Predicting Formulas of Ionic Compounds

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1 Predicting Formulas of Ionic Compounds CK12 Editor Say Thanks to the Authors Click (No sign in required)

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3 Concept 1. Predicting Formulas of Ionic Compounds CONCEPT 1 Predicting Formulas of Ionic Compounds Lesson Objectives The student will: write correct formulas for binary ionic compounds, compounds containing metals with variable oxidation numbers, and compounds containing polyatomic ions, given the elements to be combined. Introduction Ionic compounds do not exist as molecules. In the solid state, ionic compounds are in crystal lattices containing many ions each of the cation and anion. An ionic formula, like NaCl, is an empirical formula. This formula merely indicates that sodium chloride is made of an equal number of sodium and chloride ions. The three-dimensional crystal lattice structure of sodium chloride. (Source: odium-chloride-3d-ionic.png. Public Domain) Sodium sulfide, another ionic compound, has the formula Na 2 S. This formula indicates that this compound is made up of twice as many sodium ions as sulfide ions. Na 2 S will also form a crystal lattice but it won t be the same as the NaCl lattice because the organization has to have two sodium ions per sulfide ion instead of one to one. 1

4 Determining Ionic Charge The charge that will be on an ion can be predicted for most of the monatomic ions. Many of these ionic charges can be predicted for entire families of elements. There are a few ions whose charge must simply be memorized and there are also a few that have the ability to form two or more ions with different charges so the exact charge on the ion can only be determined by analyzing the ionic formula of the compound. All of the elements in family 1A have the same outer energy level electron configuration, the same number of valence electrons (one), and they are all metals with low first ionization energies. Therefore, these atoms will lose their one valence electron and form ions with a 1+ charge. This allows us to predict the ionic charges on all the 1A ions; Li +, Na +, K +, Rb +, Cs +, Fr +. Hydrogen is a special case that is usually considered by itself rather than as a member of a family. Hydrogen has the ability to form a positive ion by losing its single valence electron just as the 1A metals do. In those cases, hydrogen forms the 1+ ion, H +. In rare cases, hydrogen can take on one electron to complete its outer energy level. The compounds where this occurs are called hydrides, NaH, KH, LiH. Hydrogen also has the ability to form compounds without losing or gaining electrons and that process will be discussed in the next chapter. All of the elements in family 2A have the same outer energy level electron configuration and the same number of valence electrons (two). Be :1s 2 2s 2 Mg :1s 2 2s 2 2p 6 3s 2 Ca :1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 Sr :1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 Ba : [Xe] 6s 2 Ra : [Rn] 7s 2 Each of these atoms is a metal with a low first and second ionization energies and will therefore lose both of its valence electrons forming an ion with a 2+ charge. The ions formed in family 2A are Be 2+, Mg 2+, Ca 2+, Sr 2+, Ba 2+, and Ra 2+. There is a slight variation in family 3A. The line separating metals from non-metals on the periodic table cuts through family 3A between boron and aluminum. In family 3A, boron, 1s 2 2s 2 2p 1, behaves as a non-metal due to its higher ionization energy and higher electron affinity. In the periodic table, it is on the non-metal side of the metal/non-metal dividing line. The red line divides metals from non-metals on the periodic table. (Source: Richard Parsons. CC-BY-SA) Aluminum, on the other hand, is on the metallic side of the line and behaves as an electron donor. Aluminum, 1s 2 2s 2 2p 6 3s 2 3p 1, always loses all three of its valence electrons and forms an Al 3+ ion. Gallium and indium have the same outer energy level configuration as aluminum except they are in the next higher periods (4 and 5) and 2

