Naming Ionic Compounds

1. Naming Ionic Compounds In order to communicate conveniently compounds need to be named unambiguously. Name the cation then the anion. MonoatomicCations: Most cations are single elements and are given the same name as the element. Ex). Na+ = sodium Mg2+ = magnesium If an element can give rise to different cations use a Roman numeral to indicate the charge. This is necessary for most transition metals as well as tin and lead Ex) Fe2+ = iron (II) Fe3+ = iron (III) Au+ = gold (I) Au3+ = gold (III) Pb2+ = lead (II) Pb4+ = lead (IV)

2. Monatomic cations and the Periodic Table

3. Naming Ionic Compounds Monoatomic Anions To indicate that an element has formed an anion, we change its suffix to -ide. Ex) F = fluorine F-= Fluoride N= nitrogen N3-= Nitride O= oxygen O2-= Oxide Se= selenium Se2-= 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 littletrue ionic character. But the valence of carbon is still often 4- in many of its compounds

4. Naming Ionic Compounds 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: NH4+ Ammonium Best known H3O+ Hydronium Polyatomic Anions: ClO-Hypochlorite (Bleach) HCO3-Bicarbonate(baking Soda) Examples CN- Cyanide CrO42- Chromate OH- Hydroxide Cr2O72- Dichromate CO32- Carbonate MnO41- Permanganate When H+ is added to a polyatomic atom the prefix “bi” is placed before the ‘old’ name. Bicarbonate HS- Bisulfide HCO3- HSO4- Bisulfate

5. Naming Ionic Compounds 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. Exercise Magnesium Chloride MgCl2 CuBr2 NaNO3 (NH4)2SO4 Sn(CO3)2 Sn(HCO3)2 NaNO2 Copper(II) Bromide Sodium Nitrate Ammonium Sulfate Tin (IV) Carbonate Tin (II) Bicarbonate Sodium Nitrite

6. 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 Ionization Energy consumed. 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 Electron Affinity Energy released

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

8. Cl + Na - + Na Ionic Compounds Recall that opposite charges attract each other. A cation and an anion, experience an electrostatic force, given by: e = 1.6022 x 10-19 C unit of charge k =8.988 × 109N m2/C2 Coulomb's Constant This force pulls them together to make an ionic bond. q1 q2 r r Cl -

9. Ionic Compounds 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 q1 = +1 q2 = -1 F = ke2q1q2/r2 r = 10 nm = 1.000*10-8 m F = (8.988 × 109N m2/C2)(1.6022 *10-19 C)2(1)(-1) (1.000*10-8 m)2 F = -2.307*10-12 N

10. Ionic Compounds Exercise: Compute the energy of formation of the ionic bond in NaCl, where the bond length is 279 pm q1 = +1 q2 = -1 E = ke2q1q2/r r = 279 pm = 2.79*10-10 m E = (8.988 × 109N m2/C2)(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*1023 bonds/mol) = 498,000 J/mol = 498 kJ/mol

11. The Alkali Metals Ionic Compounds • soluble in water. Dissociation Ex) NaCl(aq) • solvation of ions: Indicated as Na(OH2)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 H2O

12. Ion-Dipole Interactions Solvation of ions by polar solvents illustrate ion-dipole forces. For water, solvation = hydration Definite enthalpies of hydration have been established for the reaction: The size of this energy is directly related to the size of the ion & the charge Z on the ion. Mg2+ has an enthalpy of -1922 kJ/mol. Why?

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

14. Formation of ionic materials ElectronAffinity 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 Ionization energy Lattice energy vaporization Formationenergy

15. Exercise Given the following data, calculate ΔLFH for NaCl(s) Enthalpy of sublimation of Na = +108 kJ/mol First ionization energy for Na = +496 kJ/mol Enthalpy of bond dissociation for Cl2(g) = +243 kJ/mol Enthalpy of electronic attraction for Cl = -349 kJ/mol Enthalpy of formation for NaCl = -411 kJ/mol

16. Enthalpy of Lattice Formation ΔLFH 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 ΔLFH . 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 ΔLFH and be more soluble. Carbonates of alkaline earth metals decompose to the metal oxide and carbon dioxide: MgCO3 undergoes this reaction at 300 °C while CaCO3 requires heating to 840 °C. Because Mg2+ is smaller than Ca2+, it is a better match for the small spherical O2- than the larger nonspherical CO32- Compounds such as Mg(ClO4)2 and CuSO4 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:

17. Ionic Lattices (NaCl) • 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?

18. 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

19. 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!

20. Structures of ZnS Lattice • ZnSadopts 2 structures: • i) zinc blende (ccp for S2-) ii) wurtzite(hcp for S2-). • What kind of holes are the cations filling? • Determine thierco-ordination number ? • Note that the two structures are very similar at the level of one ion?

21. Structures of ZnS Lattice

22. Ionic Lattices ZnSNaClCsCl 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

23. Ionic Lattices ZnSNaClCsCl 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 • 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.