1
Inorganic Chemistry (5th Edition) – Complete Solutions
Manual Author:Housecroft & Sharpe All Chapters Covered
Latest Version
FULL TEST BANK!!!
1|Page
,2 Basic concepts: atoms
1 B asic concepts: atoms
1.1 The notation:
50C
24r
shows that the atomic number, Z, is 24 and the mass number for the
isotope is 50.
Number of protons = Number of electrons = Z =
24 Number of neutrons = Mass number – Z = 50
– 24 = 26
For each isotope, Z = 24 and so there are 24 electrons and 24
protons. For mass numbers 52, 53 and 54, there are 28, 29 and
30 neutrons, respectively.
1. ‘Monotopic’ means that the element possesses only one isotope.
2
See Appendix 5 in H&S Examples other than As include P, Na and Be.
▶
1.3 (a) Al is monotopic, i.e. there is only one naturally occurring isotope.
Z = 13 Mass number = 27
Notation: ▶ Number of electrons = Number of protons =
27Al
13
13 Number of neutrons = 27 – 13 = 14
(b) Br (Z = 35) has 2 naturally occurring
isotopes.
Each isotope has 35 electrons and 35
protons.
79B 81B For the isotope with mass number 79: number of neutrons = 79
80r
35 35r
54 56 57 5 – 35 = 44 For the isotope with mass number 81: number of
FeFe Fe Fe
26 8 neutrons
= 81 – 35 = 46
(c) Fe (Z = 26) has 4 naturally occurring isotopes.
2 26 6
6 2
, 1
E he isotope with mass number 54: number of neutrons = 54 –
a
c 26 = 28 For the isotope with mass number 56: number of
h
neutrons
i
1.4 = 56 – 26 = 30 For the isotope with mass number 57:
s
o
t number of neutrons = 57 – 26 = 31 For the isotope with mass
o
p number 58: number of neutrons = 58 – 26 = 32
e
h Assume that 3H can be ignored since abundance is so low; error
a
s introduced by this assumption is negligible. The mass numbers of 1H
2 and 2H are 1 and 2 respectively. Let % 1H = x, and % 2H = 100 –
6
x
e Then:
l
e
c
t
r
o
n
s
a
n
d
2
6
p
r
o
t
o
n
s
.
F
o
r
t
3|Page
, 4 Basic concepts: atoms
x
A = 1.008 = 100 x
1 2 100
r – 2 x
+
100
100.8 = x +
200
x =
99.2
This result gives 99.2 % 1H and 0.8 % 2H. The values do not agree
with those in Appendix 5 (99.985 % 1H and 0.015 % 2H) because
we have used
Inorganic Chemistry (5th Edition) – Complete Solutions
Manual Author:Housecroft & Sharpe All Chapters Covered
Latest Version
FULL TEST BANK!!!
1|Page
,2 Basic concepts: atoms
1 B asic concepts: atoms
1.1 The notation:
50C
24r
shows that the atomic number, Z, is 24 and the mass number for the
isotope is 50.
Number of protons = Number of electrons = Z =
24 Number of neutrons = Mass number – Z = 50
– 24 = 26
For each isotope, Z = 24 and so there are 24 electrons and 24
protons. For mass numbers 52, 53 and 54, there are 28, 29 and
30 neutrons, respectively.
1. ‘Monotopic’ means that the element possesses only one isotope.
2
See Appendix 5 in H&S Examples other than As include P, Na and Be.
▶
1.3 (a) Al is monotopic, i.e. there is only one naturally occurring isotope.
Z = 13 Mass number = 27
Notation: ▶ Number of electrons = Number of protons =
27Al
13
13 Number of neutrons = 27 – 13 = 14
(b) Br (Z = 35) has 2 naturally occurring
isotopes.
Each isotope has 35 electrons and 35
protons.
79B 81B For the isotope with mass number 79: number of neutrons = 79
80r
35 35r
54 56 57 5 – 35 = 44 For the isotope with mass number 81: number of
FeFe Fe Fe
26 8 neutrons
= 81 – 35 = 46
(c) Fe (Z = 26) has 4 naturally occurring isotopes.
2 26 6
6 2
, 1
E he isotope with mass number 54: number of neutrons = 54 –
a
c 26 = 28 For the isotope with mass number 56: number of
h
neutrons
i
1.4 = 56 – 26 = 30 For the isotope with mass number 57:
s
o
t number of neutrons = 57 – 26 = 31 For the isotope with mass
o
p number 58: number of neutrons = 58 – 26 = 32
e
h Assume that 3H can be ignored since abundance is so low; error
a
s introduced by this assumption is negligible. The mass numbers of 1H
2 and 2H are 1 and 2 respectively. Let % 1H = x, and % 2H = 100 –
6
x
e Then:
l
e
c
t
r
o
n
s
a
n
d
2
6
p
r
o
t
o
n
s
.
F
o
r
t
3|Page
, 4 Basic concepts: atoms
x
A = 1.008 = 100 x
1 2 100
r – 2 x
+
100
100.8 = x +
200
x =
99.2
This result gives 99.2 % 1H and 0.8 % 2H. The values do not agree
with those in Appendix 5 (99.985 % 1H and 0.015 % 2H) because
we have used
