1. Naming Ionic Compounds In order to communicate conveniently compounds need to be named unambiguously. Name the cation then the anion. Monoatomic Cations: Most cations are single elements and are given the same name as the element. Ex). Na + = sodiumMg 2+ = magnesium If an element can give rise to different cations use a Roman numeral to indicate the charge. Ex)Fe 2+ = iron (II)Fe 3+ = iron (III) Au + = gold (I)Au 3+ = gold (III) Pb 2+ = lead (II)Pb 4+ = lead (IV) This is necessary for most transition metals as well as tin and lead

2. Monatomic cations and the Periodic Table

3. Monoatomic Anions To indicate that an element has formed an anion, we change its suffix to -ide. Ex) F = fluorine F - = Fluoride N = nitrogenN 3- = Nitride O = oxygenO 2- = Oxide Se = selenium Se 2- = Selenide Tend to form exclusively from the non-metals and one metalliod Charge = group # -18 Free ions do not exist with charges higher than 3 units. carbides have little true ionic character. But the valence of carbon is still often 4- in many of its compounds Naming Ionic Compounds

4. Polyatomic Ions Not all ions consist of just one atom. Some groups of covalently bonded atoms have overall charges. Because they are charged, they are ions (not molecules). Polyatomic cations: NH 4 + Ammonium Best known H 3 O + Hydronium Polyatomic Anions: ClO - Hypochlorite (Bleach) HCO 3 - Bicarbonate(baking Soda) Examples CN - Cyanide CrO 4 2- Chromate OH - HydroxideCr 2 O 7 2- Dichromate CO 3 2- Carbonate MnO 4 1- Permanganate When H + is added to a polyatomic atom the prefix “bi” is placed before the ‘old’ name. HCO 3 - BicarbonateHS - Bisulfide HSO 4 - Bisulfate Naming Ionic Compounds

5. Ionic Compounds To name an ionic compound, name the cation then the anion. The charge of each ion indicates how many are needed to make a neutral compound so no prefixes are necessary. MgCl 2 CuBr 2 NaNO 3 (NH 4 ) 2 SO 4 Sn(CO 3 ) 2 Sn(HCO 3 ) 2 NaNO 2 Magnesium Chloride Copper(II) Bromide Sodium Nitrate Ammonium Sulfate Tin (IV) Carbonate Exercise Tin (II) Bicarbonate Sodium Nitrite Naming Ionic Compounds

6. ClNSP Ohypochlorite (ClO - ) O2O2 chlorite (ClO 2 - ) nitrite (NO 2 - ) O3O3 chlorate (ClO 3 - ) nitrate (NO 3 - ) sulfite (SO 3 2- ) phosphite (PO 3 3- ) O4O4 perchlorate (ClO 4 - ) sulfate (SO 4 2- ) phosphate (PO 4 3- ) The polyatomic ions ending with ‘ate’ all consist of an atom surrounded by a number of oxygen atoms. Those ending in “ite” have one fewer O Naming Ionic Compounds

7. LiF formation Li loses all e’s in its valence shell Large reduction in atomic radius: Li – 152 pm Li + - 78 pm Vol = 1/7th F gains electron its valence shell Expansion in the radius because effective Z* is decreased due to extra electron. F – 71 pm F - - 133 pm Vol = 6x Ionization Energy consumed. Electron Affinity Energy released

8. LiF formation Energy is released to form the ionic bond

9. Ionic Compounds Recall that opposite charges attract each other. A cation and an anion, experience an electrostatic force, given by: This force pulls them together to make an ionic bond. Na + Cl - r q1q1 q2q2 r - Na + e = 1.6022 x 10 -19 C unit of charge k = 8.988 × 10 9 N m 2 /C 2 Coulomb's Constant

10. The energy released to form an ionic bond: Exercise Compute the force that a sodium cation and chloride anion experience when 10.00 nm apart F = ke 2 q 1 q 2 /r 2 q 1 = +1q 2 = -1 r = 10 nm = 1.000*10 -8 m F = (8.988 × 10 9 N m 2 /C 2 )(1.6022 *10 -19 C) 2 (1)(-1) (1.000*10 -8 m) 2 F = -2.307*10 -12 N Ionic Compounds

11. Exercise: Compute the energy of formation of the ionic bond in NaCl, where the bond length is 279 pm E = ke 2 q 1 q 2 /r q 1 = +1q 2 = -1 r = 279 pm = 2.79*10 -10 m E = (8.988 × 10 9 N m 2 /C 2 )(1.6022 *10 -19 C) 2 (1)(-1) (2.79*10-10 m) E = -8.27*10 -19 N m = J How energy for one mole of NaCl bonds? E(total) = E(bond) *(# of bonds) = (-8.27*10 -19 J/bonds)(6.022*10 23 bonds/mol) = 498,000 J/mol = 498 kJ/mol Ionic Compounds

