4. FORMATION OF COMPOUNDS

4. FORMATION OF COMPOUNDS The noble (inert) gases 2 He, 10 Ne, 18 Ar, 36 Kr, 54 Xe are so called because they are chemically unreactive. In science, this is indicative of stability. The noble gases are stable so they do not need to react chemically to gain stability. We know that the reason these gases are stable is because of their electron count. You can think of the electron count of each of these elements as magic numbers: 2, 10, 18, 36, 54, which confers stability on the atoms. In nature, all systems tend to move to a state having the greatest stability. Compounds are formed because atoms want to attain a maximum stability namely attaining the electron count of the nearest noble gas. Atoms can do this in two ways by forming ionic bonds or by covalent bonds. 4.1 Ionic Bonds Recall that an atom consists, in part, of protons and neutrons densely packed into a nucleus. The electrons are moving rapidly outside of, and relatively far away from the nucleus. Thus for example, if the nucleus of a hydrogen atom (located in downtown Montreal) is blown up to the size of a tennis ball, the electron will be, on the average, in Beaconsfield. Because of these distances, the electron can forget atom to which it belongs and attach itself to other atoms. The electrons of the atom can become lost. Thus, atoms are capable of losing and gaining electrons. If an atom gains an electron (which, of course, carries a negative charge), it then acquires an overall negative charge. Similarly, if an atom loses an electron, it acquires a positive charge. Substances may lose or gain more than one electron. Atoms gain charges by gaining or losing electrons. They do not change their charge by gaining or losing protons. e.g. Na Na + + e Cl + e Cl Zn Zn 2+ + 2 e Fe Fe 3+ + 3 e Fe 2+ + 2e Fe Cr 6+ + 3 e Cr 3+ O + 2 e O 2 e.g. What are the charges of the substance when the specified numbers of electrons are added or removed? 1. One electron is added to a bromine atom. 2. Three electrons are added to a nitrogen atom. 3. One electron is removed from a hydrogen atom. Chemical substances which have charges are called ions. Anions are chemical substances which have negative charges. e.g., N 3, O 2, I. Cations are chemical substances which have positive charges. e.g., H +, Na +, Ca 2+ Typically: Metals will lose electrons thereby forming cations. Nonmetals tend to gain electrons thereby forming anions. Again note that ions are NEVER formed by the addition or removal of protons. page 30

The Periodic Table is useful for determining the type of ion that will form. He Li + Be 2+ 2 Be 2 + + Be 2 + Be 2 + Be O 2 2 + Be F 2 + Ne Na + Mg 2+ Al 3+ S 2 Cl Ar K + Ca 2+ Ga 3+ Se 2 Br Kr Rb + Sr 2+ Transition metals will form cations having various charges In 3+ Te 2 I Xe Cs + Ba 2+ Rn The reason that metals tend to lose electrons is that they are to the left of the Periodic Table it is simpler for them to lose electrons to acquire the electron count to the noble gas of the preceding row. Thus, for example, it is easier for 3 Li to lose one electron to attain the electron count of He (2) than to gain 7 e to acquire the electron count of Ne (10). Likewise a nonmetal such as 9 F will prefer to gain one e to attain the Ne electron count (10) than to lose 7 e to get to He (2). Thus: To be Memorised Group IA elements: Li, Na, K, Rb, Cs will always form cations having (+1) charge. Group IIA elements: Be, Mg, Ca, Sr, Ba will always form cations having (+2) charge. Group IIIA elements: Al, Ga, In will always form cations having (+3) charge. Generally (but not always): Group VIA elements: O, S, Se, Te form anions having ( 2) charge. Group VIIA elements: F, Cl, Br, I form anions having ( 1) charge. In general, substances will not give up electrons unless there are other atoms ready to receive those electrons. Likewise, a substance cannot accept electrons without other substances ready to supply those electrons. Thus, for example, Na Na + + e Cl + e Cl Na + ------Cl The attraction of positive to negative results in the formation of an ionic bond. The resulting neutrally charged compound is known as an ionic compound. Ionic compounds are crystalline. Ionic compounds can be recognised because only they have metal atoms. page 31

