JJC 2008 H2 Chem - Chemical Bonding

JURONG JUNIOR COLLEGE

H2 CHEMISTRYCHEMICAL BONDING

JJC 2008Page 1 of 42

(2008 JCI)

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References:1. R'dmsden; "A Level Chemistry", 4th edition.2" Hill and Holman; "Chemistry in Context", 4th edition.

Gontent

. lonic (electrovalent) bondingo Gslebnt bonding and co-ordinate (dative covalent) bonding. The shapes of simple molecules. Bond energies, bond lengths and bond polarities. k*smolecular forces, including hydrogen bonding. He{aHic bonding. Bonding and physical ProPertiesr The solid state

Assessment Obiectives

Students should be able to:

describe ionic (electrovalent) bonding, as in sodium chloride and magnesium oxide, including the use of

'dot-and-cross' diagramsdescribe, including the use of 'dot-and-cross' diagrams'

z O covale-nt bonding, as in hydrogen; oxygen; nitrogen; chlorine; hydrogen chloride; carbon

dioxide; methane; ethene

,/(ii) co-ordinate (dative covalent) bonding, as in formation of the ammonium ion and in the Al.Clu

molecule.

lc) explain the shapes of, and bond angles in, molecules such as BF. (trigonal planar); CO, (linear); CHo

(tetrahedral); NH. (trigonal pyramidal); HrO (non-linear); SF. (octahedral) by using the Valence Shell

Electron Pair RePulsion theory

7 (d) describe covalent bonding in terms of orbital overlap' giving o and n bonds' (ei preOict the shapes of, and bond angles in, molecules analogous to those specified in (c)

7 if;' OescriOe hydrogen bonding, using immonia and water as examples of molecules containing -NH and -OH

9roups,19; 6xptiin the terms bond energy, bond length and bond polarity and use them to compare the reactivities of

covalent bonds,, (h) describe intermolecular forces (van der Waals' forces), based on permanent and induced dipoles, as in

CHC/3(I); Brr(l) and the liquid noble gases

,(i\ describe metallic bonding in terms of a lattice of positive ions surrounded by mobile electrons

' iJi JescriO", interpret and/oi predict the effect of different types of bonding (ionic bonding; covalent bonding;!. hydrogen bonding; other intermolecular interactions; metallic bonding) on the physical properties of

substances(k) deduce the type of bonding present from given information

,,iti show understanding of chimical reaction-s in terms of energy transfers associated with the breaking and' making of chemical bonds

(m) describe, in simple terms, the lattice structure of a crystalline solid which is:

0 ionic, as in sodium chloride, magnesium oxide(iD simple molecular, as in iodine(iiD giant molecular, as in graphite; diamond(iv) hydrogen-bonded, as in ice

iui metallic, as in copper [the concept of the 'unit cell' is not required]

(n) ouline the importance of hydrogen bonding'to the physical properties of substances, including ice and

water(o) suggest from quoted physical data the type of structure and bonding present in a substance

ioi reCoonise that materials are a finite resource and the_importance of recycling processes

iii O"r"iiU" spt hybridisation, as in ethane molecule, spt hybridisation, as in ethene and benzene molecules,

and sp hyliridiiation, as in ethyne molecule. [from 10.1 Organic Chemistry] , ,

(r) explain the shapes ol and bond angles in, molecules analogous to those specified in (e)

[from 10.1 Organic Chemistry]

(a)

(b)

1. INTRODUCTION

x As we examine the world around us, we find it to be composed mainly of elements, compounds and

mixtures. The manner in which the atoms are bound together has a profound effect on the chemical andphysical properties of these substances. For example, graphite is soft and slippery while diamond is thehardest substance known. Such differences in physical properties are best explained by the bondingwithin these substances.

x A e:frcmfesfS@d is the daceda6re"@ieg,- typically between a ffl*I}Jhere are 4 main types of chemical bond.

Positive charge Negative charge Type of chemical bond

Electrostatic

attractionbetween

!ELn6

and

*ils ffi**theno$lil{e rmbi ofthe 2 atoms

the bonding electrons

shared between them

nnffiilss naHdnns rlH&Sond

Electrostatic attraction between r* due to the setting ddir*g %es

Because chemical bond refers to the electrostatic attraction between point charges of opposite charge,the Wrld can be predicted by Coulomb's Law.

