Ionic Compound Formulas Init 10/28/2010 by Daniel R. Barnes.

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Presentation transcript:

Ionic Compound Formulas Init 10/28/2010 by Daniel R. Barnes

. . . write the correct formula SWBAT . . . . . . write the correct formula for an ionic compound if given its name.

Cl- Cl- Cl- Cl- Cl- Na+ Na+ Na+ Na+ Na+ Na+ Cl- Cl- Cl- Cl- Cl- Cl- Na+ Na+ Na+ Na+ Na+ Cl- Cl- Cl- Cl- Cl- Na+ Na+ Na+ Na+ Na+ Na+ Cl- Cl- Cl- Cl- Cl- Cl- Na+ Na+ Na+ Na+ Na+ Cl- Cl- Cl- Cl- Cl- Na+ Na+ Na+ Na+ Na+ Na+ Cl- Cl- Cl- Cl- Cl- Cl- Na+ Na+ Na+ Na+ Na+ Cl- Cl- Cl- Cl- Cl- Na+ Na+ Na+ Na+ Na+ Na+ Cl- Cl- Cl- Cl- Cl- Cl- Na+ Na+ Na+ Na+ Na+ Cl- Cl- Cl- Cl- Cl- Na+ Na+ Na+ Na+ Na+ Na+

+ IONIC COMPOUND RE-CAP F Cs Ionic compounds typically consist of a metal and a nonmetal. Nonmetals generally have much higher electronegativities than metals, so the nonmetal will steal electrons from the metal. F + Cs

+ IONIC COMPOUND RE-CAP Ionic compounds typically consist of a metal and a nonmetal. Nonmetals generally have much higher electronegativities than metals, so the nonmetal will steal electrons from the metal. As a result, the nonmetal atom becomes a negatively-charged ion, and the metal atom becomes a positively-charged ion. Because they are now oppositely-charged, the negative nonmetal ion and the positive metal ion are attracted to each other. + This electrostatic attraction constitutes an ionic bond between the metal ion and nonmetal ion. Because of their attraction for each other, the positive metal ions and the negative nonmetal ions will form an orderly crystal structure.

IONIC COMPOUND RE-CAP Although ionic compounds, like salt, are made of positive ions and negative ions, the amount of positive charge always equals the amount of negative charge. Even though it’s made of plusses and minuses, an ionic compound, overall, is neutral. Each ion in an ionic compound will have an electric charge that depends upon its location on the periodic table.

Please note the numbers at the tops of the columns. The ones that matter are the ones with the letter “A” after them These numbers tell you . . . . . . how many valence electrons each element has. This number is super-important, because it tells you how many electrons the atom is going to lose or gain. Remember that atoms like to either completely empty or completely fill their outer valence shells. How many electrons an atom loses or gains determines the atom’s electric charge when it becomes an ion.

So let's DO this!

BINARY IONIC COMPOUND EXAMPLES 1. What is the chemical formula of sodium chloride? Sodium = Na = from group 1A  loses 1 e-  Na+ Step one: determine the symbol and charge of each ion. Sodium has one valence electron, so it has two options. It can lose the electron to empty its valence shell, or it can gain seven more electrons to fill its valence shell. Since losing one electron involves fewer electrons than gaining seven, sodium loses one electron. Remember that because electrons are negative, losing electrons makes an atom positive.

BINARY IONIC COMPOUND EXAMPLES 1. What is the chemical formula of sodium chloride? Sodium = Na = from group 1A  loses 1 e-  Na+ Chlorine = Cl = from group 7A  gains 1 e-  Cl- Chlorine has seven valence electrons, so it has two options. It can lose all seven valence electrons to empty its valence shell, or it can gain one extra electron to fill its valence shell. Since gaining one electron involves fewer electrons than losing seven, chlorine gains one electron. Remember that because electrons are negative, gaining electrons makes an atom negative. Now we know the symbols and the charges of the two ions in sodium chloride.

