Inorganic Nomenclature (Naming and Writing Formulas)
Introduction questions
1)A neutral atom has no overall charge, and ion is a chemical species that has an overall
charge.
2)a) anion: negatively charged ion (i.e. Cl-)
b)cation: positively charged ion (i.e. NH4+ , Na+)
c)monatomic species: a species made up of only one atom (i.e. Ne, Li+)
d) diatomic species: a species made up of two atoms (i.e. O2 , NO , ClO-)
e)triatomic species: a species made up of three atoms (i.e. O3, NO2, I3-)
f)polyatomic species: species made up of many atoms (more that one)
(i.e. H3PO4, NO3-, NO, HS-)
Ion naming questions
1a) strontium ion
e) carbonate
b) nitride
c) nickel (III) ion
d) gold (I) ion
f) hydrogen sulphite ion ( bisulphite)
2) cyanide and hydroxide
3a) Na+
b) Cu2+
c)F-
d) PO43-
e)HC2O4- f) NH4+
Creating ionic formula questions
1a)Sn(SO4)2
b) (NH4)2C2O4
c) Li2O
d) Cu3N
e)Hg2(NO2)2
f)Fe(OH)3
g)Ag2SO4
h) Pb(ClO4)2
i) Cr2O3
j) MnO
k)KH2PO4
l)U(SO4)2
m)(NH4)2Cr2O7
n)Cu3PO4
o)Ca(ClO)2
p) NaHSO3
q) Mg(MnO4)2
r)WBr5
s)(NH4)3PO4
t)Hg(CH3COO)2
Creating ionic compound name question
1a)silver phosphate
b)aluminum sulphate
c)iron(III) sulphide
d)copper(I) chloride
e)ammonium carbonate
f)vanadium(III) chloride
g)mercury(I) carbonate
h)copper(II) sulphate
i)ammonium sulphide
j) ammomium hydrogen carbonate
(ammonium bicarbonate)
k)iron(II) oxalate
l)magnesium hydrogen sulphite
m)lithium chlorite
n)sodium monohydrogen phosphate
o)aluminum hydroxide
p)chromium (III) iodide
q)tin(IV) oxide
r)zinc dichromate
s)vanadium(V) oxide
t)strontium nitride
Kailley/Baverstock 2009 Page 1
Naming and writing formulas for hydrates questions
1a) iron(III) bromide hexahydrate
b)lithium dichromate dihydrate
c)aluminum oxide trihydrate
d) cobalt(II) fluoride tetrahydrate
e)sodium carbonate monohydrate
f)sodium sulphide nonahydrate
g)sodium sulphate decahydrate
h)nickel(II) phosphate octahydrate
i)magnesium monohydrogen phosphate heptahydrate
1a) FePO4●8H2O
b) Cd(NO3)2 ●4H2O
c) Cu3(PO4)2 ●3H2O
d) CrC2O4 ●H2O
e)NiCl2●6H2O
f)Al(NO3)3●9H2O
Naming and writing formulas of covalent compounds questions
1a)nitrogen dioxide
b)chlorine trifluoride
c)tetrasulphur dinitride
d)diphosphorus hexaoxide
e)dinitrogen trioxide
f)sulphur tetrafluoride
g)bromine monofluoride
h)sulphur hexafluoride
2a)SO3
b)PCl5
c)XeF6
d)OF2
e)CO
f)CCl4
g)P4S3
h)N2S5
i)Si3N4
Kailley/Baverstock 2009 Page 2
Summary questions
1)magnesium oxide
2)copper(II) sulphate
3)sodium acetate
4)ammonium nitrite
5)molybdenum(V) chloride
6)lithium hydroxide monohydrate
7)platinum(IV) choride
8)ammonium perchlorate
9)aluminum nitride
10)potassium permanganate
11)copper(I) sulphate
12)sulphuric acid
13)sodium carbonate decahydrate
14)sodium sulphite
15)lead(IV) hydrogen sulphate
16)tungsten (VI) fluoride
17)sodium dihydrogen phosphate
18)barium sulphide
19)ammonium chlorite
20)iron(II) hypochlorite
21)tin(II) cyanide
22)krypton difluoride
23)sodium phosphate
24)calcium sulphide
25)magnesium thiocyanate
26)silver permanganate
27)platinum(III) oxide trihydrate
28)phosphorus pentabromide
29)copper(II) acetate
30)aluminum perchlorate
31)ammonia
32)aluminum sulphide
33)sodium hydroxide
34)barium hydrogen sulphide tetrahydrate
35)dinitrogen monoxide
36)hydrogen nitrate (nitric acid)
37)cesium hydrogen carbonate
38)copper(I) sulphide
39)tricarbon disulphide
40)copper(II) nitrate hexahydrate
41)cobalt(II) chlorate
42)manganese(III) oxide
43)zinc acetate
44)acetic acid (ethanoic acid,vinegar)
45)manganese(III) phosphate
46)chromium(III) nitrate nonahydrate
47)strontium hypochlorite
48)vanadium(III) nitride
49)lead(IV) oxalate
50)cobalt(III) fluoride
51)barium sulphite
52)copper(II) dichromate
53)nitrogen triiodide
54)chromium(II) bromide
55)magnesium phosphide
56)iron(II) sulphate pentahydrate
57)calcium hydroxide
58)hydrogen phosphate(phosphoric acid)
59)radium sulphate
60)potassium hydrogen oxalate
61)dichlorine monoxide
62)titanium(IV) oxide
63)nickel(II) sulphate heptahydrate
64)magnesium chlorite
65)lead(IV) chloride
66)iron(III) hydrogen oxalate
67)diiodine pentaoxide
68)mercury(II) nitrate
69)zinc hydroxide
70)hydrogen sulphide
71)xenon trioxide
72)titanium(II) chloride
73)hydrogen fluoride (hydrofluoric acid)
74)tin(IV) chromate
75)cobalt(II) phosphate octahydrate
76)platinum(IV) sulphide
77)AgCl
78)SO2
79)Fe2(C2O4)3
80)BeO
81)Pb(CH3COO)2•10H2O
82)K2CrO4
83)Hg2(CH3COO)2
84)MoCl3
85)NH3
86)Au2S3
87)Ag2Cr2O7
88)Ca(CH3COO)2
89)Cr2(C2O4)3
90)Ca(NO2)2
91)F2O2
92)Mo2O5
93)SiF4
Kailley/Baverstock 2009 Page 3
94)Cd(CH3COO)2
95)HgCl2
96)LiHSO3
97)CH3COOH
98)Mg(ClO3)2•6H2O
99)PF3
100)CuI2
101)Ca3N2
102)Mg(OH)2
103)Mo2S5•3H2O
104)Fe(H2PO4)2
105)CI4
106)ZnSO4
107)Hg2S
108)H2SO3
109)FeF2•8H2O
110)Mg(HSO4)2
111)Al2S3
112)RaCO3
113)XeF4
114)Na2O
115)Ba3(PO4)2
116)Hg2(NO3)2•2H2O
117)NaClO
118)AuCN
119)SnBr4
120)HI
121)S4N4
122)Fe(OH)2
123)CuF
124)Sn(HCO3)2
125)N2O5
126)Zn(HSO3)2
127)Zn(ClO4)2•6H2O
128)Au(NO3)3
129)Mn2(SO4)3
130)HCl
131)CrO
132)Zn(HS)2
133)MoS3
134)Fe2(CO3)3
135)IF5
136)MnO2
137)HCN
138)Fe2(SO4)3•9H2O
139)KNO2
140)CrP
141)Ni(OH)2
142)ClO4
143)Hg(SCN)2
144)HNO2
145)PbCO3
146)NaHC2O4
147)AlBr3•6H2O
148)PbI2
149)Ag2O
150)Mn(HPO4)2
Naming and formula practice for naming and formula review questions
Naming and Formula Practice
1.
Practice for “ates”. Ate is the suffix used for the most common polyatomic ion.
Write the correct name:
a.
b.
c.
d.
e.
Ca(ClO3)2 calcium chlorate
NaBrO3 sodium bromate
Zn(IO3)2 zinc iodate
Fe(NO3)3 iron (III) nitrate
H3PO4 hydrogen phosphate
f.
g.
h.
i.