5 Concept 1. Predicting Formulas of Ionic Compounds they also lose all three of their valence electrons and form the 3+ ions, Ga 3+ and In 3+. Thallium, whose electron configuration ends with 6s 2 6p 1 should also form a 3+ ion, but for reasons not yet determined, is more stable as a 1+ ion, Tl +. The electron configurations for the members of family 6A are as follows: O :1s 2 2s 2 2p 4 S :1s 2 2s 2 2p 6 3s 2 3p 4 Se :1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 4 Te :1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 4 Each of these atoms has six valence electrons and they all have high electron affinities. These atoms will, therefore, take on additional electrons to complete the octet of electrons in their outer energy levels. Since each atom will take on two electrons to complete their octet, they will form ions. The ions formed will be O, S, Se, and Te. Family 7A elements have the outer energy level electron configuration ns 2 np 5. These atoms have the highest electron affinities in their periods and will each take on one more electron to complete the octet of electrons in their outer energy levels. Therefore, these atoms will form 1 ions; F, Cl, Br, I, and At. Family 8A elements have exactly filled outer energy levels, have very high ionization energy, and very low electron affinities. Therefore, these atoms do not give up electrons and do not take on electrons. The atoms of family 8A do not form ions. Transition Elements with Fixed Oxidation Numbers Many of the transition elements have variable oxidation states so they can form ions with different charges. We will consider some of those later in this chapter. There are also some transition elements that only form one ion and you will need to know a few of those. The electron configuration of silver, Ag, is 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 1 4d 10. You should note that this electron configuration does not follow the regular rules for the filling of orbitals. A lower energy level, the 5s orbital, is not full and yet electrons are placed in the 4d orbital. It happens that the electron configuration is slightly lower in energy when the d orbitals are exactly half full (5 electrons) or completely full (10 electrons). This also happens with chromium, copper, molybdenum, and gold. There are a few anomalies like this in the electron configurations of the transition elements and chemists do not agree as to why they occur. Whatever the reason, it gives silver an outer energy level that contains only one electron, and therefore, the silver ion is Ag +. The electron configuration for Zn is [Ar]4s 2 3d 10. Like most metals, zinc loses all of its valence electrons when it forms an ion and therefore, Zn forms a Zn 2+ ion. Cadmium is similar. The electron configuration for Cd is [Kr]5s 2 3d 10 and so it will form a Cd 2+ ion. Writing Basic Ionic Formulas In writing formulas for binary ionic compounds (binary refers to two elements, NOT two single atoms), the more electropositive element is always written first. [Even though the rule above states that the more electropositive element is always written first, there are a couple of compounds that have been commonly known and used for several decades before the rules were established. These compounds continue to be known by their common name and have their formulas written as they were for many 3

6 years. Ammonia, NH 3, for example, continues to be written as NH 3 rather than H 3 N, even though hydrogen is the more electropositive of the two elements.] Chemists use subscripts following the symbol of each element to indicate the number of atoms of that element present in the formula. For example, the formula Na 2 O indicates that the formula for the compound that forms between sodium and oxygen contains two atoms of sodium for each one of oxygen. When the subscript for an element is 1, the subscript is omitted. The number of atoms of an element with no indicated subscript is always read as 1. When an ionic compound forms, the number of electrons given off by the cations must be exactly the same as the number of electrons taken on by the anions. Therefore, if calcium, which gives off two electrons, is to combine with fluorine, which takes on one electron, then one calcium atom must combine with two fluorine atoms. The formula would be CaF 2. Suppose we wish to write the formula for the compound that forms between aluminum and chlorine. To write the formula, we must first determine the oxidation numbers of the ions that would be formed Al Then, we determine the simplest whole numbers with which to multiply these charges so they will balance (add to zero). In this case, we would multiply the 3+ by 1 and the 1 by 3. Cl 3+ 1 Al (3+)(1)=3+ (1 )(3)=3 You should note that we could multiply the 3+ by 2 and the 1 by 6 and get 6+ and 6 which would also balance, but this is NOT acceptable because we are writing empirical formulas and empirical formulas must have the lowest whole number multipliers. Once we have the lowest whole number multipliers, those multipliers become the subscripts for the symbols. Cl AlCl 3 Here s the process for writing the formula for the compound formed between aluminum and sulfur. 3+ Al (3+)(2)=6+ ()(3)=6 Therefore, the formula for this compound would be Al 2 S 3. Another method used to write formulas is called the criss-cross method. It is a quick method but it often produces errors if the user doesn t pay attention to the results. Here s the criss-cross method for writing the formula for the compound formed between aluminum and oxygen. In the criss-cross method, the oxidation numbers are placed over the symbols for the elements just as before. S 3+ Al O 4

