$$$ Quiz $$$ Ionic and Metallic Bonding/Naming Ionic Compounds
Naming Ionic Compounds
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Transcript of Naming Ionic Compounds
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.