Chemistry Pre-u Chemistry Sem 1 Chap 3

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CHEMISTRY SEMESTER 1CHEMISTRY SEMESTER 1CHAPTER 3CHAPTER 3



CHAPTER 3 CHEMICAL BONDING

31 Ionic bonding

32 Covalent bonding

33 Metallic bonding

34 Intermolecular forces Van der Waals forces andhydrogen bonding

Topic

2007 2008 2009 2010 2011 20122013

Sem 12014

Sem 1

P1 P2 P1 P2 P1 P2 P1 P2 P1 P2 P1 P2 AB

CA

B

C

3

Chemical

Bonding

43c

6a4 5a 3

3b

c1 2

2

5b2

1b

5c

7a

319

b3 19



INTERACTION

BETWEEN ELEMENTS

Metal and

non-metalNon - metal and

non-metal

Metal and

metal



bull A Lewis dot symbol consists of the symbol of an element and

one dot for each valence electron in an atom of the element



31 Ionic Bonding

bull The central idea of the ionic bonding model is the transfer of

electrons from metal atoms to non-metal atoms to form

ions that come together in a solid ionic compound where

ionic bond is formed in between oppositely charged ions

by electrostatic attraction forcesbull For example in the formation of sodium fluoride NaF

Sodium atom Fluorine atom Sodium fluoride

Electronicconfiguration

Na (1s22s22p63s1) F (1s22s22p5) Na+ F-

(1s22s22p6) (1s22s22p6)

Orbitaldiagrams Na F Na+ F-

Lewis

diagram

1s 2s 2p 3s



bull The interaction between sodium atom and fluorine atom

occur where sodium atom (with low ionisation energy)

donates electron to fluorine (with high electron affinity) to formsodium ion Na+ and fluoride ion F- respectively Note that

both ions have achieved octet arrangement of ns2np6 as it is

the most stable form of an ion formedbull The oppositely charged Na+ and F- form a giant ionic crystal

lattice with very high melting point via electrostatic attraction

i Magnesium chloride MgCl2 ii potassium oxide K2O

iii Calcium sulphide CaS iv aluminium oxide Al2O3



bull According to Coulumbs Law electrostatic energy between

two oppositely charged substance (A and B) is directly

proportional to the charge carried by each ions yet inversely

proportional to the distance between them

bull This relationship helps us predict trends in lattice energy and

explain the effects of ionic size and charge

( ) minus+

minus+

minus

times

prop r r

QQ

energylatticeor EnergyticElectrosta

nn

a) Effect of ionic size As we move down a group in theperiodic table the ionic radius increases Therefore the

electrostatic energy between cations and anions decreases

because the inter-ionic distance is greater thus the lattice

energies of their compounds should decrease as well This

prediction is borne out by the alkali-metal halides note the

regular decrease in lattice energy down a group whether we

hold the cation constant (LiF to LiI)



b) Effect of ionic charge When we compare lithium fluoride

with magnesium oxide we find cations of about equal radii

(Li+ = 76 pm and Mg2+ = 72 pm) and anions of about equalradii (F- = 133 pm and O2- = 140 pm) Thus the only

significant difference is the ionic charge LiF contains the

singly charged Li+ and F- ions whereas MgO contains the

doubly charged Mg2+ and O2- ions The difference in their

lattice energies is

∆H of LiF = - 1050 kJ mol-1

∆Hlattice of MgO = - 3923 kJ mol-1

This nearly fourfold increase in ∆Hlattice reflects the fourfold

increase in the product of the charges (1 x 1 vs 2 x 2) in the

numerator of equation above





3 Properties of Ionic Compound

a) Melting point - Ionic compound has giant ionic crystal lattice

which are hold by strong electrostatic attraction forces by

repeating of oppositely charged ions

Therefore the melting point of ionic compounds are usuallyvery high as a lot of energies are required to overcome the

electrostatic attraction forces and melted to form free moving

ions Therefore ionic compounds have very high melting

point



b) Conductivity of electricity - Most ionic compounds do not

conduct electricity (insulator) in the solid state but do

conduct it when melted or when dissolved in water (exceptsome super-ionic conductors such as AgI which have

remarkable conductivity in the solid state) Solid ionic salt

consists of immobilized ions When it melts or dissolveshowever the ions are free to move and carry an electric

current

-

are hard (does not dent) rigid (does not bend) and brittle(cracks without deforming) These properties are due to the

powerful attractive forces that hold the ions in specific

positions throughout the crystal Moving the ions out of

position requires overcoming these forces so the sample

resists denting and bending If enough pressure is applied

ions of like charge are brought next to each other and

repulsive forces crack the ionic solid suddenly







32 Covalent bondbull Studies of covalent bond was widely developed ever since

Lewis suggested that a chemical bond exist in a hydrogen gas

occur by sharing en electron between two hydrogen atoms

bull Electron pair that connect the 2 hydrogen atoms is called

covalent bond a bond in which two electrons are shared by

wo a oms an e e ec ron pa r a on e ween e wohydrogen atoms is also called as bonding pair electrons

bull In a covalent bond each electron in a shared pair is attracted to

the nuclei of both atoms This attraction holds the two atoms in

H2 together and is responsible for the formation of covalentbonds in other molecules



2 A Lewis structure is a representation of covalent bonding

in which shared electron pairs are shown either as lines

or as pairs of dots between two atoms and lone pairsare shown as pairs of dots on individual atoms Only

valence electrons are shown in a Lewis structure

a) Consider the fluorine molecule F2 The electronconfiguration of F is 1s22s22p5 The 1s electrons are in

inner shell which is nearest to the nucleus For this reason

the do not artici ate in bond formation Thus each F

atom has seven valence electrons (2s22p5) Thereforeeach fluorine atom needed one electron to achieve octet

configuration (ns2np6)

F 1 s22s22p5 F 1 s22s22p5 1s2 2s2 2p6 1s2 2s2 2p6



bull Oxygen atom has electronic configuration of 1s22s22p4 To

achieve stable octet configuration (ns2np6) each oxygen

atom need 2 electrons Hence when 2 oxygen atomsinteract they shared two electrons in between each other as

described in diagram below

O 1s2

2s2

2p4

O 1s2

2s2

2p4

1s2

2s2

2p6

1s2

2s2

2p6

bull From the structure of oxygen molecule formed each oxygen

atom shared two electrons from each other to form a double

bond in order to achieve octet configuration among each

oxygen atom



bull Nitrogen has electronic configuration of 1s22s22p3 and

required 3 electrons to achieve octet configuration (ns2np6)

In this case each nitrogen atom shared 3 electrons fromeach of its atom to form triple bond

N 1 s22s22p3 N 1 s22s22p3 1s2 2s2 2p6 1s2 2s2 2p6



Water H2O

H 1 s1 O 2 s22p4

Carbon dioxide CO2

C 2 s22p2 O 2 s22p4

Ammonia NH3

H 1 s1 N 2 s22p3Ethene C2H4

H 1 s1 C 2 s22p2



Hydrogen cyanide HCN

H 1 s1 N 2 s22p3 C 2 s22p2

Ethanoic acid CH3COOH

H 1 s1 O 2 s22p4 C 2 s22p2

Tetrachloromethane CCl4C 2 s22p2 Cl 3s23p5

Ethyne C2H2

H 1 s1 C 2 s22p2



a) Note that in ethene hydrogen cyanide and ethanoic acid all

the valence electrons are used in bonding there are no lone

pairs on the carbon atoms In fact most of the stablemolecules containing carbon do not have lone pairs on the

carbon atoms

b) Multiple bonds (double bond or triple bond) are shorter than single covalent bonds Bond length is defined as the

distance between the nuclei of two covalently bonded atoms

nitrogen triple bonds are shorter than double bonds whichin turn are shorter than single bonds The shorter multiple

bonds are also more stable than single bonds

c) Covalent bond not only exist in neutral molecule but also insome molecular ions Table below shows a few example of

molecular ions which have covalent bonds in its molecule



Carbonate ion CO32-

C 2 s22p2 O 2 s22p4

Cyanide ion CN-

N 2 s22p3 C 2 s22p2

u p a e on 4-

S 3 s23p4 O 2 s22p4

ra e on 3-

N 2 s22p3 O 2 s22p4



321 Exception of Octet Rules

1 From the Lewis structure sketched for sulphate ion SO42- we can

see that the center metal atom (sulphur) has more than 8electrons For this the molecule is described as molecule that can

expanded octet These molecules that have more than 8

electrons are located at Period 3 and below as these center

atoms have empty d-orbital available to expand the number ofelectrons positioned in the center atom Those center atom from

Period 2 such as C N O and F can only allocate a maximum

Example Both phosphorous and nitrogen are elements from Group15 with the valence electron of ns2np3 They can both form NCl3and PCl3 respectively when react with limited amount of chlorine

however under excess chlorine only PCl 5 can be formed but

not NCl 5 Explain the statement bolded

Solution This is due to phosphorous which is from period

3 have empty d-orbital to expand the octet However

nitrogen which is from Period 2 do not have emptyorbital and can only allocate 8 electrons in its shell



