Topic4_Chemicalbonding

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 BONDING4.1 Lewis Structure 4.2 Molecular Shape and Polarity4.3 Orbital Overlap and Hybridization4.4 Intermolecular Forces4.5 Metallic Bond

4.1Lewis Structure4.1.1Lewis Symbol

A Lewis symbol consists of:the symbol of an elementdots 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.

4.1.2Octet RuleOctet rule states that atoms tend to form bonds to obtain 8 electrons in the valence shell

Atoms combine to achieve stablilityto have the same electronic configuration as a noble gas

Atoms achieve noble gas configuration through:i)transferring electronsii)sharing electron

Bond formation involve transferring or sharing of only valence electrons

Electronic Configuration of Cations and Anions1)Noble gas configurationGroup 1, 2 and 13 elements donate valence electrons to form cations with noble gas configurationsExample:Na: 1s22s22p63s1Na+: 1s22s22p6 (isoelectronic with Ne)

Ca: 1s22s22p63s23p64s2Ca2+: 1s22s22p63s23p6 (isoelectronic with Ar)

Group 15, 16 and 17 elements accept electrons to form anions with noble gas configurationsExample:O: 1s22s22p4O2: 1s22s22p6 (isoelectronic with neon)Cl: 1s22s22p63s23p5Cl: 1s22s22p63s23p6 (isoelectronic with Ar)

2)Pseudonoble gas configurationd block elements donate electrons from 4s orbitals to form cations with pseudonoble gas configuration.Example:Zn: 1s22s22p63s23p64s23d10Zn2+ : 1s22s22p63s23p63d10 (pseudonoble gas configuration )

3) Stability of the half-filled orbitalsd block element can also donate electrons to achieve the stability of half-filled orbitalsExample:Mn: 1s22s22p63s23p64s23d5Mn2+: 1s22s22p63s23p63d5 (stability of half-filled 3d orbital )

Fe: 1s22s22p63s23p64s23d6Fe3+: 1s22s22p63s23p63d5 (stability of half-filled 3d orbital)

4.1.3 Formation of the bonds using Lewis SymbolsIonic (electrovalent) bondCovalent bondDative (coordinate) bond

4.1.3.1Ionic 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 ChlorideIf 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) CaCl222) MgO+

3) CaBr2++

Ionic bond is very strong, therefore ionic compounds:Have very high melting and boiling pointsHard and brittleCan 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

4.1.3.2 Covalent BondDefinition 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)

Lewis structure of F29.4Example

Covalent compounds: Compounds may have these covalent bonds:i.Single bondii.Double bondiii.Triple bond.

++orLewis structure of watersingle covalent bonds

Double bond two atoms share two pairs of electronsor

+Triple bond two atoms share three pairs of electronsor

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 electronsii.The atoms that accepts these electrons should have empty orbitals.

Single bond

Double bond

Triple bond

Steps in Writing Lewis StructuresCount 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 hydrogenIf structure contains too many electrons, form double and triple bonds on central atom as needed.

ExampleDraw the Lewis structure for each of the following compounds:i.HFii.CH4iii.CHCl3iv.NH3v.H2O

Total no. of valence electrons

H:1eF:7eTotal :8e

Number of electrons

C:4e4H:4e Total :8e

lCount electrons:

C:4eH:1e 3Cl: 21e Total:26 e

Center atom: N

4.1.5 Bond LengthCompare the bond length between single, double and triple bondBond length : The distance between nuclei of the atoms involves in the bondC CC CC C1.54 1.34 1.20

As the number of bonds between the carbon increase, the bond length decreases because C are held more closely and tightly togetherAs 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

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

SOLUTION 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.

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

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

HCN

CO2

SCNEXERCISE 3

lThree 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, AlThis 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 orbitalsORNon-metals of the 3rd, 4th, 5th.rows in the periodic table

lNitrogen 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:

CCl4CO32-HCNPCl3HNO3PO43-C2H4C2H2CH2Cl2IClNH4+NF3H2SN2H4PH3CS2NO2-XeF4NH3HCOOHSO42-ICl4-SF6O3NO2

4.2 MOLECULAR SHAPE AND POLARITY

*4.2 MOLECULAR SHAPE AND POLARITYi.VSEPR theoryii. 5 basic shapes iii. polarity

*Molecular shape: Introductionshows the 3-dimensional arrangement of atoms in a moleculePredicted by using Valence Shell Electron Pair Repulsion (VSEPR) theory

*4.2.1 VSEPRThe 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:

Decrease of the repulsion forceNote:The electron pairs repulsion will determine the orientation of atoms in space

*4.2.2 Shape of a moleculeBasic shapes are based on the repulsion between the bonding pairs.

