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8–1 Chapter 9 and 10 Bonding and Molecular Structure

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Chapter 9 and 10. Bonding and Molecular Structure. Valence and Core Electrons. Valence Electrons – electrons that participate in bonding Core Electrons – electrons in an atom that do not participate in bonding Main Group Elements – s and p orbitals - PowerPoint PPT Presentation

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Page 1: Chapter 9 and 10

8–1

Chapter 9 and 10

Bonding and Molecular Structure

Page 2: Chapter 9 and 10

8–2

Valence and Core Electrons

• Valence Electrons – electrons that participate in bonding

• Core Electrons – electrons in an atom that do not participate in bonding

• Main Group Elements – s and p orbitals

• Transition Elements – ns and (n-1) d orbitals

Page 3: Chapter 9 and 10

8–3

Lewis Dot Symbols for Elements

• Element's symbol represents the atomic nucleus together with the core electrons.

• Valence electrons are represented by dots that are placed around the symbol.

• These symbols emphasize the ns2np2 “octet” that all noble gases except helium possess.

Page 4: Chapter 9 and 10

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Lewis Dot Symbols for Main Group Elements

Page 5: Chapter 9 and 10

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Chemical Bond Formation Ionic Bonds

• Electrons are strongly displaced toward one atom and away from the other.

• Generally involve metals from the left side of the periodic table interacting with nonmetals form the far right side.

Page 6: Chapter 9 and 10

8–6

Lew Symbols and the Octet Rule for Ionic Compounds

• The electron configuration of many substances after ion formation is that of an noble gas octet rule.

• Octet rule: Main-group elements gain, lose, or share in chemical bonding so that they attain a valence octet (eight electrons in an atoms valence shell).

Page 7: Chapter 9 and 10

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Example 1

• The electron configuration of each reactant in the formation of KCl gives:

– K+ is that of [Ar] – Cl is also that of [Ar].

• The other electrons in the atom are not as important in determining the reactivity of that substance.

• The octet rule is particularly important in compounds involving nonmetals.

Page 8: Chapter 9 and 10

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Energy in Ionic Bonding

• When potassium and chlorine atoms approach each other we have:

K(g) K+(g) + e Ei = +418 kJ

Cl(g)+ e Cl(g) Eea = 349 kJ

K(g)+Cl(g) K+(g) + Cl(g) E = + 69 kJ

Page 9: Chapter 9 and 10

8–9

Driving Force of Ionic

• Positive E = energy absorbed energetically not allowed.

• Driving force must be the formation of the crystalline solid.

K+(g) + Cl(g) KCl(s)

Page 10: Chapter 9 and 10

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Born-Haber Cycle and Lattice Energies

• Overall energetics for the formation of crystalline solids can be determined from a Born-Haber cycle which accounts for all of the steps towards the formation of solid salts from the elements. For the formation of KCl from its elements we have:

kJ. KCl(s) (g)Cl/ K(s) nergies ions and e Sum react kJ KCl(s) (g)Cl(g) K KCln of solid. Formatio

kJ. Cl eCl )eaEide ions (n of chlor. Formatio

kJ e K K(g) ssium on of pota. Ionizati

kJ Cl(g) (g) Cl/ lorine tion of ch. Dissocia kJ. K(g) (s) K he metal . Sub of t

4434 21 715 5

6348 4

418 3

122 212289 1

2

2

Page 11: Chapter 9 and 10

8–11

Born-Haber Cycle

• Net energy change of 434 kJ/mol indicates energetically favored.

• Energy for the fifth step is the negative of the • lattice energy: energy required to break ionic

bonds and sublime (always positive).

Page 12: Chapter 9 and 10

8–12

Example 2

Determine the lattice energy of BaCl2 if the heat of sublimation of Ba is 150.9 kJ/mol and the 1st and 2nd ionization energies are 502 and 966 kJ/mol, respectively. The heat for the synthesis of BaCl2(s) from its elements is 806.06 kJ/mol.

Page 13: Chapter 9 and 10

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Page 14: Chapter 9 and 10

8–14

LATTICE ENERGIES AND PERIODICITY

• Lattice energy can also be determined from Coulomb’s law:– Directly proportional to charge on each

ion. – Inversely proportional to size of compound

(sum of ionic radii).

Page 15: Chapter 9 and 10

8–15

Lattice Energies

• Table (right) presents the lattice energies for alkali and alkaline earth ionic compounds. The lattice energies– decrease for compounds of

a particular cation with atomic number of the anion.

– decrease for compounds of a particular anion with atomic number of the cation.

