Chemistry NYA Class Notes and Exercises Part 4

8/10/2019 Chemistry NYA Class Notes and Exercises Part 4

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Email: [email protected]: f-228Phone: x311

CHAMPLAIN COLLEGE

SAINT-LAMBERT

GENERAL CHEMISTRY

202-NYA-205Winter 2012

Joel Robichaud
mailto:[email protected]://www.champlainonline.com/mailto:[email protected]


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Course Outline

nomenclature empirical & molecularformulas stoichiometry gas laws

molarity

UNIT 1:Basics

UNIT 2:

Atomic Theory

history of atomic theory the Bohr atom the modern approach(quantum theory) quantum numbers

electron configurations electron affinity

UNIT 3:Periodicity &

Chemical Reactions

electron configuration & chemicalproperties of elements ionization energy atomic and ionic size electronegativity & electron affinity

reactions of the main group elementswriting molecular & net ionic equations

UNIT 4:Chemical Bonding

analysis of ionic & covalent bonding

writing Lewis structures, resonancestructures formal charges shapes of molecules bond angles bond polarity dipole moments hybridization theory orbital diagrams

UNIT 5:Intermolecular Forces

intermolecular forces intramolecular

bonds dispersion forces, dipole-dipole forces hydrogen bonding relationship of melting & boiling point& solubility to intermolecular forcesclassification of substances

UNIT 6:Liquids & Solutions

properties of solids

phase changes & phasediagrams physical properties ofsolutions concentration units colligative properties
http://www.champlainonline.com/


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describe the formation of a covalent bond between two atoms using the total interactive energy of the system. (9.4) interpret the information given in a potential energy vs. internuclear distance diagram (10.3) use Lewis symbols of the elements to explain formation of ionic compounds. (9.2) apply the octet rule and know its limitations. (9.6 & 9.9) define and understand the concept of covalent bonding. (9.4) compare and contrast covalent and ionic bonding. (9.2 & 9.4) define electronegativity and explain trends within the representative elements of the periodic table. (9.5) recognize polar and non-polar covalent bonds. (9.5) draw Lewis structures for covalent compounds and polyatomic ions. (9.6 & 9.7) define terms and recognize examples of structural and geometric isomers. (22.4 & Molecular Geometry experiment) draw resonance structures with arrow pushing. (9.8) determine formal charge. (9.7) explain how the concept of resonance can be used to explain bond energies, bond lengths, and stabilities of somemolecules & ions. (9.8) learn the basic geometric shapes and the angles associated with them. (10.1)

use the valence shell electron pair repulsion theory (VSEPR) for predicting molecular geometries. (10.1) predict the molecular geometry and bond angles for species with lone pairs on the central atom. (10.1) predict the polarity of molecules based on the overall geometry and bond polarities. (10.2) use the valence bond theory to explain covalent bonding. (10.3) apply hybridization theory to show how atoms maximize their bonding possibilities. (10.4) use orbital (box) diagrams to show how atoms form a particular set of hybrid orbitals. (10.4 & Molecular Geometryexperiment) determine the geometry & bond angles of any species using the concept of hybridization of the central atom. (10.4)

predict the hybridization of carbon in its compounds, indicating those containing multiple bonds. (10.4) draw an orbital overlap diagram for the following molecules: CH4, C2H4, C2H2, H2, X2, N2, O2 & CO2 (10.4)

Unit IV: Chemical Bonding (Chang, Ch. 9, 10 & 22)Objectives:


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Unit IV: Chemical Bonding (Chang, Ch. 9, 10 & 22)

4.1 Lewis Structures(Chang, 9.1, Textbook Problems, )

Lewis Dot symbol: Consists of the symbol of an element and one dot for each valenceelectron in an atom of the element.

Br

Lewis Structures: Shows the bonding arrangement of covalent compounds &polyatomic ions in a 2-dimensional picture (based on the conceptthat atoms search to complete their octet).

Br

Br

Ex.

Bonded electrons (a covalent bond) are represented by a dash.

A pair of electron (belonging only to one atom) are represented by a lone pair.

Ex.