NiSO4 nickel (II) sulphate
CdSeO4 cadmium selenate
Na2CO3 sodium carbonate
Cs2SiO3 cesium silicate
f.
g.
h.
i.
Nickel (III) chlorate Ni(ClO3)3
Radium phosphate Ra3(PO4)2
Lithium selenate Li2SeO4
Lead (II) nitrate Pb(NO3)2
Write the correct formula:
a.
b.
c.
d.
e.
2.
Palladium (II) iodate Pd(IO3)2
Barium carbonate BaCO3
Hydrogen silicate H2SiO3
Gold (III) sulphate Au2(SO4)3
Potassium bromate KBrO3
Practice for “ites”. What does “ite” tell you about the charge and number of oxygen?
Kailley/Baverstock 2009 Page 4
One less oxygen than the ate form, but the same ion charge.
Write the correct name:
a. NaClO2 sodium chlorite
b. Al2(SO3)3 aluminum sulphite
c. H2CO2 hydrogen carbonite
d. Cu3PO3 copper (I) phosphite
e. Cd(NO2)2 cadmium nitrite
f. Hg2(IO2)2 mercury (I) iodite
g. Sr(BrO2)2 strontium bromite
Write the correct formula:
a.
b.
c.
d.
e.
3.
Platinum (IV) iodite Pt(IO2)4
Magnesium chlorite Mg(ClO2)2
Nickel (III) sulphite Ni2(SO3)3
Potassium phosphite K3PO3
Lead (II) nitrite Pb(NO2)2
f. Mercury (II) bromite Hg(BrO2)2
g. Cesium carbonite CsCO2
Practice for “per...ates”. What does “per” tell you about the charge and number of oxygen?
Per with ate means one more oxygen than the ate form with the same ion charge.
Write the correct name:
a. Au(BrO4)3 gold (III) perbromate
b. Sr(ClO4)2 strontium perchlorate
c. Na2O2 sodium peroxide
d. H2SO5 hydrogen persulphate
e. Fe(IO4)2 iron (II) periodate
f. H2O2 hydrogen peroxide
Write the correct formula:
a. Mercury (II) perchlorate Hg(ClO4)2
b. Tungensten (V) perbromate W(BrO4)5
c. Potassium peroxide K2O2
4.
d. Zirconium periodate Zr(IO4)4
e. Nickel (III) perfluorate Ni(FO4)3
Practice for “hypo...ites”. What does “hypo” tell you about the charge and number of oxygen?
Hypo with ite means two less oxygen than the ate form with the same charge
Write the correct name:
a. CsBrO cesium hypobromite
b. HClO hydrogen hypochlorite
c. NaIO sodium hypoiodite
d. Sc(NO)3 scandium hyponitrite
e. Mn(SO2)2 manganese (IV) hyposulphite
Write the correct formula:
a. Calcium hypoiodite Ca(IO)2
b. Palladium (IV) hypochlorite Pd(ClO)4
c. Cobalt (III) hyponitrite Co(NO)3
5.
d. Titanium (III) hypobromite Ti(BrO)3
e. Francium hyposulphite Fr2SO2
Practice for elements. What are the seven diatomic elements? What are the two polyatomic
elements?
Write the correct name:
a. Al aluminum
b. S8 sulphur
c. N2 nitrogen
d. I2 iodine
e. Au gold
Write the correct formula
a. Silver Ag
b. Oxygen O2
c. Chlorine Cl2
d. Phosphorous P4
e. Tin Sn
Kailley/Baverstock 2009 Page 5
6.
Practice for the acids. You must be able to write formulas from common names. This means you
must memorize the translation to the IUPAC name.
Common Name
IUPAC translation
Hydro_____ic acid becomes
hydrogen ____ide
_____ic acid becomes
hydrogen ____ate
Per_____ic acid becomes
hydrogen per ____ate
_____ous acid beomes
hydrogen ___ ite
Hypo____ous acid becomes
hydrogen hypo ____ ite
7.Write the IUPAC and common name:
a. HBrO2 hydrogen bromite
bromous acid
b. HBr hydrogen bromide
hydrobromic acid
c. HClO4 hydrogen perchlorate perchloric acid
d. H2SO4 hydrogen sulphate
sulphuric acid
e. H3PO4 hydrogen phosphate phosphoric acid
f.
H3PO3 hydrogen phosphite
g. HBrO3 hydrogen bromate
h. HIO
phosphorous acid
bromic acid
hydrogen hypoiodite hypoiodous acid
i.
HClO2 hydrogen chlorite
chlorous acid
j.
HNO3 hydrogen nitrate
nitric acid
k. HNO2 hydrogen nitrite
nitrous acid
8.Write the formula:
a. Iodous acid HIO2
b. Hydrosulphuric acid H2S
c. Perbromic acid HBrO4
d. Hypophosphorous acid H3PO2
e. Sulphurous acid H2SO3
f. Nitrous acid HNO2
g. Hydrofluoric acid HF
Kailley/Baverstock 2009 Page 6
Reading Scales, Uncertainty and Significant Figures
Measurement and uncertainty questions
NOTE: only the final answers are given, you can determine the smallest increment and
number of decimals places read to, from the final answer given. Remember the last number
is the uncertain digit so if answer states 15.25 ± 0.01 cm then answers of 15.24 ± 0.01 cm or
15.23 ± 0.01 cm would also be correct.
Reading Scales Rulers
a) A=15.25 + 0.01cm
B=16.59+ 0.01cm
b)A= 10.5 + 0.2cm
B= 14.9 + 0.2cm
c)A=5.71 + 0.05cm
B= 7.13+0.05cm
d) A= 113.6 A= 113.6+ 0.1cm B= 121.3 + 0.1cm
e)A=0.427 + 0.001cm
B= 0.555 + 0.001c
f)
A= 2.00+ 0.02cm
B= 3.31+ 0.02cm
C= 3.80 + 0.02cm
D= 4.80+0.02cm
g)
A= 99.15+ 0.01cm
C= 100.00 + 0.01cm
B= 99.46+ 0.01cm
D= 100.50+0.01cm
Reading burettes
h) 30.60 + 0.01mL
k) 24.5 + 0.1mL
n) 48.40 +0.01mL
i) 18.7 + 0.1mL
j) 0.32 + 0.01mL
l) 17.31 + 0.01mL m) 6.35 + 0.02mL
o) 19.28 + 0.02mL p) 6.51 + 0.05mL
Reading graduated cylinders
q) 30.39 + 0.01mL
t) 4.43 + 0.02mL
r) 3.08 + 0.01mL
u) 7.49 + 0.05 cm
s) 16.3 + 0.1mL
v) 20.41 + 0.02mL
2)a) 51.32 +0.01 g
d) 0.5130 + 0.0002g
b) 55 + 1mL
e) 98.9 + 0.7s
c) 455 + 3g
f) 49.8 + 0.9mL
3 a) 15.24 mL 15.26 mL
b) 109.8 mL 110.2 mL
c) 1.523 x 10-6 s 1.533 x 10-6s
4)Table to be gone over in class
Significant figure questions
1)a)6.3
d)1.3 x 102
g) 202
j) 0.0001 (1 x 10-4)
b) 0.00024 (or 2.4x10-4)
e) 3 x 1014
h) 9 x 101
k) 2
c) 1.33
f) 5.11 x 105
i) 2 x 101
l) 2.2 x 10-6
2)a) 90.4
d) 4.0076
g) 7.002 x 105
j) 6.2055 x 10-9
b) 53.0991
e) 1.864 x 104
h) -35.55
c) 7.7 x 10-5
f) –0.000769
i) 1.368 x 10-7
Kailley/Baverstock 2009 Page 7
3) a) 8.53
d) 4.0 x 102
g) 5.6 x 102
j)4.000 x 10-3 (0.004000)
b) 0.64
e) 1.67 x 104
h) –8.72 x10-3
c) –29.7
f) 30.9
I) 3.1 x 102
4)a) 0.86
d) 6.1 x102
g) 1.1
b) 102.1
e) –23.9
h) 0.109
c) 0.69
f)96
Atom Introduction Notes
Atom introduction questions
1)a)56
b)92
c)25
2)a)6
b)26
c)79
3)a)10
b)10
c)20
d)10
e)18
f)10
4)a)S2f)Mn
b)Ca2+
4+
c)Clg)V
5)a)+12
g)54
h)23
d)Ni2+
5+
h)Sb
b)+18
5-
c)+20
I)2
j)36
e)Au1+
I)S
1-
j)Fe1+
d)+7
6)proton= 1p , neutron= 1n , electron= 0e
1
0
-1
7)
Particle
Atomic
Number
Atomic
Mass
Number of
Protons
Number of
Neutrons
Number of
Electrons
56
26
30
26
207
82
125
82
56
26
26
Fe
207
82
Pb
82
70
31
Ga
31
70
31
39
31
Al
13
27
13
14
13
Au3+
79
197
79
118
76
As3-
33
75
33
42
36
27
13
197
79
75
33
Kailley/Baverstock 2009 Page 8
209
83
Bi5+
83
209
83
126
78
Number of
Protons
30
Number of
Neutrons
35
Number of
electrons
30
8)
Symbol
65
30
1-
Cd2+
88
38
2+
Te2-
103
45
75
33
36
127
53
53
74
54
59
27
27
32
27
66
30
30
36
30
112
48
48
64
46
88
38
38
50
36
127
52
52
75
54
103
45
45
58
42
75
33
33
42
36
Sr
127
52
45
Zn
112
48
35
Co
66
30
35
I
59
27
80
Br
127
53
Atomic
Number
30
Zn
80
35
Atomic
Mass
65
3+
Rh
As3-
A. Isotope Mass
1.
