Chapter 4 Chemical Chapter 4 Chemical bondingbonding4.1 Types of chemical bonding4.2 Ionic bonding4.3 Covalent bonding

Hybridisation – sp, sp2, sp34.4 Shapes of molecules4.5 Metallic bonding4.6 Intermolecular forces4.7 Bonding and physical properties

4.1 Types of chemical 4.1 Types of chemical bondingbonding Ionic bondingcovalent bonding – atoms WITHIN molecules

kept by strong covalent bonds INTERmolecular forces – weak forces

BETWEEN molecules

Few types of INTERmolecular forces:◦ Van der Waal’s forces (VDW) (dispersion forces/

temporary dipole-induced forces)◦ Permanent dipole-dipole forces◦ Hydrogen bond (H-bond)

Important to explain the structure & physical ppt of element & compounds based on the types of chemical bondings

4.2 Ionic bonding4.2 Ionic bonding interaction between metal and non-metal

through electron transfer

Electron transfer forms ion; metal loses electron form cation; non-metal gain electron form anion

Attracted each other by strong ELECTROSTATIC FORCE (between +ve and –ve ions) to form ionic compound

Sometimes also called as Electrovalent bond

Using dot-and cross diagrams◦ It shows: ◦ Outer e- shells only (VALENCE ELECTRON)◦ The charge of the ion is spread evenly, by using

square brackets◦ The charge of each ion

Examples (MUST KNOW):◦ NaCl (Figure 4.4 pg 51)◦ MgO (Figure 4.5 pg 51)◦ CaCl2(Figure 4.6 pg 51)

4.3 Covalent bond4.3 Covalent bondSingle covalent bondsharing of electrons between two or more

non-metals atomsRepresented by a SINGLE LINEPAIRS of outer shell e- NOT USED in bonding

– Lone pair

Single covalent bond

Multiple covalent bondSharing 2 pairs of electrons (double covalent bond)

=

Oxygen ethene

Carbon dioxide

Triple covalent bond

Nitrogen

N Nxx xxx

4.4 Shapes of molecules4.4 Shapes of moleculesElectron pair repulsion theoryAll e- have –ve charge and will repel each

other when they are close together

Pair of electrons in the bonds surrounding the central atom in a molecule will repel other e- pair

The repulsion forces of the e- pairs FORCES the pairs of electrons to part as further as possible until the repulsive forces are minimised.

Valence shell electron pair repulsion Valence shell electron pair repulsion (VSEPR) theory(VSEPR) theory

In the VSEPR model it is assumed that molecular geometries occur because of the influence of electron pairs within the system, more so than the ways in which atoms are bonded together. This is a qualitative theory, which therefore does not involve much in the way of mathematics.

In essence, the VSEPR theory states that electron pairs of electrons repel each other, therefore for a molecule to be at it’s most stable – electron pairs must be as far from each other as possible.

• The influence of lone electron pairs• It is important to recognise the influence of

lone electron pairs as well as those that are involved in bonding.

• Concentration of e- charge cloud: Lone > bonded

• Lone pairs e- cloud charges are wider and slightly closer to the atom

• Lone pairs of electrons are under the influence of a single atom (rather than the whole molecule), they are therefore held tightly to that particular atom.

• This is the reason that lone pair repulsions are greater than those that are bonded.

• The repulsion sequence (starting with the strongest repulsion) is shown below:• Lone Pair – Lone Pair > Lone Pair –

Bonded Pair > Bonded Pair – Bonded Pair

• Figure 4.15 pg 56.

• Consequently lone pairs of electrons have a major effect on the shape of a molecule.

• Shape and bond angles of covalently bonded molecule depends on • No. of e- pairs around each atom• Whether these e- are lone pairs or bonding

pairs

• Molecular shapes similar to those discussed in reference to Hybridization theory are generally adopted in VSEPR theory as well.

