Chapters 9 & 10 - Molecular Bonding, Geometry and Orbitals

Chapters 9 & 10: HW 9 & HW 10 -

Are due before Thursday pm, 12/6/2012

Final Exam on Tuesday 12/11/2012 8-10 am

Homework: Chapter 9 pp 366 to 372

6 8 11 12 13 16 18 19 21 22 29 39

45 48 52 55 61 63 65 67 71 75

85 97 110 117

Homework: Chapter 10 pp 411 to 417

4 5 7 8 9 10 17 - 19 (BeF2 is molecular) 22 24

Use molecular geometries when they call for geometries: 33 - 35 37

40&45(use pg 413) 49 53(hint: count pairs of Ve) 67 69

I. Chemical Bonding - A. Introduction• Atoms combine to produce new larger units, & the outermost e -

(valence e-) are mainly responsible for this chemical bonding. • Valence electrons can be:

– 1) Completely transferred to produce ions = Ionic Bonds– 2) Shared equally between atoms = Covalent Bonds– 3) Unequally shared = Polar Covalent Bonds.

Transfer of e- Unequal Sharing of e- Equal Sharing of e-

+ & - Ions ∂+ ∂- No Charges

Ionic Bonds Polar Covalent Bonds Covalent Bonds

Metal & Nonmetal Nonmetals of Nonmetals of Different electronegativ. Similar electronegativ.

NaCl, LiF HCl CO H2 F2 IBr CH4

I. Ionic Bonds B. Review

- Metals become + ions by losing electrons to have the electron configuration of the nearest inert gas. The charges for groups 1, 2 & 13 are +1, +2 & +3.

- Since some transition metals form multi charges, you need to memorize some of the more common cations:

Fe+2 Fe+3 Cu+1 Cu+2 Hg+2 Hg2+2 Zn+2 Ag+1

Cr+3 Cr+6 Pb+2 Pb+4 Sn+2 Sn+4

- Nonmetals gain electrons to mimic the nearest inert gas, & the general charges for groups 15, 16 & 17 are -3, -2 & -1.

- Molecular compounds have no charges & they share electrons.

I. Ionic Bonds B. Review

- the resulting anions & cations attract each other in such a ratio that the charges cancel out.

- Note: Do not show the charges in the final product. Example: KI NOT K+I-

- Example: Ba+2 & F- - Need two negatives to neutralize +2 charge on barium ion: Ba+2 F-1 F-1 = BaF2

- Nonmetals have too high of an ionization energy so they do not form stable positive ions (cations).

- Ionic Bond Definition: bond formed by electrostatic attraction between + & - ions.

I. Ionic Bonds C. Examples

• Examples:

Na Na+ O O-2 Na+ + Na+ + O-2 Na2OCa Ca+2 F F- Ca+2 + F- + F- CaF2

Mg Mg+2 S S-2 Mg+2 + S-2 MgS

Give the formulas for the following:

Na & Cl Na & N Ba & ILi & O Al & F Mg & NFe2+ & S Fe3+ & S

AnswersNaCl Na3N BaI2 Li2O AlF3 Mg3N2 FeS Fe2S3

I. Ionic Bonds D. Electron Dot (Lewis) Structures

- A Lewis electron dot structure is a symbol in which the valence electrons are shown as dots.

- Examples:

Na. Na+ Mg: Mg+2 Al3+

H+ H:1- (Called Hydride) :Se::::I::-1

- How many valence electrons (dots) would:

N3- O2- F- & Ne have? What about Ba+2 ?

Answers = 8 for first four & 0 for Ba+2

I. Ionic Bonds D. Electron Dot (Lewis) Structures for neutral atoms

II. Covalent Bond A. Introduction EN = electronegativity

- When two atoms of very different EN combine (metal & nonmetal), electrons are transferred, charges are formed and ionic bonds result.

- When two atoms of similar EN combine, neither has “pull” to take electrons away, and a sharing of electrons results. This occurs when a nonmetal combines with nonmetals.

- H. + .H ---) H—H equals H2 Note: H is a nonmetal

- Definition of a covalent bond: a bond formed by the sharing of two electrons. One covalent bond represented with a line; a double bond with two lines; a triple bond with three lines.