a. 10.81 amu
b. 69.8 amu
d. 72.7 amu
e. 65.5 amu
c. 108.0 amu
f. 91.3 amu
g. 95.9 amu
Kailley/Baverstock 2009 Page 9
Electron configuration questions
1a)i) ↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓
3s
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑
3s
3p
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑
3s
3p
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
ii) 1s22s22p63s2
iii) [Ne]3s2
iv) 2
b)i)
ii) 1s22s22p63s23p64s23d104p65s1
iii)[Kr]5s1
iv)1
c)i)
ii) 1s22s22p63s23p1
iii)[Ne]3s23p1
iv)3
d)i)
ii) 1s22s22p63s23p64s23d104p2
iii)[Ar]4s23d104p2
iv)4
e)i)
ii) 1s22s22p63s23p5
iii)[Ne] 3s23p5
iv)7
f)i)
ii) 1s22s22p63s23p64s23d104p4
iii)[Ar] 4s23d104p4
iv)6
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
4s
3d
4p
↑
5s
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑
4s
3d
4p
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑_ ↑
4s
3d
4p
2)We add electrons to the next available orbital space, once that is full then onto the next
orbital (keep going in order)
3)You remove electrons by following the given rule:
Rule: write the core notation, then from orbitals outside the core, remove electrons in
order from p orbitals first, then s, then d.
Remove from 4s before 3d and remove from 3p before 3s.
4)a)[He]2s22p6
Kailley/Baverstock 2009 Page 10
b) [Ar]4s23d104p6
c) [He]2s22p6
d) [Ar]4s23d10
e)[Ne]3s23p6
f)[Ar]4s23d104p6
5)
a)i)
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑
4s
3d
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
↑↓ ↑ ↑ ↑
4s
3d
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
4s
3d
ii) 1s22s22p63s23p64s23d8
iii)[Ar] 4s23d8
b)i)
ii) 1s22s22p63s23p64s23d3
iii)[Ar] 4s23d3
c)i)
ii)1s22s22p63s23p64s23d10
iii)[Ar]4s23d10
6a) [Ar]3d8
b)[Ar]3d7
c)[Ar]3d2
d)[Ne]3s23p6
e)[Ar]3d10
7)See chart on page 20
8) a. (orbital diagram) ↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑ ↑ ↑_
3s
3p
(full electron configuration)1s22s22p63s23p3 (core/abbreviated electron configuration)
[Ne]3s23p3
b. ↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
1s22s22p63s23p64s23d2
c.
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑ ↑ __ __ __
4s
3d
[Ar]4s23d2
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
↑↓ ↑↓ ↑↓ ↑ ↑ ↑_
4s
3d
1s22s22p63s23p64s23d7 [Ar]4s23d7
d.
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑_
4s
3d
4p
Kailley/Baverstock 2009 Page 11
1s22s22p63s23p64s23d104p5 [Ar]4s23d104p5
e.
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
4s
3d
4p
↑↓
5s
1s22s22p63s23p64s23d104p65s2 [Kr]5s2
f.
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
1s22s22p63s23p6 [Ne]3s23p6
g.
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
↑_
4s
1s22s22p63s23p64s1 [Ar]4s1
h.
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
4s
3d
4p
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
5s
4d
1s22s22p63s23p64s23d104p65s24d10 [Kr]5s24d10
i.
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
↑↓
4s
1s22s22p63s23p64s2 [Ar]4s2
j.
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
4s
3d
4p
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
5s
4d
5p
1s22s22p63s23p64s23d104p65s24d105p6[Kr]5s24d105p6
k.
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑_
4s
3d
4p
5s
4d
5p
6s
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
4s
3d
4p
1s22s22p63s23p64s23d104p65s24d105p66s1 [Xe]6s1
l.
↑↓
6s
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
4f
↑↓ ↑↓ ↑↓ ↑↓ ↑↓
5d
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
5s
4d
5p
↑ ↑ __
6p
1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p2 [Xe] 6s24f145d106p2
m.
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑_ __ __
4s
3d
4p
1s22s22p63s23p64s23d104p1 [Ar] 4s23d104p1
n.
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
↑↓ ↑ ↑ ↑ ↑ ↑_
4s
3d
1s22s22p63s23p64s23d5 [Ar]4s23d5
Kailley/Baverstock 2009 Page 12
o.
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
4s
3d
4p
1s22s22p63s23p64s23d104p65s24d2 [Kr] 5s24d2
9) a. ↑↓
1s
1s2
b.
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
4s
3d
4p
1s22s22p63s23p64s23d104p6 [Ar] 4s23d104p6
c.
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
4s
3d
4p
1s22s22p63s23p64s23d104p6 [Ar] 4s23d104p6
d.
↑↓
1s
↑↓
2s
1s22s2 [He]2s2
e.
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
↑ ↑ __ __ __
3d
1s22s22p63s23p63d2 [Ar]3d2
f.
↑↓
1s
↑↓ ↑↓ ↑↓ ↑
2s
2p
1s22s22p5 [He]2s22p5
g.
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
↑ ↑ ↑ ↑ ↑_
3d
1s22s22p63s23p63d5 [Ar]3d5
h.
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
↑↓ ↑↓ ↑↓ ↑↓ ↑↓
3d
1s22s22p63s23p63d10 [Ar]3d10
i.
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
↑ ↑ ↑ ↑ ↑_
3d
1s22s22p63s23p63d5 [Ar]3d5
Kailley/Baverstock 2009 Page 13
↑↓ ↑ ↑ __ __ __
5s
4d
j.
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
↑↓
4s
↑↓ ↑↓ ↑↓ ↑↓ ↑↓
3d
1s22s22p63s23p64s23d10 [Ar]4s23d10
k.
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
4s
3d
4p
↑ ↑ ↑ ↑ ↑_
4d
↑↓
1s
↑↓ ↑↓ ↑↓ ↑↓
2s
2p
↑↓ ↑↓ ↑↓ ↑↓
3s
3p
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
4s
3d
4p
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
5s
4d
1s22s22p63s23p63d104p64d5 [Kr]4d5
l.
1s22s22p63s23p63d104p65s24d10 [Kr] 5s24d10
Periodic Table
Metals/nonmetals/semi-metals questions
1)a)non-metal
f)non-metal
b)metal c)non-metal
g)semi-conductor
d)semiconductor
e)metal
2)on the right side
3)a)Ga
b)Ge
c)Sn
d)Mg
e)Bi
4)Ca>Ge>Si>P>F
5)a)Sb
b)K
c)Ge
6)P=iii
Ba=ii
Sb=iv
d)Al
Ar=v
e)Tl
f)Sb
As=I
Ionization energy questions
1a)The distance between nucleus and valence electrons increases going down a chemical
family.
b)The nuclear force of attraction decreases going down a chemical family.
c)The ionization energy decreases going down a family.
