CHAPTER 4 CHEMICAL BONDING

CHAPTER 4 CHEMICAL BONDING

Chemical bond is the force that holds two atoms together in a molecule or compound
Valence electrons play an important role in the formation of chemical bonds

CHAPTER 4 CHEMICAL BONDING
4.1 Lewis Structure 4.2 Molecular Shape and Polarity 4.3 Orbital Overlap and Hybridization 4.4 Intermolecular Forces 4.5 Metallic Bond

4.1 Lewis Structure 4.1.1 Lewis Symbol A Lewis symbol consists of: the symbol of an element dots or cross is used to represent the valence electrons in an atom of the element.

Example The Lewis symbol of atom

Elements in the same group have the same valence electronic configurations
 similar Lewis symbols.

4.1.2 Octet Rule Octet rule states that atoms tend to form bonds to obtain 8 electrons in the valence shell Atoms combine to achieve stablility  to have the same electronic configuration as a noble gas

Atoms achieve noble gas configuration through:
i) transferring electrons ii) sharing electron Bond formation involve transferring or sharing of only valence electrons

Electronic Configuration of Cations and Anions
1) Noble gas configuration Group 1, 2 and 13 elements donate valence electrons to form cations with noble gas configurations Example: Na : 1s22s22p63s1 Na+ : 1s22s22p6 (isoelectronic with Ne) Ca : 1s22s22p63s23p64s2 Ca2+ : 1s22s22p63s23p6 (isoelectronic with Ar)

Group 15, 16 and 17 elements accept electrons to form anions with noble gas configurations
Example: O : 1s22s22p4 O2 : 1s22s22p6 (isoelectronic with neon) Cl : 1s22s22p63s23p5 Cl : 1s22s22p63s23p6 (isoelectronic with Ar)

2) Pseudonoble gas configuration
d block elements donate electrons from 4s orbitals to form cations with pseudonoble gas configuration. Example: Zn : 1s22s22p63s23p64s23d10 Zn2+ : 1s22s22p63s23p63d10 (pseudonoble gas configuration )

3) Stability of the half-filled orbitals
d block element can also donate electrons to achieve the stability of half-filled orbitals Example: Mn : 1s22s22p63s23p64s23d5 Mn2+ : 1s22s22p63s23p63d5 (stability of half-filled 3d orbital ) Fe : 1s22s22p63s23p64s23d6 Fe3+ : 1s22s22p63s23p63d5 (stability of half-filled 3d orbital)

4.1.3 Formation of the bonds using Lewis Symbols
Ionic (electrovalent) bond Covalent bond Dative (coordinate) bond

4.1.3.1 Ionic bond (Electrovalent bond)
Ionic bond (electrovalent bond) is an electrostatic attraction between positively and negatively charged ions. Ionic compounds are formed when electrons are transferred between atoms (metal to nonmetal) to give electrically charged particles that attract each other .

Example 1: NaCl Sodium, an electropositive metal, tends to remove its valence electron to obtain noble gas electronic configuration (Ne) Chlorine, an electronegative element, tend to accept electron from Na to obtain noble gas electronic configuration (Ar)

+ The electrostatic forces between Na+ and Cl- produce ionic bond
These two processes occur simultaneously +

Example 2: CaCl2 Ca: 1s2 2s2 2p6 3s2 3p6 4s2
(Has two electrons in its outer shell) Cl: 1s2 2s2 2p6 3s2 3p5 (Has seven outer electrons)

Calcium Chloride If Ca atom transfer 2 electrons, one to each chlorine atom, it become a Ca2+ ion with the stable configuration of noble gas. At the same time each chlorine atom to achieve noble gas configuration gained one electron becomes a Cl- ion to achieve noble gas configuration. The electrostatic attraction formed ionic bond between the ions.

Ionic bond (Formed by transfer of electrons) Calcium Chloride
+ + 2

Example 3: LiF +

Lewis structure and formation of ionic compounds
1) CaCl2 + + 2 2) MgO +

3) CaBr2 + +

Ionic bond is very strong, therefore ionic compounds:
Have very high melting and boiling points Hard and brittle Can conduct electricity when they are in molten form or aqueous solution because of the mobile ions

Exercises: By using Lewis structure, show how the ionic bond is formed in the compounds below. ( a ) KF ( b ) BaO ( c ) Na2O

To gain stability by having noble gas configuration (octet)
Covalent Bond Definition of covalent bond Chemical bond in which two or more electrons are shared by two atoms. The electrostatic force between the electrons being shared the nuclei of the atoms. Why should two atoms share electrons? To gain stability by having noble gas configuration (octet)

Example 7e- 7e- 8e- 8e- F F + F Lewis structure of F2 F F lone pairs
single covalent bond single covalent bond F 9.4

Covalent compounds: Compounds may have these covalent bonds:
i. Single bond ii. Double bond iii. Triple bond.