4.2 Covalent Bonds We have already said that nonmetals tend to gain e to attain the electron count of the nearest noble gas (to the right). This is fine when there is a nearby atom willing to supply e s. If, however, no such supplier exists in the container, another method of acquiring e s must be found. Consider the case of a sample consisting only of 9 F atoms. 9 F requires one e to acquire the electron count of the nearest noble gas, 10 Ne. One can think of F as having an electron hole that needs to be filled. F..F Each fluorine atom seeks to seize one electron from the other fluorine atom while retaining all of its own 9 electrons. The result is that the two atoms are drawn together with the two electrons, one from each fluorine atom, located between the two nuclei. F x F This type of bond is known as a covalent bond. In covalent bonds, the electrons are shared between two atoms. Covalent bonds are formed between nonmetal atoms. page 32

4.3 Nomenclature Nomenclature is a system of naming compounds so that: given the name of the compound, you can write the formula of the compound uniquely; or given the formula of the compound, you can write the name uniquely. Before giving the system for naming compounds, it is necessary to know something about the nature of the compound. 4.3.1 BINARY IONIC COMPOUNDS The term binary compound means that the compound consists only of two different types of atoms. Exercise: State whether the following compounds are binary. 4. CO 5. CO 2 6. C 2 H 2 Cl 2 7. CH 3 CH 3 8. HCN Binary ionic compounds are formed from single atom (metal) cation and single atom (non-metal) anions. Binary ionic compound = metal + non-metal Whenever one of the atoms is a metal the compound is an ionic compound. e.g. Which of the following compounds are binary ionic compounds? CO 2 CaO CaCO 3 FeCl 3 Review: You must be able to recognize a metal from its location in the Periodic Table NaNO 2 CCl 4 4.3.1.1 Simple Binary Ionic Compounds e.g. Name each of the following compounds. metal + non-metal + ide The suffix ide is very important. It is the signature of binary compounds. Thus there are two things to look out for: 1. Is it binary? look for the suffix ide. 2. Is it ionic? look to see if there is a metal. page 33

Exercise: NaBr Determine whether each of the following is a binary ionic or binary covalent compound. CaCl 2 BaO Li 3 N NaNO 3 4.3.1.2 Chemical Formulas from Names In the previous section, we have been given the formula and from the formula we provided the name. A functional system for naming compounds requires that the reverse also be true. That is, given the name of the compound, we must be able to unambiguously write the chemical formula. Name Formula Rule: To put together the formula of a binary ionic compound you must: 1. Identify the charge of the cation and the anion. 2. Multiply each cation and each anion by small integers so that the total charge supplied by all the cations equals the total charge supplied by all the anions. 3. Write the formula of the compound with the cation + number of cations (subscripted) + anion+ number of anions (subscripted) e.g. Write the formula for each of the following compounds. 1. Potassium iodide a. Identify the type of compound This is ionic potassium is a metal. It is binary it has the suffix ide. b. Write the charges of each ion: K belongs to Group IA so it has a charge of +1: K + I belongs to Group VIIA so it has a charge of 1: I c. Write an equation showing the gain or loss of electrons by each type of ion: K K + + e I + e I d. Multiply each half equation by small integers so that the number of electrons produced in the 1st equation equals the number of electrons absorbed in the 2nd equation This one is easy. One electron is produced to form K + and one electron is absorbed to form I. e. Put together the formula KI 2. Calcium chloride When we put together a formula from the name, the final formula must have an overall charge of zero (0). a. Identify the type of compound This is ionic calcium is a metal. It is binary it has a the suffix ide. b. Write the charges of each ion: Ca belongs to Group IIA so it has a charge of +2: Ca 2+ Cl belongs to Group VIIA so it has a charge of 1: Cl c. Write an equation showing the gain or loss of electrons by each type of ion: page 34

Ca Ca 2+ + 2 e Cl + e Cl d. Multiply each half equation by small integers so that the number of electrons produced in the 1st equation equals the number of electrons absorbed in the 2nd equation Each Cl absorbs only one e. Thus we need 2 Cl s to absorb both electrons produced by the Ca. Ca Ca 2+ + 2 e 2 x [Cl + e Cl ] e. Put together the formula CaCl 2 Exercise: Write the formula for each of the following compounds. magnesium chloride barium sulphide aluminium selenide rubidium oxide strontium fluoride calcium bromide indium oxide page 35