Tips to determine the molecular shape :Step 1Draw Lewis structure of the moleculeStep 2Consider the number of bonding pairsStep 3Place bonding pairs as far as possible to minimize repulsion.

*A.Molecules with 2 bonding pairsExample:BeCl2Lewis structureBe : 2e 2Cl :14e Total : 16 e

*B.Molecules with 3 bonding-pairsExample: BCl3Lewis structure

*C. Molecules with 4 bonding pairsExample: CH4Lewis structure

*D. Molecules with 5 bonding pairsExample: PCl5Lewis structure

*E. Molecules with 6 bonding pairsExample: SF6Lewis structure S : 6e 6F : 42eTotal : 48e90o90o

*2 electron pairs in the valence shell of central atom:

Class of moleculesNumber of bonding pairsNumber of lone pairsShapeAB220

Linear



trigonal planar

*4 electron pairs in the valence shell of central atom:109.5o


Tetrahedral



Trigonal pyramidal



Octahedral

*4.2.3 Effect of lone pairs on molecular shapeThe 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: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 : 3Example : SO2

Class of molecules : AB2EMolecular shape : Bent / V-shaped

* Number of electron pair : 4Example : NH3

Class of molecules : AB3EMolecular shape : Trigonal pyramidal

* Number of electron pair : 4Example : H2O

Class of molecules : AB2E2Molecular shape : Bent / V-shaped

* Number of electron pair : 5Example : SF4

Class of molecules : AB4EMolecular shape : Distorted tetrahedron / seesaw

* Number of electron pair : 5Example : ClF3

Class of molecules : AB3E2

Molecular shape : T-shaped

* Number of electron pair : 5Example : I3-


Molecular shape : Linear

* Number of electron pair : 6Example : BrF5

Class of molecules : AB5E

Molecular shape : Square pyramidal

* Number of electron pair : 6Example : XeF4


Molecular shape : Square planar

*Shape of molecules which the central atom has one or more lone pairs

Class of moleculesNumber of bonding pairsNumber of lone pairsShapeAB2E21

Bent / V-shaped

Bond angle : < 120o



Trigonal pyramidal Bond angle : < 109.5o



Bent / V-shaped Bond angle : < 109.5o



Distorted tetrahedral(see-saw)Bond angle : < 90o



T-shaped Bond angle : < 90o



LinearBond angle : 180o



Square pyramidalBond angle :90o and 180o



Square planar Bond angle : 90o

*COMPARISON OF BOND ANGLE IN CH4, NH3 AND H2O109.5o107.3o104.5o

* a) CH4

Has 4 bonding pairs electrons.

The repulsion between the bonding pairs electrons are equal.

The bond angles are all 109.5o

*

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. b) NH3

*

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. c) H2O

*4.2.4 POLAR AND NONPOLAR MOLECULES A quantitative measure of the polarity of a bond is its dipole moment ( ). = Qr Where : = dipole moment Q = the product of the charge from electronegativity r = distance between the charges.

Dipole moments are usually expressed in debye units(D)

*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.

E.g : Polarity of HF

*

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

*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.

*ExamplePredict the polarity of the following molecules:Carbon dioxide, CO2Carbon tetrachloride, CCl4Chloromethane, CH3ClAmmonia, 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 ( = 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 ( = 0)- therefore CCl4 is a nonpolar molecule.

*

- 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 ( 0)- therefore CH3Cl is a polar molecule. ( c)Chloromethane, CH3Cl

*(d )Ammonia, NH3

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

*Factors that affected the polarity of molecules

molecular geometry

electronegativity of the bonded atoms.

*SO2 ; HBr ; SO3 ; CH2Cl2 ; ClF3 ; CF4 ; H2O ; XeF4 ; NF3 ; Cis-C2H2Cl2 ; trans-C2H2Cl2 Exercises :

Predict the polarity of the following molecules:

*4.3 ORBITAL OVERLAP AND HYBRIDIZATIONFormation Covalent BondFormation Hybrid orbitalsOrbital Overlapping

*ObjectivesAt 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.

*4.3.1 Valence Bond theoryexplains 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 electronsb) an orbital with paired electrons and another empty orbitals (dative bond)

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

*FORMATION OF COVALENT BONDValence bond theory - Covalent bond is formed when two neighbouring atomic half-filled orbitals overlap.

Two types of covalent bonds area) sigma bond ()b) pi bond ()

*+a) bondformed when orbitals overlap along its internuclear axis (end to end overlapping)Example:i.overlapping s orbitals

bond

*ii.Overlapping of s and p orbitals+Px orbital

*iii. Overlapping of p orbitals+

*b) bondFormed when two p-orbitals of the same orientation overlap sideways +

*+ bond

*Formation of bonds in a moleculeCovalent bonds may form by:a) overlapping of pure orbitalsb) overlapping of hybrid orbitals

*Overlapping of pure orbitalsExample :i.O2ii.N2

*O2Consider the ground state configuration:O :1s2 2s2 2p42pOverlapping occurs between the p-orbitals of each atom

*4.3.2 Formation Hybrid orbitalsOverlapping 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.

