10-19 joules. If an electron in the n = 1 energy level gained this amount
of energy, the value of n for its new energy level would be 2. If another hydrogen atom,
with its electron in the n = 2 energy level, emits 16.32 x 10-19 joules, the new value of n
will be 1.
Question 21
What is the maximum number of electrons that can be contained in (a) all the 3d orbitals
(b) all the 2p orbitals (c) all the orbitals with n = 3 (d) all the 3p orbitals?
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Answer:
(a) There are five 3d orbitals each of which can hold two electrons, giving a maximum of
ten electrons.
(b) There are three 2p orbitals each of which can hold two electrons, giving a maximum
of six electrons.
(c) The orbitals with n = 3 are the 3s, the 3p and the 3d. There is one 3s orbital, three 3p
orbitals, and five 3d orbitals, giving a total of nine orbitals. Each can hold two
electrons giving a maximum of eighteen electrons.
(d) There are three 3p orbitals each of which can hold two electrons, giving a maximum
of six electrons.
Question 26
How many electrons can be accommodated in the n = 4 energy level? How many orbitals are
needed to accommodate these electrons?
Answer:
The orbitals with n = 4 are the 4s, the 4p, the 4d and the 4f. There is one 4s orbital, three 4p
orbitals, five 4d orbitals and seven 4f orbitals, giving a total of sixteen orbitals. Each can hold
two electrons giving a maximum of thirty-two electrons.
Question 27
Write the electronic configurations (s, p, d) for each of the following: Be, N, C, Co, Ni, Zn,
Br, Ca, O, O2-, Na+, Cu, Al3+, V, Cr, S2-, H-, Sc.
Answer:
Beryllium has 4 electrons. Its electronic configuration is:
Be 1s2 2s2.
Nitrogen has 7 electrons. Its electronic configuration is:
N 1s2 2s2 2p3.
Carbon has 6 electrons. Its electronic configuration is:
C 1s2 2s2 2p2.
Cobalt has 27 electrons. Its electronic configuration is:
Co 1s2 2s2 2p63s2 3p64s2 3d7.
Nickel has 28 electrons. Its electronic configuration is:
Ni 1s2 2s2 2p63s2 3p64s2 3d8.
Zinc has 30 electrons. Its electronic configuration is:
Zn 1s2 2s2 2p63s2 3p64s2 3d10.
Bromine has 35 electrons. Its electronic configuration is:
Br 1s2 2s2 2p63s2 3p64s2 3d10 4p5.
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Calcium has 20 electrons. Its electronic configuration is:
Ca 1s2 2s2 2p63s2 3p64s2.
Oxygen has 8 electrons. Its electronic configuration is:
O 1s2 2s2 2p4.
The oxide ion O2- has 10 electrons. Its electronic configuration is:
O2- 1s2 2s2 2p6.
The sodium ion Na+ has 10 electrons. Its electronic configuration is:
Na+ 1s2 2s2 2p6.
Copper has 29 electrons. Its electronic configuration is:
Cu 1s2 2s2 2p63s2 3p64s1 3d10.
The aluminium ion Al3+ has 10 electrons. Its electronic configuration is:
Al3+ 1s2 2s2 2p6.
Vanadium has 23 electrons. Its electronic configuration is:
V 1s2 2s2 2p63s2 3p64s2 3d3.
Chromium has 24 electrons. Its electronic configuration is:
Cr 1s2 2s2 2p63s2 3p64s1 3d5.
The sulphide ion S2- has 18 electrons. Its electronic configuration is:
S2- 1s2 2s2 2p63s2 3p6.
The hydride ion H- has 2 electrons. Its electronic configuration is:
H- 1s2.
Scandium has 21 electrons. Its electronic configuration is:
V 1s2 2s2 2p63s2 3p64s2 3d1.
Question 28
Which atoms have the following electronic configurations? (a) 1s22s22p63s23p3(b) 1s22s22p3
(c) 1s22s22p63s23p64s23d7 (d) 1s22s22p63s2
Answer:
(a) There are fifteen electrons so the atomic number Z = 15, and the element is
phosphorous.
(b) There are seven electrons so the atomic number Z = 7, and the element is nitrogen.