4.3.2 BINARY IONIC COMPOUNDS: CATIONS WITH VARIOUS CHARGES Most metal atoms can show several charges in different compounds. Thus, for example, iron is capable of forming two types of compounds when combined with oxygen having completely different physical and chemical properties. FeO Fe 2 O 3 grey colour reddish brown (rust) According to the rules learned in the previous section, the name of both of these compounds would be: iron oxide (the compound is binary). This is not adequate however, because the name would not give us a unique chemical formula. The nomenclature rule for these types of compounds is as follows: Thus in the examples cited above: FeO would be named iron (II) oxide. Fe 2 O 3 would be named iron (III) oxide. metal (charge of the metal in roman numerals) anion-ide The difficulty with this system is that you have to either know or calculate the charges. The charges carried by atoms in a compound are known as oxidation numbers (or oxidation states). 4.3.2.1 Rules for Assigning Oxidation States to Atoms These are to be memorised in the sequence provided. 1. The sum of the oxidation numbers of all atoms in a chemical substance must add up to the overall charge of the substance. 2. Group IA elements: Li, Na, K, Rb, Cs will always form cations having (+1) charge. 3. Group IIA elements: Be, Mg, Ca, Sr, Ba will always form cations having (+2) charge. 4. Group IIIA elements: Al, Ga, In will always form cations having (+3) charge. 5. Ag always has (+1) Zn always has (+2) Al 3+ Cd always has (+2) Zn 2+ Ga 3+ Ag + Cd 2+ In 3+ 6. H will usually have (+1) charge. The exception is when H is bonded to a metal; in this case it will have an oxidation number of ( 1). 7. Usually, but not always the Group A nonmetals will have a charge associated with the group number. Thus: Group VA elements (Nitrogen group) will have ( 3). Group VIA elements (Oxygen group) will have ( 2). Group VIIA elements (Halogens) will have ( 1). Care should be utilised here. There are many exceptions. 8. The oxidation number of all other atoms must be calculated. ************ Once the oxidation states of all atoms in the compound have been calculated, you can use them to name the compound. page 36

e.g., Write the name of each of the following compounds. 1. Cr 2 O 3 a. Determine the oxidation state of each element in the compound Use Rule 1 to set up an equation 2(Cr) + 3 (O) = 0 This is an equation with two unknowns. You must choose a value of oxidation number for one of the atoms. The priority is given to Group A atoms. Choose oxygen according to its group number. (O) = 2 Substitute this into the equation above. 2(Cr) + 3 ( 2) = 0 2(Cr) = +6 (Cr) = +3 b. Write the name of the compound. It is a binary, ionic compound. chromium (III) oxide Cr has a charge of +3 2. CoBr 6 a. Determine the oxidation state of each element in the compound Use Rule 1 to set up an equation 2(Co) + 6 (Br) = 0 This is an equation with two unknowns. You must choose a value of oxidation number for one of the atoms. The priority is given to Group A atoms. Choose bromine according to its group number. (Br) = 1 Substitute this into the equation above. (Co) + 6 ( 1) = 0 (Co) = +6 b. Write the name of the compound. It is a binary, ionic compound. cobalt (VI) bromide Co has a charge of +6 3. SnS 2 a. Determine the oxidation state of each element in the compound Use Rule 1 to set up an equation (Sn) + 2 (S) = 0 This is an equation with two unknowns. You must choose a value of oxidation number for one of the atoms. The priority is given to Group A atoms. Choose sulphur according to its group number. (S) = 2 Substitute this into the equation above. (Sn) + 2 ( 2) = 0 (Sn) = +4 b. Write the name of the compound. It is a binary, ionic compound. tin (IV) sulphide Sn has a charge of +4 page 37