Lat. E, kJ/mol

Lat. E, kJ/mol

LiF 1030 MgCl2 2326

LiCl 834 SrCl2 2127

LiI 730 MgO 3795

NaF 910 CaO 3414

NaCl 788 SrO 3217

NaBr 732 ScN 7547

NaI 682

KF 808

KCl 701

KBr 671

CsCl 657

CsI 600

Page 16: Chapter 9 and 10

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Covalent Bond Formation

• Electrons involved in the bond are more or less evenly distributed between the atoms, and electrons are shared by two nuclei.

• Covalent bonding generally occurs between nonmetals, elements that lie in the upper right corner of the periodic table.

Page 17: Chapter 9 and 10

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The Covalent Bond

• Repulsive forces of the electrons offset by the attractive forces between the electrons and the two nuclei.

• Bonds are characterized in terms of energy and bond distance

Page 18: Chapter 9 and 10

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Strengths of Covalent Bonds:

• Bonds form because their formation produces lower energy state than when atoms are separated.

• Breaking bonds increases the overall energy of the system. Energy for breaking bonds has a positive sign (negative means that energy is given off).

H - H (g) 2H(g) H = 436 kJ.

Page 19: Chapter 9 and 10

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Ionic vs Covalent Bond

– Ionic compounds have high melting and boiling points and tend to be crystalline

– Covalently bound compounds tend to have lower melting points since the attractive forces between the molecules are relatively weak.

Page 20: Chapter 9 and 10

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Lewis Structures

• Lewis structures are used to indicate valence electrons and bonds

Page 21: Chapter 9 and 10

8–21

Drawing Lewis Structures

• Determine the arrangement of the atoms in the compound with respect to each other and draw a skeletal structure.

• If the compound is binary, the first element written down is usually the central atom (hydrogen is an exception to this).

• With a ternary compound (one with three kinds of elements) the middles atom in the formula is usually the central one.

Page 22: Chapter 9 and 10

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Lewis Structures Cont.

• Determine the total number of valence electrons.

• Subtract two electrons from the valence total for each bond in the skeletal structure.

Page 23: Chapter 9 and 10

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Lewis Structures Cont.

• Determine how many electrons are required for each element to have a total of eight (there are several exceptions to this rule).

• If a sufficient number of electrons are available, distribute the remaining electrons around the element symbols.

• If a sufficient number is not available, add additional bonds making certain to subtract the electrons used. (Multiple bonds often occur among the atoms C, N, and O.

Page 24: Chapter 9 and 10

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Example 3

• Draw Lewis Electron Dot Structures for the following

Page 25: Chapter 9 and 10

8–25

Lewis Structures and Resonance

• Quantum theory indicates that any position is possible for an electron.

• Equivalent electron positions often possible:

• E.g. SO2 : O=S-O and :O-S=O.

– Each structure equally likely. – the true form of the molecule is a hybrid of these

and is called resonance and the hybrid form is called a resonance hybrid.

Page 26: Chapter 9 and 10

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Example 4

• Draw Lewis Electron Dot Structures for the following

Page 27: Chapter 9 and 10

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Exceptions to the Octet Rule

• Although many molecules obey the octet rule, there are exceptions where the central atom has fewer or more than eight electrons.

• Compounds in which an atom has fewer than eight valence electrons

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• Compounds in which an atom has more than eight valence electrons

– Generally, if a nonmetal is in the third period or greater it can accommodate as many as twelve electrons, if it is the central atom.

– These elements have unfilled “d” subshells that can be used for bonding.

Page 29: Chapter 9 and 10

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Free Radicals

• Molecules with an odd number of electrons

Page 30: Chapter 9 and 10

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Example 5

Draw Lewis Electron Dot Structures for the following XeF4, ICl3, and SF4

Page 31: Chapter 9 and 10

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FORMAL CHARGES

• Formal Charge (of an atom in a Lewis formula) the hypothetical charge obtained by assuming that bonding electrons are equally shared between the two atoms involved in the bond. Lone pair electrons belong only to the atom to which they are bound.

• Formal Charge = group number – number of lone pair electrons – ½ bonding electrons

Page 32: Chapter 9 and 10

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Example 6

Determine the formal charge on all elements: PCl3, PCl5, and HNO3.

Page 33: Chapter 9 and 10

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• Formal charge (FC) allows the prediction of the more likely resonance structure.

• To determine the more likely resonance structure:– FC should be as close to zero as possible.– Negative charge should reside on the most

electronegative and positive charge on the least electronegative element.

Page 34: Chapter 9 and 10

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Example 7

Draw the resonance structures of H2SO4; determine the formal charge on each element and decide which is the most likely structure.