(c) There are twenty-seven electrons so the atomic number Z = 27, and the element is
cobalt.
(d) There are twelve electrons so the atomic number Z = 12, and the element is
magnesium.
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Question 29
Which ions have the following electronic configurations? (a) [1s2]+
(c) [1s22s22p63s23p6]2-
(b) [1s22s22p6]-
(d) [1s22s22p63s23p6]2+
Answer:
(a) This ion has two electrons and a positive charge. Thus the neutral atom would have
three electrons, Z = 3, and the element is lithium. The ion is Li+.
(b) This ion has ten electrons and a negative charge. Thus the neutral atom would have
nine electrons, Z = 9, and the element is fluorine. The ion is F-.
(c) This ion has eighteen electrons and a double negative charge. Thus the neutral atom
would have sixteen electrons, Z = 16, and the element is sulfur. The ion is S2-.
(d) This ion has eighteen electrons and a double positive charge. Thus the neutral atom
would have twenty electrons, Z = 20, and the element is calcium. The ion is Ca2+.
Question 30
Write down the total number of (i) s electrons (ii) p electrons in a chromium atom. Where are
the remaining electrons in this atom accommodated?
Answer:
The electronic configuration of chromium is:
Cr 1s2 2s2 2p6 3s2 3p6 4s1 3d5.
(i)
There are seven s electrons.
(ii)
There are twelve p electrons.
The remaining five electrons are in the 3d sublevel.
Question 31
How are the electrons in the 3p sublevel arranged in orbitals in (a) a phosphorus atom (b) a
silicon atom?
Answer:
(a) The electronic configuration of phosphorous is:
P 1s2 2s2 2px2 2py2 2pz2 3s2 3px1 3py1 3pz1. Each 3p orbital is singly occupied.
(b) The electronic configuration of silicon is:
Si 1s2 2s2 2px2 2py2 2pz2 3s2 3px1 3py1. Two 3p orbitals are singly occupied, while the third one
is empty.
Question 51
The electronic configuration of a certain element is 1s2 2s2 2p63s2.
(a) In which group of the periodic table is the element?
(b) What is the atomic number of the element?
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(c) What is the charge on the ion that an atom of the element forms on reaction with
chlorine?
Answer:
(a) The element has a total of 12 electrons in each of its atoms. The element is therefore
magnesium, and is in group II.
(b) 12.
(c) +2, since the element is in group II.
Question 61
Write the electronic configurations of each of the following ions: Ca2+, F-, K+, Al3+, Li+, Br -,
Se2-, Ga3+.
Answer:
The calcium atom has twenty electrons, so the calcium ion Ca2+ has eighteen electrons. Its
electronic configuration is:
Ca2+ 1s2 2s2 2p63s2 3p6.
The fluorine atom has nine electrons, so the fluoride ion F- has ten electrons. Its electronic
configuration is:
F- 1s2 2s2 2p6.
The potassium atom has nineteen electrons, so the potassium ion K+ has eighteen
electrons. Its electronic configuration is:
K+ 1s2 2s2 2p63s2 3p6.
The aluminium atom has thirteen electrons, so the aluminium ion Al3+ has ten electrons.
Its electronic configuration is:
Al3+ 1s2 2s2 2p6.
The lithium atom has three electrons, so the lithium ion Li+ has two electrons. Its
electronic configuration is:
Li+ 1s2.
The bromine atom has thirty-five electrons, so the bromide ion Br- has thirty-six
electrons. Its electronic configuration is:
Br- 1s2 2s2 2p63s2 3p64s2 3d10 4p6.
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The selenium atom has thirty-four electrons, so the selenium ion Se2- has thirty-six
electrons. Its electronic configuration is:
Se2- 1s2 2s2 2p63s2 3p64s2 3d10 4p6.
The gallium atom has thirty-one electrons, so the gallium ion Ga3+- has twenty-eight
electrons. The electronic configuration of the gallium atom is: Ga 1s2 2s2 2p63s2 3p64s2
3d10 4px1. Although the 4s is filled before the 3d, when forming ions electrons leave the
4s before they leave the 3d. The electronic configuration of the gallium ion is:
Ga3+ 1s2 2s2 2p63s2 3p6 3d10.
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