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Read Sections 4.1, 4.2, 4.3, 4.4, and 4.5 before viewing the slide show.

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Presentation on theme: "Read Sections 4.1, 4.2, 4.3, 4.4, and 4.5 before viewing the slide show."— Presentation transcript:

1 Read Sections 4.1, 4.2, 4.3, 4.4, and 4.5 before viewing the slide show.

2 Unit 14 Binary Ionic Compounds Criteria for Stability of Ions (4.1) Lewis Symbols for Atoms (4.2) Octet Rule (4.4) Formula Writing for Binary Ionic Compounds (4.5) Nomenclature for Binary Ionic Compounds (4.5)

3 Criterion for Stability of Ions (4.1) Group 18, the noble gases, is a particularly unreactive group The noble gases each have eight valence electrons (except for helium with two) There must be a special stability associated with eight valence electrons. The driving force in electron exchange or electron sharing to form compounds is the quest for eight valence electrons Noble gases

4 The Octet Rule (4.4) The observations on the previous slide may be summarized as the Octet Rule: Atoms tend to gain, lose, or share electrons until they are surrounded by eight valence electrons (As you might expect, there will be exceptions to this. Notably hydrogen and helium can have two valence electrons at the most.)

5 Practical Aspects of the Octet Rule Ionic Compounds Ionic compounds are made by the transfer of electrons from one atom to another Ionic compounds are formed when a metal (left of the blue line) gives up electrons to a nonmetal (right of the blue line).

6 Practical Aspects of the Octet Rule Molecular (or Covalent) Compounds (4.6) Molecular (or Covalent) compounds are made by sharing of electrons between atoms Molecular compounds are formed when two or more nonmetal atoms share electrons (Terminology note: The terms molecular or covalent may be used. Molecular is the more current term and the one we will use throughout the semester.)

7 Focus on Ionic Compounds The remainder of this unit will focus on the ionic compounds. Molecular compounds will be covered in the next unit. An abbreviated chart is provided on the next slide that will help focus on the elements of interest in our early discussion.

8 Abbreviated Periodic Chart The abbreviated periodic table here leaves out the transition and inner transition elements The red stair-stepped line separates metals and nonmetals The gray elements are metals The green elements are nonmetals In reality, there is degree of being metal or nonmetal – for now we aren’t concerned with that.

9 The Octet Rule and the Formation of Ions The octet rule states that atoms tend to lose, gain, or share electrons to attain eight valence electrons, i.e. the same number as the noble gases (Group 18) Group 1 metals have one extra electron compared to their nearest noble gas. Thus, they will tend to lose one electron to have the same number as the nearest noble gas. Losing one electron leads to the formation of a 1+ cation.

10 The Octet Rule and the Formation of Ions (continued) Similarly, Group 2 metals will tend to lose 2 electrons to become 2+ cations. Group metals will tend to also lose valence electrons (remember the number of valence electrons is the last digit in the group number) to become cations. These are a little less of a sure thing than the Groups 1 and 2

11 The Octet Rule and the Formation of Ions (continued) Consider the Group 17 nonmetals. They are one electron short of having the same number as their nearest noble gas. As a result, they will tend to form 1- anions. The same approach is used to consider Groups nonmetals, though things become a little less absolute in Groups

12 Summary of Ionic Charges Some of the charges suggested on the previous slides are virtually certain things – some are a little more flexible. The table here summarizes the charges that you can be certain of for these elements when combining metals and nonmetals.

13 Lewis Symbols (4.2) Lewis symbols provide a method of writing the symbol for an element and representing its valence electrons. One dot is placed around the symbol for each valence electron. Examples: Group 1(1 valence electron): Li· Na· K· Rb· Cs· Group 2(2 valence electrons): ·Mg· ·Ca· ·Sr· ·Ba· Group 16(6 valence electrons): Group 17(7 valence electrons): The actual location around the symbol is not important – once there are more than four electrons we usually start pairing them.