No e- pairs

No. bonding e- pairs

No. lone e- pairs

Geometry Angle Examples

2 2 0 Linear 180 BeCl2/BeH2/HCN/Ethyne

3 3 0 Trigonal planar

120 BF3, Ethene

3 2 1 Bent/V-shape

<120

4 4 0 Tetrahedral 109.5 CH4, CCl4, SO42-

Methane

4 3 1 Triangular pyramidal

<109.5 NH3, PCl3

4 2 2 Bent/V-shape

<109.5 H20

6 6 0 Octahedral 90 SF6

No e- pairs

No. bonding e- pairs

No. lone e- pairs

Geometry Angle Examples

5 5 0 Trigonal bypyramidal

90, 120 PCl5,

5 4 1 Seesaw 90, <120

SF4

5 3 2 T-shaped 90 ClF3

5 2 3 Linear 180 XeF2 , I3-

No e- pairs

No. bonding e- pairs

No. lone e- pairs

Geometry Angle Examples

6 6 0 Octahedral 90 SF6

6 1 5 Square pyramidal

90 BrF5

6 2 4 Square planar

90 XeF4 , ICl4-

Hybrid orbitals and molecular Hybrid orbitals and molecular geometrygeometry

Num of bonded e- groups

2 3 4

Composition AB2 AB3 AB4

Geometry of e- groups

Linear Trigonal planar

Bond angles 1800 1200 109.50

Hybridisation

sp sp2 sp3

Example

BeCl2/BeH2/EthyneHCN

BF3Ethene

CCl4SO4

2-

Methane

Sigma bond, Sigma bond, σσ formation of covalent bond when two atomic

orbitals on adjacent atoms overlaps

Involves the overlapping of one end of an atomic orbital from one atom to another one end of an atomic orbital from another atom; end-to-end overlapping

The electron density at the overlapping area would be the greatest, thus the σ bond is rather strong.

S-S σ bond H H H2

1s1 1s1

S-P σ bond H F HF 1s1 2p5

P-P σ bond Cl Cl Cl2

3p5 3p5

Pi bond, Pi bond, ππ formation of covalent bond when two p atomic

orbitals are overlapping sideways

Formation of pi bond are weaker than sigma bond as the bond is loosely held by the atom nuclei, and the overlapping of orbitals is smaller.

Pi bonds are only formed after σ bond is formed.

p p sideways overlapping one π bond

HybridisationHybridisationAtomic orbitals can be combined together to

form hybrid orbitals, and bonds involving such orbitals are called hybrid orbitals

The electrons rearrange themselves again in a process called hybridisation. 

3 main types of hybridisation, sp, sp2 and sp3. (Others such as sp3d, sp3d2)

In hybridisation, the no. of hybrid orbitals produced equals the total no. of atomic orbitals that are combined.

For eg: sp3 hybridisation produces (1+3) orbitals.

Only atomic orbitals that are fairly close in energy can be combined to form hybrid orbitals. Hybrid between 2s and 4s?

Example:Ground state e- configuration of the valence shell of Carbon is 2s2 2p2.

If only valence shell atomic orbitals containing single, unpaired e can be overlapped and used in covalent bonding, it is thus predicted that the molecule CH2 is linear and the bond angle of H-C-H is 90 degree. (p-orbitals are at 90 degrees to each other)

Therefore hybridisation is suggested to rationalize the observed shape of a molecule

spsp33 hybridisation hybridisation

There is a serious mis-match between this structure and the modern electronic structure of carbon, 1s22s22px

12py1.

When bonds are formed, energy is released and the system becomes more stable. If carbon forms 4 bonds rather than 2, twice as much energy is released and so the resulting molecule becomes even more stable.

What is this called?

There is only a small energy gap between the 2s and 2p orbitals, and so it pays the carbon to provide a small amount of energy to promote an electron from the 2s to the empty 2p to give 4 unpaired electrons.

The extra energy released when the bonds form more than compensates for the initial input.

The carbon atom is now said to be in an excited state.

 tetrahedron (a triangularly based pyramid) 

Other sp3 examples : CCl4, PCl3

4 C-H σ bonds

Example : Methane CH4

spsp22 Hybridisation Hybridisationonly hybridise 3 of the orbitals rather than all

four. They use the 2s electron and two of the 2p electrons, but leave the other 2p electron unchanged.

Hybrid orbitals are shorter and fatter

The three sp2 hybrid orbitals arrange themselves as far apart as possible - which is at 120° to each other in a plane.