Phosphorous pentachloride PCl5P 3 s23p3 Cl 3s23p5

Sulphur hexafluoride SF6

S 3 s23p4 F 2 s22p5

Bromine pentachloride BrCl5Cl 3s23p5 Br 4s24p5

Xenon tetrafluoride XeF4

Xe 5s25p6 F 2 s22p5



2 There are also some stable covalent compounds which have

less than 8 electrons in their center atom (incomplete octet)

The center atoms are usually metals with great number ofvalence electrons with small atomic radius such as

beryllium boron and aluminium

BeCl2Be 2s2 Cl 3s23p5 BF3B 2 s22p1 F 2 s22p5 AlCl3Al 3s23p1 F 3 s23p5

These compound possessed stability due to their short bondlength between center atom with surround atom Furthermore

they can form resonance structure between the center atom

and surrounding atoms



3 There are also some molecules which contain an odd

number of electrons Among the most common compounds

are nitrogen monoxide (NO) and nitrogen dioxide (NO2)

Odd-electron molecules are sometimes called radicals

Nitrogen monoxide NO

N 2 s22p3 O 2 s22p4

Nitrogen dioxide NO2

N 2 s22p3 O 2 s22p4

any ra ca s are g y reac ve e reason s t at t ere

is a tendency for the unpaired electron to form a

covalent bond with an unpaired electron on another

molecule For example when two nitrogen dioxide

molecules collide they form dinitrogen tetroxide



322 Dative bond

1 As shown in the Lewis structure of nitrate ion one of the N-Obond is drawn as rarr The bond rarr placed is called as dative

bond (also known as coordinative bond) where dative

bond is defined as a covalent bond in which one of the atoms

donate the lone pair electrons available Although theproperties of a coordinate covalent bond do not differ from

those of a normal covalent bond because all electrons are

a e no ma er w a e r source

2 Dative bond is usually applied for these few circumstances

below

a) To assist atom molecule ion that not yet achieved octet

configuration Making use of atom which has lone pair

electrons to those which are lack of electrons



Sulphur dioxide SO2Sulphur trioxide SO3 Carbon monoxide CO

Ozone molecule O3 Water with hydrogen ion

S

OO

O

Ammonia with hydrogen ion Ammonia with boron trifluoride



b) Formation of dimer - In order for some compounds which

have incomplete octet to achieve stability they tend to form

dimer or polymer among themselves by using dative bondTwo of the most common examples are aluminium trichloride

and beryllium dichloride

Monomer of AlCl3Dimer of aluminium chloride Al

2

Cl6

Monomer of BeCl2 Dimer of Be2Cl4 Polymer of (BeCl2)n



c) Formation of coordination compounds - Coordination

compounds are substances that contain at least one complex

ion a species consisting of a central metal cation (either atransition metal or a main-group metal) that is bonded to

molecules andor anions called ligands via dative

(coordinative) bond

hexaaquacopper (II) ion [Cu(H2O)6]2+ tetraamminenickel (II) ion [Ni(NH3)4]

2+

Hexacyanoferrate (III) ion [Fe(CN)6]3- Trioxalatocobaltate(III)ion [Co(C2O4)3]

3-

3 2 3 Hybridisation Theory



323 Hybridisation Theory

Valence bond theory

bull The basic principle of valence bond theory is that a covalentbond is formed when orbitals of two atoms overlap and the

overlapped region which is between the nuclei is occupied by

a pair of electrons The central themes of valence bond theory

derive from this principle

ndash Opposing spins of the electron pair ~ Stated in Paulis

orbitals has a maximum capacity of two electrons that musthave opposite spins For example when a covalent bond is

formed in molecule of hydrogen H2 the two 1s electrons of

two H atoms occupy the overlapping 1s orbitals and have

opposite spins



bull Maximum overlap of bonding orbitals ~ The bond strength

depends on attraction of the nuclei for the shared electrons

so the greater the orbital overlap the stronger (morestable) the bond The extent of overlap depends on the

shapes and directions of the orbitals An s orbital is spherical

but p and d orbitals have more electron density in one

direction than in another Thus whenever possible a bond

involving p or d orbitals will be oriented in the direction that

maximizes overlap For example in hydrogen fluoride (HF)

bond the 1s orbital of H overlaps the half-filled 2p orbital of Falong the long axis of that orbital Any other direction would

result in less overlap and thus a weaker bond

H b idi ti f t i bit l T t f th



bull Hybridisation of atomic orbitals ~ To account for the

bonding in simple diatomic molecules like HF we picture the

direct overlap of s and p orbitals of isolated atoms But howcan we account for the shapes of so many molecules and

polyatomic ions through the overlap of spherical s orbitals

dumbbell-shaped p orbitals and cloverleaf-shaped d orbitals

Linus Pauling proposed that the valence atomic orbitals in the

molecule are different from those in the isolated atoms The

spatial orientations of these new orbitals lead to more stable

bonds and are consistent with observed molecular shapesThe process of orbital mixing is called hybridisation and

the new atomic orbitals are called hybrid orbitals Two key

points about the number and type of hybrid orbitals are that

i The number of hybrid orbitals obtained equals the number ofatomic orbitals mixed

ii The type of hybrid orbitals obtained varies with the types of

atomic orbitals mixed

3 2 3 1 T f h b idi ti



3231 Type of hybridisation

1 sp3 hybridisation ~ When four electron groups surround

the central atom the center atom involved must prepare fourorbitals with equal energies to overlap with the four

surrounding electron groups Valence Bond theory uses

hypothetical hybrid orbitals which are atomic orbitals

obtained when two or more non-equivalent orbitals of the

same atom combine in preparation for covalent bond

formation H bridisation is the term a lied to the mixin of

atomic orbitals in an atom (usually a central atom) togenerate a set of hybrid orbitals In the case of sp3 We can

generate four equivalent hybrid orbitals from the center

atom by mixing the s orbital and the three p orbitals

Explanation Energy level diagram Diagram of orbitals




a) Ground state

Carbon which act as thecenter atom has the valence

electron of 2s22p2

b) Excited state One of the electron from 2s

is promote to 2p orbitals -

equa energy eve

c) Hybridised state

One orbital of 2s and three

orbitals of 2p combined(hybrid) and rearrange

themselves to the shape

and orientation of a

tetrahedral shape

Molecular shape Tetrahedral

Angle between bond pair 10950



2 sp2 hybridisation ~ When three electron groups surround

the central atom the center atom involved must prepare three

orbitals with equal energies to overlap with the three

surrounding electron groups In sp2 hybridisation three

equivalent (in terms of energy level) from the center atom by

mixing the one s orbital and the two p orbitals Using boron

trifluoride (BF3) as example sp2

hybridisation is explained





3 sp hybridisation ~ When two electron groups surround

the central atom we observe a linear shape which

means that the bonding orbitals must have a linear

orientation VB theory explains this by proposing thatmixing two nonequivalent orbitals of a central atom one

s and one p gives rise to two equivalent sp hybrid

orbitals that lie 1800 apart




a) Ground state

Beryllium which act as the

center atom has the valence

electron of 2s2

b) Excited state

One of the electron from 2s ispromote to 2p orbitals - equal

energy level

c) Hybridised state

One orbital of 2s and one

orbitals of 2p combined (hybrid)

and rearrange themselves to

the shape and orientation of a

linear shape

Molecular shape Linear


4 The concept of hybridisation is also useful to explain molecules with



p y p

doubletriple bonds By using the concept of the direct overlapping

orbitals and side-touch lapping orbitals the formation of multiple

bonds in ethene C2H4 and ethyne C2H2 are described

a) Ethene C2H4 - Hybridisation take place for both carbon atoms in

ethene molecule is sp2 hybridisation

bull When both hybridised C is bonded together one of the hybridisedorbital overlapped directly between each other while the other two

hybridised orbitals overlapped directly with hydrogen atoms

z

atoms form a side-touch bond between each other and form anotherbond as shown in the diagram below

bull From the diagram with C=C there are two types of bond A sigma-

bond (σ-bond) is covalent bonds formed by orbitals overlapping end-

to-end with the electron density concentrated between the nuclei ofthe bonding atoms while the second type is called a pi bond (π-

bond) which is defined as a covalent bond formed by sideways

overlapping orbitals with electron density concentrated above and

below the plane of the nuclei of the bonding atoms




a) Ground state

Carbon which act as the center

atom has the valence electron

of 2s22p2

b) Excited state

One of the electron from 2s is

promote to 2p orbitals - equalenergy level

c) Hybridised state

One orbital of 2s and two



the shape and orientation of atrigonal planar shape

Note that on unhybridised pz

orbital an electron is presence

Molecular shape Trigonal planar Angle between bond pair 1200



b) Ethyne C H has the Lewis structure of H CequivC H which the



b) Ethyne C2H2 has the Lewis structure of HminusCequivCminusH which the

bonding can be explain using sp hybridisation Table below

described how the hybridisation take place on each carbonatom and how the formation of triple bond occur

c) From the diagram formation of -CequivC- is due to the formation

of one sigma-bond by direct overlapping of one of the twohybrid orbitals on each C while the other two pi-bond bonds

are formed as a result of side-lapping of the each two

y z




a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state

One orbital from 2s and 2p

orbitals combined (hybrid) and

rearrange themselves to the

shape and orientation of a

linear shapeNote that on unhybridised py amp

pz orbitals electrons are

presence to form two

π-bonds Molecular shape Linear Angle between bond pair 1800

sp

pz py

5 Other examples of applications in valence bond theory includes the



formation of nitrogen molecule N2 and hydrogen cyanide HCN

molecule

a) Nitrogen gas is Earths most abundant gas as it cover 78 of the

content of our air Nitrogen molecule is an inert gas thanks to its

short covalent bond and also its strong triple bond Therefore

a lot of heats are required to break the chemical bond of nitrogenbefore it can be applied in industries Using valence bond theory

the bonding of nitrogen molecule is explained in the diagram

b) So when the two hybridised nitrogen atom interacting amongeach other they hence formed a linear shape and two pi-bonds