2a)The distance between the nucleus and valence electrons decreases as we go from left to
right across the periodic table.
b)The nuclear force of attraction increases going across the periodic table.
c)The ionization energy increases going across the periodic table.
Kailley/Baverstock 2009 Page 14
3)
Ionization energy INCREASES
Ionization
energy
INCREASES
4)a)Cl
Periodic Table
b)Cl
c)Ne
d)Mg
e)Ne f)I
5)a)Create graph, will be gone over in class.
b)These are the smallest atoms on each of their respective rows, and electrons are being
removed from filled orbitals, which have strong stability, which takes a lot of energy to do.
c)The valence electrons experience a smaller nuclear force of attraction due to
i)the increased distance of the electron from the nucleus(electrons are in a higher
energy level), and
ii)the shielding effect is more due to the number of filled orbitals in the core before the
valence electron is reached.
Because there is a smaller nuclear force of attraction felt by the valence electron, it is easier
to remove a valence electron as we go down a family, so the ionization energy decreases
from He to Ne to Ar.
d)The valence electrons experience a larger nuclear force of attraction due to
i)increased number of protons in nucleus as move across the periodic table.
ii)no increase in shielding effect across a period, since electrons are added to the
same energy level.
The increased nuclear force of attraction felt by valence electrons makes it harder to remove
valence electrons as we go across a period, so ionization energy increases across a period.
e)Be and Mg have filled s orbitals, so their ionization energy are higher than those of the
elements immediately before and after them. Similarly, N and P have half-filled orbitals, and
their ionization energies are higher than those of the elements immediately before and after
them. The filling of the p orbital (Ne and Ar) is a special case of increased stability leading to
increased ionization energy.
6)a)As
b)N
c)5
d)3
e)N
7)a)Al
b)Cl
c)1
d)1
e)Cl
8)Li+ and F-
Kailley/Baverstock 2009 Page 15
Electronegativity questions
1)a)Li
b)F
c)F
2a)I
b)F
c)DECREASE
3)
d)F
e)INCREASE.
Electronegativity INCREASES
Electronegativity
INCREASES
Periodic Table
4)a)F
b)Fr
c)Be
d)S (non-metal whereas Pb is a metal)
Ion-ion attraction/ionic radius questions
1)a)NaCl
b)As the charge of involved ions gets smaller, the melting point gets lower(easier to melt)
c)The strength of the ion-ion attraction is greater in smaller ions. This is due to less distance
between the + and – charges. So the melting temperature will increase in smaller ions as it is
takes more energy to separate ions which have a greater attraction.
2)Although the ions are about the same size, MgO has a greater charge on its ions (Mg2+ , O2).
The greater the charge, the greater the amount of ion-ion attraction, so the higher the melting
point.
3)a)CaO
b)BN
c)LiF
d)BaS
e)KCl
4)a)amount of repulsion increases
b)volume increases
c)negative ions are larger than the corresponding neutral atom.
5) a)amount of repulsion decreases
b)volume decreases
c)positive ions are smaller than the corresponding neutral atom.
6)Na+ is smaller circles, Cl- is larger circles.
Kailley/Baverstock 2009 Page 16
f)BeO
London force/bonding/size questions
1a)increases
b)increases
2)attraction between molecules
3)London forces are broken, because it is a physical change, the formula of F2 (bonding),
stays the same during boiling, the distance between the molecules of F2 is increased, which is
the London force attraction being broken.
4) The larger the atomic number of atoms in a molecules held together by London force, the
higher the melting and boiling point of that molecule. Both melting and boiling point require
energy to stretch and break London forces. If the atomic number of atoms increases, the
number of electrons involved in creating the London force increases, which increases the
strength of the London Force.
5)a)Kr
b)I2
c)CF4
d)CBr4
6)The students answer incorrectly implies that the bonding between Br2 molecules is broken
when melting occurs, HOWEVER, the correct answer should have the same FORMULA on
the products side, only the state has changed (which is based on the distance between the
molecules).
7)a)covalent(polar covalent)
b)ionic
c)ionic
d)covalent
e)ionic
f)covalent
g)covalent
h)ionic
8)a)F
b)Na
c)F
d)Si
e)S
f)O
9)a)Na+Cl-
b)C-O
c)Ca2+O2-
d)Mg2+O2-
e)C-C
f) N≡N
g)Se2h)P3-
I)Cl-
10)ionic bonding and London forces
11)a)Nab)Br
c)As3d)Rb+
e)Sef)O2-
12)a)London b)ionic c)London
d)covalent network
e)ionic f)London
Kailley/Baverstock 2009 Page 17
Electron (Lewis) dot structures/ shape/ polarity questions
FOLLOW YOUR NOTES! Electron dot structures must have dots in every position (no lines). This is
difficult to draw on a computer so I cannot do it! Make sure you have the correct number of valence
electrons – you must write these as dotes!
For polar molecules show the net dipole and dipoles created. Follow your notes. Showing arrows is
too difficult when typing.
3-D Shape/Name
86. a. (8e-)
H∂+
∂ - Cl
Linear
86. c. (14e-)
Cl∂
∂+ I
Linear
-
Polar, Non-Polar, IFA’s
Polar Bond creates a net
dipole. Polar Molecule,
dipole-dipole and London
Forces
3-D Shape/Name
86. b.(14e-)
I
I
Linear
Polar, Non-Polar, IFA’s
Non-polar bond so the
molecule is non-polar.
London Forces
Polar Bond creates a net
dipole. dipole-dipole,
and London Forces
86. d. (14e-)
H
H
H
No functional group, nonpolar molecule, London
Forces
C
C
86. e(12e-)
H
H
C
C
No functional group, nonpolar molecule, London
Forces
H
H
Each carbon is trigonal
planar.
86. g.
Ionic Solid
86. i. (20e-)
H
H
C
H
H
carbon is tetrahedral
oxygen is angular (there
should be 2 lone pairs on
o).
No functional group. nonpolar molecule, London
Forces
H C C H
Each carbon is linear.
86 h (12e-)
O
Polar Bonds that add
together to give a polar
molecule. dipole-dipole,
and London Forces
Angular
86. k.(12e-)
O
H
H
H
Each carbon is
tetrahedral.
86. f (10e-)
Functional group
therefore polar with
dipole-dipole,H-Bond and
London Forces.
Non-polar bond, nonpolar molecule, London
Forces
O
.Linear
86. j.(10e-)
N
N
Linear
Non-polar bond so the
molecule is non-polar.
London Forces
86. L.(20e-)
H
C
C
H
C
C
H
H
The center carbon’s are
linear and the end
carbon’s are trigonal
planar.
Kailley/Baverstock 2009 Page 18
No functional group nonpolar molecule with
London Forces.
86. m (10e-)
Linear
86. 0.(18e-)
Cl
C
S
H
Carbon which is trigonal planar.
86. q. (10e-)
C-H bond is nonpolar and C-N bond
is polar resulting in
a net dipole towards
nitrogen making this
molecule polar
resulting in dipoledipole and London
Forces.
Each bond is nonpolar making the
molecule non-polar
resulting in London
Forces.
86. n.(6e-)
H
B
H
The bonds are nonpolar making this
molecule non-polar.
London Forces only.
H
Trigonal Planar
86. p.(18e-)
This is an ion.
This is an ion.
Bent
86. r. (16e-)
This is an ion.
In bottom structure oxygen is
angular, N is linear.
In top structure oxygen is bent and N
is linear.
86. t.(18e-)
Linear
86. s.(8e-)
Angular
Bent
Kailley/Baverstock 2009 Page 19
All of the bonds
are polar and
add towards
oxygen making
this molecule
polar resulting in
dipole-dipole,
London Forces
and hydrogen
bonds.
The polar bonds
add together to
create a net
dipole, making
this molecule
polar resulting in
dipole-dipole and
London Forces.
86. u.(26e-)
S
Cl
S
Cl
Each sulphur is angular.