O H H O H O H Lewis structure of water + + or single covalent bonds

O C O C Double bond – two atoms share two pairs of electrons or
double bonds double bonds

N N Triple bond – two atoms share three pairs of electrons or
+ 8e- 8e- N or N triple bond triple bond

4.1.3.3 Coordinate Covalent Bond (Dative Bond)
Dative bond is a bond in which the pair of shared electrons is supplied by one of the two bonded atoms Involve overlapping of a full orbital and an empty orbital

Requirement for dative bonds:
i. Donor atoms should have at least one lone pair electrons ii. The atoms that accepts these electrons should have empty orbitals.

Single bond Double bond Triple bond

Steps in Writing Lewis Structures
Count total number of valence e- of atoms involved. Add 1 for each negative charge. Subtract 1 for each positive charge. Draw skeletal structure of the compound. Put least electronegative element in the center. Complete an octet for all atoms except hydrogen If structure contains too many electrons, form double and triple bonds on central atom as needed.

Example i. HF ii. CH4 iii. CHCl3 iv. NH3 v. H2O
Draw the Lewis structure for each of the following compounds: i. HF ii. CH4 iii. CHCl3 iv. NH3 v. H2O

Total no. of valence electrons
H: 1e F: 7e Total : 8e

Number of electrons C : 4e 4H : 4e Total : 8e

Center atom: N l Count electrons: C : 4e H : 1e 3Cl: 21e Total: 26 e

Compare the bond length between single, double and triple bond
The distance between nuclei of the atoms involves in the bond C C C C C C 1.54 Å Å Å As the number of bonds between the carbon increase, the bond length decreases because C are held more closely and tightly together As the number of bonds between two atoms increases, the bond grows shorter and stronger

The sum of formal charge on each atom should equal:
zero for a molecule the charge on the ion for a polyatomic ion  Formal charge is used to find the most stable Lewis structure

EXAMPLE 1) Draw all the possible Lewis structure of COCl2.
2) Predict the most plausible structure.

The most plausible structure is (2)
SOLUTION 1) 2) The most plausible structure is (2) Formal charge is determined before completing a Lewis structure to predict the most stable structure because formal charge closest to zero.

Draw the possible Lewis structures for HNO2.
EXERCISE 1 Draw the possible Lewis structures for HNO2. Determine the most plausible Lewis structures for HNO2.

Suggest the possible Lewis structure for H2SO4. Explain your answer.
EXERCISE 2 Suggest the possible Lewis structure for H2SO4. Explain your answer.

EXERCISE 3 HCN CO2 SCN

1) Incomplete octet 2) Expanded octet 3) Odd no. electron
l Three conditions: 1) Incomplete octet 2) Expanded octet 3) Odd no. electron

Occurs when central atom has less than 8 electrons.
Elements that can form incomplete octet are: Boron,B , Beryllium, Be & Aluminium, Al This is due to elements being relatively small in size but having high nuclear charge.

Occurs when central atom has more than 8 electrons.
Formed by non-metals that have d orbitals OR Non-metals of the 3rd, 4th, 5th….rows in the periodic table

l Nitrogen may form compounds that contain odd number electrons.
Example: Nitric oxide, NO Nitrogen dioxide, NO2

The use of two or more Lewis structures to represent a particular molecule.
Requirement: Molecules/ions must have multiple bonds and lone pairs electrons at the terminal atoms.

RESONANCE STRUCTURE FOR NO3-

EXERCISE: Write Lewis structures of the following compounds/ ions:
CCl4 CO32- HCN PCl3 HNO3 PO43- C2H4 C2H2 CH2Cl2 ICl NH4+ NF3 H2S N2H4 PH3 CS2 NO2- XeF4 NH3 HCOOH SO42- ICl4- SF6 O3 NO2

4.2 MOLECULAR SHAPE AND POLARITY

4.2 MOLECULAR SHAPE AND POLARITY
i. VSEPR theory ii. 5 basic shapes iii. polarity

Molecular shape: Introduction
shows the 3-dimensional arrangement of atoms in a molecule Predicted by using Valence Shell Electron Pair Repulsion (VSEPR) theory

4.2.1 VSEPR The Valence-Shell Electron Repulsion theory states that:
The valence electron pairs around the central atom are oriented as far apart as possible to minimize the repulsion between them.

The repulsion may occur either between:
a) bonding pair & another bonding pair b) bonding pair & lone pairs or c) between lone pair & another lone pairs

The strength of repulsion:
The order of repulsive force is: Bonding pair-bonding pair repulsion Lone pair- lone pair repulsion Lone pair-bonding pair repulsion > Decrease of the repulsion force Note: The electron pairs repulsion will determine the orientation of atoms in space

4.2.2 Shape of a molecule Basic shapes are based on the repulsion between the bonding pairs. Tips to determine the molecular shape : Step 1 Draw Lewis structure of the molecule Step 2 Consider the number of bonding pairs Step 3 Place bonding pairs as far as possible to minimize repulsion.