(πminusbonds) are formed as a result of side-lapping of unhybridised

py and pz orbital respectively

N N

ExplanationEnergy level

diagramDiagram of orbitals



diagram

a) Ground state

Both nitrogen whichact as the center atom

has the valence

electron of 2s22p3

b) Excited state One of the electron

from 2s is promote to 2p

orbitals - e ual ener

level

c) Hybridised state

One orbital of 2s and

one orbitals of 2p

combined (hybrid) and

rearrange themselves

to the shape and

orientation of a linear

shapeMolecular shape Linear


b) Whereas for hydrogen cyanide HCN both carbon atom and



) y g y

nitrogen atom undergoes sp hybridisation in order to form a

linear structureExplanation

Energy level diagram for

carbon atom C


nitrogen atom N

a) Ground state

Valence electronC 2s22p2

N 2s22p3

b) Excited state

One of the electron from

2s is promote to 2p

orbitals - equal energy

level

c) Hybridised state


orbitals of 2p combined

(hybrid) and form sp

hybridisation



N

σσσσ

π

σσσσ

π

6 However there are a few limitation on valence bond theory such



as when explaining the effect angle of bond-pair and bond-pair

electrons when there isare presence of lone pair electron in thecenter atom and also the difference of electronegativity

a) Bonding in ammonia NH3 and water H2O ~ Both NH3 and

H2O undergoes an arrangement similar to sp3 hybridisation

similar to that of C in methane Table below shows thehybridisation occur for nitrogen in ammonia and oxygen in water

Energy level diagram Energy level diagram

for nitrogen atom N for oxygen atom O

a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron

from 2s is promote to

2p orbitals

c) Hybridised state





three orbitals from 2p

combined (hybrid) andrearrange themselves to

the shape and

orientation similar to that

of tetrahedral shape

Similar to arrangement in

tetrahedral


tetrahedral

of ammonia NH3 andwater H2O

Shape and angleShape trigonal pyramidal

Angle 1070

Shape bent

Angle 10450

Number of bond pair amp

lone pair electrons

Bond pair electron 3

Lone pair electron 1



bull The angle of bond pair - bond pair electrons in ammonia and0 0



water are 1070 and 10450 respectively which is lesser than in

methane molecule (10950

) This can be explained by the factof the presence of lone-pair electrons in both ammonia and

water Since the lone pair - lone pair electron repulsion is

stronger than lone pair - bond pair electron repulsion than

bond pair - bond pair electron repulsion it is expected that therepulsion between lone pair - bond pair electrons in ammonia

is stronger than bond pair - bond pair electrons repulsion

hence caused the angle to squeeze to a smaller angle As

for water since there is a presence of lone pair - lone pair

electrons repulsion it results the bond pair - bond pair

electron repulsion to be much smaller hence caused the

angle to squeeze to a smaller angle Though valence bondtheory does not actually explain the hybridisation especially

when it relates to the repulsion occur involving lone pair

electron another theories shall be applied to study such

effect All these shall be discussed further in VSEPR theory

b) Effect of electronegativity and bonding angle between bond pair

- bond pair electrons ~



- bond pair electrons

i As discussed earlier in the bonding of water (H2

O) the

molecular shape and angle is described in the diagram below

Sul hur which is also an element from Grou 16 formed h dro en

sulphide H2S when sulphur react with hydrogen However unlike

water the bonding angles is much smaller compare to water This is

due to the difference of electronegativity and also the bond length

between O and S in molecule Since O is more electronegative

compare to S the O-H bond is pulled closer toward O Furthermorethe bond length of O-H is shorter compare to S-H As a result the

bonding pair-bonding pair electrons repulse greater with each other in

H-O-H and caused the angle become greater

ii Another example is between NH3 and PH3 The orbital

f



diagram for both NH3 and PH3 are described below Both

nitrogen N and phosphorous P are from the same groupwhich is Group 15

Similar to the above case P in phosphine is less

electronegative than N in ammonia and bond length of N-H

is shorter than P-H As a result H is pull closer to N and

repulsion between bonding of H-N-H is greater compare toH-P-H and caused the angle between H-N-H is greater

compare to H-P-H

324 Valence Shell Electron Pair Repulsion (VSEPR) Theory

1 M l l t i th th di i l t f t



1 Molecular geometry is the three-dimensional arrangement of atoms

in a molecule A moleculersquos geometry affects its physical andchemical properties such as melting point boiling point density

and the types of reactions it undergoes The basic concept of

VSEPR theory is based on the three general rules below

a) As far as electron-pair repulsion is concerned double bonds andtriple bonds can be treated like single bonds This approximation is

good for qualitative purposes However you should realize that in

ldquo rdquo

because there are two or three bonds between two atoms theelectron density occupies more space

b) If a molecule has two or more resonance structures we can apply

the VSEPR model to any one of them Formal charges are usually

not shown

c) The order of repulsion strength of lone pair and bond pair are

lone-pair amp lone-pair electrons repulsion are the strongest

followed by lone-pair amp bond-pair electrons repulsion while

bond-pair amp bond-pair electrons repulsion is the weakest



No of

surroun

No of

loneMolecular Diagram of the molecular Example of



Classsurroun

d

atoms

pair

electron

Molecular

geometry

Diagram of the molecular

shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-

AB2E2 2 2 Shape Bent

H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-

a) phosphorous trichloride PCl3S1 Total valence electrons

P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e

Total electrons = 26

S2 Electrons used bond = 3 x 2e-

Electrons left = 26 - 6 = 20 e-

S3 e- at surround atom = 3 x 6e-Electrons left = 20 - 18 = 2 e-

S4a ) The 2e- remain is placed atthe center atom P

Since molecule contain 3

surrounding atom and 1 lone pairelectrons hence

Arrangement tetrahedral

Shape trigonal pyramidal

b) Carbonate ion CO32-

S1 Total valence electrons

C = 4 e 3 O = 3 x 6e- + 2e accept



C = 4 e- 3 O = 3 x 6e + 2e- accept Total electrons = 24



- -


S4b) Since the center atom C not

yet achieved octet a double bondis form using any e- from O

Since molecular ion contain 3


Arrangement and

Shape trigonal planar

c) iodine tetrachloride ion ICl4-


I = 7 e- 4 Cl = 4 x 7e- + 1e- accept



I = 7 e 4 Cl = 4 x 7e + 1e accept Total electrons = 36



- = -


S4a ) The 2e- remain is placed at

the center atom P



Arrangement octahedralShape square planar

a) Iodide ion I3- b) Antimony pentachloride ion

[SbCl ]2



[SbCl5]2-

Arrangement trigonal

bipyramidalShape linear

Arrangement octahedralShape square pyramidal

3 4

Arrangement trigonalbipyramidal

Shape T-shape


Shape see-saw

325 Electronegativity and Polarity of Molecules

1 From all the chemical bonding discussed so far ionic and



o a e c e ca bo d g d scussed so a o c a d

covalent bonding models portray compounds as beingformed by either complete electron transfer or complete

electron sharing However in most real compounds the

type of bonding lies somewhere between these extremes

Thus the great majority of bonds are more accuratelythought of as ldquopolar covalentrdquo that is partially ionic and

partially covalent

Pure ionic compound Pure covalent compound Polar covalent compound

2 One of the most important concepts in chemical bonding is

electronegativity (EN) the relative ability of a bonded atom to



attract the shared electrons Electronegativity is a relative concept

meaning that an elementrsquos electronegativity can be measured only

in relation to the electronegativity of other elements Linus Pauling

devised a method for calculating relative electronegativities of most

elements

Molecule Fluorine F2 Hydrogen fluoride HF

δ+ δ



Lewis

structure

δ+ δminus

Polarity Non-polar molecule Polar molecule

∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no

different between the EN

Since F is more electronegative than H

therefore the bondin air electrons

Explanation

therefore bonding pair

electrons was not pulled to

either atom hence remain

in the middle between 2 F

atom

were pulled closer to F atom This will

caused F to have greater electron

density compare to H Therefore F has

partial negative charge (δminus) while H

carries partial positive charge (δ+)