4 C-H σ bonds, 1 C-C σ bond, 1 C-C π bond

Example : Ethene C2H4

sp hybridisationsp hybridisation

2 C-H σ bonds, 1 C-C σ bond, 2 C-C π bond

Example : Ethyne C2H2

BENZENEBENZENEBenzene has two resonance structures,

showing the placements of the bonds. 

Each of the 6-C atoms is sp2 hybridisation

The H-C-C bond is 120 degree

σ bond ◦ sp2 hybridised orbital of C atom overlap with 1s atomic orbital

of hydrogen produced single covalent σ C-H bond ◦ Each C atom forms 3 σ bond (2 for C, 1 for H)

π bond◦ 6 2p orbitals of the C atoms which are not hybridised, overlap

sideways on both sides to form 6 molecular orbitals of the π type.

◦ Each C atom contributes 1 π bond◦ However the π bond forms are not localised (DELOCALISED)

between pairs of C atoms as in an alkene.◦ The delocalised electrons from the π bond result in resonance

structure of benzene.

Pg 375.

4.3 Co-ordinate bonding (dative 4.3 Co-ordinate bonding (dative covalent bonding)covalent bonding) Is formed when 1 atom provides BOTH the e-

needed for COVALENT BOND

What is the term for both e- (pairs of e-) that is available for bonding?

Sometimes atom with lone pairs tend to share e- with other element or ion which have an empty orbital or those with an incomplete outer shell

Also known as coordinate bond with the symbol “→”

Lone pair with UNFILLED orbital (ions)

1. Ammonium

ORH +

H N HH

Compounds with an incomplete outer shell tend to bond datively and form dimer (Eg: AlCl3 , BeCl2)

1. AlCl1. AlCl3 / 3 / AlAl22ClCl6 6 dimerdimerAt high temperature, Aluminium chloride sublimes

(turns straight from a solid to a gas) at about 180°C to AlCl3

At low temperature, 2 AlCl3 molecules form 1 Al2Cl6 (Dimers of AlCl3)

Are able to combined because of the lone pairs e- on the 2 Cl atoms

Lone pair with INCOMPLETE orbital

Each Cl atom has 3 lone pairs e-

1of the 3 lone pairs e- are used to form coordinate bonds with Al

2. 2. BeClBeCl22

4.5 Metallic bonding4.5 Metallic bonding In metals, the atoms are held together by

metallic bonding to form a regular arrangement called lattice.

In a metal, the bonding of e- are delocalized over the entire crystal (delocalised electrons) by losing their outer shell e- to become cation.

Metal consist of a fixed position in a lattice of positive metal 'ions' surrounding by a 'sea' of delocalised electrons

The delocalized electrons hold the metal cations strongly since the metal cations and the electrons are oppositely charged

The metallic bond is neither covalent nor ionic, but it is a strong bond (electrostatic bond) because most metals have relatively high melting points

The delocalized electrons are moving randomly, from place to place

It do not involve in transferring or sharing electrons with the neighboring atom to form cations, but instead the outer energy levels of the metal atoms overlap

Fig 4.28 pg 60

Strength of metallic bonding increases with◦ Inc +ve charge on the ions◦ Dec size of metal ions in the lattice◦ Inc num of mobile e- per atom

Metallic bonding & the properties Metallic bonding & the properties of metalsof metals1. Most metals have high melting points (m.pt)

and high boiling points (b.pt)

2. Metals conduct electricity◦ When metal is subjected to an electrical potential, the

valence e- are free to move and hence all metals are very good conductors of electricity

◦ Conductivity across groups in the periodic table◦ What about transition elements?