A. Molecules with 2 bonding pairs
shape Linear 180° Example: BeCl2 Lewis structure Be : 2e 2Cl :14e Total : 16 e Cl .. : Be

B. Molecules with 3 bonding-pairs
Example: BCl3 Lewis structure Repulsive forces between pairs are the same 120° Trigonal planar B: 3e 3Cl : 21e Total: 24e B .. : Cl

C. Molecules with 4 bonding pairs
Example: CH4 Lewis structure Equal repulsion between bonding pairs – equal angle 109.5° Tetrahedral C H

D. Molecules with 5 bonding pairs
Example: PCl5 Lewis structure Shape: 120° Trigonal bipyramidal 90° P Cl .. :

E. Molecules with 6 bonding pairs
Example: SF6 Lewis structure Octahedral S : 6e 6F : 42e Total : 48e 90o 90o

Number of bonding pairs
2 electron pairs in the valence shell of central atom: Class of molecules Number of bonding pairs Number of lone pairs Shape AB2 2 Linear 180°

3 electron pairs in the valence shell of central atom: Class of molecules Number of bonding pairs Number of lone pairs Shape AB3 3 trigonal planar 120°

4 electron pairs in the valence shell of central atom: Class of molecules Number of bonding pairs Number of lone pairs Shape AB4 4 Tetrahedral 109.5o

5 electron pairs in the valence shell of central atom: Class of molecules Number of bonding pairs Number of lone pairs Shape AB5 5 Trigonal pyramidal 120° 90°

6 electron pairs in the valence shell of central atom: Class of molecules Number of bonding pairs Number of lone pairs Shape AB6 6 Octahedral 90°

4.2.3 Effect of lone pairs on molecular shape
The geometries of molecules and polyatomic ions, with one or more lone pairs around the central atom can be predicted using VSEPR. The molecular geometry is determined by the repulsions of electron pairs in the valence shell of the central atoms.

Repulsion between electron pairs decreases in the order of:
Bonding pair- bonding pair repulsion Lone pair- bonding pair repulsion Lone pair- lone pair repulsion > Stronger to weaker repulsion

Electrons in a bond are held by the attractive forces exerted by the nuclei of the two bonded atoms therefore, they take less space of repulsion. Lone- pair electrons in a molecule occupy more space; therefore they experience greater repulsion from neighboring lone pairs and bonding pairs

Number of electron pair : 3
Example : SO2 Class of molecules : AB2E Molecular shape : Bent / V-shaped

Example : NH3 Class of molecules : AB3E Molecular shape : Trigonal pyramidal

Example : H2O Class of molecules : AB2E2 Molecular shape : Bent / V-shaped

Example : SF4 Class of molecules : AB4E Molecular shape : Distorted tetrahedron / seesaw

Example : ClF3 Class of molecules : AB3E2 Molecular shape : T-shaped

Example : I3- Class of molecules : AB2E4 Molecular shape : Linear

Example : BrF5 Class of molecules : AB5E Molecular shape : Square pyramidal

Example : XeF4 Class of molecules : AB4E2 Molecular shape : Square planar

Shape of molecules which the central atom has one or more lone pairs Class of molecules Number of bonding pairs Number of lone pairs Shape AB2E 2 1 Bent / V-shaped Bond angle : < 120o

4 electron pairs in the valence shell of central atom: Class of molecules Number of bonding pairs Number of lone pairs Shape AB3E 3 1 Trigonal pyramidal Bond angle : < 109.5o

4 electron pairs in the valence shell of central atom: Class of molecules Number of bonding pairs Number of lone pairs Shape AB2E2 2 Bent / V-shaped Bond angle : < 109.5o

5 electron pairs in the valence shell of central atom: Class of molecules Number of bonding pairs Number of lone pairs Shape AB4E 4 1 Distorted tetrahedral (see-saw) Bond angle : < 90o

5 electron pairs in the valence shell of central atom: Class of molecules Number of bonding pairs Number of lone pairs Shape AB3E2 3 2 T-shaped Bond angle : < 90o

5 electron pairs in the valence shell of central atom: Class of molecules Number of bonding pairs Number of lone pairs Shape AB2E3 2 3 Linear Bond angle : 180o

6 electron pairs in the valence shell of central atom: Class of molecules Number of bonding pairs Number of lone pairs Shape AB5E 5 1 Square pyramidal Bond angle :90o and 180o

