1. Chemical Bonding I: Ions & Ionic Bonds Glencoe: Chapter Eight

2. 8.1: Forming Chemical Bonds Chemical bonds are the attractive forces that hold two or more atoms together. Chemical bonds are the attractive forces that hold two or more atoms together. Valence electrons are used to arrange elements into groups. They are also involved in the formation of chemical bonds. Valence electrons are used to arrange elements into groups. They are also involved in the formation of chemical bonds.

3. Why lose electrons? Sodium has one valence electron, represented by 1s 2 2s 2 2p 6 3s 1 Sodium has one valence electron, represented by 1s 2 2s 2 2p 6 3s 1 When sodium loses its valence electron, it looks like: 1s 2 2s 2 2p 6 When sodium loses its valence electron, it looks like: 1s 2 2s 2 2p 6 What element is this??? What element is this??? The octet rule states that atoms gain, lose or share electrons in order to acquire a full set of 8 valence electrons (or completely filled s and p orbitals) to get an unusually stable configuration. The octet rule states that atoms gain, lose or share electrons in order to acquire a full set of 8 valence electrons (or completely filled s and p orbitals) to get an unusually stable configuration.

4. Formation of Positive Ions Positive ions form when electrons are lost. Positive ions form when electrons are lost. Metals typically form cations. Metals typically form cations. Reactivity of metals is based on the ease at which electrons are lost. Reactivity of metals is based on the ease at which electrons are lost. Positive ions are called cations. Positive ions are called cations.

5. Transition Metals Transition metals have an outer energy level of ns 2. Going from left to right, inner d sublevels are being filled. Transition metals have an outer energy level of ns 2. Going from left to right, inner d sublevels are being filled. Transition metals commonly form +2 cations because they lose those two valence electrons. However, d orbital electrons may also be lost. Transition metals commonly form +2 cations because they lose those two valence electrons. However, d orbital electrons may also be lost.

6. Formation of Negative Ions Nonmetals form negative ions, called anions. Nonmetals have high electron affinities. Nonmetals form negative ions, called anions. Nonmetals have high electron affinities. Anions are formed through the addition of one or more electrons in the outer energy level. Anions are formed through the addition of one or more electrons in the outer energy level. Anions are designated with the –ide ending. Ex: nitrogen becomes nitride. Anions are designated with the –ide ending. Ex: nitrogen becomes nitride.

7. Ionic Radius Atoms can gain or lose electrons to form ions. Because electrons are negatively charged, atoms that gain or lose electrons acquire a net charge. Atoms can gain or lose electrons to form ions. Because electrons are negatively charged, atoms that gain or lose electrons acquire a net charge. An ion is a atom or a bonded group of atoms with a +/- charge. An ion is a atom or a bonded group of atoms with a +/- charge. When atoms lose electrons and form cations, they always become smaller. When atoms lose electrons and form cations, they always become smaller.

8. The reason for the size decrease is because the electron lost is always a valence electron. The reason for the size decrease is because the electron lost is always a valence electron. The loss of a valence electron may leave a completely empty outer orbital, which results in a smaller radius. The loss of a valence electron may leave a completely empty outer orbital, which results in a smaller radius. The resulting electrostatic repulsion between the remaining electrons causes them to be pulled closer to the nucleus. The resulting electrostatic repulsion between the remaining electrons causes them to be pulled closer to the nucleus.

9. When atoms gain electrons, they form negatively charged ions called anions. They always become larger as a result. When atoms gain electrons, they form negatively charged ions called anions. They always become larger as a result. The addition of an electron to an atom increases the electrostatic repulsion between the valence electrons, forcing them apart. The addition of an electron to an atom increases the electrostatic repulsion between the valence electrons, forcing them apart. The increased distance results in a larger radius. The increased distance results in a larger radius.

10. Trends within periods As you move across a period, cations get smaller. Then when you get to group 15, the much larger anions also get smaller. So, for ionic radii, the general trend within a period is for it to decrease from left to right. As you move across a period, cations get smaller. Then when you get to group 15, the much larger anions also get smaller. So, for ionic radii, the general trend within a period is for it to decrease from left to right.

