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TOPIC 3.
IONIC COMPOUNDS: formation, formulas and naming
Chemical bonding. When elements combine to form compounds, a chemical bond holds the atoms together. There are two basic types of chemical bonds possible in compounds, IONIC BONDS and COVALENT BONDS. Both types of bond involve a redistribution of electrons between the bonded atoms. Ionic bonding will now be examined while covalent bonding will be dealt with in Topic 4.
Formation of ions and ionic bonds. From Topic 2 it was seen that by removing electrons from the atoms of some elements (metals), their outer electron level can be made to be identical to that of the nearest noble gas, this being the energetically favourable condition which is responsible for the lack of reactivity of noble gases. When atoms gain or lose electrons to attain the same electron arrangement as the noble gas, they are said to become isoelectronic with the noble gas. However, the process of formation of cations from metal atoms requires energy input known as its ionization energy, and is not spontaneous as many books lead to believe. In this process, such atoms acquire a positive electrical charge equal in magnitude to that of the electrons removed and they are then called cations. After the required number of electrons has been removed such that the resulting cation is isoelectronic with a noble gas, the energy needed to remove further electrons is too great and so this does not occur. Hence with only a relatively small energy input required, Na atoms can form Na +cations in which state they are isoelectronic with the noble gas neon, but sodium cannot form Na 2+^ or Na 3+cations as the energy requirements would be excessive. With a few exceptions, in the process of becoming isoelectronic with a noble gas, no more than three electrons can be removed from an atom in the formation of cations as the ionization energy required is too great and the large positive charge would result in an unstable cation. To form compounds that would require the transfer of more than 3 electrons between reacting atoms, a different type of bonding - covalent bonding - is generally used.
Elements from some other groups (non-metals) can attain this energetically favourable state through their atoms gaining enough electrons to acquire the same electron arrangement as the nearest noble gas. In this process these elements gain a negative charge and are called anions. Again a limit of not more than three electrons can be transferred and once the noble gas electron arrangement has been acquired, no further electrons can be gained. Hence Cl atoms can form Cl – anions (isoelectronic with argon) but not Cl 2–^ or Cl 3–anions.
Thus an atom of sodium can lose the single electron located in the third energy level to form an Na +ion which then has 8 electrons in its outer energy level just like the neon atom. An atom of chlorine can gain an electron to form the Cl – ion which then has 8 electrons in its outer energy level just like argon. As electrons which are lost by an atom of one element must be accepted by an atom of another element, then sodium atoms and chlorine atoms can form Na +^ and Cl – ions by transferring a single electron per atom from Na to Cl. When this transfer occurs, the resulting ions of Na +^ and Cl , having opposite electrical charges, will be– attracted to each other to form the solid compound sodium chloride. This electrostatic attraction is the basis for ionic bonding.
III - 1
Note that there would be no sodium or chlorine atoms present in the compound, only the ions derived from them. While the resulting compound of formula Na Cl+^ – is also known commonly as “salt”, the term SALT is a general one which applies to any ionic compound, not just to sodium chloride.
Combining ratios and formulas of ionic compounds. The ionic compound formed must be electrically neutral, so the ratio of the number of Na +^ ions to the number of Cl – ions present in sodium chloride must be 1:1, resulting in the formula Na Cl+^ – as the simplest for this compound. Analysis of the compound sodium chloride would always show it to consist of sodium ions and chloride ions present in this ratio of 1:1. In Topic 1 it was pointed out that all the halogen elements including chlorine occur as diatomic molecules and not single atoms. Thus in forming sodium chloride, 1 molecule of chlorine, written as Cl , 2 would react with 2 atoms of sodium to form 2Cl –^ and 2Na +ions in order that the ratio of + charge to! charge be 1:1. However, in writing the formula, the simplest whole-number ratio of cation and anion is used. Thus although the formula Na 2 Cl 2 still has a Na :Cl ratio of 1:1, the formula should be written as Na Cl.
As was discussed in Topic 2, some cations may have a charge larger than +1, for example Ca 2+^ and Al 3+. Similarly, some anions such as O^ 2–^ and N 3–have more than a single negative charge. Again, when ionic compounds involving such species are formed, the overall charge on the resulting compound must be zero. This is achieved if the cations and anions are formed and therefore combine in the appropriate ratio so that the total size of the charge on the cations equals the total size of the charge on the anions. For example, the ionic compound calcium chloride which results from the reaction between calcium and chlorine consists of Ca2+ ions and Cl– ions which are present in the compound in the ratio of 1 calcium ion to 2 chloride ions. The formula for calcium chloride is therefore Ca Cl 2. 2+ –
[Note the use of the subscripted 2 to show that there are 2 chloride ions present in the formula. It is incorrect to write this as Ca 2+2Cl^ –^ or Ca 2+Cl^ 2– ].