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Continuation...In a molecule with a formulaAxn:

Ais usually the central atom to which X are attached.

Ais usually the one which can make the most bonds or the least electronegative.

When drawing a Lewis structure:

Draw a diagram connecting all the atoms with a SINGLE bond.

Add the # of valence electrons to find the total # of valence electrons.

Note: each covalent bond requires 2 electrons (subtract the electrons needed for covalent bonds).The other electrons complete the octet (each atom wants to see 8 electrons, Hydrogen 2).

4.1.1 Exercises:1. Complete the Lewis Diagram for the following molecules:

a) CH4 b) H2O c) NH3


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Exceptions to the Octet Rule

Compounds with Beryllium & Boron are content to not complete the octet:

Be is content to see 4 electrons B is content to see 6 electrons

If there are insufficient electrons to complete all octets, then you must

complete the octet of the most electronegative atoms first.

4.1.2 Exercises:

1. Complete the Lewis Diagram for the following molecules:

a) BeCl2 b) BeH2

c) BF3 d) BH3


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Charged Molecules

For every negative charge, one electron must be added to the total electron count.

For every positive charge, one electron must be subtracted to the total electron count.

4.1.3 Exercises:

1. Complete the Lewis Diagram for the following molecules:

a) H3O+ b) NH4

+

c) BCl4- d) NH2

-


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Atoms from the 3rd or Later Periods

If the central atom is from the 3rd period or later, it is allowed to go above the octet(atoms in the 2nd period NEVER go above the octet).

Complete the octet of outer atoms and if theres any extra [expanded octet]they are added to the central atom (only if it is from the 3rd period or more).


a) SCl4 b) ClF3

c) XeF2 d) BrF4+

e) SbCl5


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Double & Triple Bonds

Double bonds are createdwhen atoms dont complete their octet(most likely N, O, C & S).

Lone pairs are converted into a covalent bond.


a) O2 b) C2H4

c) H2CO d) N2

e) CO f) CN-


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4.2 Formal Charges

Formal charge: charge given to a particular atom in a Lewis diagram.

Neutral Removed 1 electron Added 1 electron

Formal charge = 0 Formal charge = +1 Formal charge = -1

Carbon 4 bonds

0 lone pair

3 bonds

0 lone pair

3 bonds

1 lone pair

Nitrogen 3 bonds

1 lone pair

4 bonds

0 lone pair

2 bonds

2 lone pairs

Oxygen 2 bonds

2 lone pairs

3 bonds

1 lone pair

1 bonds

3 lone pairs

C C C

N N N

O O O

Established by comparing the # of electrons before & after bonding(the sum of the formal charges must equal the charge of the compound).


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Continuation...

You can also use the formula:

Formal charge = Group # of Atom - # of unshared e- - # of covalent bond

Formal charge = Group # of Atom ( (# of shared electrons) + # of unshared e-)

or

4.2.1 Exercises:

1. Complete the Lewis Diagram for the following molecules and assign formal charges:

a) H2O b) H3O+

c) NH3 d) SbCl52-


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Continuation...

When there is a choice:

The Lewis structure which creates the least # of formal charge is preferred. The Lewis structure where the negative charge is on the most electronegative atomis preferred (or positive on the least electronegative).

4.2.2 Exercises:

1. Complete the Lewis Diagram for the following molecules and assign formal charges:

a) CO2 b) SCN-


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4.2.2 Exercises:2. Complete the Lewis Diagram for the following molecules and assign formal charges:

a) H2S b) AsF5-

c) BeF2 d) C2H2

e) TeF4 f) CS2

g) BrF3 h) ICl4-

i) N2H4 j) HCN

k) BrF5 l) XeF3+

m) XeF4 n) C2Cl4

o) BCl3 p) C2F2

q) N2H2


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4.3 Isomers

Isomers: compounds with the same molecular formula but different structures(atoms are connected differently, bonds have to be broken to get fromone to another).

Note: While single bonds can rotate, double bonds cannot.