- Atoms share valence e’s to get stable e- configurations; elements in rows ≥ 3 use d orbitals & can have >8 bonding e. Examples: PF5 SF6 XeF4

II. Covalent Bonds A. Introduction

- There can be one, two or three covalent bonds between two atoms:

1) A shared pair of e’s is indicated with a line — (Single Bond)

2) Two shared pairs = (Two Bonds or a Double Bond)

3) Three shared pairs (Three Bonds or a Triple Bond)

- One shared electron from each bonding atom is a normal covalent bond.

- Two from one bonding atom and none from the other is called a Coordinate Covalent Bond; both types result in sharing a pair of electrons and is one covalent bond.

- We can show the result as a Lewis Structure: covalent bonds with lines and nonbonding valence electrons as dots; any charges are also shown.

- If we can draw a Lewis Structure, then the molecule or polyatomic ion probably exists.

- Molecules & polyatomic ions use covalent bonds.

- Generally in neutral, stable structures:Group 14 forms 4 bondsGroup 13 “ 3 bonds Group 16 “ 2 bonds Group 17 & H “ 1 bond

II. Covalent Bonds A. Introduction

II. Covalent Bonds B. Examples

H. + F::: ---) H F:::

H. + O + .H ---) H O H

:N + N: ---) :N N:

:::Cl. + .O. + .Cl::: ---) :::Cl O Cl:::

::O: + :C: + :O:: -----) ::O = C = O::

II. Covalent Bonds C. Lewis Structures 1. Rules

Note: Rules 1 and 4 are double checks on your structure.

1. Calculate total # of valence electrons - take into account the charge if the sample is a polyatomic ion.

2. a) Place atom that forms most bonds at center (Closest to Group 14 & Lowest if in same group). b) If is a charge, then take this into account on the central atom.

3. Arrange other atoms around central atom & allow sharing so that each atom has stable electron configuration. Show bonding pairs as lines & nonbonding valence e- as dots.

4. Make sure that: a) each atom has a stable electron configuration & b) you still have the same number of valence electrons that you started with in step #1.

II. Covalent Bonds C. Lewis Structures 2. Examples

Notes: - on acids the acidic H’s are on O atoms.

- on ionic or charged structures show the charge.

H2O HI BaF2 Br2 BF4-

O3 CO2 CO3-2 NO2

- SO4-

2

NH4+ C2H6 C2H4 C2H2 NF3

H2CO POCl3 H3PO4 H2SO4 OH-

II. Covalent Bonds D. Miscellaneous1) Electronegativity (EN) & Polar Covalent Bonds

- Bonds formed between atoms of different EN (not different enough to form ionic bonds) - result in the bonding electrons spending more time around the more negative element. Partial charges result: ∂+ and ∂-

- Polar Covalent or Polar Bond is a covalent bond in which there is unequal sharing of the bonding electrons.

- One part of molecule will be more positive and other part will be more negative = dipole. ∂+ ∂-

- Note that ‘e tugs’ may cancel out on a symmetrical molecule yielding a nonpolar molecule.

- Examples: H-Cl F-I Se=O F-F- Which one of the above molecules is NOT polar?Answer = F-F (equal & opposite ‘tugs’ on bonding e’s)

II. Covalent Bonds D. Miscellaneous1) Electronegativity & Polar Covalent Bonds

• There may be unequal sharing with covalent bonds resulting in a slight separation of charge as with H - Cl

H Cl:::

II. Covalent Bonds D. Miscellaneous1. Electronegativity & Covalent Bonds

- Note the transition in bonding with respect to the electrons:

Transfer Unequal Sharing Equal Sharing

Ions Partial Charges No Charge

Ionic Polar Covalent Covalent

Electronegativity #

II. Covalent Bonds D. Miscellaneous2. Formal Charges

- Formal Charge Definition: Hypothetical charge assuming bonding electrons are shared & n electrons belong to the given atom.

- Useful in 1) organic chemistry & 2) when have two or more possible Lewis Structures, then LS with lowest FCs most likely.

- Draw the Lewis Structure & use following on a selected atom:

FC = # VE - 1/2 # BE - # n

VE = Valence ElectronsBE = Bonding Electronsn = Nonbonding Valence Electrons

- Examples: H3O: FC on O = 6 - 1/2(6) - 2 = +1 H2N:: FC on N = 5 - 1/2(4) - 4 = -1

II. Covalent Bonds D. Miscellaneous 3) Bond Lengths, Energy & Vibration

- Bond Length = Average distance between nuclei in a bond.

- Typical bond length ~ 100 to 200 pm.

- Bonding nuclei vibrate about average length and can cause bonds to change vibrational energy levels with Infrared EMR.