5 electron pairs in the valence shell of central atom: Class of molecules Number of bonding pairs Number of lone pairs Shape AB4E2 4 2 Square planar Bond angle : 90o

COMPARISON OF BOND ANGLE IN CH4, NH3 AND H2O

a) CH4 Has 4 bonding pairs electrons. The repulsion between the bonding pairs electrons are equal. The bond angles are all 109.5o

b) NH3 has 3 bonding pairs electron and 1 lone pair electron. according to VSEPR, lone pair - bonding pair > bonding pair - bonding pair repulsion. Lone- pair repels the bonding-pair more strongly, the three NH bonding-pair are pushed closer together, thus HNH angle in ammonia become smaller, 107.3o.

c) H2O Has 2 bonding pairs electrons and 2 lone pair electrons. According to VSEPR, lone pair – lone pair > lone pair – bonding pair > bonding pair – bonding pair repulsion. Lone-pair tend to be as far from each other as possible. Therefore, the two OH bonding-pairs are pushed toward each other. Thus, the HOH angle is 104.5o.

4.2.4 POLAR AND NONPOLAR MOLECULES µ µ = Qr Where : µ = dipole moment
A quantitative measure of the polarity of a bond is its dipole moment ( µ ). µ = Qr Where : µ = dipole moment Q = the product of the charge from electronegativity r = distance between the charges. Dipole moments are usually expressed in debye units(D)

E.g : Polarity of HF Hydrogen fluoride is a covalent molecule with a polar bond. F atom is more electronegative than H atom, so the electron density will shift from H to F. The symbol of the shifted electron can be represented by a crossed arrow to indicate the direction of the shift.

The consequent charge separation can be represented by :
δ + : partial positive charge δ - : partial negative charge

Diatomic molecules containing atoms of different elements (e. g
Diatomic molecules containing atoms of different elements (e.g. : HCl, NO and CO) have dipole moments and are called polar molecules. Diatomic molecules containing atoms of the same element (e.g. : H2, N2 and Cl2) do not have dipole moments and are called nonpolar molecules.

For polyatomic molecules, the polarity of the bond and the molecular geometry determine whether there is a dipole moment. Even if polar bond are present, the molecules will not necessarily have a dipole moment.

Example Carbon dioxide, CO2 Carbon tetrachloride, CCl4
Predict the polarity of the following molecules: Carbon dioxide, CO2 Carbon tetrachloride, CCl4 Chloromethane, CH3Cl Ammonia, NH3

(a) Carbon dioxide, CO2 - molecular geometry : linear
- oxygen is more electronegative than carbon, - Dipole moment can cancell each other - has no net dipole moment (µ = 0) - therefore CCl4 is a nonpolar molecule.

(b) Carbon tetrachloride, CCl4
- molecular geometry : tetrahedral - Chlorine is more electronegative than carbon, - Dipole moment can cancell each other - has no net dipole moment (µ = 0) - therefore CCl4 is a nonpolar molecule.

( c) Chloromethane, CH3Cl
- molecular geometry : tetrahedral - Cl is more electronegative than C, C is more electronegative than H - Dipole moment cannot cancell each other - has a net dipole moment (µ ≠ 0) - therefore CH3Cl is a polar molecule.

(d ) Ammonia, NH3 - molecular geometry : tetrahedral
- N is more electronegative than H, - Dipole moment cannot cancell each other - has a net dipole moment (µ ≠ 0) - therefore NH3 is a polar molecule.

Factors that affected the polarity of molecules
molecular geometry electronegativity of the bonded atoms.

BOND POLAR MOLECULES NON-POLAR Symetrical molecules - basic molecular shape with the same terminal atom - molecules with lone pairs linear (from trigonal bipyramidal) and square planar with the same terminal atom Non-symetrical molecules - basic molecules with different terminal atom - molecules with lone pairs except linear and square planar

Exercises : Predict the polarity of the following molecules:
SO2 ; HBr ; SO3 ; CH2Cl2 ; ClF3 ; CF4 ; H2O ; XeF4 ; NF3 ; Cis-C2H2Cl2 ; trans-C2H2Cl2

4.3 ORBITAL OVERLAP AND HYBRIDIZATION
Formation Covalent Bond Formation Hybrid orbitals Orbital Overlapping

Objectives At the end of this subtopic, students should be able to:
1. Draw and describe the formation of sigma() and pi() bonds from overlapping of orbitals. 2. Draw and explain the formation of hybrid orbitals of a central atom: sp, sp2, sp3, sp3d, sp3d2 using appropriate examples. 3. Draw orbitals overlap and label sigma() and pi() bonds of a molecule.