4.3.1 Exercises:

1. Draw different isomers for the following molecules:

a) C2H2Cl2 b) C2H4O

c) C2H4ClBr d) C3H5F


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4.4 Resonance

Resonance: when it is possible to shift electrons around (i.e. delocalization) using

lone pairs & double bonds (both Lewis structures are equally plausible).

Note: The actual molecule is a hybrid of the resonance structures.

Resonance adds a lot of stability to the molecule.

4.4.1 Exercises:1. Draw the Lewis diagram (if not given) and resonance structures for the following:

a) b) NO2-

c) d) SO2

e) SO3 f) CO32-

C

C

C

C

C

C

H

H

H

H

H

H

N

O

O O

O O O


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4.5 Bond LengthLength of bonds is affected by the # of bonds between two atoms.

Bonds between 2 carbon atoms:

C2H6 = 0.154 nm

C2H4 = 0.134 nm

C6H6 = 0.139 nm

C2H2 = 0.120 nm

The more bonds between 2 atoms, the shorter the length of that bond.

Ex.

4.5.1 Exercises:1. Arrange the following in order of increasing bond length

a) The N-N bond in N2, N2H4, N2H2

b) The C-O bond in CH3OH, HCO2-, CO3

2-, H2CO, CO

c) The N-O bond in NO+, NO2-, NO3-


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4.5 Molecular Geometry, VSEPR Theory

Valence Shell Electron Pair Repulsion Theory (VSEPR Theory):

A molecule will adopt a3-dimensional shapewhere bonds & lone pairs,are as far away as possible from one another to minimize repulsion between them.

The next table describes the shape of molecules with a repulsion # from 2-6:

Arepresents the central atomB represents bondsE represents a lone pair of electrons

Note: Single, double or triple bonds all count as one(they all occupy the same region in space).

The electronic repulsion between lone pairs (LP) & bond pairs (BP) is:LPLP > LPBP > BPBP

The repulsion number (also known as the #of electron clouds)includes: Lone Pairs (LP) & Bond Pairs (BP).


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Linear

Trigonalbipyramidal

Octahedral

Trigonalplanar

Tetrahedral

Linear

Trigonalbipyramidal

Octahedral

Trigonalplanar

Tetrahedral

Molecular Geometry: No Lone Pair (LP)


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Linear (rep. # = 2)

Octahedral (rep. # = 6)

AB6AB

5

AB4AB3

AB2

Trigonal planar (rep. # = 3)

rigonal bipyramidal (rep. # = 5)

Tetrahedral (rep. # = 4)

Molecular Geometry: No Lone Pair (LP)


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When a lone pair is added, it requires a bit more room than a covalent bondtherefore all bond angles will become smaller.

H

C

HH

H

H

N

HH

H

N

HH

H

H

O

H

109.5 109.5 107.3 104.7

Ex.

Molecular Geometry: Introduction of Lone Pairs (LP)

Note 2: while the angles for a tetrahedral geometry without LP must be known (i.e. 109.5o),simply write < 109.5o for the angles of a tetrahedral geometry where a LP is present.

Note 1: angle values are only given for two adjacent atoms connected through a centralatom (i.e. never give angle values between a LP and another atom).


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Trigonal bipyramidal

Tetrahedral

Trigonalpyramidal

Seesaw

Bent

Trigonal planar

Bent

Tetrahedral

Molecular Geometry: With Lone Pair(s) (LP)


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Octahedral



Octahedral

Squareplanar

Squarepyramidal

Linear

T-shaped


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AB2E(rep. # 2)

AB3E(rep. # 4)

AB2E2(rep. # 4)

AB4E

(rep. # 5)


Tetrahedral

Bent

Trigonal

bipyramidal

Trigonalplanar

Tetrahedral

Trigonal

pyramidal

Bent

Seesaw
http://en.wikipedia.org/wiki/File:Seesaw-3D-balls.pnghttp://en.wikipedia.org/wiki/File:AX4E1-3D-balls.pnghttp://en.wikipedia.org/wiki/File:Pyramidal-3D-balls.pnghttp://en.wikipedia.org/wiki/File:AX3E1-3D-balls.pnghttp://en.wikipedia.org/wiki/File:Bent-3D-balls.pnghttp://en.wikipedia.org/wiki/File:AX2E2-3D-balls.pnghttp://en.wikipedia.org/wiki/File:Bent-3D-balls.pnghttp://en.wikipedia.org/wiki/File:AX2E1-3D-balls.png