BINARY IONIC COMPOUND EXAMPLES 1. What is the chemical formula of sodium chloride? Sodium = Na = from group 1A  loses 1 e-  Na+ Chlorine = Cl = from group 7A  gains 1 e-  Cl- Na Cl Now, we need to figure out how many of each ion will make a single formula unit of the compound. We need to make sure that the number of plusses equals the number of minuses. This one is easy because the sodium ion is plus one and the chlorine ion is minus one. The two ions are electrically equal but opposite, so one of each ion will be fine.

BINARY IONIC COMPOUND EXAMPLES 2. What is the chemical formula of beryllium chloride? Beryllium = Be = from group 2A  loses 2 e-  Be2+ Chlorine = Cl = from group 7A  gains 1 e-  Cl- THINK PAIR SHARE! 30 seconds . . . Be2+ Cl- BeCl2 Cl- S’s: +2 -2 Now, we need to figure out how many of each ion will make a single formula unit of the compound. Beryllium chloride is going to be a little more complicated than sodium chloride because there are two plusses but only one minus. The solution is pretty simple, though. Just add another chloride ion, and you’ve got a total of two minuses.

BINARY IONIC COMPOUND EXAMPLES 3. What is the chemical formula of potassium oxide? Potassium = K = from group 1A  loses 1 e-  K+ Oxygen = O = from group 6A  gains 2 e-  O2- THINK PAIR SHARE! 30 seconds . . . K+ O2- K2O K+ S’s: +2 -2 Potassium oxide has a problem similar to beryllium chloride, but this time, there’s more minus than plus. Remember that ionic compounds form when nonmetals steal electrons from metals. A neutral oxygen atom is hungry for two electrons, but a potassium atom is only going to give up one. It’s going to take two potassium ions to balance out against a single oxide ion.

BINARY IONIC COMPOUND EXAMPLES 4. What is the formula of aluminum oxide? Aluminum = Al = from group 3A  loses 3 e-  Al3+ Oxygen = O = from group 6A  gains 2 e-  O2- THINK PAIR SHARE! 30 seconds . . . Al3+ O2- Al3+ O2- With aluminum oxide, the plusses don’t equal the minuses either. This one’s going to be a bit trickier to balance. There are two ways to go about it. We’ll do the “one ion at a time” method first. At first, there are 3 plusses but only 2 minuses, so what we need is more minuses. We get this by adding another oxide ion. Unfortunately, there is now a total of 4 minuses, which is too much. We need another aluminum ion.

BINARY IONIC COMPOUND EXAMPLES 4. What is the formula of aluminum oxide? Aluminum = Al = from group 3A  loses 3 e-  Al3+ Oxygen = O = from group 6A  gains 2 e-  O2- Al3+ O2- Al2O3 Al3+ O2- O2- S’s: +6 -6 Oh boy. Now we have more plusses than minuses again. One more oxide ion should help. Holy cow! I think it’s balanced now! Yay for correct answers! Yay! (There is a quicker way than this, though, called the “criss-cross” method. We’ll show it to you a little later.)

BINARY IONIC COMPOUND EXAMPLES 5. What is the formula of iron (III) bromide? Iron = Fe  if (III)  loses 3 e-  Fe3+ Bromine = Br = from group 7A  gains 1 e-  Br - THINK PAIR SHARE! 30 seconds . . . Fe3+ Br - FeBr3 Br - Br - S’s: +3 -3 The “(III)” means that the iron is in the +3 “oxidation state”. For now, consider Roman numerals to indicate positive charge. Now, we have the usual problem of getting plusses equal to minuses. Keep adding ions until + = - .

BINARY IONIC COMPOUND EXAMPLES 6. What is the formula of iron (II) nitride? Iron = Fe  if (II)  loses 2 e-  Fe2+ Nitrogen = N = from group 5A  gains 3 e-  N3- THINK PAIR SHARE! 30 seconds . . . Fe2+ N3- Fe3N2 Fe2+ N3- Fe2+ S’s: +6 -6 Last time, it was iron (III), and now it’s iron (II). As it turns out, iron has more than one possible oxidation state. That’s true of a lot of elements. Once again, though, the Roman numeral tells you . . . . . . the positive charge of the metal ion.