Dipole

moment

magnitude

and

vector

Since there is no

difference between EN

the dipole moment is 0

and no resultant dipole

moment nor vector


There is presence of dipole moment in

HF and the vector of resultant dipole

moment is pointed to the direction of F

(symbolised by I )

a) Comparisons above are basically the difference between an

element with compound where diatomic molecules



containing atoms of different elements (for example HClCO and NO) have dipole moments and are called polar

molecules while diatomic molecules containing atoms of

the same element (for example H2 O2 and F2) are

examples of non-polar molecules because they do nothave dipole moments However not necessarily a covalent

bond compound is guaranteed a polar molecule

b) For a molecule made up of three or more atoms both the

polarity of the bonds and the molecular geometry determine

whether there is a dipole moment Even if polar bonds are

present the molecule will not necessarily have a dipole

moment For example comparison between sulphur dioxideand sulphur trioxide

Molecules Sulphur trioxide SO3 Sulphur dioxide SO2



Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35

the dipole moment of the entire molecule

is made u of three bond moments that isthe dipole moment of the entire molecule

Bond

moment

dipole

moment

magnitude

and vector

individual dipole moments in the polar

SminusO bonds The bond moment is a vector

quantity which means that it has both

magnitude and direction The measured

dipole moment is equal to the vector sum

of the bond moments The three bondmoments in SO3 are equal in magnitude

Because they point in opposite directions

in a planar SO3 molecule the sum of

resultant dipole moment would be zero

lone pair electron Even though two bond

moments in SO2 are equal in magnitude

however the presence of the lone pair

electrons which caused the repulsion of

bond pair electrons to be lesser Because

they point in downward directions in a bent

SO2 molecule the overall vector points

downward and the sum of resultant

dipole moment would not be zero

hence a polar molecule

From the example of SO2 and SO3 used we can tell that if a

polyatomic molecule is a symmetrical molecule (molecule



with no lone pair electrons in it) it may be a non-polarmolecule However if a polyatomic molecule is an

asymmetrical molecule (molecule with lone pair electrons in

it) it may be a polar molecules

c) Even though a polyatomic molecule may be symmetrical if

polar molecule as the bonding moments are different andcaused the magnitude of dipole moment of the molecule is not

equal to zero However if the surrounding atoms are the

same bonding moments are equal in magnitude and the

resultant vector cancel-off each other causing the dipolemoment is equal to zero hence form a non-polar molecule

For example

MoleculeMethane CH4 Chloromethane

( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation

As methane is a symmetrical

molecule and the surrounding atoms

are the same the vector of bond

moment cancel off each other hencecaused the dipole moment is equal to

zero

Since a foreign element Cl is in thesymmetrical molecule and Cl is more

EN than the rest of the atoms the

vector and magnitude is heading to

the direction of Cl caused a small

dipole moment present in molecule

hence polar

Covalent molecule



Diatomic molecule

Same element Different

Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule

Same surroundatoms Different surroundatoms

Non-polarmolecule

Polar molecule

326 Electronegativity and Type of Chemical Bond



1 The type of bond that would form can be told by using the

difference of electronegativity (∆EN) Larger the difference the

more tendency of electron from low electronegativity atom to

move to the atom with higher electronegativity and form ioniccompound

a) The relationship between the ionic character and the difference

in the electronegativity of the bonded atom is shown in the

diagram and graph below



b) From the graph above the dotted line represent the arbitrary line

between ionic and covalent characteristic of a molecule To be

more specific there more likely an ionic compound may have high




covalent characteristic (exemplified by LiI) or conversely covalent

compound having high ionic characteristic (exemplified by HF)

c) The covalent characteristic of a molecule is dependent on the ability

of a cation to polarise an anion Polarisation indicates the ability ofa cation to attract the electron density of an anion when put next to

the cation involved When a cation is able to pull the electron

electron with cation hence increase the covalency of the molecule

The covalency properties of a molecule is dependent on the cation

and anion where they can be explained qualitatively via

bull Polarisation power of cation bull Polarisability of anion

A+ 983128 991251

983106983083 983129983085

3611 Polarisation Power of Cation

Polarisation Power of Cation ndash measure the ability of a cation to

polarise the electron cloud of the anion




2 factors determining the polarisation power of cation

Charge of cation Size of cation

rArr Greater the charge of ion higher the

effective nuclear charge of cationhence it will be able to attract the

neighboring electron density of anion

rArr Smaller the size of cation closer the

neighboring anion to the nucleus ofcation hence easier for the cation to

polarise the anion and result an

This will caused the polarization power

of cation increase hence increase thecovalent characteristic of cation

increment in the polarization power of

cation and increase the covalentcharacteristic of cation

diams Both factors can be explained in another term called as charge density where

Charge Density = Charge Ionic Radius

diams From the equation above Charge Density will have a greater value provided that

cation has a high charge and small cationic radius

diams Greater the charge density higher the polarization power greater the covalent

characteristic of the cation

3612 Polarisability of Anion

bull Polarisability of an anion ~ ability of the anion to allow the electron

density to be polarised by cation




bull 2 factors determining the polarisability of an anion

Charge of anion Size of anion

rArrGreater the charge of anion lower theeffective nuclear charge of anion This will

weakened the electrostatic attraction forces

rArrLarger the size of anion further theoutermost electron from the nucleus

of the anion easier for the cation to

bull Unlike cation anion does not have a term that combined bothfactors of charge and ionic radius However information of

polarisability of anion enable the prediction of the covalent

characteristic of a molecule since in order to form a covalent bond

it depend on both polarisation power of cation and polarisability ofthe anion

e ween nuc eus an e ou ermos

electron in anion and increase the polarisability of the anion hence increase

the covalent characteristic of anion

po ar se e an on an cause e

polarisability to increase henceincrease the covalent characteristic

of anion

362 Prediction of Chemical Bond Fajansrsquo Rule

bull In 1923 Kazimierz Fajans formulated an easy guidance to predict

whether a chemical bond will be covalent or ionic and depend on




the charge on the cation and the relative sizes of the cation and

anion They can be summarized in the following table

Ionic compound Low positive charge Large cation Small anion

Covalent compound High positive charge Small cation Large anion

bull Based on these guidance the bonding of a few compoundsshall be discussed to understand the application of Fajansrsquo

Rule in the chemical bonding

Lithium halide (LiX)

bull Lithium ion Li+ (1s2) has a small size due to only 1 shell



present in its ion But since it has a low charge so its chargedensity is not too high That is why all lithium halide are ionic

compound The covalency of lithium halide varies from a

highly ioniccharacteristic to highly covalency depending on

the polarisability of the anion next to Li+

bull When a group of halide F ndash Cl ndash Br ndash I ndash is put close to Li+ the

Group 17 halide LiF is highly ionic since the fluoride ion hassmall ionic size and low charge hence has low polarisability

Ionic size increase with the increasing shell when going down

to Group 17 halide hence increase the polarisability which

allowed lithium ion to polarise the anionrsquos electron densityhence increase the covalency

983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148

Aluminium halide (AlX3) and aluminium oxide (Al2O3)

bull Aluminium ion (Al3+) has high charge density due to its high



charge unit and its small ionic radius So depending on the anionaluminium has a high tendency to form covalent compound For

example when going down to Group 17 halide aluminium fluoride

(AlF3) forms ionic compound (since F- has a low polarisability)

while aluminium trichloride (AlCl3) aluminium tribromide (AlBr 3)and aluminium iodide (AlI3) form covalent compound (since

chloride bromide and iodide have high polarisability) This

(10400

C) while aluminium trichloride and tribromide are 1920

C and780C respectively

bull As for aluminium oxide (Al2O3) it is an ionic compound with high

covalent characteristic as aluminium ion has high covalent

characteristic due to its high charge density This explained thehigh melting point of Al2O3 (20500C) yet it is insoluble in water It

also explained the amphoteric properties of aluminium oxide where

aluminium oxide can act as an acid (covalent characteristic) as

well as a base (ionic characteristic)

33 Metallic Bonding

1 Metallic bonding occurs when large

numbers of metal atoms interact



Unlike the reaction of metal with

non metal which involve electrons

transfer when two metal atoms

interact they can also share their

valence electrons in a covalent

bond and form gaseous diatomic

-

model of metallic bonding

proposes that all the metal atoms

in the sample contribute their

valence electrons to form an

electron ldquoseardquo that is

delocalized throughout the

piece The metal ions are

submerged within this electron

sea in an orderly array

a) The model we will use to study metallic bonding is band theory

because it states that delocalized electrons move freely through

ldquobandsrdquo formed by overlapping molecular orbitals



b) Consider magnesium for example The electron configuration of

Mg is 1s22s22p63s2 so each atom has two valence electrons in the

3s orbital In a metallic crystal the atoms are packed closelytogether so the energy levels of each magnesium atom are

affected by the immediate neighbors of the atom as a result of

the energy scale that they are more appropriately described as aldquobandrdquo The closely spaced filled energy levels make up the

valence band The upper half of the energy levels corresponds

to the empty delocalized molecular orbitals formed by the overlap

of the 3p orbitals This set of closely spaced empty levels is calledthe conduction band As a conductor the conduction band and

valence band are overlapped hence electrons can travel freely

among the two bands hence conduct electricity



c) Theoretically greater the number of valence electrons in a

metal greater the number of electrons delocalised higher

the conductivity However the conductivity decrease withtemperature as vibration of the lattice of ion impedes the free

movement of electron in conduction band

2 Semiconductors are element that normally are not conductors but

will conduct electricity at elevated temperatures or when combined

with a small amount of certain other elements These elements are



usually metalloid such as silicon and germanium(a) The energy gap between the conduction band and valence band

of these solids is much smaller than that for insulator If the energy

needed to excite electrons from the valence band into the conduction

band is provided the solid becomes a conductor Note that thisbehavior is opposite that of the metals

the conductivity of the semiconductors Doping can be done by

adding one of the following

i dopant atoms containing fewer valence electrons Hence the

semiconductor formed is positive p - type semiconductor

ii dopant atoms with extra valence electrons Hence the semiconductorformed is negative n - type semiconductor