- λ of EMR absorbed characteristic of the bond, ie C-H absorbs

at one λ and C=O absorbs at another λ. Can tabulate λ’s and use for ID.

- Can determine the λ of EMR absorbed with IR spectrophotometer. From result can readily tell which chemical bonds are present or absent.

II. Covalent Bonds D. Miscellaneous 3) Typical Infrared Spectrum & ID of possible compound

C-H stretch

C-H bend

(Related to λ or ν)

stretch

stretch

II. Covalent Bonds D. Miscellaneous 3) Bond Lengths, Energy & Vibration

- Bond Energy = Energy needed to break a bond in the gaseous phase. ∆H (enthalpy) = + value for breaking bond (adding energy) & - for forming bond (releasing energy).

- Typical bond energies ~ 150 to 1000 kJ/mol

- Use bond energies (Table 9.5, pg 361) to calculate other bond energies and heats of reaction (∆Hr).

- ∆Hr = ∑ ∆Hbonds broken - ∑ ∆Hbonds formed

II. Covalent Bonds D. Miscellaneous 3) Bond Energy Calculations

∆Hr = the change in heat energy for a reaction ∆Hr = ∑ [∆Hbonds broken] - ∑ [∆Hbonds formed]

Example 1: Calculate the ∆H(rxn) for following: 1 H2 + 1 Cl2 ------) 2 HCl

- use Table 9.5; p 360; the bond energies/mol are:[ H-H = 432 ; Cl-Cl = 240; H-Cl = 428 kj/mol ]

1 H-H 1 Cl-Cl 2 H-Cl

∆Hr = [432 + 240] - [2 x 428] = -184 kJ

- 184 kJ Heat released = exothermic rxn

II. Covalent Bonds Example 2: Calculate the ∆Hr

1CH4 + 4 Cl2 -----) 1CCl4 + 4 HCl

C-H = 411 kJ Cl-Cl = 240 kJ C-Cl = 327 kJ H-Cl = 428 kJ

1CH4 + 4 Cl2 -----) 1CCl4 + 4 HCl

4x411 4x240 4x327 4x428

1644 960 1308 1712

∆Hr = ∑ [∆Hbonds broken] - ∑ [∆Hbonds formed]

∆Hr = [1644 + 960] - [1308 + 1712]

∆Hr = [ 2604 ] - [ 3020 ] = - 416 kJ

II. Covalent Bonds D. Miscellaneous4. Resonance

Resonance - Situation in which there are two or more structures which only differ in the placement of the double bonds. Have extra stability when this occurs.

- Example: O3

::O = O - O::: :::O - O = O::

- Note the delocalized double bond.

- Real structure is a hybrid between the two, and the bond lengths are part way between O=O & O-O

O - O – O

II. Covalent Bonds D. Miscellaneous4. Resonance

Kekule

II. Covalent Bonds D. Miscellaneous5. Exceptions to Octet Rule

- There are exceptions to the “closest inert gas” rule:

1) Covalent compounds containing Group 13 atoms are satisfied with 6 valence electrons (not 8).

- Example: BI3 (is covalent & 6 electrons around B)

2) Elements ≥ row 3 can use s, p & d orbitals and have > 8 VE.

- Covalent Examples: SF6 PF5 XeF4

- Ionic Examples: Transition metals: Many use d subshells in

addition to their Valence e- . Mn+2 & Mn+5 as examples:

Mn0 = [Ar] 4s2 3d5 Mn+5 = [Ar] 4s2 Mn+2 = [Ar] 3d5

III. Shapes of Molecules A. Introduction

- Molecular Shapes play a major role in:

1) Physical Properties

2) Chemical Properties

3) Biological Properties

Examples: - Benzene vs cyclohexane- Cis verses Trans fats

- Two 3-D forms (R & S) of ibuprofen

III. Shapes of Molecules A. Introduction

• Two classes of molecular shapes:

1) Electron Geometry - the general shape of the molecule or polyatomic ion as determined by the positions of the bonding and nonbonding valence electron pairs. There are five general shapes.

2) Molecular Geometry - the general shape of the molecule or polyatomic ion as determined by the positions of the bonding atoms. There are over 10 different shapes/names.

III. Shapes of Molecules B. The Valence Shell Electron Pair Repulsion (VSEPR) Model

- VSEPR predicts overall shapes of molecules by assuming that pairs of valence electrons try to get as far apart as possible.