In this formula, 2 Cl ions carry a total charge = 2–^! while the single Ca 2+ion carries a charge = 2+. Thus electrical neutrality is preserved in the compound. The ionic compound lithium oxide contains Li +^ ions and O 2–ions in the ratio of 2 lithium ions to 1 oxide ion so that the total charge on the compound is zero. The formula of lithium oxide then must be Li 2 O , the subscripted 2 being used to
indicate that there are two Li +ions in the formula.
Likewise, the ionic compound aluminium oxide containing Al 3+^ ions and O 2–ions has the formula Al 2 O 3 so that the total positive charge (6 +) exactly equals the 3+ 2–
total negative charge (6 !).
What will the charge be on a cation in a compound? There is a very simple rule that gives the number of electrons that can be removed from an atom of an element to form a cation without requiring an excessive amount of energy. Most elements of the first three groups of elements shown in Table 2 on Page I-21 of Topic 1 mainly form ionic compounds. The first group, the alkali metals, has 1 outer level electron, the second group has 2 outer electrons and the third group has 3 outer electrons. These are therefore also the number of electrons which must be removed from atoms of elements of each of these groups in order to obtain the same outer shell structure as the noble gas of closest atomic number. Thus when forming ions, the first group of elements form 1+ cations in ionic compounds, the second group of elements form 2+ cations in ionic compounds, and when elements of the third group form ionic compounds, the cations usually have a 3+ charge.
The following list of common cations formed by elements from other groups should be committed to memory. In some cases a Roman numeral I, II, III or IV is written as part of the name of the cation. This is needed for those elements which can have more than a single ionic state - for example, the element tin (Sn) can form ions with a 2+ charge [tin(II), Sn2+ ] or a 4+ charge [tin(IV), Sn^ 4+]. Note that very few cations with a 4+ charge exist and Sn 4+is unusual. Where there is only one possible ionic charge for a cation, the Roman numerals are not used.
COMMON CATIONS FROM FAMILIES OTHER THAN 1, 2 OR 3.
Ion Symbol Ion Symbol Ion Symbol
silver(I) Ag +^ zinc Zn 2+^ iron(II) Fe2+
copper(I) Cu +^ lead(II) Pb 2+^ iron(III) Fe3+
copper(II) Cu 2+^ cobalt(II) Co 2+^ cadmium Cd2+
tin(II) Sn 2+^ chromium(III) Cr 3+^ mercury(II) Hg2+
tin(IV) Sn 4+^ manganese(II) Mn 2+^ nickel(II) Ni2+
bismuth(III) Bi 3+^ gold(III) Au 3+^ platinum(II) Pt2+
What will be the charge on an anion in a compound? Again, the grouping of elements helps to answer this question. All the elements of the seventh group (halogens) require just one more electron to obtain the electron structure (8 outer level electrons) of the nearest noble gas. Consequently, when these elements form anions they all do so by gaining one electron and thus carry a 1! charge. Similarly, all elements of the sixth group are two electrons short of the nearest noble gas electron structure and so they all form 2! charged anions when in ionic compounds. Although less commonly observed, when elements from the fifth group starting with nitrogen (all 3 electrons short of the noble gas structure) form anions they do so by gaining 3 electrons and thus carry a 3! charge. No stable anions form from single atoms gaining more than three electrons. Thus knowing to which group an element belongs allows one to deduce its likely charge in an ionic compound.
Check your understanding of this section. State the rules that allow one to deduce the likely charge on the cation formed from a metal atom and an anion formed from a non-metal atom. How is the charge on a cation that can have more than one charge indicated? State the rules for naming a binary compound. How would one know that in the compound Fe O , the iron is present as Fe 2 3? 3+
How are ionic compounds named? Compounds of two elements such as all those discussed in this Topic are called BINARY COMPOUNDS. Rules for naming binary ionic compounds are very simple and are as follows: The compound is named as two separate words, the cation being named first and the anion last. The cation name is the same as the element but with the charge appended in brackets as Roman numerals where necessary. Anions formed from an element take the stem from the name of the element and the ending "ide" is attached. No special ending is needed for the cation as it is obvious that a compound is being named when two words are used in the name. Lower case letters are used for the name, including the first letter. The following examples illustrate the correct naming of binary ionic compounds.