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AB3E2

AB2E3

AB5E

AB4E2


Trigonalbipyramidal Linear

Octahedral

Octahedral

Trigonalbipyramidal

Squareplanar

Square

pyramidal

T-shaped
http://en.wikipedia.org/wiki/File:Square-planar-3D-balls.pnghttp://en.wikipedia.org/wiki/File:AX4E2-3D-balls.pnghttp://en.wikipedia.org/wiki/File:Square-pyramidal-3D-balls.pnghttp://en.wikipedia.org/wiki/File:AX5E1-3D-balls.pnghttp://en.wikipedia.org/wiki/File:Linear-3D-balls.pnghttp://en.wikipedia.org/wiki/File:AX2E3-3D-balls.pnghttp://en.wikipedia.org/wiki/File:T-shaped-3D-balls.pnghttp://en.wikipedia.org/wiki/File:AX3E2-3D-balls.png


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4.6 Polarity of MoleculesA molecule is considered to be polar if it has a net dipole.

Steps to follow:

4.6.1 Exercise:1. Determine if the following molecules are polar or non-polar, if they are polar then determine the

direction of the resultant.

a) H2O b) PF3c) BF3 d) HF

e) CO2 f) ClF3g) CH4

Avector is drawn for all bonds which have dipole, the arrow going towards the mostelectronegative atom (if the electronegativity difference is larger, the vector is also larger).

If after the addition of the bond dipole vectors, there is a net vector then the molecule ispolar, otherwise the molecule is non-polar.

Note: a molecule may be non-polar even though it has polar bonds.First draw the Lewis diagram of the molecule,Then do the VSEPR sketch,Then find all vectors,Then find the resultant.


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4.6.1 Exercises:2. For the following determine or draw: Lewis Structure

Repulsion number

Name of geometry without lone pairsVSEPR sketchName of actual geometryBond anglesIf the molecule is polar or non-polar

a) OF2 b) CH4

c) BeCl2 d) CH2F2

e) BF3 f) NH3

g) PCl5 h) ClF3

i) SF6 j) IF5

k) BrF4-

l) XeF2

m) XeF4 n) ClF4+

o) SO2 p) H2CO

q) HCN r) N2H4

s) CO2 t) NH2-

u) I3-


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4.7 HybridizationOnly electrons that are alone in an orbital can bind, all others form lone pairs.

How to explain the following molecule: CH4?

Hybridization splits electron pairs to other orbitals close

in energy in order to have the desired # of single electrons.

2p2

2s2C :

C* :

C* :

2p

2s

sp3

Excited state: promotion of an electron from s to p.

[He]

[He]

[He]

Hybridization: the half-filled orbitals are blended to obtain

degenerate orbitals (i.e. all of the same energy).

Result: formation of 4 equivalent orbitals.

Problem: as written, only 2 bonds can be made.

Problem: different orbitals different types of bonds.


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2p3

2s2 N : [He]

N* : [He]

In this case the electron pair does not need to be split.

sp3

MolecularGeometry

(excludes LP):

Electron pairArrangement(includes LP):

Many molecules however, possess not only bond pairs (BP), but also lone-pair s(LP).

Ex. NH3

Question: Is hybridization is still required?

< 109.5o

Note: a lone pair is represented by 2 electrons in an orbital.

4.7 Hybridization

Solution: hybridization affords 4 equivalent sp3 orbitalsand explains the tetrahedral geometry.

Note: one of the sp3

hybridized orbitals contains a Lone Pair (LP).


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4.7.1 Exercises:

1. Determine the hybridization of the orbitals in the following central atom in order to

obtain the molecules below.For each, give: # of covalent bonds

# of electron pairsOrbital box diagram:Excited State:Hybridized State:

a) BeH2 b) BH3

c) CH4 d) PCl5

e) SF4 f) SF6

g) XeF2 h) XeF4

i) KrF4 j) ClF3


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4.8 Valence Theory of BondingA bond is not actually a line, it is an overlap between 2 electron clouds.