3 Insulators are substances that do not conduct electricity

no matter how high the temperature is applied to the



substances involved The energy gaps between theconduction band and valence band of these element is very

large hence regardless how much energies were applied to

these insulator it will not be able to conduct electricity nor

heat Glass and woods are good examples of insulator Inwood and glass the gap between the valence band and the

Consequently much more energy is needed to excite anelectron into the conduction band Lacking this energy

electrons cannot move freely Therefore glass and wood

are insulators ineffective conductors of electricity



34 Intermolecular forces Van der Waals forces and hydrogen

bonding

1 The nature of the state of matter of substances and their changes



are due primarily to forces among the molecules Both bonding

(intramolecular) forces and intermolecular forces arise from

electrostatic attractions between opposite charges Bonding forces

are due to the attraction between cations and anions (ionic

bonding) nuclei and electron pairs (covalent bonding) or metal

cations and delocalized valence electrons (metallic bonding)

between molecules as a result of partial charges or the attraction

between ions and molecules The two types of forces differ in

magnitude and forces explains why

a) Bonding forces are relatively strong because they involve

larger charges that are closer togetherb) Intermolecular forces are relatively weak because they typically

involve smaller charges that are farther apart





3 Induced dipole Forces (Dispersion forces) - Consider how a helium

atom (monoatomic gas which have dipole moment = 0) interact

with the following species



Helium with cationHelium with polar

moleculeHelium with Helium

He

He

a) From the diagram we can tell that if an ion or a polar

molecule is placed near an atom or a non-polar molecule the

electron distribution of the atom (or molecule) is distorted by

th f t d b th i th l l l lti i



the force exerted by the ion or the polar molecule resulting in

a kind of dipole The dipole in the atom (or non-polar

molecule) is said to be an induced dipole because the

separation of positive and negative charges in the atom(or non-polar molecule) is due to the proximity of an ion or a

polar molecule

b) However the weak attractive interaction between a non-polar

molecule to another non-polar molecule are unlike when

placed near an ion or polar molecule Between two non-polar

atom they form among themselves a short induced dipole

hence attract each other temporary Therefore the forcesformed between them are very weak and can be broken

easily Such interaction is also known as London forces

Alkane CH4 C2H6 C3H8 C4H10 C5H12 C6H14 C7H16 C8H18




RMM 16 30 44 58 72 86 100 114

Boiling ndash ndash ndash ndash

point oC

TrendRelative molecular mass increased

weak Van Der Waals forces increased

d) For molecules especially organic compounds which have the

same molecular mass and functioning group they may have

different boiling point depend on the molecular structure For

example pentane C5H12 with molecular mass 72 has 3 isomers



example pentane C5H12 with molecular mass 72 has 3 isomersas shown in table below

Molecule 22-dimethylpropane 2-methylbutane n-pentane

structure

Boiling point

oC54 218 363

Total surface area increased


Boiling point increased

4 Dipole-Dipole Forces ~ When polar molecules lie near one

another as in liquids and solids their partial charges act as

tiny electric fields that orient them and give rise to dipole-

dipole forces the positive pole of one molecule attracts the




negative pole of another These are the forces that give polar

compound a higher boiling point than the non-polar

compound

Molecule FormulaMolecular

mass

Dipole

moment

Boiling point

(K)

Propane CH CH CH 44 0 08 231



Propane CH3CH2CH3 44 008 231

Dimethyl

ether CH3OCH3 46 13 248

Methyl

chlorideCH3Cl 505 187 249

Ethanal CH3CHO 44 269 294

Acetonitrile CH3CN 41 392 355

Almost the same molecular massHowever greater the dipole moment

Stronger the dipole-dipole forces

Higher the boiling point

5 Hydrogen bond ~ a special type of dipole-dipole interaction

between the hydrogen atom in a polar bond as in NminusH OminusH

or FminusH with an electronegative O N or F atom Diagram

below shows a few example of interaction between molecules




using hydrogen bond

Hydrogen bond

StF is more electronegative than N

Even though F is more electronegative than O amp

Hydrogen bond between FndashH is stronger than OndashH

However water form more hydrogen bond



Strong

hydrogen

bond

F is more electronegative than NHydrogen bond between FndashH is stronger than NndashH


b) The factors of hydrogen bonding can also use to explain the

solubility of some organic compound in water like example

ethane cannot dissolve in water but ethanol can dissolve in

water due to the hydrogen bonding




c) Some organic compound form dimer using hydrogen bond

For example when glacial ethanoic acid is dissolved inorganic solvent it form a dimer using hydrogen bond via the

interaction between O-H and C=O in each of the molecule

d) In some case hydrogen bond can also be used to form which

is the intermolecular forces and intramolecular forces For

example in 2-nitrophenol and 4-nitrophenol the boiling point

of the 2 compounds can be explain below




Since 2-nitrophenol form strong hydrogen bond asintramolecular forces the interaction between 2-nitrophenol

molecules are weaker among each other compare to 4-

nitrophenol which used hydrogen bond as theirintermolecular forces With stronger hydrogen bond which act

as the intermolecular forces the boiling point of 4-nitrophenol

is expected to be higher than 2-nitrophenol



CHAPTER 3 CHEMICAL BONDING

31 Ionic bonding

32 Covalent bonding

33 Metallic bonding

34 Intermolecular forces Van der Waals forces andhydrogen bonding

Topic

2007 2008 2009 2010 2011 20122013

Sem 12014

Sem 1

P1 P2 P1 P2 P1 P2 P1 P2 P1 P2 P1 P2 AB

CA

B

C

3

Chemical

Bonding

43c

6a4 5a 3

3b

c1 2

2

5b2

1b

5c

7a

319

b3 19



INTERACTION

BETWEEN ELEMENTS

Metal and


non-metal

Metal and

metal







31 Ionic Bonding








Na (1s22s22p63s1) F (1s22s22p5) Na+ F-

(1s22s22p6) (1s22s22p6)


Lewis

diagram

1s 2s 2p 3s



















( ) minus+

minus+

minus

times

prop r r

QQ


nn
















lattice energies is

















point








current

-


































F 1 s22s22p5 F 1 s22s22p5 1s2 2s2 2p6 1s2 2s2 2p6







O 1s2

2s2

2p4

O 1s2

2s2

2p4

1s2

2s2

2p6

1s2

2s2

2p6




oxygen atom






N 1 s22s22p3 N 1 s22s22p3 1s2 2s2 2p6 1s2 2s2 2p6



Water H2O

H 1 s1 O 2 s22p4

Carbon dioxide CO2

C 2 s22p2 O 2 s22p4

Ammonia NH3


H 1 s1 C 2 s22p2




H 1 s1 N 2 s22p3 C 2 s22p2


H 1 s1 O 2 s22p4 C 2 s22p2


Ethyne C2H2

H 1 s1 C 2 s22p2






carbon atoms









Carbonate ion CO32-

C 2 s22p2 O 2 s22p4

Cyanide ion CN-

N 2 s22p3 C 2 s22p2

u p a e on 4-

S 3 s23p4 O 2 s22p4

ra e on 3-

N 2 s22p3 O 2 s22p4




















S 3 s23p4 F 2 s22p5



Xe 5s25p6 F 2 s22p5


















N 2 s22p3 O 2 s22p4


N 2 s22p3 O 2 s22p4








322 Dative bond








below








S

OO

O









2

Cl6








(coordinative) bond


2+


3-





Valence bond theory









opposite spins
















































a) Ground state


electron of 2s22p2



equa energy eve

c) Hybridised state





tetrahedral shape


























a) Ground state



electron of 2s2

b) Excited state


energy level

c) Hybridised state





linear shape






p y p








z











a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state



















y z




a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state









sp

pz py





molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond



























INTERACTION

BETWEEN ELEMENTS

Metal and


non-metal

Metal and

metal







31 Ionic Bonding








Na (1s22s22p63s1) F (1s22s22p5) Na+ F-

(1s22s22p6) (1s22s22p6)