*10.4HybridizationHybrid orbitals have different shapes from original atomic orbitalsTypes of hybridisation reflects the shape/geometry of a moleculeOnly the central atoms will be involved in hybridisation

*Hybridization of orbitalsspsp2sp3 sp3dsp3d2

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

*sp3 hybridMixing of s and three p orbitals

*Example:1)CH4

Lewis structure : Valence orbital diagram ;H :C ground state :C excited :C hybrid :

Orbital Overlap :

Molecular Geometry :

*Example : Methane, CH4Ground state : C : 1s2 2s2 2p2Lewis StructureExcitation: to have 4 unpaired electronsExcited state :sp3 hybrid shape: tetrahedralC

*Fig. 10.8sp3-Hybridized C atom in CH4sp3sp3sp3sp31s1s1s

*Example 2 : NH3 Lewis structure : Valence orbital diagram ;H :N ground state :N excited :N hybrid :

Orbital Overlap :

Molecular Geometry :

*Fig. 10.9sp3sp3sp3sp31s1s1s

*Example:3)H2OLewis structure:

Valence orbital diagram;O ground state :O hybrid :

Orbitals overlap:

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

*Fig. 10.12spxpysp2sp2sp2one s orbital + two p orbitalsthree sp2 orbitals

*simplified drawing of sp2 orbitals:

Shown together (large lobes only)

*Example:1)BF3Lewis structure:

Valence orbital diagram;F :B ground state :B excited :B hybrid :

Orbital overlap:

*Example: BF3sp2sp2sp2F : 1s22s22p5Shape : trigonal planarPure p orbital

*Example:2)C2H4Lewis structure:

Valence orbital diagram;C ground state :C excited :C hybrid :

Orbital overlap:

*Fig. 10.16a-c bond bonds

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

*Formation of sp Hybrid Orbitals10.4Types of hybrid orbitalsProduces linear shape

*Example:1)BeCl2Lewis structure:

Valence orbital diagram;Cl :Be ground state :Be excited :Be hybrid :

Orbital overlap:

*Fig. 10.11

*Example:2)C2H2Lewis structure:

Valence orbital diagram;C ground state :C excited :C hybrid :

Orbital overlap:

*Fig. 10.19a-c

*Example:3)CO2Lewis structure:

Valence orbital diagram;O:C ground state :C excited :C hybrid :

Orbital overlap:

*sp3d hybridizationone 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)PCl5Lewis structure:

Valence orbital diagram;Cl :P ground state :P excited :P hybrid :

Orbital overlap:

*Example:2)ClF3Lewis structure:

Valence orbital diagram;F :Cl ground state :Cl excited :Cl hybrid :

Orbital overlap:

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

*Simplified drawing of sp3d2 orbitals:

*Example:1)SF6Lewis structure:

Valence orbital diagram;F :S ground state :S excited :S hybrid :

Orbital overlap:

*Example:2)ICl5Lewis structure:

Valence orbital diagram;Cl :I ground state :I excited :I hybrid :

Orbital overlap:

*No of Lone Pairs+No of Bonded AtomsHybridizationExamples23456spsp2sp3sp3dsp3d2BeCl2BF3CH4, NH3, H2OPCl5SF6Count the number of lone pairs AND the numberof atoms bonded to the central atom10.4

*Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

*Exercise:For each of the following, draw the orbital overlap to show the formation of covalent bonda)XeF2b)O3c)ICl4d) OF2

4.4 Intermolecular forces

Describe intermolecular forcesi. van der Waals forces :- dipole-dipole interactions or permanent dipole- London forces or dispersion forcesii. Hydrogen bonding2.Explain factors that influence the strength of van der Waals forces3.Explain the effects of hydrogen bonding oni. boiling pointii. Solubilityiii. Density of water compared to iceExplain the relationship between :i. intermolecular forces and vapour pressureii. Vapour pressure and boiling pointLEARNING OUTCOMES At the end of the lesson, students should be able to;

*4.4 Intermolecular forces4.4.1Types of intermolecular forces4.4.2The effect of intermolecular forces on the physical properties.