7 Concept 1. Predicting Formulas of Ionic Compounds But in this method, the oxidation numbers are then criss-crossed and used as the subscripts for the other atom (ignoring sign). This produces the formula Al 2 O 3 for the compound and this is the correct formula. The errors that occur with this method would occur when you are writing the formula for a compound in which the two atoms have oxidation numbers of 2+ and. Obviously, the two subscripts should both be 1, but the criss-cross method would produce subscripts of 2 for each atom. Here s another example of a criss-cross error. If you used the original method of finding the lowest multipliers to balance the charges, you would get the correct formula PbO 2, but the criss-cross method produces an incorrect formula Pb 2 O 4 (both subscripts could be divided by 2 to produce the lowest whole number ratio). If you are alert to the final result and check you make sure you do not have double or triple the correct value, then the criss-cross method works. Otherwise, stick to the original method of writing formulas. Metals with Variable Oxidation Number There are a few metals that form ions with different charges. For some of them, the reason is apparent from their electron configuration but for others, the reason is not yet determined. It is necessary for you to memorize some of these metals and their different oxidation numbers. When writing formulas, you are given the oxidation number, so there is no problem if a metal forms more than one ion. When we get to naming ionic compounds, however, it is absolutely vital that you recognize metals that can have more than one oxidation number. A partial list of variable oxidation number metals includes iron, copper, tin, lead, nickel, and gold. We find, in the case of iron, that it can form both Fe 2+ and Fe 3+ ions. The electron configuration for neutral Fe is 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 6. It is fairly obvious why iron forms the 2+ ion, it simply loses all its valence electrons just as other metals do. The electron configuration for the Fe 2+ ion is [Ar]3d 6. What is it about this electron configuration that would allow the iron ion to lose one more electron? Just a few paragraphs back, we were talking about the electron configuration for silver and it was mentioned that the d orbitals have slightly lower energy when they are exactly half full or completely full. If this Fe 2+ ion were to lose one more electron, the 3d orbitals would be exactly half full with five electrons. When iron reacts, if there is normal pull on its electrons from the non-metal, it will form Fe 2+ and if the pull on its electrons is particularly strong, it will form Fe 3+. Copper can form ions with either a +1 or a +2 charge, Cu + and Cu 2+. The Cu 2+ is more common and more stable. Tin forms both Sn 2+ and Sn 4+. Lead also forms both Pb 2+ and Pb 4+. Nickel forms both Ni + and Ni 2+ and finally, gold forms both Au + and Au 3+ ions. There are more variable oxidation number metals and you will meet some of them but you should memorize iron, copper, tin, lead, nickel, and gold. 5

8 Polyatomic Ions Thus far, we have been dealing with ions made from single atoms. Such ions are called monatomic ions. There also exists a group of polyatomic ions, ions composed of a group of atoms that are covalently bonded and behave as if they were a single ion. Almost all the common polyatomic ions are negative ions. The only common positive polyatomic ion is ammonium, NH 4 +. The name and formula of ammonium ion is similar to ammonia, BUT it is not ammonia and you should not confuse the two. These polyatomic ions are another group of names and formulas that you need to memorize. Ammonium ion, NH 4 + Acetate ion, C 2 H 3 O 2 Carbonate ion, CO 3 Chromate ion, CrO 4 Dichromate ion, Cr 2 O 7 Hydroxide ion, OH Nitrate ion, NO 3 Phosphate ion, PO 4 3 Sulfate ion, SO 4 Sulfite ion, SO 3 You should know these well enough that when someone says the name, you can respond with the formula and charge, and if someone shows you the formula and charge, you can respond with the name. These polyatomic ions require additional consideration when you write formulas involving them. Suppose we are asked to write the formula for the compound that would form between calcium and the nitrate ion. We begin by putting the charges above the symbols just as before Ca NO 3 (2+)(1)=2+ (1 )(2)= The multipliers needed to balance these ions are 1 for calcium and 2 for nitrate. We wish to write a formula that tells our readers that there are two nitrate ions in the formula for every calcium ion. When we put the subscript 2 beside the nitrate ion in the same fashion as before, we get something strange - Ca(NO 3 ) 2. With this formula, we are indicating 32 oxygen atoms which is VERY wrong. Chemists decided that the solution to this problem is to put parentheses around the nitrate ion before the subscript is added. Therefore, the correct formula looks like this, Ca(NO 3 ) 2. Calcium phosphate would look like this Ca 3 (PO 4 ) 2. Please note that parentheses are only used when they are needed! If a polyatomic ion does not need a subscript other than an omitted 1, then parentheses are not needed. When students make an error with parentheses, it is almost never a case of leaving off parentheses when they are needed. The most common errors with parentheses is students put them around everything. The goal is to have every single chemist in the world write exactly the same formula for the same compound. If you put parentheses around something where they are not needed, the reader of your formula knows you made a mistake but he doesn t know whether you put on parentheses you didn t need or whether you left off a subscript. Practicing with Parentheses 1 Write the formula for the compound that will form from aluminum and acetale. 6