11. Trends within groups As you move down a group, an ion’s valence electrons are in higher principal energy levels, resulting in a gradual increase in ionic size. As you move down a group, an ion’s valence electrons are in higher principal energy levels, resulting in a gradual increase in ionic size. Thus, the ionic radii for both cations and anions increase as you move down groups. Thus, the ionic radii for both cations and anions increase as you move down groups.

13. Introduction of the Green Sheet

14. Ionic Compounds- Give and Take Attraction between + ions and - ions Attraction between + ions and - ions Electrons go from metals to nonmetals Electrons go from metals to nonmetals electron transfer metal nonmetal ion + ion – metal nonmetal ion + ion – Electrons lost = Electrons gain

15. Formulas of Ionic Compounds Formulas of ionic compounds are determined from the charges on the ions atoms ions atoms ions     –     – Na  +  F :  Na + : F :  NaF         sodium fluorine sodium fluoride formula Charge balance: 1+ 1- = 0 Charge balance: 1+ 1- = 0

16. Writing a Formula Write the formula for the ionic compound that will form between Ba 2+ and Cl . Solution: 1. Balance charge with + and – ions 2. Write the positive ion of metal first, and then the negative ion Ba 2+ Cl  Cl  negative ion Ba 2+ Cl  Cl  3. Write the number of ions needed as subscripts BaCl 2 BaCl 2

17. Learning Check Write the correct formula for the compounds containing the following ions: A. Na +, S 2- 1) NaS 2) Na 2 S3) NaS 2 1) NaS 2) Na 2 S3) NaS 2 B. Al 3+, Cl - 1) AlCl 3 2) AlCl 3) Al 3 Cl 1) AlCl 3 2) AlCl 3) Al 3 Cl C. Mg 2+, N 3- 1) MgN 2) Mg 2 N 3 3) Mg 3 N 2 1) MgN 2) Mg 2 N 3 3) Mg 3 N 2

18. Naming Binary Ionic Compounds Contain 2 different elements Contain 2 different elements Name the metal first, then the nonmetal as -ide. Name the metal first, then the nonmetal as -ide. Use name of a metal with a fixed charge Use name of a metal with a fixed charge Groups 1A, 2A, 3A and Ag, Zn, and Cd Examples: NaClsodium chloride ZnI 2 zinc iodide Al 2 O 3 aluminum oxide

19. Learning Check Complete the names of the following binary compounds: Na 3 Nsodium ________________ KBrpotassium________________ Al 2 O 3 aluminum ________________ MgS_________________________

20. Transition Metals Transition Metals Many form 2 or more positive ions 1+ 2+ 1+ or 2+ 2+ or 3+ Ag + Cd 2+ Cu +, Cu 2+ Fe 2+, Fe 3+ Ag + Cd 2+ Cu +, Cu 2+ Fe 2+, Fe 3+ silver cadmium copper(I) ion iron(II) ion silver cadmium copper(I) ion iron(II) ion ion ion copper (II) ion iron(III) ion ion ion copper (II) ion iron(III) ion Zn 2+ zinc ion zinc ion

21. Names of Variable Ions Use a roman number after the name of a metal that forms two or more ions Transition metals and the metals in groups 4A and 5A FeCl 3 (Fe 3+ ) iron (III) chloride CuCl (Cu + ) copper (I) chloride SnF 4 (Sn 4+ ) tin (IV) fluoride PbCl 2 (Pb 2+ )lead (II) chloride Fe 2 S 3 (Fe 3+ )iron (III) sulfide Fe 2 S 3 (Fe 3+ )iron (III) sulfide

22. Learning Check Complete the names of the following binary compounds with variable metal ions: FeBr 2 iron (_____) bromide Cu 2 Ocopper (_____) oxide SnCl 4 ___(_____ ) ______________ Fe 2 O 3 ________________________ Fe 2 O 3 ________________________ CuS________________________

23. Complete p. 236, #47 – 56 & #60 - 78 even. This will help you better understand ionic bond formation. Complete p. 236, #47 – 56 & #60 - 78 even. This will help you better understand ionic bond formation.