86. w.(48e-)
F
F
F
S
F
F
F
Octahedral
86. y.(24e-)
Each sulphurchlorine bond is
polar and add
toward chlorine.
As a result the
dipoles cancel
making this
molecule nonpolar yielding
London Forces
only.
The polar bonds
in the axial
positions cancel
out and so do
the polar bonds
in the plane
cancel out which
makes this
molecule nonpolar yielding
London Forces
only.
This is an ion.
86. v.(14e-)
H
N
N
H
H
H
Each nitrogen is trigonal pyramidal.
86. x.(16e-)
The nitrogenhydrogen bonds
are polar
towards
nitrogen. As a
result the dipoles
cancel making
this molecule
non-polar
yielding London
Forces only.
This is an ion.
N is linear.
86. z.(20e-)
H
H
H
H
Each carbon is trigonal planar.
Trigonal planar.
Kailley/Baverstock 2009 Page 20
No functional group
therefore non-polar
resulting in London
Forces.
86. aa.(28e-)
Top left N – trigonal pyramidal,
top right N – linear.
2nd structure left N – trigonal
planar, right N – bent
3rd structure left N – trigonal
planar, right N bent
4th structure left N – trigonal
planar, right N linear
86. cc.(34e-)
Br
Br
Se
Br
Br
See-saw
First structure – polar
Second structure –
polar.
Third structure –
polar
Fourth structure –
non-polar
Most resonance
structures are polar
yielding both London
Forces and dipoledipole attractions.
86. bb.(24e-)
O
O
H
C
C
H
H
H
Top carbon is trigonal
planar. Right oxygen is
angular and bottom carbon
is tetrahedral.
The bonds are nonpolar making this
molecules non-polar
resulting in London
Forces.
Kailley/Baverstock 2009 Page 21
Functional group
therefore polar.
London Forces, and
hydrogen bonds.
Review of attraction questions
1.
a.
b.
c.
d.
Ion charge and ion size
Higher charge and smaller ion size
Charge
When charge is tied
2. Rank the following from lowest to highest boiling point. Be able to explain your answer.
a. KCl < NaF < PbO < PbO2
b. CrCl2 < CrS < CrO < CrN
c. Cal2 < CaBr2 < CaCI2 < CaF2
3.
a. A London Force is the weak attraction between temporary dipoles. A London Forces
can form because electrons are moving in atoms and molecules. At any instant in time
the distribution is uneven. Where there is an area of more electrons a slight negative
charge forms and the area with less electrons has a slight positive charge. The
attraction occurs between two molecules or atoms.
b. Number of electrons, contact area
c. More electrons, greater contact area (unbranched)
d. The number of electrons
e. When electrons are tied
4. Rank the following non-polar molecules from lowest to highest boiling point. Be able to
explain your answer.
a. Ne < Kr < Xe < Rn
b. Ne < N2 < O2 < F2
c. Cl2 < P4 < I2 < S8
d. CH4 < N2 < CO2 < CCl4
e. C2H6 < C3H8 < C5H12 < C7H16 < C9H20
f. Propyne < cyclopropane < propane < butane
g. Butane < 2,2-dimethylbutane < 2-methylpentane < decane
5.
a. O=O
b.
c.
d.
e.
f.
= non-polar covalent bond, …London Force
O=O
London forces change
Bonds change
O2(l) + energy → O2(g)
O2(g) + energy → 2 O(g)
“e” requires more energy. Bonds are stronger than London Forces. Therefore more
energy is required to break a bond.
6.
a. Electronegativity and Molecule size
b. Bigger electronegativity difference and smaller molecule size
c. Hydrogen Bond
Kailley/Baverstock 2009 Page 22
7. Rank the polar molecules from strongest to weakest dipole-dipole attraction. Be able to
explain your answer. These are all considered to be small covalent molecules.
a. HF > HCl > HBr > HI
b. H2O > SO2 > ClO2 > NO2
c. H2O> NH3 > HCl > CHCl3
d. H2O > CH3OH > CH3NH2 > C2H5NH2
8.
a. To melt or boil carbon we must break the network of covalent bonds. A network of
covalent bonds is much stronger than an attraction. As a result lots of energy is required
to break these bonds.
b. The layers of graphite are held together by weak London Forces. When we write on a
piece of paper we create friction which is energy. This is enough energy to break the
London Forces and a layer of graphite is left on our paper.
c. Lead has the same basic structure as graphite (weak London forces between the layers)
so we could write with lead. Lead is not used because it is toxic. We stop using lead in
pencils in the 1500s.
9. Covalent networks > ion-ion attractions > London Forces in large molecules – would be
solid > Hydrogen Bonds > dipole-dipole attractions in small molecules and London Forces
in small molecules (depend on the molecule)
10. Organize the following from lowest to highest melting points. *Polar
a. N2 < O2 < ethane < methylpropane < butane
b. CH4 < Ar < CO2 < *HCl < *H2O
c. F2 < NH3 < Br2 < KCl < SrO < CaO < C
d. Cl2 < H2O < P4 < I2 < Si
11. HF boils at 19oC so on a cold day it is a liquid and a hot day it is a gas. The rest of these
are gases at room temperature. HF has the highest boiling point because it has hydrogen
bonds which with relatively small molecules are a significant attraction. This means more
enegy is required to break the hydrogen bonds giving a higher boiling point. The rest of
these molecules are polar, but they do not have hydrogen bonds. This means HCl, HBr
and HI have dipole-dipole attractions which are weaker than hydrogen bonds. HCl with
the highest electronegativity difference would have a stronger dipole-dipole attraction
followed by HBr and then HI. These molecules also have London Forces which get
stronger with an increasing number of electrons. Since HI and HBr have a higher boiling
point than HCl the London Forces must be stronger than the dipole-dipole attractions.