NaCl sodium chloride (contains Na +^ and Cl ions)–
KI potassium iodide (contains K +^ and I ions)–
CaF 2 calcium fluoride (contains Ca and F ions) 2+ –
Rb O 2 rubidium oxide (contains Rb and O ions)
BaS barium sulfide (contains Ba 2+^ and S 2–ions)
Mg N 3 2 magnesium nitride (contains Mg and N ions) 2+ 3–
CuCl 2 copper(II) chloride (contains Cu and Cl ions) 2+ –
SnS tin(II) sulfide (contains Sn 2+^ and S 2–ions)
Fe O 2 3 iron(III) oxide (contains Fe and O ions) 3+ 2–
AuCl 3 gold(III) chloride (contains Au and Cl ions) 3+ –
Objectives of this Topic. When you have completed this Topic including the tutorial questions, you should have achieved the following goals:
- Understand the process whereby ionic bonds form as a result of electron transfer from one atom to another with consequent electrostatic attraction between the resultant ions.
- Understand that the formula for an ionic compound must contain an equal number of +ve and !ve charges and therefore be electrically neutral.
- Be able to write formulas and names for binary ionic compounds.
(iv) sodium and sulfur
(v) magnesium and nitrogen
(vi) rubidium and chlorine
(vii) caesium and phosphorus
(viii) potassium and iodine
(ix) calcium and selenium
(x) strontium and chlorine
(xi) lithium and oxygen
(xii) magnesium and bromine
(xiii) rubidium and nitrogen
(xiv) calcium and fluorine
(xv) aluminium and sulfur
(xvi) caesium and selenium
(xvii) barium and phosphorus
(xviii) sodium and nitrogen
(xix) potassium and chlorine
(xx) strontium and iodine
- Give the formula for each of the following binary compounds.
(i) silver(I) iodide
(ii) magnesium chloride
(iii) copper(II) oxide
(iv) copper(I) oxide
(v) barium nitride
(vi) manganese(II) sulfide
(vii) mercury(II) oxide
(viii) iron(II) bromide
(ix) aluminium oxide
(x) iron(III) chloride
- Write the name for each of the following compounds.
(i) AgCl
(ii) Mg N 3 2
(iii) CaBr 2
(iv) Al O 2 3
(v) CuCl 2
(vi) PbO
(vii) MnS
(viii) ZnI 2
(ix) KCl
(x) Ca P 3 2
(xi) CrCl 3
(xii) BaSe
(xiii) CoCl 2
(xiv) Fe O 2 3
(xv) FeCl 2
(xvi) SrI 2
(xvii) SnBr 2
(xviii) MgO
(xix) Rb N 3
(xx) LiF
(xxi) PtBr 2
(xxii) Bi O 2 3
(xxiii) AuCl 3
- Give the formula for each of the following binary compounds.
(i) cadmium fluoride (ii) strontium chloride
(iii) cobalt(II) sulfide (iv) lead(II) iodide
(v) tin(II) oxide (vi) iron(III) oxide
(vii) chromium(III) nitride (viii) calcium bromide
(ix) potassium oxide (x) sodium phosphide
- Complete the chemical crossword puzzle on the following pages.