Lewis

diagram

1s 2s 2p 3s



















( ) minus+

minus+

minus

times

prop r r

QQ


nn
















lattice energies is

















point








current

-


































F 1 s22s22p5 F 1 s22s22p5 1s2 2s2 2p6 1s2 2s2 2p6







O 1s2

2s2

2p4

O 1s2

2s2

2p4

1s2

2s2

2p6

1s2

2s2

2p6




oxygen atom






N 1 s22s22p3 N 1 s22s22p3 1s2 2s2 2p6 1s2 2s2 2p6



Water H2O

H 1 s1 O 2 s22p4

Carbon dioxide CO2

C 2 s22p2 O 2 s22p4

Ammonia NH3


H 1 s1 C 2 s22p2




H 1 s1 N 2 s22p3 C 2 s22p2


H 1 s1 O 2 s22p4 C 2 s22p2


Ethyne C2H2

H 1 s1 C 2 s22p2






carbon atoms









Carbonate ion CO32-

C 2 s22p2 O 2 s22p4

Cyanide ion CN-

N 2 s22p3 C 2 s22p2

u p a e on 4-

S 3 s23p4 O 2 s22p4

ra e on 3-

N 2 s22p3 O 2 s22p4




















S 3 s23p4 F 2 s22p5



Xe 5s25p6 F 2 s22p5


















N 2 s22p3 O 2 s22p4


N 2 s22p3 O 2 s22p4








322 Dative bond








below








S

OO

O









2

Cl6








(coordinative) bond


2+


3-





Valence bond theory









opposite spins
















































a) Ground state


electron of 2s22p2



equa energy eve

c) Hybridised state





tetrahedral shape


























a) Ground state



electron of 2s2

b) Excited state


energy level

c) Hybridised state





linear shape






p y p








z











a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state



















y z




a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state









sp

pz py





molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond































31 Ionic Bonding








Na (1s22s22p63s1) F (1s22s22p5) Na+ F-

(1s22s22p6) (1s22s22p6)


