introductory inorganic chemistry

Chem 59-250

Introductory Inorganic Chemistry

What is Inorganic Chemistry?

Chem 59-250

Chem 59-250

Chem 59-250

Chem 59-250

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As: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3

Chem 59-250

For more information about these periodic tables visit the site where I obtained the pictures: http://chemlab.pc.maricopa.edu/periodic/default.html

Chem 59-250

Chem 59-250

Classes of Inorganic Substances

Elements Ionic Compounds Covalent Compounds

Atomic/Molecular Gases

Ar, N2

Simple (binary)

NaCl

Simple (binary)

NH3, H2O, SO2

Molecular Solids

P4, S8, C60

Complex (polyatomic ions)

Na2(SO4)

Complex (polyatomic)

As(C6H5)3, organometallic compounds

Network Solids

diamond, graphite (C)

“red” phosphorus (P)

Network ions

Mg3(Si2O5)(OH)2 (talc)

Network Solids

SiO2, polymers

Solid/Liquid Metals

Hg, Ga, Na, Fe, Mg

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Elements

Atomic/Molecular Gases

Ar, N2, O2 , Br2

Molecular Solids

P4, S8, C60

Network Solids

diamond, graphite (C)

“red” phosphorus (P)

Solid/Liquid Metals

Hg, Ga, Fe, Na, Mg

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Ionic Compounds

Simple (binary)

NaCl

Complex (polyatomic ions)

Na2(SO4), Na2Mg(SO4)2

Network ions

Mg3(Si4O10)(OH)2 (talc)

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Covalent Compounds

Simple Molecular (binary)

NH3, H2O, CO2, SO2

Complex Molecular

As(C6H5)3, organometallic compounds

Network Solids

SiO2, polymers

F

F

P F

F

F

HH

N

H

H H

O

Chem 59-250

Review of Concepts

Thermochemistry:Standard state: 298.15 K, 1 atm, unit concentration

Enthalpy Change, H°H° = H°products - H°reactants

Entropy Change, S°

Free Energy Change, GG = H - TSAt STP:G° = H° - (298.15 K)S°

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Standard Enthalpy of Formation, H°f

H° for the formation of a substance from its constituent elements

Standard Enthalpy of Fusion, H°fus Na(s) Na(l)

Standard Enthalpy of Vapourization, H°vap Br2(l) Br2(g)

Standard Enthalpy of Sublimation, H°sub P4(s) P4(g)

Standard Enthalpy of Dissociation, H°d ½ Cl2(g) Cl(g)

Standard Enthalpy of Solvation, H°sol Na+(g) Na+

(aq)

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Ionization Enthalpy, H°ie

The enthalpy change for ionization by loss of electron(s)

Na(g) Na+(g) + e- H°ie = 502 kJ/mol

Al(g) Al+(g) + e- H°ie = 578 kJ/mol

Al+(g) Al2+(g) + e- H°ie = 1817 kJ/mol

Al2+(g) Al3+

(g) + e- H°ie = 2745 kJ/mol

Thus:Al(g) Al3+

(g) + e- H°ie = 5140 kJ/mol

Chem 59-250

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Chem 59-250

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Electron Attachment Enthalpy, H°ea

The enthalpy change for the gain of an electron

Cl(g) + e- Cl-(g) H°ea = -349 kJ/mol

O(g) + e- O-(g) H°ea = -142 kJ/mol

O-(g) + e- O2-

(g) H°ea = 844 kJ/mol

Electron Affinity, EA = -H°ea + 5/2 RT

EA = -H°ea

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Not easy to measure so many are missing

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Why should we care about these enthalpies?

They will provide us information about the strength ofbonding in solids.

NaCl(s)

Na(s) Na(g) Na+(g)

½ Cl2(g) Cl(g) Cl-(g)H°eaH°d

H°ieH°sub

H°f

Lattice Energy, U

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Bond Energy, EA-B

Diatomic:H-Cl(g) H(g) + Cl(g) H = 431 kJ/mol

Polyatomic:H-O-H(g) H(g) + O-H(g) H = 497 kJ/mol

O-H(g) H(g) + O(g) H = 421 kJ/mol

Thus:H-O-H(g) 2 H(g) + O(g) H = 918 kJ/mol

Average O-H bond energy = 918 / 2 EO-H = 459 kJ/mol

Chem 59-250H

H

NN

H

H

H2N-NH2(g) 4 H(g) + 2 N(g) H = 1724 kJ/mol

NH3(g) 3 H(g) + N(g) H = 1172 kJ/mol

Thus average N-H bond energy = 1172 / 3 EN-H = 391 kJ/mol

Since 1724 = 4 EN-H + EN-N

We can estimate N-N bond energy to be:

1724 – 4(391) = 160 kJ/mol

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MeC

Me

OH H+ Me C

O

Me

H

H

EH-H = 436 kJ/mol

EC=O = 745 kJ/mol

EC-H = 414 kJ/mol

EC-O = 351 kJ/mol

EO-H = 464 kJ/mol

Hrxn = E(bonds broken) – E(bonds formed)

Hrxn = (436 + 745) – (414 + 351+ 464) kJ/mol

Hrxn = -48 kJ/mol

Chem 59-250

Remember that such calculated bond energies can change

For H2N-NH2(g): EN-N = 160 kJ/mol

For F2N-NF2(g): EN-N = 88 kJ/mol

For O2N-NO2(g): EN-N = 57 kJ/mol

They are only a rough approximation and predictions must bemade cautiously.

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Free Energy Change, G = H - TS

At STP:G° = H° - (298.15 K) S°

The two factors that determine if a reaction is favourable:

If it gives off energy (exothermic)H = Hproducts - Hreactants

H < 0

If the system becomes “more disordered”S = Sproducts - Sreactants

S > 0

If G < 0, then reaction is thermodynamically favourable

Chem 59-250

G lets us predict where an equilibrium will lie throughthe relationship:G = -RT ln K

aA + bB + cC + … hH + iI + jJ + …

K [ ] [ ] [ ]

[ ] [ ] [C ]

H I J

A B

h i j

a b c

So if G < 0, then K > 1 and equilibrium lies to the right.

There are three possible ways that this can happen with respect to H and S.

Chem 59-250If both enthalpy and entropy favour the reaction:i.e. H < 0 and S > 0 then G < 0.

S(s) + O2(g) SO2(g) H° = -292.9 kJ/molTS° = 7.5 kJ/mol

G° = -300.4 kJ/mol

If enthalpy drives the reaction:i.e. H < 0 and S < 0, but |H| > |TS|, then G < 0.

N2(g) + 3 H2(g) 2 NH3(g) H° = -46.2 kJ/molTS° = -29.5 kJ/mol

G° = -16.7 kJ/mol

If entropy drives the reaction:i.e. H > 0 and S > 0, but |H| < |TS|, then G < 0.

NaCl(s) Na+(aq) + Cl-(aq) H° = 1.9 kJ/mol

TS° = 4.6 kJ/molG° = -2.7 kJ/mol

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ln ln lnK

K

H

R

1

T

1

T1 2

1

2

o

1 2K K

How do people obtain these values?

Measure change in equilibrium constants with temperatureto get H° using the relationship:

Measure the equilibrium constant for the equilibrium,then determine G° using the relationship ? :

G° = -RT ln K

Often not that easy…

Chem 59-250Reduction-Oxidation (RedOx) reactions:

Reduction – gain of electronsOxidation – loss of electrons

E°, the standard potential for an equilibrium,gives access to G° through the following relationship:

G° = - nF E°where,n = number of electrons involvedF = Faraday’s constant = 96.4867 kJ mol-1 V-1 (e-)-1

Note: if G° < 0, then must be E° > 0

So favourable reactions must have E° > 0

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Half-Cell Reduction Potentials

Al3+(aq) + 3 e- Al(s) E° = -1.67 V

Sn4+(aq) + 2 e- Sn2+

(aq) E° = 0.15 V

thus for:

2 Al(s) + 3 Sn4+(aq) 2 Al3+

(aq) + 3 Sn2+(aq)

E° = -(-1.67 V) + (0.15 V) = 1.82 V for 6 electrons

So: G° = - nF E° = - (6 e-)F (1.82 V) = -1054 kJ/mol

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Oxidation state diagrams (Frost Diagrams)

Relative Energy vs. Oxidation State (under certain conditions)

Provides:- Relative stability of oxidation states

-Energies available orrequired for RedOx reactions(the slope between reactantand product)

Chem 59-250 Oxidation state diagrams (Frost Diagrams)Some important information provided by Frost diagrams:

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The diagram for Mn displays many of these features.

Oxidation state diagrams (Frost Diagrams)

The most useful aspect of Frost diagrams is that they allow us to predict whether a RedOx reaction will occur for a given pair of reagents and what the outcome of the reaction will be. This is described in the handout.