- To predict either the Electron or Molecular Geometry, draw the Lewis Structure and note how many “things” (atoms & pairs of non-bonding valence e- = n) are about the central atom.

- Example: H - O - H Four “things” around O = 2 atoms & 2 n

Electron Geometries

- Two things = Linear, Bond Angles of 180o

- Three things = Trigonal Planar (Trigonal), B Angles of 120o

- Four things = Tetrahedral, Bond Angles of 105-109o

- Five things = Trigonal Bipyramidal, B A 90 & 1200

- Six things = Octahedral, Bond Angles of 900

III. Molecular Shapes B. VSEPR 1. Electron geometry Pg 374 - 10.2 (pairs = things)

III. Molecular Shapes B. VSEPR 2. Molecular Geometry Pg 376 - 10.4 (my terminology)

Electron Geometry Molecular Geometry

(Pyramidal)

(Trigonal)

(Bent)

(Bent)

III. Molecular Shapes B. VSEPR 2. Molecular Geometry Pg 381 - 10.8

Electron Geometry Molecular Geometry

III. Molecular Shapes B. VSEPR 2. Molecular Geometry Pg 381 - 10.8

Electron Geometry Molecular Geometry

III. Molecular Shapes B. VSEPR 3. Polarity of Molecules

- To have a polar molecule one must have BOTH:

1. One or more polar bonds

2. A geometry that does not cancel

- Polarity is like a vector; opposite pulls can cancel separation of charge.

- Examples:

O=C=O - Nonpolar due to opposite pulls (linear).

CF4 - Nonpolar due to opposite pulls (tetrahedral).

:NF3 - Polar; e- pulled to bottom of the pyramid.

IV. Hybrid Atomic Orbitals (HAO)A. Introduction

- There are two theories to account for bonds:

1. Valence Bond Theory - Assumes that bonds involve only valence electrons.

2. Molecular Orbital Theory - Assumes that all electrons are involved in bonding.

- We use both theories.

- The Valence Bond Theory allows atoms to rearrange the valence orbitals to form Hybrid Atomic Orbitals (HAO) in order to best fit the bonding situation.

- The # of HAO equals the AOs involved, & all new HAO have similar energies & shapes.

HAO - Introduction

H

CH H

HC

H

H

H

H

C:2s 2p

1s

4 x H:

+

C forms 4 bonds to H

Electron configuration of C shows only 2 places to bond

and these are 90o apart.

but . . .

1) How does carbon form 4 equal bonds?2) Why are bond angles109.5o ?

IV. Hybrid Atomic OrbitalsB. SP3 C, O, N Examples

1) SP3 - An s combines with three p AOs to yield four sp3 HAOs. C, O, N use SP3 HAOs when four “things” are attached or when forming only single bonds. Example with carbon:

. .

2p __ __ __ . . . .

. . ------) ___ ___ ___ ___

2s ___ sp3

- These four SP3 HAO are 109o apart, and each can form 1 covalent bond.

- C, N & O use SP3 HAO when forming only single bonds.

sp3 Hybridization of Atomic Orbitals

+

+ +

4 atomic orbitals2s + 2px + 2py + 2pz

yield4 hybrid atomic orbitals

4 sp3 orbitals

Four SP3 HAOs directed to the corners of a tetrahedron

C:s p 4 sp3

109.5o

IV. Hybrid Atomic OrbitalsB. SP2 C, O, N Examples

2) sp2 - an s combines with two p AO to form three sp2 HAO. This frequently leaves one p AO left over to form a double bond. The overall shape is trigonal planar. Carbon Example:

. .

2p __ __ __ . . . .

. . ------) ___ ___ ___ ___2s ___ sp2 p

- The three sp2 HAO are 120o apart. The p AO is 90o from plane of sp2 HAOs and is used to form a double bond.

- C, O & N use SP2 when forming one double bond.

sp2 Hybridization of Atomic Orbitals

4 atomic orbitals{2s + 2px + 2pz }, 2py

yield 3 sp2 hybrid orbitals+ 1 py

sp2 HAO directed to the corners of a triangle (trigonal

planar)

C:s p psp2

+

+

+

120o

IV. Hybrid Atomic OrbitalsB. SP1 C, O, N Examples

3) sp or sp1 - an s combines with one p AO to form two sp HAO. This leaves two p AO left over to form a triple bond or two double bonds. The overall shape is linear. Example with Carbon:

. .

2p __ __ __ . . . .

