1 / 22

Lecture 7: Ionic Compounds

Lecture 7: Ionic Compounds. Dr Harris Suggested HW : ( Ch 6 ) 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)

ula
Download Presentation

Lecture 7: Ionic Compounds

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lecture 7: Ionic Compounds Dr Harris Suggested HW: (Ch 6)4, 9, 16, 24, 25a., 29, 51*, 52*, 60 *Use pg 186 for electron affinities

  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 ionsare 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 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

  4. Forming Ionic Compounds 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+ Na+ cation Na atom

  5. Forming Ionic Compounds 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]) • Unlike ionization energy, which describes the energy input needed to extract an electron, electron affinitydescribes the energy released an electron is added. 18 e- 17 e- 1st Electron Affinity 17 p+ 17 p+ Cl atom Cl- anion

  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] 3s1) + Cl ( [Ne] 3s2 3p5) ---> Na+ Cl- [Ne] [Ar] IONIC COMPOUND Lewis dot structure of the product Cl- Na+

  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 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)

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

  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 + 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

  10. Dissociation • 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

  11. Dissociation of a Salt 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-

  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 + e- - e- Cation, X+ Neutral X Anion, 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 energyof 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 • Form Na+ (ionization energy) • Form Cl- (electron affinity) • Join them together (coulombic energy)

  17. Calculating The Energy Change Due To Formation of Ionic Compounds 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) ? Now we must calculate the coulombic energy

  18. 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 coulombic 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)

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

  20. Group Example • Given the following data, calculate the energy of reaction to form CsCl and Cs2O 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] 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

  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. 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`

More Related