Valence Bond theory explains the formation of covalent bonds and the molecular geometry outlined by the VSEPR. States that a covalent bond is formed when the neighboring atomic orbitals overlap. Overlapping may occur between: a) orbitals with unpaired electrons b) an orbital with paired electrons and another empty orbitals (dative bond)

Change in electron density as two hydrogen atoms approach each other.
Example: H The s-orbital of the Hydrogen atom Change in electron density as two hydrogen atoms approach each other. High electron density as the orbitals overlap (covalent bond formed) 10.3 120

FORMATION OF COVALENT BOND
Valence bond theory - Covalent bond is formed when two neighbouring atomic half-filled orbitals overlap. Two types of covalent bonds are a) sigma bond (σ) b) pi bond ()

a)  bond formed when orbitals overlap along its internuclear axis (end to end overlapping) Example: i. overlapping s orbitals H H H H +  bond

ii. Overlapping of s and p orbitals
Px orbital H x x H +  bond

iii. Overlapping of p orbitals
+ x x x  bond

b) bond Formed when two p-orbitals of the same orientation overlap sideways y  bond y y +

y y y y +  bond

Formation of bonds in a molecule
Covalent bonds may form by: a) overlapping of pure orbitals b) overlapping of hybrid orbitals

Overlapping of pure orbitals
Example : i. O2 ii. N2

O2 Consider the ground state configuration: O :   O O 1s2 2s2 2p4
Two unpaired electrons to be used in bonding. O : 1s2 2s2 2p4 1s 2s y x O y O  2p Overlapping occurs between the p-orbitals of each atom σ 

y O y x  O σ 

4.3.2 Formation Hybrid orbitals
Overlapping of hybrid orbitals and the pure orbitals occur when different type of atoms are involved in the bonding. Hybridization of orbitals: mixing of two or more atomic orbitals to form a new set of hybrid orbitals The purpose of hybridisation is to produce new orbitals which have equivalent energy Number of hybrid orbitals is equal to number of pure atomic orbitals used in the hybridization process.

Hybridization Hybrid orbitals have different shapes from original atomic orbitals Types of hybridisation reflects the shape/geometry of a molecule Only the central atoms will be involved in hybridisation 10.4

Hybridization of orbitals
sp sp2 sp3 sp3d sp3d2

sp3 hybridization one s orbital and three p orbitals are mixed to form four sp3 hybrid orbitals the geometry of the four hybrid orbitals is tetrahedral with the angle of 109.5o .

sp3 hybrid Mixing of s and three p orbitals sp3 sp3 sp3 sp3

Example: 1) CH4 Lewis structure : Valence orbital diagram ; H :
C ground state : C excited : C hybrid : Orbital Overlap : Molecular Geometry :

Example : Methane, CH4 Ground state : C : 1s2 2s2 2p2 C H sp3 hybrid
Lewis Structure C H 1s 2s 2p Excitation: to have 4 unpaired electrons H sp3 Excited state : 1s 2s 2p C sp3 hybrid shape: tetrahedral

sp3-Hybridized C atom in CH4
Fig. 10.8 sp3-Hybridized C atom in CH4 sp3 sp3 sp3 1s 1s sp3 1s

Valence orbital diagram ; H : N ground state : N excited : N hybrid :
Example 2 : NH3 Lewis structure : Valence orbital diagram ; H : N ground state : N excited : N hybrid : Orbital Overlap : Molecular Geometry :

Fig. 10.9 sp3 sp3 sp3 1s 1s sp3 1s

Example: 3) H2O Lewis structure : Valence orbital diagram; O ground state : O hybrid : Orbitals overlap:

sp2 hybridization one s orbital and two p orbitals are mixed to form three sp2 hybrid orbitals the geometry of the three hybrid orbitals is trigonal planar with the angle of 120o .

one s orbital + two p orbitals three sp2 orbitals
Fig s sp2 px py sp2 sp2 one s orbital + two p orbitals three sp2 orbitals

simplified drawing of sp2 orbitals:
Shown together (large lobes only)

Example: 1) BF3 Lewis structure : Valence orbital diagram; F : B ground state : B excited : B hybrid : Orbital overlap:

Example: BF3    F : 1s22s22p5 Shape : trigonal planar
Pure p orbital   sp2 sp2 F : 1s22s22p5 sp2  Shape : trigonal planar

Example: 2) C2H4 Lewis structure : Valence orbital diagram; C ground state : C excited : C hybrid : Orbital overlap:

Fig a-c  bonds  bond

sp hybridization one s orbital and one p orbital are mixed to form two sp hybrid orbitals the geometry of the two hybrid orbitals is linear with the angle of 180o

Formation of sp Hybrid Orbitals
Types of hybrid orbitals Formation of sp Hybrid Orbitals sp Produces linear shape 10.4

Example: 1) BeCl2 Lewis structure : Valence orbital diagram; Cl : Be ground state : Be excited : Be hybrid : Orbital overlap:

Example: 2) C2H2 Lewis structure : Valence orbital diagram; C ground state : C excited : C hybrid : Orbital overlap:

Fig a-c

Example: 3) CO2 Lewis structure : Valence orbital diagram; O: C ground state : C excited : C hybrid : Orbital overlap:

sp3d hybridization one s orbital, three p orbitals and one d orbital are mixed to form five sp3d hybrid orbitals. the geometry of the five hybrid orbitals is trigonal bipyramidal with the angle of 120o and 90o

simplified drawing of sp3d orbitals:

Example: 1) PCl5 Lewis structure : Valence orbital diagram; Cl : P ground state : P excited : P hybrid : Orbital overlap:

Example: 2) ClF3 Lewis structure : Valence orbital diagram; F : Cl ground state : Cl excited : Cl hybrid : Orbital overlap:

sp3d2 hybridization one s orbital, three p orbitals and two d orbitals are mixed to form six sp3d2 hybrid orbitals the geometry of the six hybrid orbitals is octahedral with the angle of 90o

Simplified drawing of sp3d2 orbitals:

Example: 1) SF6 Lewis structure : Valence orbital diagram; F : S ground state : S excited : S hybrid : Orbital overlap:

Example: 2) ICl5 Lewis structure : Valence orbital diagram; Cl : I ground state : I excited : I hybrid : Orbital overlap:

How do I predict the hybridization of the central atom?
Count the number of lone pairs AND the number of atoms bonded to the central atom No of Lone Pairs + No of Bonded Atoms Hybridization Examples 2 sp BeCl2 3 sp2 BF3 4 sp3 CH4, NH3, H2O 5 sp3d PCl5 6 sp3d2 SF6 10.4

Exercise: For each of the following, draw the orbital overlap to show the formation of covalent bond a) XeF2 b) O3 c) ICl4 d) OF2

4.4 Intermolecular forces
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LEARNING OUTCOMES At the end of the lesson, students should be able to;
Describe intermolecular forces i. van der Waals forces : - dipole-dipole interactions or permanent dipole - London forces or dispersion forces ii. Hydrogen bonding 2. Explain factors that influence the strength of van der Waals forces 3. Explain the effects of hydrogen bonding on i. boiling point ii. Solubility iii. Density of water compared to ice Explain the relationship between : i. intermolecular forces and vapour pressure ii. Vapour pressure and boiling point yati-ab

4.4 Intermolecular forces
4.4.1 Types of intermolecular forces 4.4.2The effect of intermolecular forces on the physical properties. 171 yati-ab

Intermolecular Forces
Intermolecular forces are the attractive forces between molecules 172 yati-ab

Effects of intermolecular forces on physical properties
Have effects on these physical properties: a) boiling point b) melting point c) solubility d) density e) electrical conductivity 173 yati-ab

Between covalent molecules
Intermolecular Forces Van der Waal Forces Hydrogen Bond Between covalent molecules Between covalent molecules with H covalently bonded to F, O or N 174 yati-ab

4.4.1.1 van der Waal Forces Forces that act between covalent molecules
Three types of interaction: i. Dipole-dipole attractive forces - act between polar molecules ii. London Dispersion forces - act between non-polar molecules 175 yati-ab yati-ab 175

Dipole-dipole forces (permanent dipole forces)
Exist in polar covalent compounds Polar molecules have permanent dipole due to the uneven electron distributions Example: + - Cl H + - Cl H Chlorine is more electronegative, thus it has higher electron density Dipole-dipole forces; the partially positive end attracts the partially negative end 176 yati-ab

4.4.1.2. London Dispersion Forces
attractive forces that exist between non-polar molecules result from the temporary (instantaneous) polarization of molecules The temporary dipole molecules will be attracted to each other and these attractions is known as the London Forces or London Dispersion forces 177 yati-ab

The formation of London forces
At any instant, electron distributions in one molecule may be unsymmetrical. The end having higher electron density is partially negative and the other is partially positive. An instant dipole moment that exists in a molecule induces the neighboring molecule to be polar. 178 yati-ab

Example: London forces in Br2
Electrons in a molecule move randomly about the nucleus Br At any instant, the electron density might be higher on one side The temporary dipole molecule induce the neighboring atom to be partially polar - + - + Br Br Temporary dipole molecule London forces 179 yati-ab

Factors that influence the strength of the van der Waals forces.
The molecular size/molecular mass Molecules with higher molar mass have stronger van der Waals forces as they tend to have more electrons involved in the London forces. Example: CH4 has lower boiling point than C2H6 Note: However if two molecules have similar molecular mass, the dipole-dipole interaction will be more dominant. Example: H2S has higher boiling point than CH3CH3 180 yati-ab