3. Metals conduct heat◦ Conduction of heat is due the vibrations in the crystal

lattice by the delocalised e-

◦ 4. metallic lustre and opaque because it has the shiny coating or covering and have light-reflective qualities on its surface

The lustre, or shine, of metals is caused by the electrons reflecting light

All pure metals reflect well

However some metals do not seem to do so, like lead, iron, etc. which coated with a thin layer of oxide (rust). If this layer is scraped off, the reflective metal can be seen underneath

4.6 Intermolecular forces4.6 Intermolecular forces3 types of INTERmolecular forces:

◦ Van der Waal’s forces (VDW) (dispersion forces/ temporary dipole-induced forces)

◦ Permanent dipole-dipole forces◦ Hydrogen bond (H-bond)

General trend of the strengths (Table 4.2 pg 61.):Ionic > covalent > H-bond > permanent dipole-dipole> VDW

In order to understand how intermolecular forces work, we have to know about ELECTRONEGATIVITY and bond POLARITY

4.6.1 Electronegativity 4.6.1 Electronegativity (EN)(EN)Electronegativity is the ability of a particular atom,

which is covalently bonded to another atom, to attract the bond pair of e- towards itself.

It is a measure of the tendency of an atom to attract electrons towards itself

The Pauling scale - use in quantifying electronegativity

Think: Group1 or Group 7 has higher electronegativity?Going down a group, electronegativity inc/dec?

So, which atom has the highest electronegativity?

Pauling scale:◦ Higher value (non-metals): more attracted; relatively

negative charge◦ Lower value (metals): less attracted; relatively positive

charged

Relating electronegativity to metalsMetallic atoms have higher/lower

electronegativity.

Most metals have EN of < 2 and most non-metals have EN >2.

Predicting the nature of the bond using EN.predict the nature of the chemical bond by looking at the difference of electronegativity value

◦ no electronegativity difference OR difference < 0.5 form non-polar bond; usually have to be the same atoms of identical electronegativity or different atoms but similar electronegativities; H2, Br2, Cl2, N–Cl

◦ electronegativity difference from 0.5-2.0 form polar bond; usually from different atoms with different electronegativities; H2O, H–Cl

◦ greater difference of electronegativity (> 2.0) leads to an ionic bond; NaCl, CaBr2

4.6.2 Polarity4.6.2 Polarity

When EN between 2 atoms are the same in covalent bonding, or the electrons density are equally share – non-polar

Cl Cl

When EN between 2 atoms are different, the more EN atom attracts the pair of e- in the bond towards it.

This result a permanent partial charge; polar bond

Bond polarity is due to difference of EN. The bigger the diff in EN between 2 atoms, the more polarized the bond. Such as ionic bonds.

H Cl

δ+ δ–

Polarity:Centre +ve charge does not coincide with the

centre –ve chargee- distribution is assymmetricThe two atoms are partially charged (δ+ / δ-)

The shift of electron density in polar molecule can be symbolized by placing a crossed arrow ( ) to indicate the direction of the shift

H – F

Dots & cross diagrams could explain polarity. Eg. C-F.

Check EN values for both. Explain why using dots & cross.◦ Nuclear charge◦ Distance◦ Shielding (net +ve charge)

The polarity of molecule is generally measured quantitatively as dipole moment, μ which is the product of charge,

μ = δ × d (unit = D, debye)

The dipole moment is defined as the product of a partial charges ( δ+ and δ–) and distance (d).

The molecule would be non-polar due to μ = 0; but if the bond is polar, the molecule would be polar where it has μ ≠ 0

Polar molecules Polar molecules compoundcompoundWhen consisting of more than 2 atoms:

◦ Polarity (dipole moment) of each bond◦ Arrangement of bonds in the molecule

Molecules with unequal distribution of atoms and whose dipoles do not cancel each other are polar molecules.

Molecular Dipole Moments are the vector sum of the individual bond dipole moments. They depend on the magnitude and direction (vector) of the bond dipoles

EXAMPLE

1. (carbonyl grp C=O)

Methanal (Formaldehyde) and CO2. Formaldehyde is highly polar while carbon dioxide is nonpolar . Since CO2 is a linear molecule, the dipoles cancel each other.

Asymmetry Symmetry

2. CH3Cl Chloromethane

Both C-Cl and C-H bonds are polar covalent.

However the C-H bond contributions are small.

Most of the dipole moment is due to the C-Cl bond.

CH

ClHH

Asymmetryμ ≠ 0

3. H2O 4. NH3 Ammonia

OH

H

:

:

μ ≠ 0Asymmetry

NH

H

H:

μ ≠ 0Asymmetry

EN of O atom > H => polar O-H bond

EN of N > H atom, causing N-H to be polar.