14 Writing Formulas for Ionic Compounds (4.4) Consider the reaction of sodium and chlorine using Lewis symbols. Overall: Things to note: The sodium (Na) begins with one valence electron but donates it to the chlorine. Thus the sodium has the same number of electrons as the noble gas neon and chlorine has the same number as argon. When the sodium gives up its electron it becomes a 1+ cation, Na + When the chlorine accepts the electron it becomes a 1- anion, Cl - Since the sodium gives up one electron and the chlorine takes on one electron, they react in a 1:1 ratio – all electrons are accounted for. The formula for the compound becomes NaCl, indicating that one Na reacts with one Cl All of the Group 1 elements react with the Group 17 elements in the same ratio.

15 Writing Formulas for Ionic Compounds (4.4) What if it’s not a 1:1 ratio? Consider the reaction of barium and fluorine using Lewis symbols. Overall: Things to note: The barium begins with two valence electrons but donates them to two fluorine atoms. Thus the barium has the same number of electrons as the noble gas xenon and fluorine has the same number as neon. When the barium gives up its electrons it becomes a 2+ cation, Ba 2+ When the fluorine accepts the electron it becomes a 1- anion, F - Since the barium gives up two electrons and the each fluorine takes one electron, the atoms react in a barium:fluorine ratio of 1:2 – all electrons are accounted for. The formula for the compound becomes BaF 2 indicating that one Ba reacts with two F All of the Group 2 elements react with the Group 17 elements in the same ratio.

16 Writing Formulas for Ionic Compounds (4.4) The key to formula writing for ionic compounds is to recognize that all electrons must be accounted for. Every electron given up by the metal must be accepted by the nonmetal. A relatively easy way of doing this is to consider a crossover approach. Write the element with its ionic charge and place the charge on one as the subscript on the other and vice versa Examples: Sr and Br as ions Sr 2+ and Br - crossover charges Sr 2+ and Br - gives SrBr 2 Al and O as ions Al 3+ and O 2- crossover charges Al 3+ and O 2- gives Al 2 O 3 K and S as ions K + and S 2- crossover charges K + and S 2- gives K 2 O

17 Nomenclature of Binary Ionic Compounds(4.5) Naming binary ionic compounds involves two key rules The metal retains its name The nonmetal ending changes to –ide For example: BaO → barium oxide K 2 S → potassium sulfide SrCl 2 → strontium chloride Al 2 S 3 → aluminum sulfide Ca 3 N 2 → calcium nitride

18 Metals with Multiple Possible Charges Besides the charges illustrated in the abbreviated periodic table below, several elements can take more than one charge in ionic form. For example, iron (Fe) exists both as Fe 2+ and Fe 3+. This does not change the formula writing approach, but does complicate nomenclature a little. For example, iron could form both FeCl 2 and FeCl 3. The name iron chloride would not distinguish between them so another approach, outlined on the next slide, is necessary.

19 Naming Compounds Containing Metals with Multiple Possible Charges Some of the key elements that show this trait include: Fe 2+ Fe 3+ Cu + Cu 2+ Pb 2+ Pb 4+ Au + Au 3+ The simplest approach to naming compounds with these ions in them is to write the name of the metal followed by its charge in Roman numerals in parentheses and then the nonmetal ending in –ide. The names of the ions above would become: Fe 2+ iron (II)Cu + copper(I) Pb 2+ lead(II) Au + gold(I) Fe 3+ iron (III)Cu 2+ copper(II) Pb 4+ lead(IV) Au 3+ gold(III) The compounds below would be named as indicated: FeCl 2 iron (II) chlorideFeCl 3 iron (III) chlorideCu 3 N 2 copper(II) nitride PbOlead (II) oxidePbO 2 lead (IV) oxideAu 2 O 3 gold (III) oxide Notice in the bottom rows that the charge on the metal had to be figured out by working backwards. For PbO, O is 2- so Pb must be 2+ for the sum of the charges to be zero.

20 The End


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