An overlap between orbital can lead to various types of bond:

bond & bond (others are possible but will not be discussed)

The Bond:

a stronger bond is obtained when there is a greater overlap between electron clouds. the atoms cant get too close because at a certain point repulsion between nuclei occurs.

A bond can result from the overlap of:2 s orbitals, s & p orbital,

2 p orbital (head-to-head), 2 hybrid orbitals,s & hybrid orbital, p & hybrid orbital.

The 1st bond between 2 atoms is always a s bond. The s bond allows free rotation between the atoms.


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4.8 Valence Theory of Bonding

All other bonds between atoms (for this course).

The Bond:

The bond is rigid & no rotation is possible. Is the result from 2 p orbitals side by side.

Hybridization hints:

Repulsion Number Hybridization

2 sp

3 sp2

4 sp3

5 sp3d

6 sp3d2


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4.9 Hybridization for More Complex Molecules

If the central atom has a formal charge, the # of electrons must reflect the charge.

4.9.1 Exercises:1. Determine the hybridization of the orbitals in the following central atom in order to obtain the

following molecules:

a) NH4+ # of bonds

# of electron pairs# of bondsOrbital box diagram:Excited State:

Hybridized State:

b) AsF52- # of bonds

# of electron pairs# of bondsOrbital box diagram:Excited State:Hybridized State:

If the charge is positive: there are less electrons than a typical neutral atom.If the charge is negative: there are more electrons than a typical neutral atom.


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A bond

is higher in energy than a bond

: meaning thatelectrons which form a bond should not be hybridized.

1. Determine the hybridization of the orbitals in the following central atom in order to obtain the

following molecules:a) C2H4 # of bonds

# of electron pairs# of bondsOrbital box diagram:Excited State:

Hybridized State:

b) C2H2 # of bonds# of electron pairs# of bondsOrbital box diagram:Excited State:

Hybridized State:

4.9 Hybridization for More Complex Molecules

4.9.2 Exercises:


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4.10 Orbital Overlap Diagram

The Orbital Overlap Diagram represents molecules as electron clouds

(closer to reality, rather than representing bonds with straight lines).

First find the VSEPR sketch,Then the hybridization of each atom,Then bonds are replaced by electron clouds,

Indicate the type of bond ( or ),Electron clouds must be labelled according to their name/hybridization.

H2 Lewis Diagram

Orbital Box Diagram:

VSEPR Sketch Orbital Overlap Diagram

Steps to follow:


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Continuation...

CH4 Lewis Diagram

Repulsion Number (C) : Hybridization (C) :

VSEPR sketch Orbital Overlap Diagram

Br2 Lewis Diagram

Repulsion Number (Br1) : Hybridization (Br1) :

Repulsion Number (Br2) : Hybridization (Br2) :



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O2 Lewis Diagram

Repulsion Number (O1) : Hybridization (O1) :

Repulsion Number (O2) : Hybridization (O2) :


C2H4 Lewis Diagram

Repulsion Number (C1) : Hybridization (C1) :



Continuation...


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Continuation...

N2 Lewis Diagram

Repulsion Number (N1) : Hybridization (N1) :

Repulsion Number (N2) : Hybridization (N2) :


C2H2 Lewis Diagram





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Continuation...

CO2 Lewis Diagram

Repulsion Number (C) : Hybridization (C) :

Repulsion Number (O) : Hybridization (O) :



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End of unit exercises:1. For the following molecules give: Box diagram of the central atom

Box diagram of the excited central atom

Box diagram of the hybridized atomVSEPR sketchBond AnglesName of geometry

a) BeH2 b) BH3

c) CH4 d) NH3

e) H2O f) HF

g) PCl5 h) SF4

i) ClF3 j) XeF2

k) KrF4 l) SiF42-

m) ClF4+ n) AsF5

-2

o) BrF4- p) IF6

+

q) XeF3+ r) CN-

s) CH2O t) CO2

Chemistry NYA Class Notes and Exercises Part 4

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