. . ------) ___ ___ ___ ___2s ___ sp1 p p

- C, O & N use sp1 HAO when they form two double bonds or a triple bond.

IV. Hybrid Atomic OrbitalsB. Examples: C, N, O Summary

sp1, sp2, sp3 HAO. C, N, O always use these HAO when they form bonds.

Can predict which HAO used by C, N or O by drawing the Lewis Structure (LS) & using the following rules.

Rules applied to a given C, N or O atom in a NEUTRAL, STABLE molecule. Draw LS, then:

sp3 all single bonds (elec geom = tetrahedral; BA=105-109o)

sp2 one double bond (elec geom = trigonal planar; BA = 120o)

sp1 two double bonds (elec geom = linear; BA = 180o)

IV. Hybrid Atomic OrbitalsB. Examples: C, N, O Summary

Predicting HAO for neutral, stable compounds:- Draw Lewis structure.- Note how many double bonds- Use the rules given on previous slide.

Examples: Predict HAO on C, N & O

1) CH3-CH=O:: 2) CH2=C=CH2

..

3) CH3-NH–CH3 4) H – C N:

Answers: 1) sp3 sp2 sp2 2) sp2 sp1 sp2

(H uses 1s) 3) sp3 sp3 sp3 4) sp1 sp1

N C

H

CH

C

H O

OH

CC

CC

C

C

HH

O

H

H

H

H

H

Assigning HybridizationAssign hybridization to the C, O & N atoms

sp3

sp3 sp2sp2

sp3

sp3

sp3

sp2

sp2sp2

sp2

sp2

H C

H

H

I

sp3

C C C

H

H

H

H

C C C

H

H

H

H

sp sp sp3

sp2

sp

sp2

Summary of HAO for N, O, C Atoms in neutral, stable moleculesUseful for Organic Molecules

Draw Lewis Structure, and note the # of double bonds

# Double Bonds HAO Used Bond Angle

0 sp3 109 – 104o

1 sp2 120o

2 (or 1 triple) sp1 180o

sp3d and sp3d2 are formed when an atom needs 5 or 6 new orbitals. Draw Lewis Structure & count # of “Things” attached to atom of interest [“Thing” = a) pairs of nonbonding valence e & b) atoms]. Then use this chart to predict HAO used.

Examples: SeCl6 XeF2

Answers: sp3d2 sp3d1

Summary of HAO for All Atoms

# “Things” Hybridization

5 sp3d

6 sp3d2

IV. Hybrid Atomic OrbitalsC. Multiple Bonds

- There are two general categories of covalent bonds:

1) Sigma - a bond in which the e density is cylindrically symmetrical. Formed from all AO & HAO overlaps except a p atomic orbital overlapping with a p atomic orbital sideways.

2) Pi - a bond in which the e density is not cylindrically symmetrical. Only formed from sideways overlap of two p AOs.

-Note: Sigma bonds are stronger than Pi bonds.

- A single bond = 1 sigma

- A double bond = 1 sigma & 1 pi (used sp2 HAO)

- A triple bond = 1 sigma & 2 pi (used sp1 HAO)

V. Molecular Orbital Theory

- A theory in which one assumes that all electrons are influenced during bonding and that the new molecular orbitals are spread over the entire molecule.

- AO overlap to produce MO, and the new MO are classified as bonding and antibonding sigma (σ) and pi (π) bonds. # MO produced = # AO used from both bonding units.

- The MO have unique energy levels.

Molecular Orbital Theory

Two ways to describe bonding:1) Valence Bond Theory (Already Discussed)2) Molecular Orbital Theory (MO Theory)

MO Theory results from a rigorous, mathematical description of bonding

VB Theory - overlap of atomic orbitals, e- density localized in bondMO Theory - combination of atomic orbitals to form molecular orbitals, e-

density spread throughout entire molecule

Two atomic orbitals on different atoms combine to form two molecular orbitals:1. A low-energy bonding orbital2. A high-energy antibonding orbital

H 1s orbital H 1s orbital

*

Simple Molecular Orbital Diagram: H2

H-H Bonding MO(filled)

H-H Antibonding MO(unfilled)

node

Constructive combination of atomic orbitals; lower

energy

En

erg

yDestructive combination of atomic orbitals; higher

energy

In stable bonding situations, usually only the bonding orbitals (, ) are filled

(draw MO diagram for He2 & He2+ Which is possible; ~ stable)

s + s sigma () bond