Water is a bent molecule NH3 Ammonia is a trigonal pyramidal

oxygen end has a partial –ve charge; whereas the H end has a partial + charge.

N end has a partial –ve charge; whereas the H end has a partial + charge.

Are the bonds polar?

0.5 < Eneg diff < 2

YES – POLAR BONDS NO – NON POLAR BONDS

NON-POLARMOLECULEIs the molecule symmetrical?

NO YES

NON-POLARMOLECULE

POLARMOLECULE

EN differences present?

Type of molecule

Type of bond

Dipole moment Shape Type of

molecule

Diatomic molecule, A2

Non – polar μ = 0 Linear, symmetry

Non – polare.g. H2, Cl2

Polar μ ≠ 0 Linear, asymmetry

Polare.g. HCl

Triatomic molecule,

AY2

Polar μ ≠ 0 Linear, asymmetry

Polare.g. HCN

Polar μ = 0 Linear, symmetry

Non – polare.g. CO2,

C2H2

Tetra-atomic

molecule, AY3

Polar μ = 0Trigonal planar,

symmetryNon – polar

e.g. BCl3

Polar μ ≠ 0Trigonal planar,

AsymmetryPolar

e.g. BFCl2

Type of molecule

Type of bond

Dipole moment Shape Type of

molecule

AY2E Polar μ ≠ 0 Bent, asymmetry Polar

Penta-atomic

molecule, AY4

Polar μ = 0 Tetrahedral, symmetry

Non – polare.g. CH4

Polar μ ≠ 0 Tetrahedral, asymmetry

Polare.g. CH3Cl

AY3E Polar μ ≠ 0 Pyramidal, asymmetry

Polare.g. NH3

AY2E2 Polar μ ≠ 0 Bent, asymmetry

Polare.g. H2O

Hexa-atomic

molecule, AY5

Polar μ = 0Trigonal

bypyramidal, symmetry

Non – polare.g. PCl5

Polar μ ≠ 0Trigonal

bypyramidal, asymmetry

Polare.g. PCl4F

Type of molecule

Type of bond

Dipole moment Shape Type of

moleculeHepta-atomic

molecule, AY6

Polar μ = 0 Octahedral, symmetry

Non – polare.g. SF6

Polar μ ≠ 0 Octahedral, asymmetry

Polare.g. SF5Cl

Polarity and chemical Polarity and chemical reactivityreactivityUsing polarity for chemical reaction

Example: triple bonds

Are they polar molecules?CO is use as a reducing agent (element who

donate e-)

Example: Nucleophilic substitution of Bromoethane with OH- ions (pg 233)

Intermolecular forcesIntermolecular forcesDeviation from the ideal gas law is due to weak

forces of attraction between molecules

These forces are responsible for the formation of the condensed phases, that are liquid and gases

They are collectively known as Van der Waals’ forces

There are 3 types: Van der Waals’ forces (dispersion forces/ temporary dipole-induced

forces/London forces) Dipole-dipole forces Hydrogen bonding

Van der Waals’s forces Van der Waals’s forces (VDW)(VDW)Also knows as dispersion forces / London forces/

All molecules experience VDW which result from the motion of electrons

Average over time, e- distribution around an atom/molecule is symmetrical. But at a particular instant, by chance, the e- are concentrated in 1 region of the atom/molecule. A temporary dipole was set up.

This dipole could induce a dipole on neighbouring molecules. Hence the name temporary-induced dipole forces. (Figure 4.33 pg 63)

The strength of the dispersion forces depends on how easily the e- cloud of the atom/molecule could be induced/polarised

The further away the e- are from the nucleus, the easier it is to polarised the e- cloud.

Factors affecting VDW:1. Higher no. of e-/p; the higher/lower the VDW???

2. Molecular shape / contact points bet molecules. ◦ The more contact point, the higher/lower the VDW???◦ Elongated/linear molecules compared to branched

molecules?

◦ Elongated molecules are more spread out, increasing the contact point bet molecules and are more easily polarised compared to small, compact molecules.

◦ What about surface area of molecules?◦ What about increasing molecular weight?

So, when VDW is high? What does it mean/cause?

When VDW is high, B.pt / enthalphy of vaporisation (kJ/mol-1) is high.