9 Concept 1. Predicting Formulas of Ionic Compounds The charge on an aluminum ion is 3+ and the charge on an acetate ion is 1. Therefore, three acetate icons are required to combine with one aluminum ion. This is also apparent by the criss-cross method. However, we cannot place a subscript of 3 beside the oxygen subscript of 2 without inserting parentheses first. Therefore, the formula will be Practicing with Parentheses 2 Al(C 2 H 3 O 2 ) 3 Write the formula for the compound that will form from ammonium and phosphate. The charge on an ammonium ion is 1+ and the charge on a phosphate ion is 3. Therefore, three ammonium ions are required to combine with one phosphate ion. The criss-cross procedure will place a subscript of 3 next to the subscript 4. This can only be carried out if the ammonium ion is first placed in parentheses. Therefore, the proper formula is Practicing with Parentheses 3 (NH 4 ) 3 PO 4 Write the formula for the compound that will form from aluminum and phophate. Al 3+ PO 4 3 Since the charge on an aluminum ion is 3+ and the charge on a phosphate ion is 3, these ions will combine in a one-to-one ratio. In this case, the criss-cross method would produce an incorrect answer. Since it is not necessary to write the subscripts of one, no parentheses are needed in this formula. Since parentheses are not needed, is actually INCORRECT to use them. AlPO 4 More Examples Magnesium and hydroxide...mg(oh) 2 Sodium carbonate...na 2 CO 3 Barium acetate...ba(c 2 H 3 O 2 ) 2 Hydrogen phosphate...h 3 PO 4 Ammonium dichromate...(nh 4 ) 2 Cr 2 O 7 Practice! 7

10 Lesson Summary The oxidation number for ions of the representative elements can usually be determined by the number of valence electrons. The number of valence electrons for the representative elements is equal to the number in the family name. Some transition elements have fixed oxidation numbers and some have variable oxidation numbers. Some metals, such as iron, copper, tin, lead, nickel, and gold, have variable oxidation numbers. There are groups of atoms formed into an ion and called polyatomic ions that behave as if they were a single ion. Formulas for ionic compounds contain the lowest whole number ratio of subscripts such that the sum of the subscript of the more electropositive element times its oxidation number plus the subscripts of the more electronegative element times its oxidation number equals zero. Review Questions 1. Fill in the chart by writing formulas for the compounds that might form between the ions in the columns and rows. Some of these compounds don t exist but you can still write formulas for them. TABLE 1.1: NO 3 SO 4 Cl S PO 4 3 OH Cr 2 O 7 CO 3 Na + Ca 2+ Fe 3+ NH 4 + Sn 4+ Further Reading / Supplemental Links Website with lessons, worksheets, and quizzes on various high school chemistry topics. Lesson 5-3 is on Writing Chemical Formulas. Vocabulary oxidation number The charge or apparent charge that an atom in a compound or ion would have if all the electrons in its bonds belonged entirely to the more electropositive atom. 8

11 Concept 1. Predicting Formulas of Ionic Compounds polyatomic ion An electrically charged species formed by covalent bonding of atoms of two or more different elements, usually non-metals. 9