e.g., Write the formula for each of the following compounds 1. Iron (III) chloride a. Identify the charges of the ions (Fe) = +3 (Cl) = 1 b. Combine the two types of atoms so there is a net charge of zero Fe Fe 3+ + 3 e 3 x [ Cl + e Cl ] c. Write the formula: FeCl 3 2. Manganese (V) oxide a. Identify the charges of the ions (Mn) = +5 (O) = 2 b. Combine the two types of atoms so there is a net charge of zero 2 x [ Mn Mn 5+ + 5 e ] 5 x [ O + 2 e O 2 ] c. Write the formula: Mn 2 O 5 3. Mercury (II) nitride a. Identify the charges of the ions (Hg) = +2 (N) = 3 b. Combine the two types of atoms so there is a net charge of zero 3 x [ Hg Hg 2+ + 2 e ] 2 x [ N + 3 e N 3 ] c. Write the formula: Hg 3 N 2 *********************************** General Rule for the Nomenclature of all Binary Ionic Compounds FCharge ofi Metal G J anionide metal H K Do not put this in if the metal has only one oxidation state. See Rules 2, 3, 4 and 5. e.g., Write the names of each of the following compounds K 2 S W 2 O 5 CdO ZnCl 2 AgCl page 38

4.3.3 BINARY COVALENT COMPOUNDS These can be recognised because they contain no metal atoms. For our purposes, only nonmetals form covalent bonds. Binary covalent compounds: nonmetal + nonmetal Binary covalent compounds are named quite differently from binary ionic compounds. In the case of binary ionic compounds, once the oxidation numbers of all species are known, there is only one way to put the compound together. In the case of binary covalent compounds, the situation is completely different. The same pair of atoms may form several covalent compounds. The case of the reactions of nitrogen (N 2 ) with oxygen (O 2 ) demonstrates this. Possible combinations include: NO N 2 O NO 2 N 2 O 3 N 2 O 4 N 2 O 5 Since the products are not predictable when nonmetals react, we must use a different system of nomenclature. We use numerical prefixes to state explicitly the number of each type of atom present in the molecule. Memorise the following list of numerical prefixes 1 mono- 2 di- 3 tri- 4 tetra- 5 penta- 6 hexa- 7 hepta- 8 octa- 9 nona- 10 deca- Nomenclature: (prefix) nonmetal (prefix) nonmetal -ide Number of first nonmetal atom (omit if it is mono) First nonmetal is named. It is the one closest to the metal side Signature of binary compounds page 39

1. NO a. Is it ionic or covalent? Is it binary? (it is covalent and binary) b. Number of N-atoms = 1 prefix = mono Number of O-atoms = 1 prefix = mono c. Name the compound mononitrogen monoxide nitrogen monoxide 2. N 2 O a. Is it ionic or covalent? Is it binary? (it is covalent and binary) b. Number of N-atoms = 2 prefix = di Number of O-atoms = 1 prefix = mono c. Name the compound dinitrogen monoxide 3. NO 2 a. Is it ionic or covalent? Is it binary? (it is covalent and binary) b. Number of N-atoms = 1 prefix = mono Number of O-atoms = 2 prefix = di c. Name the compound mononitrogen dioxide nitrogen dioxide 4. N 2 O 3 a. Is it ionic or covalent? Is it binary? (it is covalent and binary) b. Number of N-atoms = 2 prefix = di Number of O-atoms = 3 prefix = tri c. Name the compound dinitrogen trioxide 5. N 2 O 4 a. Is it ionic or covalent? Is it binary? (it is covalent and binary) b. Number of N-atoms = 2 prefix = di Number of O-atoms = 4 prefix = tetra c. Name the compound dinitrogen tetroxide 6. N 2 O 5 a. Is it ionic or covalent? Is it binary? (it is covalent and binary) b. Number of N-atoms = 2 prefix = di Number of O-atoms = 5 prefix = penta c. Name the compound dinitrogen pentoxide 7. P 4 O 6 a. Is it ionic or covalent? Is it binary? (it is covalent and binary) b. Number of P-atoms = 4 prefix = tetra Number of O-atoms = 6 prefix = hexa c. Name the compound tetraphosphorous hexoxide page 40