Group 8

M.pt and B.pt cannot be correlated simply to molecular weight or to the no. of e- in a molecule

Chemical symbol and name Atomic number Electron

arrangement Melting point Boiling point

Atomic radius pm (10-12m) and

nanometres nm (10-9m)

He helium 2 2 -272oC , 1K -269oC , 4K 49 and 0.049

Ne neon 10 2.8 -249oC , 24K -246oC , 27K 51 and 0.051

Ar argon 18 2.8.8 -189oC , 84K -186oC , 87K 94 and 0.094

Kr krypton 36 2.8.18.8 -157oC , 116K -152oC , 121K 109 and 0.109

Xe xenon 54 2.8.18.18.8 -112oC , 161K -108oC , 165K 130 and 0.130

Rn radon 86 2.8.18.32.18.8 -71oC , 202K -62oC , 211K 136 and 0.136

Generally when comparing 2 molecules: Dealing with 2 factor (molecular weight and polar)

Case 1Estimate the B. pt for Silane SiH4and Hydrogen

sulphide H2S

With approximately the same molecular weight, more polar molecule has higher m.pt and b.pt.

This is due to the addition of dipole-dipole attraction to the VDW.

Compound Silane Hydrogen sulphide

Molecular formula SiH4 H2SDipole moment (D) 0 0.97Relative molecular mass Mr

32.1 34.1

B. Pt (K) 161.2 212.3

Case 2Estimate the B.pt for Chloromethane CH3Cl and

Fluoromethane CH3F

When both molecules are polar, higher molecular weight molecule has higher m.pt and b.pt.

This is due to the greater no. of e-, the stronger the VDW attraction.

Compound Chloromethane

Fluoromethane

Molecular formula CH3Cl CH3FDipole moment (D) 1.87 1.85Relative molecular mass Mr

50.5 34

B. Pt (K) 248.8 194.6

Permanent dipole-dipole Permanent dipole-dipole forcesforcesexist between polar covalent molecule which due

to the unequally distribution of the electrons; polarized

have a permanent dipole where each element having permanent partial charge; dipole molecule

Is the forces between 2 permanent dipole molecules

A result of electrical interactions among dipole neighbouring molecules.

These forces can either be attractive or repulsive, depending on the orientation of the molecules

The additional partial ordering of molecules can cause a substance to persist as a solid or liquid state at temperatures than otherwise expected

The net force in a large collection of molecules results from many individual interactions or both types.

These forces are significant only when molecules are in close contact

ExampleDipole moment of HCl molecules, because of the

force of attraction between oppositely charged particles, there is a small dipole-dipole force of attraction between adjacent HCl molecules

Strength of dipole-dipole interaction between HCl?

Bond strength between H and Cl atom? What bond?

3.3 kJ/mol B.Pt = -85.0oC

HCl exist in what form at room temp?

The larger/smaller the dipole moment, the greater/lesser the dipole interaction strength??

Therefore, when dipole interaction strength increase, m.pt/B.pt ? What state?

When dealing with 2 factor (molecular weight and dipole-dipole forces) between molecules

ExampleEstimate the B.pt for Propane C3H8, Dimethyl

ether CH3OCH3, Ethanenitrile CH3CN

The larger the dipole moment, the greater the strength of the dipole-dipole interaction.

Compound Propane Dimethyl ether

Ethanenitrile

Molecular formula C3H8 CH3OCH3 CH3CNDipole moment (D) 0.1 1.3 3.9Relative molecular mass Mr

44.1 46.07 41.05

B. Pt (K) 231 248 355

Hydrogen bonds (-------)Hydrogen bonds (-------)Extra strong intermolecular, permanent dipole-

permanent dipole attraction between molecules.

Is form when 1 molecule whose H atom is covalently bonded tightly to a high EN atom (N, O, F)

BETWEEN1 molecule having an available lone pair e- from N, O, F

H-bond arise because the N-H / O-H / F-H bonds are HIGHLY POLAR.

In addition, the H atom (small and no shielding e- to shield its nucleus) can be closely approach.