Page 35: Chapter 9 and 10

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Bond Properties

• Bonds Order• Bond Length• Bond Energy• Bond Polarity

Page 36: Chapter 9 and 10

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Bond Order

• The order of a bond is the number of bonding electrons pairs shared by two atoms in a molecule.

• A fractional bond order is possible in molecules and ions having resonance structures

Page 37: Chapter 9 and 10

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Bond Length

• Bond length is the distance between the nuclei of two bonded atoms. Bond length is determined in a large part by the size of the atoms.

• Bond length becomes shorter as bond order increases

Page 38: Chapter 9 and 10

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Bond Dissociation Enthalpies

• Bond dissociation energy, BE – the energy required to break one mole of a type of bond in an isolated molecule in the gas phase.

• Useful for estimation of heat of unknown reactions.

Page 39: Chapter 9 and 10

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• Bond energies are all positive• The energies are average bond energies• Bond energies are defined in terms of

gaseous atoms of molecular fragments• Bond energies increase with bond order

Page 40: Chapter 9 and 10

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Page 41: Chapter 9 and 10

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• Hess’s law can be used with bond dissociation energies to estimate the enthalpy change of a reaction.

• The breaking in a C – H bond would be C – H(g) C(g) + H(g) H = BE = 410 kJ.

– Sign always positive since energy must be supplied to break bond.

Page 42: Chapter 9 and 10

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Using Bond Dissociation Enthalpies

• Estimate the heat of formation of H2O(g) from bond dissociation energies. Thus determine:

• H2(g) + ½ O2(g) H2O(g) = ?

  H – H (g) 2H(g) H = BE = 436 kJ ½ O=O O(g) H = BE = 494/2 = 247 kJ2H(g) + O(g) H – O – H (g) H = 2BE = 2*459 kJ  H2(g) + ½ O2(g) H2O(g) = 235 kJ

Actual = 241.8 KJ

• Can be determined by summing all the energies for the bonds broken and subtract from it the sum of the energies for the bonds formed.

Page 43: Chapter 9 and 10

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Example 8

Estimate the energy change for the chlorination of ethylene:– CH2=CH2(g) + Cl2(g) CH2ClCH2Cl

Page 44: Chapter 9 and 10

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Bond Polarity and Electronegativity

• Electronegativities

– increase from bottom to top of periodic table and – increase to a maximum towards the top right.

• can provide an insight as to the type of bond that would be expected.

Page 45: Chapter 9 and 10

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Electronegativities9_12

Li1.0

Na0.9

K0.8

Rb0.8

Cs0.7

Fr0.7

Be1.5

Mg1.2

B2.0

Al1.5

C2.5

Si1.8

N3.0

P2.1

O3.5

S2.5

F4.0

Cl3.0

Ca1.0

Sr1.0

Ba0.9

Ra0.9

Sc1.3

Y1.2

La–Lu1.1–1.2

Ti1.5

Zr1.4

Hf1.3

V1.6

Nb1.6

Ta1.5

Cr1.6

Mo1.8

W1.7

Mn1.5

Tc1.9

Re1.9

Fe1.8

Ru2.2

Os2.2

Co1.8

Rh2.2

Ir2.2

Ni1.8

Pd2.2

Pt2.2

Cu1.9

Ag1.9

Au2.4

Zn1.6

Cd1.7

Hg1.9

Ga1.6

In1.7

Tl1.8

Ge1.8

Sn1.8

Pb1.8

As2.0

Sb1.9

Bi1.9

Se2.4

Te2.1

Po2.0

Br2.8

I2.5

At2.2

IA IIA

IIIB IVB VB VIB VIIB IB IIB

IIIA IVA VA VIA VIIA

VIIIB

H2.1

Ac–No1.1–1.7

Page 46: Chapter 9 and 10

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Bond Polarity

• Ionic bonds formed when displacement of electrons is essentially complete

• Covalent bonds forms when no displacement of electrons occurs

• Polar covalent forms when bond pair is not equally shared between two atoms and the electrons are displace toward one of the atoms from a point midway between them

Page 47: Chapter 9 and 10

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Polar Bonds

With a polar bond exists between two atoms, a small charge on the atom due to that bond develops. + and designates which is the positive and negative side respectively

Page 48: Chapter 9 and 10

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Example 9

Determine the relative polarities of HF, HCl, HBr and HI.

Page 49: Chapter 9 and 10

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Molecular Geometry and Directional Bonding

• Atoms oriented in very well defined relative positions in the molecule.

• Molecular Geometry = general shape of the molecule as determined by the relative positions of the atomic nuclei.