\\^c\rtGMb'o.Law: describes the mrunihsde"r#eedd@db*oreWxeem*** ct€r,ges

Xdow\ nA?J tui..ei,w;tik- f c.lnotrqeJ

t

x

x

Me*ry where Q* & q- are charges of positive & negative points;

d is distance between the two point charges

Note that Coulomb's Law is covered under Physics buf NOT in the Chemistry syllabus"

ln this topic, we shall study the different types of chemical bond and understand how they affect thephysical properties of materials ll

JJC 2008Page 2 of 42

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Figure 1.1 An Overview of Structure and Bonding

IONIC BONDINGI

II

I

Giant lonicStructure

COVALENT BONDING(INCLUDTNG DATIVE COVALENT BOND)

I

Giant M ol ec u Ia r/C ovalentStructure -"i r' 1

-I

ti''-"jl'-

Limited examples: graPhite,diamond, silicon, SiOz

METALLIC BONDINGI

I

I

Giant MetallicStructure

i ( :; t, :1"-,--- \

i"' I\r-:i\f1 i

i t'vi' d .,=1-- it-.--*... ,-r,,.., i

l'!r'i\'ri\---'

Sr'mple Mol ec u I a r/C ova I e ntStructure

F,-JE-; .,f"' . i

contains simplemolecules whichare held by

Hydrogen Bonding

I

I

between polarmolecules thal

contain H directlybonded toF,OorN

II

HF, NH3, H2O,

cH3cH20H,cH3cooH,

CHrCONHr...etc

van derWaals'forces

Pe rm anent Dipole-Pe rm an ent Di poleAttracti on (P D -P D aft racti on )

I

I

between polar molecules

II

HC/, CHC/3, HCHO

cH3cocH3,CH.OCH....etc

I n du ced Dipole-l nduced Di PoleAftraction (ID-ID aftraction)

I

I

between non-polar molecules/atoms

II

Hz, Clz, Brr, COr, CHa

CC/., He, Ne, Ar...etc

INTERMOLECULAR FORCES

Sorne Useful Concepts

. 1" Atomic radius/size

-: Across a period, atomic radius decreases gradually because nuclear charge increases as the number

of protons in the nucleus increases while the valence electrons occupy the same outermost shell.

Thus, electrons are drawn closer to the nucleus.

Down a group, atomic radius increases because the number of quantum shells that contain electrons

increases.

**This concept is useful in determining the strength of chemical bond.

2" Electronegativity - ':i t ' '

-''f\;r r" tr-' ''

Electronegativity refers to the ability of an atom to attract fdditional electronstto itself.

The higher the electronegativity value, the more electronegative the atom is.

Table 1.1 Electronegativity values for main group elements (based on Pau

H 2.1 He

Li 1.0 Be 'l.5 B 2.0 c 2.5 N 3.0 o 3.5 F 4.0 Ne-Na 0.9 Mq 1.2 A/ 1.5 si 1.8 P 2.1 s 2.5 c, 3.0 Ar

K 0.8 Ca 1.0 Ga 1.6 Ge 1.8 As 2.0 Se 2.4 Br 2.8 Kr

Rb 0.8 Sr 1.0 ln 1.7 Sn 1.8 sb 1.9 fe 2.1 | 2.5 Xe

Across a period, electronegativity increases.Down a group, electronegativity decreases.

JJC 2008 Page3ol42

-.:.> Therefore, metals have low electronegativity and they tend to form cations.Non-metals have high electronegativity and they tend to form anions.lmplications:Efements in a compound with large electronegativity difference tend to form ionic bonds.Elements in a compound with small electronegativity difference tend to form covalent bonds...r. I :

The concept of electronegativity and difference in electronegativity values is useful in predicting the typeof bonding and structure present in an elemenVcompound.

3. Metals and non-metals.

4. Noble Gas Configurations

All noble gases are. chemically stable or inert. They are extremely unreactive. Their stabilities arerelated to their electronic co_nfigurations. The number of electrons in the outer shell of a nbble gas is8 (ns2npo) except for He 1ts2;. inus, stability is gained when the atom is in the octet state. This is knownas the octet rule.

ln 1916, w' Kossel and G.N. Lewis explained the formation of chemical bond. Atoms can achieve thestable octet structure by

(a) transferring electrons from one atom to another(b) sharing electrons between atoms

i.e. forming ionic bondingi.e. forming covalent bonding

5' Heat is released (evolved) during bond formation (AH is negative). +.ri -r.\'i

Heat is absorbed during bond breaking (AH is positive).-; .,,. 1.,'^,,1r',,, ,

Exercise 1: with reference to Figure 1.1, predict the bonding and structure of the following substances.

H

Li Be HeB c N o F Ne

Na Mo Sc Ti Cr Mn Fe Co Ni Cu Zn AI Si P S cl ArK Ca Zr Nb Mo Tc Ru Rh Pd Aq ni Ga Ge As Se Br Kr

Rb Sr La Hf Ta W Re Os' lr Pt Au Hq In Sn sb Te I XeCs Ba Ac Rf Db So Bh Hs Mt TI Pb Bi Po At Rn

Substances Bonding Structure

Rubidium i 'l * ''t -l ,-,_ t,\...\ .+- J ; .- _\/, *,. )

Silicon,. \* ia\:

Bromine i:, \

Mgo il' n: r'r l ' r

Hzo r.i-). );-\.iir. 1, .,2 , _,r,\i.-r,' ,.,,'1r i

l:! : )'1.. ii]' r-v.;l;'i t, ,,, . '., .-'f

si02 rr .\ \,.. t,/\"f J ' r''- l

JJC 2008 Page 4 of 42

2.