12. 12 The Alkali Metals Ionic Compounds –soluble in water. Dissociation Ex) NaCl (aq) –solvation of ions: Indicated as Na(OH 2 ) 6 + (aq) or Na + (aq). –Solvated ions must be more stable (lower energy) than the initial crystal lattice so that energy is released. –Alkali metals have large hydration enthalpies. ex)The beaker gets hot when NaOH is dissolved in H 2 O

13. Ion-Dipole Interactions Mg 2+ has an enthalpy of -1922 kJ/mol. Why? The size of this energy is directly related to the size of the ion & the charge Z on the ion. Definite enthalpies of hydration have been established for the reaction: For water, solvation = hydration Solvation of ions by polar solvents illustrate ion-dipole forces.

14. Ionic Materials Lattice - 3-D pattern of ions - Minimize repulsive forces -Maximize attractive forces - Large lattice energy released - Brittle?? - Charge Balance - High melting point

15. Formation of ionic materials Whether element from ionic compounds depends on the balance between: 1) Ionization energy 2) Electron affinity 3) Lattice energy 4) Phase transition energies 5) Bond energies Dissociation energy Electron Affinity Ionization energy Lattice energy Formation energy vaporization

16. 16 Exercise Given the following data, calculate Δ LF H for NaCl (s) Enthalpy of sublimation of Na = +108 kJ/mol First ionization energy for Na = +496 kJ/mol Enthalpy of bond dissociation for Cl 2(g) = +243 kJ/mol Enthalpy of electronic attraction for Cl = -349 kJ/mol Enthalpy of formation for NaCl = -411 kJ/mol

17. 17 Enthalpy of Lattice Formation Δ LF H is determines properties such as solubility, thermal stability and hydration: If similar in charge and in size, they pack closely, resulting in a large negative Δ LF H. Thus it takes a lot of energy to break up the lattice, hence and the compounds are not very soluble. Ionic compounds with mismatched ions tend to have less negative Δ LF H and be more soluble. Carbonates of alkaline earth metals decompose to the metal oxide and carbon dioxide: MgCO 3 undergoes this reaction at 300 °C while CaCO 3 requires heating to 840 °C. Because Mg 2+ is smaller than Ca 2+, it is a better match for the small spherical O 2- than the larger nonspherical CO 3 2- Compounds such as Mg(ClO 4 ) 2 and CuSO 4 have small cations relative to their anions. As such, they can literally pull water out of the air (and are therefore useful as drying agents) to surround the small cation with the relatively small water molecules. This gives a hydrated solid:

18. 18 Count the number of anions (green) and cations (silver) in a unit cell. Consider that the anions form a lattice and the cations fill holes in the lattice. Three types of “holes” can be found: Octahedral - surrounded by six atoms. Cubic - surrounded by eight atoms. Tetrahedral - surrounded by four atoms. What type of holes are the cations filling? Ionic Lattices (NaCl)

19. 19 Rock Salt NaCl Lattice Structure This is based on the face-centred cubic unit cell, as the Cl – ions occupy all the sites of the fcc lattice Smaller sodium cations fit into remaining gaps to maximize Coulombic attraction This is the so-called lattice energy

20. 20 Structure of CsCl Lattice Count the number of anions (teal) and cations (gold) in a unit cell. What kind of lattice is formed by the anions in this picture? What type of “hole” is found in this lattice? Note that this lattice is not body-centered cubic (bcc)! The lattice is named for the anions only!

21. 21 ZnS adopts 2 structures: i) zinc blende (ccp for S 2- ) ii) wurtzite (hcp for S 2- ). What kind of holes are the cations filling? Determine thier co-ordination number ? Note that the two structures are very similar at the level of one ion? Structures of ZnS Lattice

22. 22 Structures of ZnS Lattice

23. 23 Ionic Lattices Anion radius (r - ) 184 pm 181 pm 181 pm Cation radius (r + ) 75 pm 99 pm 169 pm Co-ordination # anions 4 6 8 cations 4 6 8 Anion lattice type FCC FCC SC Holes for cations Tetra. Octa. Cubic ZnS NaCl CsCl

24. 24 Ionic Lattices Anion radius (r - ) 184 pm 181 pm 181 pm Cation radius (r + ) 75 pm 99 pm 169 pm r + /r - 0.408 0.547 0.934 ZnS NaCl CsCl The radius ratio indicates which packing is most stable: If r + /r – is between 0.225 and 0.414, we get a structure like ZnS If r + /r – is between 0.414 and 0.732, we get a structure like NaCl If r + /r – is above 0.732, we get a structure like CsCl These threshold number can be derived geometrically.