*Intermolecular Forces Intermolecular forces are the attractive forces between molecules

*Effects of intermolecular forces on physical propertiesHave effects on these physical properties:a)boiling pointb)melting pointc)solubilityd)densitye)electrical conductivity

*Intermolecular ForcesVan der Waal ForcesHydrogen BondBetween covalent moleculesBetween covalent molecules with H covalently bonded to F, O or N

*4.4.1.1 van der Waal ForcesForces that act between covalent moleculesThree types of interaction:i.Dipole-dipole attractive forces- act between polar moleculesii.London Dispersion forces- act between non-polar molecules

*Dipole-dipole forces (permanent dipole forces)Exist in polar covalent compoundsPolar molecules have permanent dipole due to the uneven electron distributionsExample:Chlorine is more electronegative, thus it has higher electron densityDipole-dipole forces; the partially positive end attracts the partially negative end

*4.4.1.2.London Dispersion Forcesattractive forces that exist between non-polar moleculesresult from the temporary (instantaneous) polarization of moleculesThe temporary dipole molecules will be attracted to each other and these attractions is known as the London Forces or London Dispersion forces

*The formation of London forcesAt 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.

*Example: London forces in Br2Electrons in a molecule move randomly about the nucleus At any instant, the electron density might be higher on one sideThe temporary dipole molecule induce the neighboring atom to be partially polarTemporary dipole moleculeLondon forces

*Factors that influence the strength of the van der Waals forces.The molecular size/molecular massMolecules 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 C2H6Note:However if two molecules have similar molecular mass, the dipole-dipole interaction will be more dominant.Example: H2S has higher boiling point than CH3CH3

*4.4.1.3 Hydrogen intermolecular bondDipole-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:Hydrogen intermolecular bond

*Other examples:NH3 liquidCovalent bondHydrogen intermolecular bondHydrogen intermolecular bondHydrogen intermolecular bond

*inCH3OHConsider ethanol, CH3OHNot a hydrogen bondH is not bonded to either F, O or N

*Example: H2O ___ covalent bond----- hydrogen bond

*Properties of compounds with Hydrogen intermolecular forcesHave 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.Boiling point

*Solubility A. Dissolve in polar solventThe 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

*ExampleNH3 dissolves in water because it can form Hydrogen intermolecular bond with water.Hydrogen bond

*Problem:Explain the trend of boiling point given by the graph below:HFHClHBrHIT/oCMolecular mass

*AnswerHF 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.

*The effect of Hydrogen bond on water moleculesThe 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.

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

*Hydrogen bond takes one of the tetrahedral orientation and occupy some spaceIce form tetrahedral arrangement

*H2O(l) is denser than H2O(s) becausethe hydrogen bond in ice arrange the H2O molecules in open hexagonal crystalH2O molecules in water have higher kinetic energy and can overcome the hydrogen bond V-shaped water molecules slide between each other.

*Fig. 11.13

*The boiling points of these substances are affected by:a)the number of hydrogen bonds per moleculeb)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 > CH4Boiling points of substance with Hydrogen intermolecular bonds

*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.

Answer:

*Effects of intermolecular forces on physical properties1)Boiling pointFor molecules with similar size, the order of intermolecular strength:Hydrogen bond > dipole-dipole forces > London dispersion forcesStrength of intermolecular forces boiling point

*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.

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

*Boiling point HF > NH3Fluorine is more electronegative than nitrogen ,thus the hydrogen bonding in H-F is stronger than H-N.

Vapour PressureMolecules 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 systemthe molecules strike the container wall and exert some pressure

Fig. 11.34

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.

In a close system .Liquid molecules vapouriseVolume of liquid becomes lessVapour molecules are trapped in the close containerSome of the vapour molecules may collide and lose their energy. They re-enter the liquid surfaceSystem reaches equilibrium dynamic equilibriumVolume of liquid remains constantMolecules have enough energy to overcome intermolecular forcesRate of vaporisation is faster than the rate of condensation Rate of vapourisation is equal to the rate of condensationPressure exerted by the vapour molecules is known as the vapour pressure

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

Factors that affects vapour pressurei.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.

Fig. 11.35

Boiling the processIncreasing 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.

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)

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.

4.5 Metallic bond

Metallic bondAn 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.*

*Metallic bonds

*Electrostatic force in a metalMetallic Bond (Electron-sea Model)Metals form giant metallic structureEach 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.

*metals have high melting pointhigh energy is required to overcome these strong electrostatic forces between the positive ions and the electron sea in the metallic bond

Physical properties of metalsMetallic bonds formed by the electrostatic forces exist between positive ions and the free moving electrons

*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 electronsThe strength of the metallic bonds

*Boiling points in metalseeeeeeeeeeeeeeNaMgHas one valence electronthe electrostatic force acting between positive ions and free moving electrons form metallic bonds Has 2 valence electronsStronger 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,Mg has higher boiling point than Na

*Example: Explain the difference in the boiling point of the two metals given:Magnesium 11300 oCAluminum 24500 oC

*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

Answer

The strength of metallic bond is directly proportional to the boiling point.The stronger metallic bond,the higher the boiling point.*

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Topic4_Chemicalbonding

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