Presentation is loading. Please wait.

Presentation is loading. Please wait.

Suggested HW: 4, 9, 16, 24, 25a., 29, 51*, 52*, 60 *Use pg 186 for electron affinities.

Similar presentations


Presentation on theme: "Suggested HW: 4, 9, 16, 24, 25a., 29, 51*, 52*, 60 *Use pg 186 for electron affinities."— Presentation transcript:

1

2 Suggested HW: 4, 9, 16, 24, 25a., 29, 51*, 52*, 60 *Use pg 186 for electron affinities

3 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 attract. The result is an ionic compound. Let’s consider the formation of a very common ionic compound, NaCl (s)

4 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 Ionic Compounds

5 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 + Neutral Na atom Na + cation 10 e - Mechanism of an Ionic Reaction

6 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]) Electron affinity describes the energy released when an electron is added. 17 p + 1 st Electron Affinity 17 p + Neutral Cl atom Cl - anion 17 e- 18 e- Mechanism of an Ionic Reaction

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

8 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 complete 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) Predicting and Balancing Charge

9 metals nonmetals Predicting and Balancing Charge

10 Write the chemical formulas and Lewis structures of the following ionic compounds: – Calcium oxide – Magnesium Chloride – Sodium Sulfide – Potassium Phosphide 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 Group Examples

11 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 Dissolving Ionic Compounds in Water

12 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 - Na + Cl - Dissolving Ionic Compounds in Water

13 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 Electrolytes

14 Cations tend to be smaller than their neutral atom counterparts, and anions seem to be larger Anions have large electron clouds because the excess of 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 + + e - - e - Ionic Radii

15

16 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 Energy Changes In Reactions

17 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) Energy Changes In Reactions

18 1.(Ionization of Na) Na(g)  Na + (g) + e - Δ E I = aJ *Positive sign means energy is added. 2.(Ionization of Cl) Cl(g) + e -  Cl - (g) Δ E EA = aJ * Negative sign means energy is released. 3.(Coulombic energy) Na + (g) + Cl - (g)  Na + Cl - (s) ? Energy Changes In Reactions

19 The third step is to join the two ions, 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 energy change (E c ) that results when two ions come together. Coulombic Energy

20 Negative energy change indicates a favorable process Solve….

21 Given the following data, calculate the energy of reaction to form CsCl given that the first ionization energy of Cesium is aJ Example

22 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 Electron Configurations of Transition Metals

23 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 Electron Configurations of Transition Metals


Download ppt "Suggested HW: 4, 9, 16, 24, 25a., 29, 51*, 52*, 60 *Use pg 186 for electron affinities."

Similar presentations


Ads by Google