Lewis

diagram

1s 2s 2p 3s



















( ) minus+

minus+

minus

times

prop r r

QQ


nn
















lattice energies is

















point








current

-


































F 1 s22s22p5 F 1 s22s22p5 1s2 2s2 2p6 1s2 2s2 2p6







O 1s2

2s2

2p4

O 1s2

2s2

2p4

1s2

2s2

2p6

1s2

2s2

2p6




oxygen atom






N 1 s22s22p3 N 1 s22s22p3 1s2 2s2 2p6 1s2 2s2 2p6



Water H2O

H 1 s1 O 2 s22p4

Carbon dioxide CO2

C 2 s22p2 O 2 s22p4

Ammonia NH3


H 1 s1 C 2 s22p2




H 1 s1 N 2 s22p3 C 2 s22p2


H 1 s1 O 2 s22p4 C 2 s22p2


Ethyne C2H2

H 1 s1 C 2 s22p2






carbon atoms









Carbonate ion CO32-

C 2 s22p2 O 2 s22p4

Cyanide ion CN-

N 2 s22p3 C 2 s22p2

u p a e on 4-

S 3 s23p4 O 2 s22p4

ra e on 3-

N 2 s22p3 O 2 s22p4




















S 3 s23p4 F 2 s22p5



Xe 5s25p6 F 2 s22p5


















N 2 s22p3 O 2 s22p4


N 2 s22p3 O 2 s22p4








322 Dative bond








below








S

OO

O









2

Cl6








(coordinative) bond


2+


3-





Valence bond theory









opposite spins
















































a) Ground state


electron of 2s22p2



equa energy eve

c) Hybridised state





tetrahedral shape


























a) Ground state



electron of 2s2

b) Excited state


energy level

c) Hybridised state





linear shape






p y p








z











a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state



















y z




a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state









sp

pz py





molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond











































( ) minus+

minus+

minus

times

prop r r

QQ


nn
















lattice energies is

















point








current

-


































F 1 s22s22p5 F 1 s22s22p5 1s2 2s2 2p6 1s2 2s2 2p6







O 1s2

2s2

2p4

O 1s2

2s2

2p4

1s2

2s2

2p6

1s2

2s2

2p6




oxygen atom






N 1 s22s22p3 N 1 s22s22p3 1s2 2s2 2p6 1s2 2s2 2p6



Water H2O

H 1 s1 O 2 s22p4

Carbon dioxide CO2

C 2 s22p2 O 2 s22p4

Ammonia NH3


H 1 s1 C 2 s22p2




H 1 s1 N 2 s22p3 C 2 s22p2


H 1 s1 O 2 s22p4 C 2 s22p2


Ethyne C2H2

H 1 s1 C 2 s22p2






carbon atoms









Carbonate ion CO32-

C 2 s22p2 O 2 s22p4

Cyanide ion CN-

N 2 s22p3 C 2 s22p2

u p a e on 4-

S 3 s23p4 O 2 s22p4

ra e on 3-

N 2 s22p3 O 2 s22p4




















S 3 s23p4 F 2 s22p5



Xe 5s25p6 F 2 s22p5


















N 2 s22p3 O 2 s22p4


N 2 s22p3 O 2 s22p4








322 Dative bond








below








S

OO

O









2

Cl6








(coordinative) bond


2+


3-





Valence bond theory









opposite spins
















































a) Ground state


electron of 2s22p2



equa energy eve

c) Hybridised state





tetrahedral shape


























a) Ground state



electron of 2s2

b) Excited state


energy level

c) Hybridised state





linear shape






p y p








z











a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state



















y z




a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state









sp

pz py





molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond

































lattice energies is

















point








current

-


































F 1 s22s22p5 F 1 s22s22p5 1s2 2s2 2p6 1s2 2s2 2p6







O 1s2

2s2

2p4

O 1s2

2s2

2p4

1s2

2s2

2p6

1s2

2s2

2p6




oxygen atom






N 1 s22s22p3 N 1 s22s22p3 1s2 2s2 2p6 1s2 2s2 2p6



Water H2O

H 1 s1 O 2 s22p4

Carbon dioxide CO2

C 2 s22p2 O 2 s22p4

Ammonia NH3


H 1 s1 C 2 s22p2




H 1 s1 N 2 s22p3 C 2 s22p2


H 1 s1 O 2 s22p4 C 2 s22p2


Ethyne C2H2

H 1 s1 C 2 s22p2






carbon atoms









Carbonate ion CO32-

C 2 s22p2 O 2 s22p4

Cyanide ion CN-

N 2 s22p3 C 2 s22p2

u p a e on 4-

S 3 s23p4 O 2 s22p4

ra e on 3-

N 2 s22p3 O 2 s22p4




















S 3 s23p4 F 2 s22p5



Xe 5s25p6 F 2 s22p5


















N 2 s22p3 O 2 s22p4


N 2 s22p3 O 2 s22p4








322 Dative bond








below








S

OO

O









2

Cl6








(coordinative) bond


2+


3-





Valence bond theory









opposite spins
















































a) Ground state


electron of 2s22p2



equa energy eve

c) Hybridised state





tetrahedral shape


























a) Ground state



electron of 2s2

b) Excited state


energy level

c) Hybridised state





linear shape






p y p








z











a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state



















y z




a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state









sp

pz py





molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond




































point








current

-


































F 1 s22s22p5 F 1 s22s22p5 1s2 2s2 2p6 1s2 2s2 2p6







O 1s2

2s2

2p4

O 1s2

2s2

2p4

1s2

2s2

2p6

1s2

2s2

2p6




oxygen atom






N 1 s22s22p3 N 1 s22s22p3 1s2 2s2 2p6 1s2 2s2 2p6



Water H2O

H 1 s1 O 2 s22p4

Carbon dioxide CO2

C 2 s22p2 O 2 s22p4

Ammonia NH3


H 1 s1 C 2 s22p2




H 1 s1 N 2 s22p3 C 2 s22p2


H 1 s1 O 2 s22p4 C 2 s22p2


Ethyne C2H2

H 1 s1 C 2 s22p2






carbon atoms









Carbonate ion CO32-

C 2 s22p2 O 2 s22p4

Cyanide ion CN-

N 2 s22p3 C 2 s22p2

u p a e on 4-

S 3 s23p4 O 2 s22p4

ra e on 3-

N 2 s22p3 O 2 s22p4




















S 3 s23p4 F 2 s22p5



Xe 5s25p6 F 2 s22p5


















N 2 s22p3 O 2 s22p4


N 2 s22p3 O 2 s22p4








322 Dative bond








below








S

OO

O









2

Cl6








(coordinative) bond


2+


3-





Valence bond theory









opposite spins
















































a) Ground state


electron of 2s22p2



equa energy eve

c) Hybridised state





tetrahedral shape


























a) Ground state



electron of 2s2

b) Excited state


energy level

c) Hybridised state





linear shape






p y p








z











a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state



















y z




a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state









sp

pz py





molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond
































current

-


































F 1 s22s22p5 F 1 s22s22p5 1s2 2s2 2p6 1s2 2s2 2p6







O 1s2

2s2

2p4

O 1s2

2s2

2p4

1s2

2s2

2p6

1s2

2s2

2p6




oxygen atom






N 1 s22s22p3 N 1 s22s22p3 1s2 2s2 2p6 1s2 2s2 2p6



Water H2O

H 1 s1 O 2 s22p4

Carbon dioxide CO2

C 2 s22p2 O 2 s22p4

Ammonia NH3


H 1 s1 C 2 s22p2




H 1 s1 N 2 s22p3 C 2 s22p2


H 1 s1 O 2 s22p4 C 2 s22p2


Ethyne C2H2

H 1 s1 C 2 s22p2






carbon atoms









Carbonate ion CO32-

C 2 s22p2 O 2 s22p4

Cyanide ion CN-

N 2 s22p3 C 2 s22p2

u p a e on 4-

S 3 s23p4 O 2 s22p4

ra e on 3-

N 2 s22p3 O 2 s22p4




















S 3 s23p4 F 2 s22p5



Xe 5s25p6 F 2 s22p5


















N 2 s22p3 O 2 s22p4


N 2 s22p3 O 2 s22p4








322 Dative bond








below








S

OO

O









2

Cl6








(coordinative) bond


2+


3-





Valence bond theory









opposite spins
















































a) Ground state


electron of 2s22p2



equa energy eve

c) Hybridised state





tetrahedral shape


























a) Ground state



electron of 2s2

b) Excited state


energy level

c) Hybridised state





linear shape






p y p








z











a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state



















y z




a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state









sp

pz py





molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond



















































F 1 s22s22p5 F 1 s22s22p5 1s2 2s2 2p6 1s2 2s2 2p6







O 1s2

2s2

2p4

O 1s2

2s2

2p4

1s2

2s2

2p6

1s2

2s2

2p6




oxygen atom






N 1 s22s22p3 N 1 s22s22p3 1s2 2s2 2p6 1s2 2s2 2p6



Water H2O

H 1 s1 O 2 s22p4

Carbon dioxide CO2

C 2 s22p2 O 2 s22p4

Ammonia NH3


H 1 s1 C 2 s22p2




H 1 s1 N 2 s22p3 C 2 s22p2


H 1 s1 O 2 s22p4 C 2 s22p2


Ethyne C2H2

H 1 s1 C 2 s22p2






carbon atoms









Carbonate ion CO32-

C 2 s22p2 O 2 s22p4

Cyanide ion CN-

N 2 s22p3 C 2 s22p2

u p a e on 4-

S 3 s23p4 O 2 s22p4

ra e on 3-

N 2 s22p3 O 2 s22p4




















S 3 s23p4 F 2 s22p5



Xe 5s25p6 F 2 s22p5


















N 2 s22p3 O 2 s22p4


N 2 s22p3 O 2 s22p4








322 Dative bond








below








S

OO

O









2

Cl6








(coordinative) bond


2+


3-





Valence bond theory









opposite spins
















































a) Ground state


electron of 2s22p2



equa energy eve

c) Hybridised state





tetrahedral shape


























a) Ground state



electron of 2s2

b) Excited state


energy level

c) Hybridised state





linear shape






p y p








z











a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state



















y z




a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state









sp

pz py





molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond































O 1s2

2s2

2p4

O 1s2

2s2

2p4

1s2

2s2

2p6

1s2

2s2

2p6




oxygen atom






N 1 s22s22p3 N 1 s22s22p3 1s2 2s2 2p6 1s2 2s2 2p6



Water H2O

H 1 s1 O 2 s22p4

Carbon dioxide CO2

C 2 s22p2 O 2 s22p4

Ammonia NH3


H 1 s1 C 2 s22p2




H 1 s1 N 2 s22p3 C 2 s22p2


H 1 s1 O 2 s22p4 C 2 s22p2


Ethyne C2H2

H 1 s1 C 2 s22p2






carbon atoms









Carbonate ion CO32-

C 2 s22p2 O 2 s22p4

Cyanide ion CN-

N 2 s22p3 C 2 s22p2

u p a e on 4-

S 3 s23p4 O 2 s22p4

ra e on 3-

N 2 s22p3 O 2 s22p4




















S 3 s23p4 F 2 s22p5



Xe 5s25p6 F 2 s22p5


















N 2 s22p3 O 2 s22p4


N 2 s22p3 O 2 s22p4








322 Dative bond








below








S

OO

O









2

Cl6








(coordinative) bond


2+


3-





Valence bond theory









opposite spins
















































a) Ground state


electron of 2s22p2



equa energy eve

c) Hybridised state





tetrahedral shape


























a) Ground state



electron of 2s2

b) Excited state


energy level

c) Hybridised state





linear shape






p y p








z











a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state



















y z




a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state









sp

pz py





molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond






























N 1 s22s22p3 N 1 s22s22p3 1s2 2s2 2p6 1s2 2s2 2p6



Water H2O

H 1 s1 O 2 s22p4

Carbon dioxide CO2

C 2 s22p2 O 2 s22p4

Ammonia NH3


H 1 s1 C 2 s22p2




H 1 s1 N 2 s22p3 C 2 s22p2


H 1 s1 O 2 s22p4 C 2 s22p2


Ethyne C2H2

H 1 s1 C 2 s22p2






carbon atoms









Carbonate ion CO32-

C 2 s22p2 O 2 s22p4

Cyanide ion CN-

N 2 s22p3 C 2 s22p2

u p a e on 4-

S 3 s23p4 O 2 s22p4

ra e on 3-

N 2 s22p3 O 2 s22p4




















S 3 s23p4 F 2 s22p5



Xe 5s25p6 F 2 s22p5


















N 2 s22p3 O 2 s22p4


N 2 s22p3 O 2 s22p4








322 Dative bond








below








S

OO

O









2

Cl6








(coordinative) bond


2+


3-





Valence bond theory









opposite spins
















































a) Ground state


electron of 2s22p2



equa energy eve

c) Hybridised state





tetrahedral shape


























a) Ground state



electron of 2s2

b) Excited state


energy level

c) Hybridised state





linear shape






p y p








z











a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state



















y z




a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state









sp

pz py





molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond



























Water H2O

H 1 s1 O 2 s22p4

Carbon dioxide CO2

C 2 s22p2 O 2 s22p4

Ammonia NH3


H 1 s1 C 2 s22p2




H 1 s1 N 2 s22p3 C 2 s22p2


H 1 s1 O 2 s22p4 C 2 s22p2


Ethyne C2H2

H 1 s1 C 2 s22p2






carbon atoms









Carbonate ion CO32-

C 2 s22p2 O 2 s22p4

Cyanide ion CN-

N 2 s22p3 C 2 s22p2

u p a e on 4-

S 3 s23p4 O 2 s22p4

ra e on 3-

N 2 s22p3 O 2 s22p4




















S 3 s23p4 F 2 s22p5



Xe 5s25p6 F 2 s22p5


















N 2 s22p3 O 2 s22p4


N 2 s22p3 O 2 s22p4








322 Dative bond








below








S

OO

O









2

Cl6








(coordinative) bond


2+


3-





Valence bond theory









opposite spins
















































a) Ground state


electron of 2s22p2



equa energy eve

c) Hybridised state





tetrahedral shape


























a) Ground state



electron of 2s2

b) Excited state


energy level

c) Hybridised state





linear shape






p y p








z











a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state



















y z




a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state









sp

pz py





molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond




























H 1 s1 N 2 s22p3 C 2 s22p2


H 1 s1 O 2 s22p4 C 2 s22p2


Ethyne C2H2

H 1 s1 C 2 s22p2






carbon atoms









Carbonate ion CO32-

C 2 s22p2 O 2 s22p4

Cyanide ion CN-

N 2 s22p3 C 2 s22p2

u p a e on 4-

S 3 s23p4 O 2 s22p4

ra e on 3-

N 2 s22p3 O 2 s22p4




















S 3 s23p4 F 2 s22p5



Xe 5s25p6 F 2 s22p5


















N 2 s22p3 O 2 s22p4


N 2 s22p3 O 2 s22p4








322 Dative bond








below








S

OO

O









2

Cl6








(coordinative) bond


2+


3-





Valence bond theory









opposite spins
















































a) Ground state


electron of 2s22p2



equa energy eve

c) Hybridised state





tetrahedral shape


























a) Ground state



electron of 2s2

b) Excited state


energy level

c) Hybridised state





linear shape






p y p








z











a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state



















y z




a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state









sp

pz py





molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond






























carbon atoms









Carbonate ion CO32-

C 2 s22p2 O 2 s22p4

Cyanide ion CN-

N 2 s22p3 C 2 s22p2

u p a e on 4-

S 3 s23p4 O 2 s22p4

ra e on 3-

N 2 s22p3 O 2 s22p4




















S 3 s23p4 F 2 s22p5



Xe 5s25p6 F 2 s22p5


















N 2 s22p3 O 2 s22p4


N 2 s22p3 O 2 s22p4








322 Dative bond








below








S

OO

O









2

Cl6








(coordinative) bond


2+


3-





Valence bond theory









opposite spins
















































a) Ground state


electron of 2s22p2



equa energy eve

c) Hybridised state





tetrahedral shape


























a) Ground state



electron of 2s2

b) Excited state


energy level

c) Hybridised state





linear shape






p y p








z











a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state



















y z