Page 50: Chapter 9 and 10

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• Theories describing the structure and bonding of molecules are:– VSEPR = considers mostly electrostatics in

determining the geometry of the molecule.– Valence Bond Theory = considers quantum

mechanics and hybridization of atomic orbitals.– Molecular Orbital Theory = claims that upon bond

formation new orbitals that are linear combinations of the atomic orbitals are formed.

Page 51: Chapter 9 and 10

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Valence Share Electron Pair Repulsion (VSEPR) Theory

• Valence Share Electron-Pair Repulsion (VSEPR) model allows us to predict the molecular shape by assuming that the repulsive forces of electron pairs cause them to be as far apart as possible from each other.

Page 52: Chapter 9 and 10

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PREDICTING EXPECTED GEOMETRY ACCORDING TO

VSEPR THEORY.

• Lewis dot structure determines the total # of electrons around the central atom. Multiple bonds (double and triple) count as one.

• The number of bonding and nonbonding electron pairs determines the geometry of electron pairs and the molecular geometry.

• Only the valence electron pairs are considered in determining the geometry.

Page 53: Chapter 9 and 10

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Page 55: Chapter 9 and 10

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Page 57: Chapter 9 and 10

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Page 58: Chapter 9 and 10

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Page 59: Chapter 9 and 10

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Procedure

• Count the number of valence electrons and write the Lewis Structure

• Determine the number of unshared electron pairs, E, and the number of shared pairs, X.

• Determine the arrangement of bonds and unshared pairs that minimizes electron pair repulsions

• Describe the shape of the molecule in terms of the positions of the atoms

• Note: Multiple bonds are treated as single bonds• Lone e Pairs affect geometry more than bonding

Page 60: Chapter 9 and 10

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Page 61: Chapter 9 and 10

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Example 10

– BeCl2.and CO2

Page 62: Chapter 9 and 10

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Example 11

– BF3, COCl2, O3, SO2

Page 63: Chapter 9 and 10

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Example 12

– CH4, PCl3, H2O

Page 64: Chapter 9 and 10

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Example 13

– PCl5, SF4, ClF3, XeF2 (lone pair in axial position for a trigonal bipyramidal structure).

Page 65: Chapter 9 and 10

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Example 14

– SF6, IF5, XeF4

Page 66: Chapter 9 and 10

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Polarity of Molecules

• Bond dipole a positive charge next to a negative charge.

• Dipole moment, the magnitude of the net bond dipole of a molecule = Qxr Q = the net charge separation; r = the separation distance. Units: debyes (D) where 1 D = 33.36x1030 Cm.

Page 67: Chapter 9 and 10

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The DipoleThe Dipole    A dipole arises when two electrical charges of equal A dipole arises when two electrical charges of equal

magnitude but opposite sign are separated by distance.magnitude but opposite sign are separated by distance.

Page 68: Chapter 9 and 10

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• A polar bond forms when two atoms of between two atoms involved in a bond have significantly different electronegativities. – Most electronegative substance will have a slight negative

charge (represented as )– The positive (electron poor) side of the bond is

represented as + or points in direction of the negative charge.

• Net polarity (dipole moment) of a molecule is obtained using the vector sum of polarities of individual bonds.

Page 69: Chapter 9 and 10

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The sum of these vectors will give us the dipole for the The sum of these vectors will give us the dipole for the moleculemolecule

For a polyatomic molecule we treat the dipoles as 3D vectors

Page 70: Chapter 9 and 10

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Example 14

Determine if NH3, H2O, CO2 have dipole moments.

Page 71: Chapter 9 and 10

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MOLECULAR SHAPES:VALENCE BOND THEORY (VBT)

• Valence Bond Theory: a quantum mechanical description of bonding that pictures covalent bond formation as the overlap of two singly occupied atomic orbitals.

Page 72: Chapter 9 and 10

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• VSEPR effective but ignores the orbital concepts discussed in quantum mechanics.

• H2 forms due to overlap of two 1s orbitals.

• Electron densities from p-subshell electrons overlap to produce a bond in F2.

Page 73: Chapter 9 and 10

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Overlap of OrbitalsOverlap of Orbitals

Page 74: Chapter 9 and 10

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The point at which the potential energy is a minimum is called the equilibrium bond distance

The degree of overlap is determined by the system’s potential energy

equilibrium bond distance

Page 75: Chapter 9 and 10

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Hybrid Orbitals

• In molecules, the orbitals occupied by electron pairs

are seldom “pure” s or p orbitals. Instead, they are “hybrid” orbitals formed by combining s and p or s,p and d orbitals.