4.4.1.3 Hydrogen intermolecular bond
Dipole-dipole interaction that acts between a Hydrogen atom that is covalently bonded to a highly electronegative atom ; F, O ,N in one molecule and F,O or N of another molecule. Example: + - F H + - F H Hydrogen intermolecular bond 181 yati-ab

Other examples: NH3 liquid .. .. .. .. .. H2O
Hydrogen intermolecular bond O H .. O H .. Hydrogen intermolecular bond N H .. Covalent bond O H .. Hydrogen intermolecular bond 182 yati-ab

Consider ethanol, CH3OH CH3OH H is not bonded to either F, O or N
CH3OH and methane in CH3OH C H Hydrogen bond Not a hydrogen bond C H O .. C H O .. H is not bonded to either F, O or N 183 yati-ab

Example: H2O ___ covalent bond hydrogen bond 184 yati-ab

185 yati-ab

Properties of compounds with Hydrogen intermolecular forces
Boiling point Have relatively high boiling point than compounds having dipole-dipole forces or London forces - the Hydrogen bond is the strongest attraction force compared to the dipole- dipole or the London forces. 186 yati-ab

Solubility A. Dissolve in polar solvent
The molecules that posses Hydrogen bonds are highly polar. They may form interaction with any polar molecules that act as solvent. B. Dissolve in any solvent that can form Hydrogen bonds 187 yati-ab

Example NH3 dissolves in water because it can form Hydrogen intermolecular bond with water. N H .. O Hydrogen bond 188 yati-ab

Problem: Explain the trend of boiling point given by the graph below:
T/oC HF HI HBr HCl Molecular mass 189 yati-ab

Answer HF can form hydrogen bonds between molecules while HCl, HBr and HI have van der Waals forces acting between molecules. Hydrogen intermolecular bond is stronger that the van der Waals forces. More energy is required to break the Hydrogen bond. Boiling point increases from HCl to HI. The strength of van der Waals forces increases with molecular mass. Since molecular mass increases from HCl to HI, thus the boiling point will also increase in the same pattern. 190 yati-ab

The effect of Hydrogen bond on water molecules
The density of water is relatively high compared to other molecules with similar molar mass. Reason: Hydrogen intermolecular bonds are stronger than the dipole-dipole or the London forces. Thus the water molecules are drawn closer to one another and occupy a smaller volume. 191 yati-ab

Density Ice (solid H2O) has lower density compared to its liquid. Refer to the structure of ice yati-ab

Ice form tetrahedral arrangement
Hydrogen bond takes one of the tetrahedral orientation and occupy some space 193 yati-ab

H2O(l) is denser than H2O(s) because
 the hydrogen bond in ice arrange the H2O molecules in open hexagonal crystal  H2O molecules in water have higher kinetic energy and can overcome the hydrogen bond  V-shaped water molecules slide between each other. 194 yati-ab

195 yati-ab

196 yati-ab

Fig 197 yati-ab

Boiling points of substance with Hydrogen intermolecular bonds
The boiling points of these substances are affected by: a) the number of hydrogen bonds per molecule b) the strength of H intermolecular forces which directly depends on the polarity of the hydrogen bond Example: Explain the trend of boiling points given below: The order of the increase in boiling point is: H2O > HF > NH3 > CH4 198 yati-ab

Answer: by looking at the polarity of the bond, we have
(Order of polarity: HF > H2O > NH3) but H2O has the highest boiling point. For H2O, the number of hydrogen bonds per molecule affects the boiling point. Each water molecule can form 4 hydrogen bonds with other water molecules. More energy is required to break the 4 Hydrogen bonds. HF has higher boiling point than NH3 because F is more electronegative than Nitrogen. CH4 is the lowest - it is a non polar compound and has weak van der Waals forces acting between molecules. 199 yati-ab

Effects of intermolecular forces on physical properties
1) Boiling point For molecules with similar size, the order of intermolecular strength: Hydrogen bond > dipole-dipole forces > London dispersion forces Strength of intermolecular forces ↑  boiling point ↑ 200 yati-ab

201 yati-ab

Why boiling point H2O > HF and HF > NH3?
Fluorine is more electronegative than oxygen, therefore stronger hydrogen bonding is expected to exist in HF liquid than in H2O. However, the boiling point of H2O is higher than HF because each H2O molecules has 4 hydrogen bonds. 202 yati-ab

On the other hand, H-F has only 2 hydrogen bonds.
Therefore the hydrogen bonds are stronger in H2O rather than in H-F. 203 yati-ab

Boiling point HF > NH3
Fluorine is more electronegative than nitrogen ,thus the hydrogen bonding in H-F is stronger than H-N. 204 yati-ab

Vapour Pressure Molecules can escape from the surface of liquid at any temperature by evaporation in a closed system : vapour molecules which leaves the surface cannot escape from the system the molecules strike the container wall and exert some pressure yati-ab