POLYATOMIC IONS NH 4 + Ammonium CHO 2 3 2 Acetate Hg 2 2+ Mercury (I) 2 CO 2 4 Oxalate MnO 4 Permanganate OH Hydroxide 2 Cr 2 O 7 Dichromate CN Cyanide CrO 4 2 Chromate SCN Thiocyanate SO 4 2 Sulfate ClO 4 Perchlorate SO 3 2 Sulfite ClO 3 Chlorate PO 4 3 Phosphate ClO 2 Chlorite PO 3 3 Phosphite ClO Hypochlorite NO 3 Nitrate CO 3 2 Carbonate NO 2 Nitrite 4.3.4 Naming Acids Acids (Arhennius definition) substances which, when dissolved in water, produce H + ions. By implication then, acids are ionic compounds in which the cations are all protons, H +. Nomenclature of Acids is based on the nomenclature of the anion as there is no need to reference the cation which is always H +. Anion Corresponding Acid Anion Example Corresponding Acid ide hydroic acid Cl, chloride HCl, hydrochloric acid ate ic acid ClO, chlorate HClO, chloric acid 3 3 ite ous acid ClO, hypochlorite HClO, hypochlorous acid Examples: HI hydroiodic acid HCN hydrocyanic acid HIO 4 periodic acid HBrO 2 bromous acid page 41

4.3.5 Acid anions For anions having multiple charges, there is no reason for the cations to be identical. It is perfectly legitimate to have an ionic compound with a formula that combines several different cations. For example, the combination of lithium and ammonium cations with phosphate can have: (NH 4 )Li 2 PO 4 Ammonium dilithium phosphate (NH 4 ) 2 LiPO 4 diammonium lithium phosphate Note that the sequence in which the cations are named is alphabetical. As you would expect, ionic compounds having one or more H + cations, but not all, are special cases. Thus for example, compounds such as: Na 2 HPO 4 and NaH 2 PO 4 Ba(HCO 2 ) 2 are possible. The nomenclature is based on the anion root. Cation (Cation Charge) + (prefix for the number of hydrogen omit mono) hydrogen + root anion. Thus Na 2 HPO 4 sodium hydrohen phosphate NaH 2 PO 4 sodium dihydrogen phosphate Ba(HCO 2 ) 2 barium hydrogen carbonite Note that although the name is based on the anion root, the anion in these hydrogenated anions includes the hydrogens. Thus the following substances decompose as follow: Na HPO 2Na + HPO Anion = HPO not PO NaH PO Na + H PO Anion = H PO not PO + 2 2 3 2 4 4 4 4 + 3 2 4 2 4 2 4 4 page 42

Exercises on Nomenclature For each substance whose name is given, write the formula, if the formula is given, provide the name. Use the Periodic Table as your main reference. If you encounter ions with which you are unfamiliar, you may look these up in your textbook. This assignment will count as part of the quiz grade. It is due on Thursday, March 18, 2004. 1. carbon dioxide 3. calcium hypochlorite 5. barium hydroxide 7. cobalt (II) acetate 9. calcium fluoride 11. mercury (II) chlorate 13. Ammonia 15. manganese (II) oxalate 17. sodium carbonate 19. mercury (I) chloride 21. aluminium oxide 23. gallium selenate 25. iron (III) sulphide 27. iron (II) sulphate heptahydrate 29. barium bromate 31. lead (II) iodide 33. carbonic acid 35. sodium sulphite 37. periodic acid 39. magnesium hydrogen sulphate 2. BaCl 2 4. KMnO 4 6. FePO 4 8. SnS 10. CBr 4 12. Al 2 (Cr 2 O 7 ) 3 14. Al 2 (SO 4 ) 3 16. CdCl 2 18. KH 2 PO 4 20. Al(ClO 2 ) 3 22. Zn(NO 3 ) 2 24. CaCl. 2 6H 2 O 26. NH 4 NO 2 28. H 3 PO 3 (aq) 30. H 2 O 2 32. Ca(OH) 2 34. K 2 SO 3 36. Hg(BrO 3 ) 2 38. HCN (aq) 40. Cr 2 O 3 page 43