The "Criss-Cross" Method PATTERN? DO SEE YOU Sodium chloride is made of the ions Na+ & Cl- The formula of sodium chloride is NaCl A Beryllium chloride is made of the ions Be2+ & Cl- The formula of beryllium chloride is BeCl2 Potassium oxide is made of the ions K+ & O2- The formula of potassium oxide is K2O Aluminum oxide is made of the ions Al3+ & O2- The formula of aluminum oxide is Al2O3 Iron (III) bromide is made of the ions Fe3+ & Br - The formula of iron (III) bromide is FeBr3 Iron (II) nitride is made of the ions Fe2+ & N3- The formula of iron (II) nitride is Fe3N2

The "Criss-Cross" Method Sodium chloride is made of the ions Na+ & Cl- The formula of sodium chloride is NaCl The “Criss-Cross” short cut usually works just fine. Beryllium chloride is made of the ions Be2+ & Cl- The formula of beryllium chloride is BeCl2 Potassium oxide is made of the ions K+ & O2- The formula of potassium oxide is K2O Usually . . . Aluminum oxide is made of the ions Al3+ & O2- The formula of aluminum oxide is Al2O3 Iron (III) bromide is made of the ions Fe3+ & Br - The formula of iron (III) bromide is FeBr3 Iron (II) nitride is made of the ions Fe2+ & N3- The formula of iron (II) nitride is Fe3N2

BINARY IONIC COMPOUND EXAMPLES 7. What is the formula of tin (IV) oxide? Tin = Sn  if (IV)  loses 4 e-  Sn4+ Oxygen = O = from group 6A  gains 2 e-  O2- THINK PAIR SHARE! 30 seconds . . . Sn4+ O2- SnO2 O2- S’s: +4 -4 The one-ion-at-a-time method gives us . . . Sn4+ O2- The criss-cross method gives us . . . Since you can divide both two and four by two, do so, and you get our original answer. If you do the criss-cross method and you get subscripts that all can be divided by the same number, you should divide them all by the same number. Sn2O4

2 1 = 4 2 In math, if you get 2/4 as your answer, you’re supposed to . . . . . . reduce the fraction to . . . ½. You do this by dividing the top and bottom of the fraction by the same number. (In this case, dividing the top and the bottom by two) Sn2O4 If, in chemistry, you get . . . SnO2 You should reduce it to . . . Remember this if you decide to use the “criss-cross” method instead of the “one ion at a time” method.

Rules for Determining an Ionic Compound’s Formula * If the cation is an element with a Roman # in parentheses, the Roman numeral indicates how positively-charged the cation is. * If the cation has no Roman #, use the octet rule to determine how many e- it should lose  size of positive charge. * If the anion is polyatomic, you’ll probably have to either memorize its formula or look it up in a chart. * If the anion is monatomic, use the octet rule to figure out how many e- it wants to gain  size of negative charge * Once you know the charge of the cation and the anion, use the criss-cross method to determine the subscripts for each ion. * If the subscripts in the formula can be divided by the same whole number, then divide them by that number.

Cl- Cl- Cl- Cl- Cl- Na+ Na+ Na+ Na+ Na+ Na+ Cl- Cl- Cl- Cl- Cl- Cl- Na+ Na+ Na+ Na+ Na+ Cl- Cl- Cl- Cl- Cl- Na+ Na+ Na+ Na+ Na+ Na+ Cl- Cl- Cl- Cl- Cl- Cl- Na+ Na+ Na+ Na+ Na+ A salt crystal may go on and on, but this presentation is over. Cl- Cl- Cl- Cl- Cl- Na+ Na+ Na+ Na+ Na+ Na+ Cl- Cl- Cl- Cl- Cl- Cl- Na+ Na+ Na+ Na+ Na+ Cl- Cl- Cl- Cl- Cl- Na+ Na+ Na+ Na+ Na+ Na+ Cl- Cl- Cl- Cl- Cl- Cl- Na+ Na+ Na+ Na+ Na+ Cl- Cl- Cl- Cl- Cl- Na+ Na+ Na+ Na+ Na+ Na+