- platinum(II) oxide 49. platinum(II) sulfide
- manganese(II) iodide 51. copper(II) phosphide
- copper(II) nitride 53. aluminium phosphide
- calcium sulfide 54. lead(II) nitride
- lead(II) phosphide 55. gold(III) oxide
- iron(III) phosphide 57. antimony
- tin(II) phosphide 58. tin(II) nitride
- bismuth(III) oxide 59. bismuth(III) sulfide
- mercury(II) nitride 60. ozone
- chromium(III) oxide 62. chromium(III) selenide
- bismuth 63. lithium phosphide
- nitrogen (molecule) 66. aluminium oxide
- aluminium sulfide 67. copper(I) nitride
- copper(I) phosphide 69. silver(I) nitride
- iron(III) selenide 72. caesium phosphide
- silver(I) phosphide 73. manganese(II) sulfide
- gold(III) nitride 74. bismuth(III) iodide
- ozone 75. rubidium phosphide
- caesium nitride 76. iron(II) oxide
- bismuth(III) bromide 77. manganese
- rubidium nitride 78. sodium fluoride
- iron(II) sulfide 79. potassium nitride
- sodium iodide 80. nickel(II) sulfide
- potassium phosphide 81. aluminium chloride
- nickel(II) oxide 82. sodium nitride
- aluminium fluoride 83. copper(II) oxide
- sodium phosphide 84. mercury(II) oxide
- copper(II) sulfide 87. magnesium selenide
- lithium chloride 88. lead(II) sulfide
- lithium nitride 89. lead(IV) fluoride
- magnesium oxide 90. barium sulfide
- lead(II) oxide 91. tin(II) oxide
- barium selenide 92. lead
- tin(II) sulfide 93. rubidium chloride
- rubidium fluoride 94. strontium oxide
- strontium sulfide 95. cadmium sulfide
- cadmium oxide 96. gold(III) nitride
- tin(IV) chloride
- calcium oxide
- zinc sulfide
- bismuth(III) nitride
CHEMICAL CROSSWORD No. 2
ELEMENTS AND BINARY COMPOUNDS OF METALS WITH NON-
METALS
1 2 3 4 5 6
7 8 9 10
11 12 13 14
15 16 17
18 19 20 21 22
23 24 25 26
27 28 29 30
31 32 33 34
35 36 37 38
39 40 41 42
43 44 45 46
47 48 49
50 51 52 53
54 55 56
57 58 59 60
61 62 63
64 65 66 67
68 69
70 71 72 73
74 75 76
77 78 79 80
81 82 83 84
85 86 87 88
89 90 91
92 93 94 95 96
97 98 99 100
- (i) silver(I) chloride (ii) magnesium nitride
(iii) calcium bromide (iv) aluminium oxide
(v) copper(II) chloride (vi) lead(II) oxide
(vi) manganese(II) sulfide (viii) zinc iodide
(ix) potassium chloride (x) calcium phosphide
(xi) chromium(III) chloride (xii) barium selenide
(xiii) cobalt(II) chloride (xiv) iron(III) oxide
(xv) iron(II) chloride (xvi) strontium iodide
(xvii) tin(II) bromide (xviii) magnesium oxide
(xix) rubidium nitride (xx) lithium fluoride
(xxi) platinum(II) bromide (xxii) bismuth(III) oxide
(xxiii) gold(III) chloride
- (i) CdF 2 (ii) SrCl 2
(iii) CoS (iv) PbI 2
(v) SnO (vi) Fe O 2 3
(vii) CrN (viii) CaBr 2
(ix) K O 2 (x) Na P 3
CHEMICAL CROSSWORD No. 2
ELEMENTS AND BINARY COMPOUNDS OF METALS WITH NON-
METALS
1
Ni
2 Fe Br 3
3 Li
4 Te
5 Ge
6 Sn 7 K Cl
8 Na 2 S
9 Si
10 Ca O 11
Cr I 3
12 K 2 O
13 Tl
14 Sr F 2 15 Rb 2 Se
16 In
17 Ba Cl 2 18
Pt
19 Cs 2 S
20 Ga
21 Cu Br 2
22 Al 23 Ag 2 O
24 Rn
25 Ni I 2
26 Cr N 27
Cu 2 Se
28 Xe
29 Fe F 2
30 Sr
S
31 Kr
32 Zn Cl 2
33 Ba 3
34 N 2 35
As
36 Ar
37 Cd Br 2
38 Mg 3 N 2 39 Ne
40 Pb I 2
41 Ni 3 P 2
42 Hg 43
He
44 Sn F 2
45 Fe 3 P 2
46 Zn Se 47 Hg Cl 2
48 Zn 3 N 2
49 Pt O 50
Mn I 2
51 Cu 3 N 2
52 Ca S
53 Al
2
54 Pb 3 P 2
55 Au
56 Fe P 57
Sb
58 Sn 3 P 2
59 Bi 2
60 O 3 61 Hg 3 N 2
62 Cr 2 O 3
63 Li 64
Bi
65 N 2
66 Al 2 S 3
67 Cu 3 P
2
68 Fe 2 Se 3
69 Ag 3 P 70
Au N
71 O 3
72 Cs 3 N
73 Mn 74 Bi Br 3
75 Rb 3 N
76 Fe S 77
Mn
78 Na I
79 K 3 P
80 Ni O 81 Al F 3
82 Na 3 P
83 Cu S
84 Hg 85
Li Cl
86 Li 3 N
87 Mg O
88 Pb O
3
89 Pb N
90 Ba Se
91 Sn S 92
Pb
93 Rb F
94 Sr S
95 Cd O
96 Au 97 Sn Cl 4
98 Ca O
99 Zn S
100 Bi N