a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state









sp

pz py





molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond



























Carbonate ion CO32-

C 2 s22p2 O 2 s22p4

Cyanide ion CN-

N 2 s22p3 C 2 s22p2

u p a e on 4-

S 3 s23p4 O 2 s22p4

ra e on 3-

N 2 s22p3 O 2 s22p4




















S 3 s23p4 F 2 s22p5



Xe 5s25p6 F 2 s22p5


















N 2 s22p3 O 2 s22p4


N 2 s22p3 O 2 s22p4








322 Dative bond








below








S

OO

O









2

Cl6








(coordinative) bond


2+


3-





Valence bond theory









opposite spins
















































a) Ground state


electron of 2s22p2



equa energy eve

c) Hybridised state





tetrahedral shape


























a) Ground state



electron of 2s2

b) Excited state


energy level

c) Hybridised state





linear shape






p y p








z











a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state



















y z




a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state









sp

pz py





molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond












































S 3 s23p4 F 2 s22p5



Xe 5s25p6 F 2 s22p5


















N 2 s22p3 O 2 s22p4


N 2 s22p3 O 2 s22p4








322 Dative bond








below








S

OO

O









2

Cl6








(coordinative) bond


2+


3-





Valence bond theory









opposite spins
















































a) Ground state


electron of 2s22p2



equa energy eve

c) Hybridised state





tetrahedral shape


























a) Ground state



electron of 2s2

b) Excited state


energy level

c) Hybridised state





linear shape






p y p








z











a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state



















y z




a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state









sp

pz py





molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond










































N 2 s22p3 O 2 s22p4


N 2 s22p3 O 2 s22p4








322 Dative bond








below








S

OO

O









2

Cl6








(coordinative) bond


2+


3-





Valence bond theory









opposite spins
















































a) Ground state


electron of 2s22p2



equa energy eve

c) Hybridised state





tetrahedral shape


























a) Ground state



electron of 2s2

b) Excited state


energy level

c) Hybridised state





linear shape






p y p








z











a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state



















y z




a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state









sp

pz py





molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond



























322 Dative bond








below








S

OO

O









2

Cl6








(coordinative) bond


2+


3-





Valence bond theory









opposite spins
















































a) Ground state


electron of 2s22p2



equa energy eve

c) Hybridised state





tetrahedral shape


























a) Ground state



electron of 2s2

b) Excited state


energy level

c) Hybridised state





linear shape






p y p








z











a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state



















y z




a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state









sp

pz py





molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond





























S

OO

O









2

Cl6








(coordinative) bond


2+


3-





Valence bond theory









opposite spins
















































a) Ground state


electron of 2s22p2



equa energy eve

c) Hybridised state





tetrahedral shape


























a) Ground state



electron of 2s2

b) Excited state


energy level

c) Hybridised state





linear shape






p y p








z











a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state



















y z




a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state









sp

pz py





molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond
































2

Cl6








(coordinative) bond


2+


3-





Valence bond theory









opposite spins
















































a) Ground state


electron of 2s22p2



equa energy eve

c) Hybridised state





tetrahedral shape


























a) Ground state



electron of 2s2

b) Excited state


energy level

c) Hybridised state





linear shape






p y p








z











a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state



















y z




a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state









sp

pz py





molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond































(coordinative) bond


2+


3-





Valence bond theory









opposite spins
















































a) Ground state


electron of 2s22p2



equa energy eve

c) Hybridised state





tetrahedral shape


























a) Ground state



electron of 2s2

b) Excited state


energy level

c) Hybridised state





linear shape






p y p








z











a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state



















y z




a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state









sp

pz py





molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond




























Valence bond theory









opposite spins
















































a) Ground state


electron of 2s22p2



equa energy eve

c) Hybridised state





tetrahedral shape


























a) Ground state



electron of 2s2

b) Excited state


energy level

c) Hybridised state





linear shape






p y p








z











a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state



















y z




a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state









sp

pz py





molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond








































































a) Ground state


electron of 2s22p2



equa energy eve

c) Hybridised state





tetrahedral shape


























a) Ground state



electron of 2s2

b) Excited state


energy level

c) Hybridised state





linear shape






p y p








z











a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state



















y z




a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state









sp

pz py





molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond
















































a) Ground state



electron of 2s2

b) Excited state


energy level

c) Hybridised state





linear shape






p y p








z











a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state



















y z




a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state









sp

pz py





molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond



























p y p








z











a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state



















y z




a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state









sp

pz py





molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond



























a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state



















y z




a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state









sp

pz py





molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond




































y z




a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state









sp

pz py





molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond



























a) Ground state



of 2s22p2

b) Excited state



c) Hybridised state









sp

pz py





molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond




























molecule









N N





diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond



























diagram

a) Ground state


has the valence

electron of 2s22p3




level

c) Hybridised state


one orbitals of 2p



to the shape and







) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond



























) y g y




carbon atom C


nitrogen atom N

a) Ground state


N 2s22p3

b) Excited state


2s is promote to 2p


level

c) Hybridised state




hybridisation



N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond



























N

σσσσ

π

σσσσ

π











a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond


































a) Ground state

Valence electron

N 2s22p3

O 2s2

2p4

b) Excited state

One of the electron


2p orbitals

c) Hybridised state







the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond






























the shape and




tetrahedral


tetrahedral



Angle 1070

Shape bent

Angle 10450


lone pair electrons





























O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond
















































O) the











f









compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond

































compare to H-P-H











ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond

































ldquo rdquo




not shown







No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond



























No of

surroun

No of




Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond



























Classsurroun

d

atoms

pair

electron

Molecular

geometry


shape

Example of

molecules

AB2 2 0 Linear CO2

BeCl2

AB3 3 0Trigonal

lanar

AlCl3BF3

NO3-

AB2E 2 1Shape

Bent

SO2

O3

NO2-

AB4 4 0Shape

Tetrahedral

CH4

SiCl4SO

4

2-

Shape NH3



AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond



























AB3E 3 1 Trigonal

pyramidal

PCl3

SO32-


H2O

SCl2H2O2

AB5 5 0

Shape

Trigonal

bipyramidal

PCl5SbF5

AB4E 4 1Shape

See-Saw

SCl4PF4

-



AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond



























AB3E2 3 2

Arrangeme

nt

Trigonal

bipyramidalShape

T-shape

ICl3

BrF3

AB2E3 2 3

Arrangement

Trigonal

bipyramidal

Shape

linear

I3-

BrCl2-

A



AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond



























AB6 6 0

Arrangeme

ntamp Shape

Octahedral

SF6

AB5E 5 1

Arrangement

Octahedral

Sha e

SbCl52-

IF

Square

pyramidal

AB4E2 4 2

Arrangeme

nt

OctahedralShape

Square

planar

XeF4

BrF4-


P = 5 e- 3 Cl = 3 x 7e-



P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond



























P = 5 e- 3 Cl = 3 x 7e












C = 4 e 3 O = 3 x 6e- + 2e accept






- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond






























- -






Arrangement and




I = 7 e- 4 Cl = 4 x 7e- + 1e- accept






- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond






























- = -



the center atom P





[SbCl ]2



[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond



























[SbCl5]2-




3 4


Shape T-shape


Shape see-saw










partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond
































partially covalent










elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond































elements


δ+ δ



Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond



























Lewis

structure

δ+ δminus


∆EN 40 - 40 = 0 40 - 21 = 19

Since there are no




Explanation





atom






Dipole

moment

magnitude

and

vector

Since there is no




moment nor vector





(symbolised by I )


















Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond








































Lewis

structure

and shape


EN S = 25 O = 35 S = 25 O = 35



Bond

moment

dipole

moment

magnitude

and vector
































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond



































For example


( hl f ) CH Cl



(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond



























(chloroform) CH3Cl

Lewis

structure

-

Explanation





zero






hence polar

Covalent molecule



Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond



























Diatomic molecule


Polyatomic molecule

As mmetric element

Non-polarmolecule

Polar molecule

al Symmetrical

Polarmolecule


Non-polarmolecule

Polar molecule




























A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond



















































A+ 983128 991251

983106983083 983129983085














































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond



































































of anion


























983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond















































983107983148 991251



983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond



























983116983145983083

983110 991251

983106983154 991251

983107983148










(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond
































(10400









33 Metallic Bonding











-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond

































-





























































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond









































































bonding



























He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond


















































He

He











polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond































polar molecule











RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond




























RMM 16 30 44 58 72 86 100 114


point oC











structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond





























structure

Boiling point

oC54 218 363













compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond






























compound


mass

Dipole

moment

Boiling point

(K)





Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond




























Dimethyl


Methyl














using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond




























using hydrogen bond

Hydrogen bond







Strong

hydrogen

bond



























Strong

hydrogen

bond

























Chemistry Pre-u Chemistry Sem 1 Chap 3

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Transcript of Chemistry Pre-u Chemistry Sem 1 Chap 3