LECTURE 7: IONIC COMPOUNDS (Ch. 6)

LECTURE 7: IONIC COMPOUNDS (Ch. 6)

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

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 ionsare 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)

Ionic Compounds • We know that Na(s) and Cl2(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

Mechanism of an Ionic Reaction Na: [Ne] 3s1 Cl: [Ne] 3s2 3p5 • For Na, the nearest noble gas is Ne. To reach the Ne configuration, it needs to lose a single electron. Na ( [Ne] 3s1 ) ---> Na+ ([Ne]) + e- 11 e- 10 e- 1st Ionization Energy 11 p+ 11 p+ Neutral Na atom Na+cation

Mechanism of an Ionic Reaction Na: [Ne] 3s1 Cl: [Ne] 3s2 3p5 • For Cl, the nearest noble gas is Ar. To reach the Ar configuration, it needs to gaina single electron. Cl ([Ne] 3s2 3p5) + e- ---> Cl- ([Ar]) • Electron affinitydescribes the energy released an electron is added. 18 e- 17 e- 1st Electron Affinity 17 p+ 17 p+ Cl- anion Neutral Cl atom

Mechanism of an Ionic Reaction • Na and Cl can simultaneously achieve a noble gas configuration if an electron is transferred from the metal (Na) to the nonmetal (Cl) [Ne] 3s1+ [Ne] 3s23p5---> Na+Cl- [Ar] [Ne] IONIC COMPOUND Lewis dot structure of the product Cl- Na+

Predicting and Balancing Charge • So now, we understand that ionic compounds form when metal and nonmetal ions interact • We also see why sodium chloride is NaCl, not NaCl2 or Na2Cl, 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)

Predicting and Balancing Charge metals nonmetals 1+ 2+ 3+ 3- 2- 1-

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 + O2 ? • Na + N2 ? • Show the electron transfer process in the formation of calcium oxide using the noble gas electron configuration, as was shown for NaCl

Dissolving Ionic Compounds in Water • Ionic compounds completely dissociate in water, forming individual ions. Ions become completely ‘hydrated’. H2O (L) Na+ Cl- Na+(aq) + Cl- (aq) • Here, NaCl is the solute, water is the solvent

Dissolving Ionic Compounds in Water Na+ Cl- • 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-

Electrolytes • 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

Ionic Radii • 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 + e- - e- Cation, X+ Neutral X Anion, X-

Ionic Radii

Energy Changes In Reactions • 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 energyof attraction between the cation and anion

Energy Changes In Reactions • Lets revisit the reaction: Na(g) + Cl(g)  NaCl(s) • Ignore the monatomic chlorine • To form NaCl, there are 3 steps • Form Na+ (ionization energy) • Form Cl- (electron affinity) • Join them together (coulombic energy)

Energy Changes In Reactions 1. (Ionization of Na) Na(g)  Na+(g) + e-ΔEI = +0.824 aJ *Positive sign means energy is added. 2. (Ionization of Cl) Cl(g) + e- Cl-(g)ΔEEA = -0.580 aJ *Negative sign means energy is released. 3. (Coulombic energy) Na+(g) + Cl-(g)  Na+Cl-(s) ?

Coulombic Energy • The third step is to join the two ions, as shown below. rNa = 102 pm rCl = 181 pm • The equation shown above is Coulomb’s Law, which gives the energy change (Ec) that results when two ions come together. • Q1 and Q2 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 aJ•pm)

Solve…. • Negative energy change indicates a favorableprocess

Example • Given the following data, calculate the energy of reaction to form CsCl given that the first ionization energy of Cesium is 0.624 aJ

Electron Configurations of Transition Metals • When a transition metal forms an ion, electrons are first removed from the preceding s-orbital. Fe: [Ar] 4s2 3d6 Fe2+: [Ar] 3d6 Fe3+: [Ar] 3d5 • If the ionization of a transition metal results in an unpaired s-electron, that electron will move into the valence d orbital • Ni: [Ar] 4s2 3d8 • Ni+: [Ar] 4s1 3d8 ---> [Ar] 3d9

Electron Configurations of 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. FeCl2 ---> Here, Fe is 2+. So, we name this compound: Iron (II) chloride FeCl3 ---> Here, Fe is 3+. Iron (III) chloride • Name the following: TiO2, WCl6 Titanium (IV) oxide, Tungsten (VI) chloride

LECTURE 7: IONIC COMPOUNDS (Ch. 6)