To determine the average no. of H-bonds formed per molecule depends on: ◦ No. of H atom attached to N,O, F◦ No. of lone pairs present

Students to draw H-bond bet (HF, NH3, H2O) molecules

B.Pt = -33OC

B.Pt = 19.5OC

When dealing with 2 factor (molecular weight and H-bond) between molecules

Estimate the B.pt between Ethane and MethanolCompound Ethane MethanolMolecular formula C2H6 CH3OHRelative Molecular Mass

30 32

B.Pt (K) 184.6 337.2

The unique properties of The unique properties of waterwaterWater is the best example of hydrogen bond

with several unique properties

Has the capability in dissolving many ionic compounds◦ Known as hydration; the compound was

said to be hydrated

The peculiar properties of The peculiar properties of waterwater1. Enthalphy change of

vapourisation and B.pt.

2. Surface tension and viscosity

3. Ice is less dense than water

1. Enthalphy change of vapourisation 1. Enthalphy change of vapourisation and B.pt.and B.pt. In general, higher molecular weight

increases B.pt due to the dispersion (VDW) forces.

Evident by Group 14 hydrides.H2O, HF, NH3 have 10e- each. This shows that the H-bond strength does not depend on the polarisability of the molecule.

2. Surface tension and 2. Surface tension and viscosityviscosity

It has comparatively high surface tension and viscosity◦ Surface tension: property of the surface of a liquid

that allows it to resist an external force, due to the cohesive nature of its molecules. Due to H-bond exert a significant downward force at liquid surface

◦ Viscosity: measure of a fluid's resistance to flow. High due the H-bonding reduces water’s ability to slide over each other.

3. Ice is less dense than 3. Ice is less dense than waterwater Ice (solid of water) is less dense than water

UNLIKE OTHER SOLID◦ This is because the water molecules are

bonded in 3D network where 1 water molecule is H-bonded with 4 other water molecule forming a tetrahedral lattice

Density = mass/volume

The repeating units of the tetrahedral lattice forming a hexagonal network and finally ice where the structure is more ‘open’, allowing the molecules to stay slightly apart

Trends in physical Trends in physical propertiespropertiesGenerally, when the molecular mass increases,

melting and boiling point increases; e.g. alkane and alkene

However when the structure of the molecules become more branched, the boiling point decreases

When the polarity of molecule increases, it will cause an increase in the boiling point; comparison between alcohol and alkane

The presents of hydrogen bond will have effects on the boiling temperature

summarysummaryInteracting molecules or ions

Are ionsinvolve?

Van derWaals onlye.g. H2, Cl2

Are H atoms bondedto N, O, F?

Are polarmoleculesinvolve?

Are polarmolecules andions are both

present?

Dipole-dipoleForces

e.g. HCl, HCN

HydrogenBond

e.g. H2O,NH3

Ion-dipoleForces

e.g. NaCl inH2O

IonicBond

e.g. NH4NO3

Yes

Yes

NoNo

No

No

Yes

Yes

Increase in strength

Bond order, Bond order, Bond length, Bond Bond length, Bond energy,energy,Bond orderbonding electron pairs between two bonded atoms

Bond lengthDue to double bonds have greater quantity of –ve

charge between 2 atom, the double bonds are shorter than single bonds

C—O 143 pm > C=O 122 pm > C≡O 113 pm

Bond energyBond energy is the energy needed to break 1 mole of a given bond

in GASEOUS MOLECULE.

It is always a +ve in value because energy is needed to break the bond

Example

bond lengths increase with increasing atom size, and bond energy decreases.

Bond Length increases => M.pt/B.pt ???

Bond energy VS. B.pt/M.pt ????

Bond energy B.pt/M.pt ????Bond energy is the energy needed to break 1 mole of a given bond in GASEOUS MOLECULE.

B.Pt is the temp where both liquid and gaseous phase exists in equilibrium at a particular external pressure.

INTRAMOLECULAR FORCES/bondCovalent (single, double, triple bond)

What about Ionic and metallic bonding?

INTERMOLECULAR FORCESVDW, Dipole-dipole, H-bond

The higher the bond energy, the higher the strength of the intramolecular forces; or vice versa

The stronger the intermolecular forces, the higher the B.pt/Mpt. When the forces are stronger, it take more energy (heat) to break them apart