• Hybridization determined by using VSEPR to establish the geometry, i.e., the number of electron clouds around the central atom. The number of electron clouds = the number of hybrid orbitals.

Page 76: Chapter 9 and 10

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sp

Page 77: Chapter 9 and 10

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2s

Formation of sp hybrid orbitalsFormation of sp hybrid orbitals

The combination of an s orbital and a p orbital produces 2 new orbitals called sp orbitals.

Page 78: Chapter 9 and 10

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sp2

Page 79: Chapter 9 and 10

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Formation of spFormation of sp22 hybrid orbitalshybrid orbitals

Page 80: Chapter 9 and 10

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sp3

Page 81: Chapter 9 and 10

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Formation of spFormation of sp33 hybrid orbitalshybrid orbitals

Page 82: Chapter 9 and 10

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sp3d

Page 83: Chapter 9 and 10

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sp3d2

Page 84: Chapter 9 and 10

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Electron Geometry and Hybrid Orbitals

Page 85: Chapter 9 and 10

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Valence Bond Theory and Multiple Bonds Sigma bondSigma bond () A bond where the line of electron A bond where the line of electron density is concentrated symmetrically along the line density is concentrated symmetrically along the line connecting the two atoms. Only 1 sigma bond can exist connecting the two atoms. Only 1 sigma bond can exist between two atomsbetween two atoms

Page 86: Chapter 9 and 10

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Pi bondPi bond () A bond where the overlapping regions A bond where the overlapping regions exist above and below the internuclear axis (with a exist above and below the internuclear axis (with a nodal plane along the internuclear axis). Occur when nodal plane along the internuclear axis). Occur when multiple onds form between two atomsmultiple onds form between two atoms

Page 87: Chapter 9 and 10

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• C2H4 planar with a trigonal geometry = sp2 hybridization for each of the carbon atoms and they form bonds with hydrogen.

• Each carbon has 4 orbitals in its valence shell. This means one of the p-orbitals for each C is not hybridized.

• Proximity to each other results in overlap to give a charge distribution resembling a cloud which is above and below the plane of the molecule - a –bond

Page 88: Chapter 9 and 10

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HH22C=CHC=CH22

Page 89: Chapter 9 and 10

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HH22C=CHC=CH22

Page 90: Chapter 9 and 10

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C2H2

• sp (linear) hybridized. Leads to the existence of a bond as well as two bonds.

Page 91: Chapter 9 and 10

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Example: Example: HCHCCHCH

Page 92: Chapter 9 and 10

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Summarizing

• single bond is a bond, • double bond is a bond and a bond, • triple bond is a bond and 2 bonds.

Page 93: Chapter 9 and 10

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MO Theory of Bonding

• Molecular Orbital Theory extends quantum theory and states that electrons spread throughout the molecule in molecular orbitals = region in a molecule in which an electron is likely to be which is similar to the concept discussed in quantum theory. Molecular orbitals are considered to be the result of the combination of atomic orbitals.

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The one that is lower in energy is called the bonding orbital,

The one higher in energy is called an antibonding orbital.

These two new orbitals have different energies. 

BONDING

ANTBONDING

Molecular Orbital (MO) TheoryMolecular Orbital (MO) Theory

Page 95: Chapter 9 and 10

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Energy level diagrams / molecular Energy level diagrams / molecular orbital diagramsorbital diagrams

Page 96: Chapter 9 and 10

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MO Theory for 2nd row diatomic MO Theory for 2nd row diatomic moleculesmolecules

 Molecular Orbitals (MO’s) from Atomic Orbitals (AO’s) 1. 1. # of Molecular Orbitals = # of Atomic Orbitals# of Molecular Orbitals = # of Atomic Orbitals

2.2. The number of electrons occupying the Molecular orbitals is equal The number of electrons occupying the Molecular orbitals is equal to the sum of the valence electrons on the constituent atoms.to the sum of the valence electrons on the constituent atoms.

3.3. When filling MO’s the Pauli Exclusion Principle Applies (2 electrons When filling MO’s the Pauli Exclusion Principle Applies (2 electrons per Molecular Orbital)per Molecular Orbital)

4. 4. For degenerate MO’s, Hund's rule applies.For degenerate MO’s, Hund's rule applies.

5.5. AO’s of similar energy combine more readily than ones of different AO’s of similar energy combine more readily than ones of different energyenergy

6.6. The more overlap between AOs the lower the energy of the The more overlap between AOs the lower the energy of the bonding orbital they create and the higher the energy of the bonding orbital they create and the higher the energy of the antibonding orbital.antibonding orbital.

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Example: Example: LiLi22