Fig yati-ab

The pressure exerted by those molecules is called vapour pressure (or maximum vapour pressure)
Vapour pressure is the pressure exerted by a vapour in equilibrium with its liquid phase. yati-ab

In a close system …. Liquid molecules vapourise
Molecules have enough energy to overcome intermolecular forces Vapour molecules are trapped in the close container Rate of vaporisation is faster than the rate of condensation Volume of liquid becomes less Some of the vapour molecules may collide and lose their energy. They re-enter the liquid surface Rate of vapourisation is equal to the rate of condensation System reaches equilibrium dynamic equilibrium Pressure exerted by the vapour molecules is known as the vapour pressure Volume of liquid remains constant yati-ab

Dynamic equilibrium and vapour pressure
Number of liquid molecules leaving the surface is the same as the number of vapour molecules entering the liquid surface. The vapour pressure at this stage is constant and known as the equilibrium vapour pressure. Dynamic equilibrium is reached when: Rate of evaporation = rate of condensation Note: Equilibrium vapour pressure = saturated vapour pressure = vapour pressure yati-ab

Factors that affects vapour pressure
i. Intermolecular forces Molecules with weak intermolecular forces can easily vapourise. More vapour molecules will be present and exert higher pressure.  the weaker intermolecular forces the higher is the vapour pressure. ii. Temperature Heating causes more molecules to have high kinetic energies that are higher than their intermolecular forces. More liquid molecules will form vapour.  vapour pressure increases with temperature. yati-ab

Fig yati-ab

Boiling – the process Increasing the temperature will increase in the vapour pressure. As heat is applied, the vapour pressure of a system will increase until it reaches a point whereby the vapour pressure of the liquid system is equal to the atmospheric pressure. Boiling occurs and the temperature taken at this point is known as the boiling point. At this point, the change of state from liquid to gas occurs not only at the surface of the liquid but also in the inner part of the liquid. Bubbles form within the liquid. yati-ab

Boiling Point: the temperature at which the vapour pressure of a liquid is equal to the external atmospheric pressure. Normal Boiling Point: the temperature at which a liquid boils when the external pressure is 1 atm (that is the vapour pressure is 760 mmHg) yati-ab

Factors affecting the boiling point:
1. Intermolecular forces A substance with weak intermolecular forces can easily vapourise and the system requires less heat to achieve atmospheric pressure, thus it boils at a lower temperature. 2. Atmospheric pressure When the external atmospheric pressure is low, liquid will boil at a lower temperature. yati-ab

Metallic bond yati-ab

yati-ab

Metallic bond An electrostatic force between positive charge metallic ions and the sea of electrons. Bonding electrons are delocalized over the entire crystal which can be imagined as an array of the ions immersed in a sea of delocalized valence electron. 217 yati-ab

Metallic bonds Positive ions are immersed in the sea of electrons
Free moving electrons e 218 yati-ab

Electrostatic force in a metal
Metallic Bond (Electron-sea Model) Metals form giant metallic structure Each positive ion is attracted to the ‘sea of electrons’. These atoms are closely held by the strong electrostatic forces acting between the positive ions and the ‘sea of electrons’. These free moving electrons are responsible for the high melting point of metals and the electrical conductivity. 219 yati-ab

Physical properties of metals
metals have high melting point high energy is required to overcome these strong electrostatic forces between the positive ions and the electron sea in the metallic bond + Metallic bonds formed by the electrostatic forces exist between positive ions and the free moving electrons e 220 yati-ab

The strength of the metallic bonds
The strength of the metallic bond increases with the number of valence electrons and the size of ions. The smaller the size of positive ions the greater is the attractive force acting between the ions and the valence electrons 221 yati-ab

Boiling points in metals
Na Mg ee +1 +2 e e e e e e ee +2 +1 e ee e e e e e ee +1 +2 Has 2 valence electrons Has one valence electron Stronger metallic bond due to the size of Mg being smaller than Na and the strong electrostatic force between +2 ions and the two valence electrons, the electrostatic force acting between positive ions and free moving electrons form metallic bonds Mg has higher boiling point than Na 222 yati-ab

Example: Explain the difference in the boiling point of the two metals given: Magnesium – oC Aluminum – oC 223 yati-ab

Answer The cationic size of Al is smaller compared to magnesium and its charge is higher (+3). Mg has two valence electrons Al has three valence electrons involved in the metallic bonding. The strength of metallic bond in Aluminium is greater than that of Magnesium Al has higher boiling point 224 yati-ab

The stronger metallic bond,the higher the boiling point.
The strength of metallic bond is directly proportional to the boiling point. The stronger metallic bond,the higher the boiling point. 225 yati-ab

CHAPTER 4 CHEMICAL BONDING

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CHAPTER 4 CHEMICAL BONDING

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