Kailley/Baverstock 2009 Page 23
Chemical and Physical Changes
A. Balancing and predicting products
Balancing and predicting product questions
1. Balance the following:
a. 2 Pb + O2 → 2 PbO
b. H2 + I2 → 2 HI
c. N2 + 3 Br2 → 2 NBr3
d. 2 K + 2 H2O → 2 KOH + H2
e. 4 PH3 + 3 O2 → P4 + 6 H2O
f. C8H16 + 12 O2 → 8 CO2 + 8 H2O
g. 2 LiNO3 → 2 LiNO2 + O2
h. BaC2 + 2 O2 → Ba + 2 CO2
i. C7H16 + 11 O2 → 7 CO2 + 8 H2O
j. K2SO3 + SrCl2 → 2 KCl + SrSO3
k. 2 NaOH + H2SO3 → Na2SO3 + 2 H2O
l. Mg(OH)2 + 2 NH4Br → MgBr2 + 2 NH3 + 2 H2O
m. 5 C + 2 SO2 → CS2 + 4 CO
n. Mg3P2 + 6 H2O → 3 Mg(OH)2 + 2 PH3
o. W 2O5 + 5 Ba → 5 BaO + 2 W
p. 2 K2O2 + 2 H2O → 4 KOH + O2
q. X3O4 + 4 H2 → 3 X + 4 H2O
r. Hg + 2 H2SO4 → HgSO4 + 2 H2O + SO2
s. 2 Fe + 3 H2SO4 → Fe2(SO4)3 + 3 H2
t. 2 Si4H10 + 13 O2 → 8 SiO2 + 10 H2O
u. 4 NH3 + O2 → 2 N2H4 + 2 H2O
v. 2 C17H34 + 51 O2 → 34 CO2 + 34 H2O
w. 4 BP + 6 F2 → 4 BF3 + P4
x. SrSO4·2H2O + 2 SO3 → SrSO4 + 2 H2SO4
y. 4 C3H7N2O7 + 5 O2 → 12 CO2 + 14 H2O + 4 N2
z. C7H16O4S2 + 11 O2 → 7 CO2 + 8 H2O + 2 SO2
aa. 9 K + 4 CdI2 → 8 KI + KCd4
bb. HIO3 + 5 HI → 3 H2O + 3 I2
cc. Ga4C3 + 12 H2O → 4 Ga(OH)3 + 3 CH4
dd. 2 Mg(NO3)2 ·3H2O + 3 LaC2 → 2 Mg(NO3)2 + 3 La(OH)2 + 3 C2H2
ee. C2H5NO2 + 5 Cl2 → C2Cl5NO2 + 5 HCl
ff. Ca3(PO3)2 + 3 SiO2 + 3 C → 3 CaSiO3 + 3 CO + 2 P
gg. In2C6 + 6 H2O → 2 In(OH)3 + 3 C2H2
hh. 2 RbF + BaO + H2O → BaF2 + 2 RbOH
ii. 4 CsH + AlF3 → CsAlH4 + 3 CsF
jj. 2 BaF2 + 2 H2SO4 + SiO2 → 2 BaSO4 + SiF4 + 2 H2O
kk. 3 MgSi2 + 2 AsBr3 → 6 Si + 2 As + 3 MgBr2
ll. 2 ZrO2 + B4C + 3 C → 2 ZrB2 + 4 CO
mm. 4 NH3 + 5 O2 → 4 NO + 6 H2O
nn. SiF4 + 8 KOH → K4SiO4 + 4 KF + 4 H2O
oo. 2 NH4I + BaO → 2 NH3 + BaI2 + H2O
pp. 4 KPb + 4 C2H5Cl → Pb(C2H5)4 + 3 Pb + 4 KCl
qq. Mg2C + 4 H2O → 2 Mg(OH)2 + CH4
rr. 4 NoF3 + O2 + 4 HF → 4 NoF4 + 2 H2O
ss. 3 NO2 + H2O → 2 HNO3 + NO
tt. 3 KAlH4 + 4 BCl3 → 3 KCl + 3 AlCl3 + 2 B2H6
Kailley/Baverstock 2009 Page 24
2. Predict the products and write the balanced reaction with states.
a. H2SO3 (aq) + 2 NaOH (aq) → Na2SO3 (aq) + 2 H2O (l)
b. H3PO4 (aq) + 3 KOH (aq) → K3PO4 (aq) + 3 H2O (l)
c. 3 H2SO4 (aq) + 2 Fe(OH)3 (s) → Fe2(SO4)3 (aq) + 6 H2O (l)
d. H4P2O7 (aq) + 2 Ca(OH)2 (s) → Ca2P2O7 (s) + 4 H2O (l)
(Ca2P2O7 is not on the chart but Ca3(PO4)2 is low solubility so a common sense guess would
be to use (s))
e. H2SO4 (aq) + Ba(OH)2 (s) → BaSO4 (s) + 2 H2O (l)
3. Predict the products and write the balanced reaction with states.
a. 2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(g)
b. Mg(s) + CuSO4(aq) → MgSO4(aq) + Cu(s)
c. 4 Na(s) + O2(g) → 2 Na2O(s)
d. 2 Fe(NO3)3(aq) + 3 MgS(aq) → Fe2S3(s) + 3 Mg(NO3)2(aq)
e. 2 N2O(g) → 2 N2(g) + O2(g)
f. Sn(OH)4(s) + 4 HBr(aq) → SnBr4(aq) + 4 H2O(l)
g. Cl2(g) + 2 KI(aq) → 2 KCl(aq) + I2(s)
h. 16 Al(s) + 3 S8(s) → 8 Al2S3(s)
i. C6H12O6(l) + 6 O2(g) → 6 CO2(g) + 6 H2O(g)
j. 3 HF(aq) + Fe(OH)3(aq) → FeF3(aq) + 3 H2O(l)
k. 2 H2O2(l) → O2(g) + 2 H2O(l)
l. FeCl2(aq) + K2S(aq) → FeS(s)_ + 2 KCl(aq)
m. 2 Ca(s) + O2(g) → 2 CaO(s)
n. H2SO4 (aq) + 2 NaOH (aq) → Na2SO4 (aq) + 2 H2O (l)
o. C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g)
p. 4 Cr(s) + 3 SnCl4(aq) → 4 CrCl3(aq) + 3 Sn(s)
q. Pb(NO3)2(aq) + K2CrO4(aq) → PbCrO4(s) + 2 KNO3(aq)
r. 2 Fe(s) + 3 I2(s) → 2 FeI3(s)
s. C3H6OS2(l) + 6 O2(g) → 3 CO2(g) + 3 H2O(g) + 2 SO2(g)
t. MgCl2(s) → Mg(s) + Cl2(g)
u. Co(NO3)2(aq) + H2S(aq) → CoS(s) + 2 HNO3(aq)
v. H4P2O7(aq) + 4 KOH (aq) → K4P2O7 (aq) + 4 H2O (l)
w. Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)
x. 2 HI(g) → H2(g) + I2(s)
4. Predict the products and write the balanced reaction with states.
a. 2 HNO3 (aq) + Sr(OH)2 (aq) → Sr(NO3)2 (aq) + 2 H2O (l)
b. 2 C6H4(OH)2 (l) + 13 O2 (g) → 12 CO2 (g) + 6 H2O (g)
c. Zn (s) + Ni(NO3)2 (aq) → Ni (s) + Zn(NO3)2 (aq)
d. 2 AlCl3 (aq) + 3 Na2CO3 (aq) → Al2(CO3)3 (s) + 6 NaCl (aq)
e. 4 Al(s) + 3 O2 (g) → 2 Al2O3 (s)
f. Ba(OH)2 (s) + H2SO4 (aq) → BaSO4 (s) + 2 H2O (l)
g. 2 NO2 (g) → N2 (g) + 2 O2 (g)
h. Cl2 (g) + CaBr2 (aq) → Br2 (l) + CaCl2 (aq)
i. C9H20O4S2 (l) 14 O2 (g) → 9 CO2 (g) + 10 H2O (g) + 2 SO2 (g)
j. ZnSO4 (aq) + SrCl2 (aq) → SrSO4 (s) + ZnCl2 (aq)
k. 8 Zn (s) + S8 (s) → 8 ZnS (s)
l. 2 NH3 (g) → N2 (g) + 3 H2 (g)
m. HCl (aq) + KOH (aq) → KCl(aq) + H2O (l)
n. 2 ICl (l) → I2 (g) + Cl2 (g)
(ICl is polar (dipole-dipole) and has about the same number of electrons (London Forces) as Br2 therefore I
would classify it as a liquid, but solid would also be acceptable)
Kailley/Baverstock 2009 Page 25
o. 2 Na3PO4 (aq) + 3 Ca(OH)2 (s) → Ca3(PO4)2 (s) + 6 NaOH (aq)
p. C4H8S (l) + 7 O2 (g) → 4 CO2 (g) + 4 H2O (g) + SO2 (g)
q. Mg (s) + ZnSO4 (aq) → Zn (s) + MgSO4 (aq)
r. 4 Li (s) + O2 (g) → 2 Li2O (s)
B. Endothermic and exothermic reactions
Enthalpy and PE diagram questions
1.
a. Step 1 requires energy because bonds are broken.
b. Step 2 releases energy because bonds are formed.
c. Step 2 must release more energy than step 1 requires because energy is released overall
making the reaction exothermic.
d. Draw a PE diagram for each step. Label the activation energy and ∆H.
PE
(kJ)
PE
(kJ)
Rxn Progress
Rxn Progress
2. The energy needed to break an H-Br bond is 366 kJ.
a. HBr + 366 kJ → H + Br
b. 366 kJ of energy are released during the synthesis of HBr
3. Which of the following are exothermic or endothermic?
a. exothermic
b. endothermic
c. endothermic
d. exothermic
e. endothermic
f. “b” C12H22O11(s) + heat → C12H22O11(l)
“e” C25H52(l) → C25H52(s) + heat
4. The chemicals in the beaker are losing heat which makes the beaker hot and the reaction
exothermic.
5. The products have more PE than reactants in an endothermic reaction.
6. Heat must be removed or lost during an exothermic reaction.
7. Is ∆H > O for an endothermic reaction and ∆H < O for an exothermic reaction.
8.
50
40
PE
(kJ)
30
Ae
∆H
20
10
Rxn Progress
Kailley/Baverstock 2009 Page 26
9.
10. X → Y + 10kJ
11. ∆H = -35 kJ, Reactants have more PE than the products.
