1. Lecture 8: Ionic Compounds Dr Harris 9/11/12 HW: Ch 6: 4, 25, 29, 51, 58, 69

2. Ionic Compounds The nucleus of an atom is unchanged by chemical reactions (number of protons never changes) However, electrons are readily added and lost and ions are formed When a metal reacts with a nonmetal, ions form and interact. The result is an ionic compound. Let’s consider the formation of a very common ionic compound, NaCl (s)

3. Forming Ionic Compounds We know that Na(s) and Cl 2 (g) react together to form NaCl (s), but how? The most important thing to know about chemical reactions is that atoms undergoing a reaction will always seek to reach a noble gas configuration Let’s look at the electron configurations of Na and Cl

4. Forming Ionic Compounds Na: [Ne] 3s 1 Cl: [Ne] 3s 2 3p 5 For Na, the nearest noble gas is Ne. To reach the Ne configuration, it needs to lose a single electron. Na ( [Ne] 3s 1 ) ---> Na + ([Ne]) + e- 11 p + 11 e - 1 st Ionization Energy 11 p + Na atom Na + cation 10 e -

5. Forming Ionic Compounds Na: [Ne] 3s 1 Cl: [Ne] 3s 2 3p 5 For Cl, the nearest noble gas is Ar. To reach the Ar configuration, it needs to gain a single electron. Cl ([Ne] 3s 2 3p 5 ) + e - ---> Cl - ([Ar]) Unlike ionization energy, which describes the energy input needed to extract an electron, electron affinity describes the energy released an electron is added. 17 p + 1 st Electron Affinity 17 p + Cl atom Cl - anion 17 e- 18 e-

6. Forming Ionic Compounds Na and Cl can simultaneously achieve a noble gas configuration if an electron is transferred from the metal (Na) to the nonmetal (Cl) Na ( [Ne] 3s 1 ) + Cl ( [Ne] 3s 2 3p 5 ) ---> Na + Cl - [Ne] [Ar] IONIC COMPOUND Cl - Na + Lewis dot structure of the product

7. Predicting Charge So now, we understand that ionic compounds form when metal and nonmetal ions interact We also see why sodium chloride is NaCl, not NaCl 2 or Na 2 Cl, etc. The overall charge of any molecule must be zero. Since the Na loses an electron to become Na +, and Cl gains an electron to become Cl -, only one of each ion is needed to balance the charge. In ionic compounds, the metal is always positively charged (cation) and the nonmetal is always negatively charged (anion)

8. Predicting Charge 1+1+ 2+2+ 3+3+ 3-3- 2-2- 1-1- metals nonmetals

9. Group Examples Write the chemical formulas and Lewis structures of the following ionic compounds: Calcium oxide Magnesium Chloride Sodium Sulfide Determine the ionic product and balance: Mg + O 2  ? Na + N 2  ? Show the electron transfer process in the formation of calcium oxide using the noble gas electron configuration, as was shown for NaCl

10. Dissociation Ionic compounds completely dissociate in water, forming individual ions. Ions become completely ‘hydrated’. Na + Cl - Na + (aq) + Cl - (aq) H 2 O (L) Here, NaCl is the solute, water is the solvent

11. Water molecules “solvate” ionic compounds, ripping the ions apart. The negative oxygen atoms (red) attracted to the positive Na +, and the positive hydrogens are attracted to the negative Cl - Dissociation of a Salt In Water Na + Cl -

12. Conducting Electric Current Ions in solution are capable of conducting electric current (hence, the term electrolyte). Ions are able to transport charge across the water. Non-ionic solutions (covalent) do not exhibit this property because they do not dissociate

13. Ion Size Depends on Charge As you may have noticed in my drawings in previous slides, cations tend to be smaller than their neutral atom counterparts, and anions seem to be larger Anions have larger electron clouds because the excess negative charge causes repulsion, which leads to expansion of the electron cloud. The excess positive charge in cations draws the electron cloud closer to the nucleus Neutral X Anion, X - Cation, X +

14. Ionic Radii

15. Energy Is Absorbed or Released When an Ionic Compound Forms The electrostatic attraction, or the electrical attraction between positive and negative ions, is what holds an ionic compound together When two ions form an ionic compound, there is an overall change in energy. We can calculate this energy by considering: the ionization energy of the metal the electron affinity of the nonmetal the coulombic energy of attraction between the cation and anion

16. Calculating The Energy Change Due To Formation of Ionic Compounds A chemical reaction is considered favorable if the energy of the product is less than that of the reactants. In other words, the change in energy is negative. Lets revisit the reaction: Na(g) + Cl(g)  NaCl(s) Ignore the monatomic chlorine To form NaCl, there are 3 steps 1.Form Na + (ionization energy) 2.Form Cl - (electron affinity) 3.Join them together (coulombic energy)

17. 1.(Ionization of Na) Na(g)  Na + (g) + e - ΔE I = +0.824 aJ * Positive sign means energy is added. 2.(Ionization of Cl) Cl(g) + e -  Cl - (g) ΔE EA = -0.580 aJ * Negative sign means energy is released. 3.(Coulombic energy) Na + (g) + Cl - (g)  Na + Cl - (s) Now we must calculate the coulombic energy ? Calculating The Energy Change Due To Formation of Ionic Compounds

18. Coulombic Energy The third step is to join the two atoms, as shown below. r Na = 102 pm r Cl = 181 pm Q 1 and Q 2 are the charges of the metal and nonmetal d is the distance between the nuclei. This is the sum of the ionic radii. k is a constant. (231 aJpm) The equation shown above is Coulomb’s Law, which gives the coulombic energy change (E c ) that results when two ions come together.

19. Solve Negative energy change indicates a favorable process Na + Cl -

20. Group Example Given the following data, calculate the energy of reaction to form CsCl and Cs 2 O given that the first ionization energy of Cesium is 0.624 aJ

21. Electron Configurations of Transition Metal Ions When a transition metal forms an ion, electrons are first removed from the preceding s-orbital. Fe: [Ar] 4s 2 3d 6 Fe 2+ : [Ar] 3d 6 Fe 3+ : [Ar] 3d 5 If the ionization of a transition metal results in an unpaired s- electron, that electron will move into the valence d orbital Ni: [Ar] 4s 2 3d 8 Ni + : [Ar] 4s 1 3d 8 ---> [Ar] 3d 9

22. Roman Numeral Nomenclature for Transition Metals Transition metals can have multiple positive ionic charges. To distinguish, a roman numeral is placed in front of a transition metal in a compound to identify its charge. Ex. FeCl 2 ---> Here, Fe is 2 +. So, we name this compound: Iron (II) chloride FeCl 3 ---> Here, Fe is 3 +. Iron (III) chloride Name the following: TiO 2, WCl 6 Titanium (IV) oxide, Tungsten (VI) chloride`