Naming Ionic CompoundsIn 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+ = sodium Mg2+ = magnesium

If an element can give rise to different cations use a Roman numeral to indicate the charge.

Ex) Fe2+ = iron (II) Fe3+ = iron (III)Au+ = gold (I) Au3+ = gold (III)Pb2+ = lead (II) Pb4+ = lead (IV)

This is necessary for most transition metals as well as tin and lead

Monatomic cations and the Periodic Table

Monoatomic AnionsTo indicate that an element has formed an anion, we change its suffix to -ide.

Ex) F = fluorine F- = Fluoride N = nitrogen N3- = NitrideO = 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

Naming Ionic Compounds

Polyatomic IonsNot 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 MnO4

1- Permanganate

When H+ is added to a polyatomic atom the prefix “bi” is placed before the ‘old’ name.

HCO3- Bicarbonate HS- Bisulfide HSO4

- Bisulfate

Naming Ionic Compounds

Ionic CompoundsTo 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.

MgCl2

CuBr2

NaNO3

(NH4)2SO4

Sn(CO3)2

Sn(HCO3)2

NaNO2

Magnesium Chloride

Copper(II) BromideSodium Nitrate Ammonium SulfateTin (IV) Carbonate

Exercise

Tin (II) Bicarbonate

Sodium Nitrite

Naming Ionic Compounds

LiF formation

Li Li

. +

F F

::. .. .

. :. .. . -

Li loses all e’s in its valence shellLarge reduction in atomic radius:Li – 152 pm Li+ - 78 pm Vol = 1/7th

F gains electron its valence shellExpansion 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

Li Li

. +

LiF formation

Energy is released to form the ionic bond

Ionic CompoundsRecall 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.

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q eq eF kr

Na+ Cl-r

q1 q2

Clr

-Na+

e = 1.6022 x 10-19 C unit of charge

k = 8.988 × 109 N m2/C2

Coulomb's Constant

The energy released to form an ionic bond:

1 2q eq eE kr

ExerciseCompute the force that a sodium cation and chloride anion experience when 10.00 nm apart

F = ke2q1q2/r2 q1 = +1 q2 = -1

r = 10 nm = 1.000*10-8 m

F = (8.988 × 109 N m2/C2)(1.6022 *10-19 C)2(1)(-1) (1.000*10-8 m)2

F = -2.307*10-12 N

Ionic Compounds

Exercise: Compute the energy of formation of the ionic bond in NaCl, where the bond length is 279 pm

E = ke2q1q2/r q1 = +1 q2 = -1

r = 279 pm = 2.79*10-10 m

E = (8.988 × 109 N 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

Ionic Compounds

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The Alkali MetalsIonic Compounds

– soluble in water. DissociationEx) 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

2 2M H O [M(OH ) ] Z zn hydrationn H

Ion-Dipole Interactions

ion dipole 3

qF

r

Mg2+ 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 = hydrationSolvation of ions by polar solvents illustrate ion-dipole forces.

NaFNa

FNaF

FNaF

NaFNa

NaFNa

FNaF

Ionic MaterialsLattice - 3-D pattern of ions

- Minimize repulsive forces

-Maximize attractive forces

- Large lattice energy released

- Brittle??

- Charge Balance

- High melting point

Formation of ionic materials

Whether element from ionic compounds depends on the balance between:

1) Ionization energy2) Electron affinity 3) Lattice energy4) Phase transition energies5) Bond energies

Dissociation energy

Electron Affinity

Ionizationenergy Lattice

energy

Formation energy

vaporization

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ExerciseGiven the following data, calculate ΔLFH for NaCl(s)

Enthalpy of sublimation of Na = +108 kJ/molFirst 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/molEnthalpy of formation for NaCl = -411 kJ/mol

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Enthalpy of Lattice FormationΔLFH 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 CO3

2-

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:

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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 thelattice.

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)

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Rock Salt NaCl Lattice StructureThis is based on the face-centred cubic unit cell, as the Cl– ions occupy all the sites of the fcc latticeSmaller sodium cations fit into remaining gaps to maximize Coulombic attraction

This is the so-called lattice energy

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Structure of CsCl LatticeCount the number of anions (teal)and cations (gold) in a unit cell.

What kind of lattice is formed by the anionsin 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 forthe anions only!

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ZnS adopts 2 structures:

i) zinc blende (ccp for S2-) ii) wurtzite (hcp for S2-).

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

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Structures of ZnS Lattice

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Ionic Lattices

Anion radius (r -) 184 pm 181 pm 181 pmCation 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

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Ionic Lattices

Anion radius (r -) 184 pm 181 pm 181 pmCation 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 ZnSIf r+/r – is between 0.414 and 0.732, we get a structure like NaClIf r+/r – is above 0.732, we get a structure like CsCl

These threshold number can be derived geometrically.