Chemical and physical change review questions
1. What are the clues we use to determine whether a chemical or physical change has occurred?
A chemical change may have a large change in heat, a colour change, bubbles forming, a new
state forming whereas a physical change usually has a smaller change in heat and state will
only change if you are melting, boiling etc, but the composition of the substance will not
change.
2. In what types of chemical reactions do the charges of substances change? How do we
determine the charge of metals with more than one charge?
Charges change in synthesis, decomposition, combustion and single replacement reactions.
When metals have more than one charge we first look at the data booklet on page 3. If there
is a starred and unstarred version we choose the unstarred version because it is more stable.
If not then use the complex periodic table and the charge in bold is the most stable it is also the
first one written. Do not use the complex periodic table for non-metals.
3. How do we determine the states of elements?
Use the periodic table where red is gas, blue is liquid and black is solid. We assume room
temperature and standard pressure.
Hydrocarbons with 1 to 4 carbons will be gas, with 5 to 15 carbons will be liquid and more than
15 will be solid. We assume organic compounds with functional groups will be liquid; however
when they get large they will be solid.
4. In what types of reactions is water present? How do we determine the states for compounds
when water is present?
Water is present during single replacement and double replacement reactions. We use the
solubility table when water is present where soluble means (aq) and low solubility means (s)
5. When do we use the activity series? How do we use the activity series?
We only use the activity series for single replacement reactions. If the element is more
reactive than what it is replacing then a reaction will occur. If the element is less reactive than
what it should replace then no reaction will occur.
Kailley/Baverstock 2009 Page 27
6. Write balanced reactions with states:
a. Ethyne is burned
2 C2H2(g) + 3 O2(g)
→ 4 CO2(g) + 2 H2O(g)
b. Zinc reacts with copper (II) sulphate
Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)
c. Iodine is added to a solution of potassium chloride
I2(s) + KCl(aq) → no reaction
d. Copper reacts with sulphur
8 Cu(s) + S8(s) → 8 CuS(s)
e. Cadmium reacts with oxygen
2 Cd(S) + O2(g) → 2 CdO(s)
f. Nickel is added to a solution of zinc nitrate
Ni(s) + Zn(NO3)2(aq) → no reaction
g. Nitric acid reacts with strontium hydroxide.
2 HNO3(aq) + Sr(OH)2(aq) → Sr(NO3)2(aq) + 2 H2O(l)
h. Iron (II) sulphide reacts with oxygen
2FeS(s) + 3O2(g) 2FeO(s) + 2SO2(g) OR
i.
j.
k.
l.
m.
n.
o.
p.
4 FeS(s) + 7 O2(g) → 2 Fe2O3(s) + 4 SO2(g)
Ammonia decomposes
2 NH3(g) → N2(g) + 3 H2(g)
Aluminum chloride reacts with sodium carbonate
2 AlCl3(aq) + 3 Na2CO3(aq) → Al2(CO3)3(s) + 6 NaCl(aq)
1-propanamine is burned
4 C3H7NH2(l) + 25 O2(g) → 12 CO2(g) + 4 NO2(g) + 18 H2O(g)
Calcium sulphite decomposes
CaSO3(s) → CaO(s) + SO2(g)
Nickel reacts with bromine
Ni(s) + Br2(l) → NiBr2(s)
Phosphoric acid reacts with aluminum hydroxide
H3PO4(aq) + Al(OH)3(s) → AlPO4(s) + 3 H2O(l)
A solution of lead (II) chlorate reacts with potassium iodide
Pb(ClO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KClO3(aq)
Could have also put Pb(ClO3)2(s) but I did say solution☺
Ethanoic acid reacts with ammonia
CH3COOH(aq) + NH3(aq) → NH4+(aq) + CH3COO-(aq)
7. What observations or tests could be used to determine the type of reaction or products of the
following reactions?
a. Precipitation - if two clear solutions are mixed and the result is a cloudy mixture then a
ppt formed
b. Neutralization- if two clear solutions are mixed and the products remain clear then the
reactants should be tested with litmus paper. If red litmus turns blue and blue litmus
turns red then we are reacting an acid and base and this is neutralization.
c. A single replacement reaction that produces hydrogen gas.
A burning splint is used to test for hydrogen gas. If the splint goes out with a pop then
hydrogen is formed.
d. Hydrocarbon combustion products.
Carbon dioxide is present if limewater turns cloudy and water is present is blue CoCl2
paper turns pink.
e. A decomposition reaction that produces oxygen gas.
Kailley/Baverstock 2009 Page 28
A glowing splint is used to test for oxygen gas. If the splint reignites or glows brighter
then oxygen is formed.
8. What are the observations, both qualitative and quantitative, for an endothermic reaction?
An endothermic reaction has qualitative observations of feeling cold and a quantitative
observations of a temperature decrease.
9.
When a reaction releases heat what happens to the thermometer?
The thermometer absorbs the heat which speeds up particles in the thermometer and as a
result the liquid expands and the temperature increases.
10. Write a reaction for the decomposition of calcium carbonate. Include ∆H at the end of the
reaction.
Heat + CaCO3(s) → CaO(s) + CO2(g)
a. Draw a potential energy diagram for this reaction. Label the axes, reactants, products,
activation energy and ∆H.
Ae
PE
(kJ)
R
P
∆H
Rxn Progress
b. What is the sign for ∆H? ∆H = PE products – PE reactants since PE products is larger
the sign for ∆H is positive.
11. Write a reaction for the dissolving of zinc bromate in water. This process is exothermic.
Include heat in the reaction. Water is not included in this reaction. Why?
Zn(BrO3)2(s) → Zn2+(aq) + 2 BrO3-(aq) + heat
Water does not react it only uses attractions to separate the ions. The water molecules are still
interacting with each other before and after. The ions are now separated from each other and
therefore physically changed.
Kailley/Baverstock 2009 Page 29
12. Draw a diagram showing the dissolving of zinc bromate. Label the attractions that change.
13. If the dissolving of zinc bromate is exothermic what does this mean in terms of attraction and
energy changes.
An exothermic process releases heat overall. This means more energy is released by the
forming of ion-dipole attractions than is required to break the ion-ion attractions.
Kailley/Baverstock 2009 Page 30
14. Phase Change diagram
Draw a phase change diagram for the heating of hydrogen fluoride from -100oC to 30oC. The
melting point of hydrogen fluoride is -840C and the boiling point is 19oC. Label the axes and
include a title.
40 g
(g, PE constant, KE increasing, particles
speeding up
20
l-g evaporation
-**bp
no change to attractions)
(l-g PE increases, KE constant, particle speed constant, hydrogen bonds are
breaking)
0
Temp(oC)
-20 l
(l, PE constant, KE increasing, particles speeding up, no change to attractions)
-40 -60 mp**
-100 -
(s-l PE increases, KE constant, particle speed constant, hydrogen bonds are weakening.
s-l melting
s (s, PE constant, KE increasing, particles speeding up, no change to attractions
Heat Added (kJ)
15. Ammonium carbamate undergoes deposition (gas directly to solid) at 60oC. Draw a
phase change diagram for the cooling of ammonium carbamate from 100oC to 400C. Label
the axes and include a title.
100 Temp(oC)
-
g(PE constant, KE decreasing, particles slow down, no change to attractions)
(g-s PE decreasing, KE constant, particle speed is constant, attractions are forming)
g-s deposition
60 -dp
s(PE constant, KE decreasing, particles slow down, no change to attractions)
40 Heat removed (kJ)
16. A test tube of melted wax (C25H52) is placed in a beaker of water when solidification is starting.
a. How does the temperature of the melted wax change during solidification? Explain with
reference to intermolecular forces of attractions.
During solidification the attractions between the wax particles are being strengthened
and this releases energy. However the energy released is equal to that absorbed by
the water. As a result the particle speed will not change and the temperature of the
wax is constant during the phase change.
b. How will the temperature of the water change while the wax is solidifying? Explain with
reference to the heat change occurring as the wax solidifies.
The water is absorbing the heat released by the phase change as the wax solidifies.
This causes the temperature of the water to increase.
17.
௚
Density is 1.00 ± 0.02 ௠௅
௚
18. 1.00 ௠௅
Kailley/Baverstock 2009 Page 31
Moles
A. Empirical and molecular formula
1.
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
k.
l.
m.
n.
BF3
SiH4
P2O5
I4O9
FeO
Fe2O3
Fe3O4
Li2O
C3H5Cl3
K2Cr2O7
Mg2P2O7
H3PO3
C4H8O3
C4H5N3
2. C3H6
3. N2O4
4. C10H22
5. C2H2
6. CO
7. Si2F6
8. B2H6
9. C3H8
10. O3
1.
2.
3.
4.
5.
6.
7.
8.
Stoichiometry
a. 0.900 mol C4F2 b. 13.5 mol F2 c. 3.80 mol F2
a. 54.0 g H2O b. 96.0 g NO c. 84.0 L O2 d. 16.8 L NH3
a. 308.0 g CO2 b. 147 g O2 c. 57.4 g C7H16 d. 56.0 L O2 e. 75.4 L O2 f. 97.3 g H2O
a. 271.0 L O2 b. 1.49 x 1021 molecules CO2 c. 2.4 x 1012 molecules H2O d. 10.6 mL O2
a. 0.108 g CO2 b. 348 mL gas c. 18.0 L O2 d. 0.676 g CO2
18.2 g SiCl4 and 0.427 g H2
6.85 x 105 L NH3
a. P4O10(s) + 6 H2SO4(l) → 4 H3PO4(aq) + 6 SO3(g) b. 22.3 g P4O10 and 10.6 L SO3
Percent yield and percent purity questions
1. a. 3.75 g Fe2O3 b. 82.6%, 53.7 g FeCO3 c. 24.1 g Fe2O3, 107 % yield Fe2O3 d. 23.2 kg FeCO3
4 FeCO3(s) + O2(g) → 2 Fe2O3(s) + 4 CO2(g)
2. 27.3 g PbS2, 9.09 % yield PbS2
2 PbS2(s) + 5 O2(g) → 2 PbO(s) + 4 SO2(g)
3. a. 92.9 g C6H5NO2, 18.5 g C6H5NO2, 20.0 % yield C6H5NO2 b. 58.9 g C6H6 impure, 11.8 g C6H6
pure reacted, 47.1 g C6H6 unreacted
C6H6(l) + HNO3(aq) → C6H5NO2(l) + H2O(l)
4. a. 10.1 g SiF4 b. 5.84 g SiO2 reacted, 6.41 g SiO2 unreacted c. 21.22 g SiF4 expected, 47.7 %
yield SiF4
SiO2(s) + 4 HF(g) → SiF4(g) + 2 H2O(g)
5. a. 10.792 g CuO expected, 7.0 g CuO actual b. 291 g Cu2(OH)2CO3 pure, 210 g CuO expected,
7.98 kg ore
Cu2(OH)2CO3(s) → CO2(g) + 2 CuO(s) + H2O(g)
Kailley/Baverstock 2009 Page 32
6.
a. 67.5 kg Ag2S pure, 58.8 kg Ag b. 0.0358 g Ag2S pure, 0.234 % purity Ag2S c. 172.0 g Ag,
71.62 % yield Ag d. 28.7 kg Ag2S pure, 3.32 x 103 kg Ag2S impure e. 0.27 g Ag
Ag2S(s) + C(s) + 2 O2(g) → 2 Ag(s) + CO2(g) + SO2(g)
Conductivity
Conductivity and solution questions
1. Ions must be present since a bulb turns on and since only the lowest wattage bulb turns on
showing that the concentration of ions are low.
2. NaCN, HBr, LiOH, H2SO4, K3PO4, NaOH…all the bright bulbs
3. Since no bulbs turn on this shows glucose does not produce ions.
4. Non-conducting compounds are all covalent compounds and organic molecules. The exception to
this rule is acids, amines and ammonia.
5. H2SO4 and HBr are acids
6. LiOH and NaOH are bases
7. As solutions these substances yield a high concentration of ions and therefore all bulbs glow
brightly.
Conductivity and liquid/solid questions
Questions
1. NaCN, LiOH, K3PO4
2. Solids generally do not conduct electricity the only exception is pure metals. Metals are held
together by metallic attractions where valence electrons are held loosely to the nuclei of atoms.
These electrons are able to move through the metal enabling them to conduct electricity.
3. To conduct ions must be able to move. NaCl(s) cannot conduct because the ions do not move
while NaCl(aq) has free moving ions allowing for conductivity.
Conductivity and concentration questions
Questions
1. A higher concentration of ions leads to greater conductivity.
2. NaCl → Na+ + Cl- , 2 M ions
K3PO4 → 3 K+ + PO43-, 4 M ions
1 M K3PO4 has a higher concentration of ions leading to greater conductivity present in each
solution
Water and conductivity questions
1. Pure liquid molecules are not made up of ions therefore there are no ions able to move and
complete the circuit.
2. Water reacts with the acids to produce ions which cause conductivity.
3. As more water is added more ions are produced. At some point the maximum amount of ions are
produced and therefore when more water is added the concentration of ions will decrease.
Kailley/Baverstock 2009 Page 33
Summary questions
1.
Which of the following would you expect to form ionic solutions and which form molecular
solutions?
a. I
b. I
c. I
d. I
e. M
f. I
g. I
h. M
i. I
j. M
2. A solution contains water and therefore ions will exist in any solution (H3O+ and OH-) making any
solution able to conduct.
3. Which of the following will conduct electricity?
a. No
i. Yes
j. No
b. No
c. Yes
k. Yes
d. Yes
l. Yes
m. No
e. No
n. No
f. Yes
g. No
o. Yes
h. Yes
p. No
q. Yes
r. No
s. Yes
t. Yes
u. Yes
v. Yes
w. Yes
x. No
Dissolving
Dissolving questions
1. Iodine and heptane are both non-polar therefore London Forces form between heptanes and
iodine and are strong enough to break the London Forces between the iodine molecules causing
iodine to be soluble in heptane. Water is polar and can only form weak London Forces with iodine.
Therefore only a small number of iodine molecules can dissolve in water.
2. By having a long non-polar tail, soap is able to dissolve non-polar grease molecules while the polar
head can dissolve in water.
3. A non-polar solvent can only form London Forces with ionic compounds. London Forces are not
strong enough to break ion-ion attractions in an ionic compound.
4. Water can only form weak London Forces with heptane. Therefore only a small number of
heptane molecules can dissolve in water.
5. a. H2O b. butanol c. H2O
6. Non-polar solvents when larger have more electrons and can form stronger London Forces with
non-polar solutes.
7.
O=O
= non-polar covalent bond, …London Force
O=O
8. Classify the most significant attraction found between the particles (ion-ion, Hydrogen bond,
dipole-dipole or London Force).
a. London Forces
g. Dipole-dipole
b. London Forces
h. H-Bond
c. Dipole-dipole
i. Dipole-dipole
d. H-Bond
j. Ion-ion attractions
e. London Forces
k. Non-polar covalent bond
f. London Forces
l. Polar covalent bond
Kailley/Baverstock 2009 Page 34
9. Heptane, is a good solvent for which of the following?
a. I2(s) Yes
c. H2O(l) No
b. NaI(s) No
d. C8H18(l) Yes
10. You are given the following solvents: water, ethanol, ethanal and carbon tetrachloride. Which of
these solvents could you use to dissolve?
a. Br2 carbon tetrachloride, ethanal and ethanol (ethanol and ethanol have polar and non-polar
parts and Br2 is small enough to enable weak London Forces to allow Br2 to dissolve in ethanol
and ethanol)
b. NaNO2 water (ethanol and ethanol don’t have the same strength of HB to break apart ion-ion)
c. CO2 carbon tetrachloride, ethanol and ethanal (to some amount since they have a non-polar
part)
d. Methanal: water,, ethanol and ethanal
e. Sulphur: carbon tetrachloride (Sulphur is held by strong London Forces and therefore ethanol
and ethanol cannot dissolve sulphur.)
11. A is NaCl, B is benzene, C is carbon tetrachloride, X is polar, Y is non-polar and Z is water.
Kailley/Baverstock 2009 Page 35