2"1

METALLIC BONDING

Metallic Bondinq in Metals

A metal consists of a lattiee of metal cations surrounded by mobile electrons which are the valenceelectrons of the metal atoms (see Figure 2.'t).

Figure 2.1 Model of a metal: Lattice of positive ions surrounded by mobile electrons

o".o"o"o.oo"o"o"o"

o"o"o"o'o

JJC 2008 Page 5 of42

:-::-:-=.

Metal #ms bse theirvalence electrons toachieve the stable noblegas configuration.

a As shown in Figure 2.1, metallic bond exists extensively throughout the whole structure.

Thus metals are said to have giant metallic structure.

2.2 Strenqth:of Metallic B

Two main factors affecting the strength of metallic bond:

1. Number of valence electrons

As the number of valence electrons increases, the charg.e of the positive metal ions 1\Ailr'tr I \)'''

and the number of mobite electrons ' 'n i-ri'-i-*ir'!:'r . Thus, the metallic bond becoms5 :''-i"'! i "'i'\i- -, .

e.g. strength of metallic bond of Na < Mg < A/

't J]:,: ti,;i.j' =;Ji|2. Charge density of the metal ion

Charge density of an ion =charse on the ion

size of ion

For metal ions with the same charge, 16." i'rttl'\ rJ'

the size of the ions, the .t"r"l; - I

,n"charge density of the ions, ths ifl^Jl,/r l:/.,will be the electrostatic attraction for the mobile electrons.

e.g" strength of metallic bond of Na > K > Rb

Therefore, metallic bond strength increases with increasing number of valence electrons and withincreasing charge

- density of the metal ion (decreasing size and increasing charge).

Hence, metallic bond strength increases across the period and decreases down the group.

Recafl Coulomb's Law: F qql*q_^-d"

T

2.3

1.

Phvsical Properties of Metals

Volatility

Metals generally have melting and boilingamount of energy is required to overcome the strongmetal ions and mobile electrons.

points (or low volatility) because largeelectrostatic attraction between positive

2"

Melting and boiling points of metals increase as the strength of metallic bond increases"E.g- melting point of Na < Mg < At white metting point of Na > K > Rb

Thermal Conductivity

Metals are good heat conductors becEuse heat travels throughvibrationsandcoltisionsofthe ,. i:1 ir t' r;-' the metal lattice as a result of the raoid

Figure 2.2 Metals conduct heat well

hot end cotd end

ffiIXT ;,ff:'il;X:, , i, '

stror€ty tess strongly

3. ElectricalGonductivity

ltelals 1re good electricat conductors in the and , -' '

states because there are11 :' ' I ' ' ' that can carry charges through the metal lattice. (Etectric curentresulfs from a netflow of charges.) Etectricat conductivity (units: o-t cm-t; lncreases as number of mobite electronslncreases.

Figure 2.3 Electrical conductivity of metala:.:: ::

.'+Direction ofelectron flow

Mobile electrons,noldrE through Ulqtanice carryir€:negalive charges

Direction ofcurrent flow

Table 2'2 Electricar conductivities of some metars |n 196 Eg-l gm-1


Table 2.1 Melting and boiling points of some metals in oC

@=. Q-@ @ -@"',1@"o-ieJ@-".Q-n :a- =r - :a:ra : 3l@-@_@ _@ =.@

Li 0.108 Be 0.313Na 0.210 Mq 0.226 At 0.377K 0.139 Ca 0.298

4. Solubility

Metals are generally insoluble in polar solvents such as water and non-polar solvents such as

hexane, benzene.

However, some reactive metals such as K, Na, Ca can react, not dissolve (strictly speaking), with water to

give a solution. e.g. Na(s) + Hzo(t) -> NaoH(aq) + %Hz@)

Metals may be soluble in liquid metals forming alloys'

Hardness

Metals are fairly hard. lts hardness depends on the strength of the metallic bond. However, metals are

;;;\J;rlJ '@an be bent or pressed into different shape) sni ct.i c;'ri p (can be pulled into wires)

because the layers in the lattice can slide over each other without breaking the strong metallic

bond.

",";",;ffi g=;a;

ry'uct tv

lrmetars

9;qffi-

5.

nWhen a force is applied, metallic bond is not broken. Thus, metals will not break when it is bent.

Exersise2: (a) Describe the structure and bonding present in aluminium"

(b) Exptainwhy the electrical conductivity of aluminium is higher than that of sodium.

qlo \^-.* +^qrr'$l r- ruott'+-)\t

(a)

(b)

Structure:

Bonding: ' -.'.}'(.|{.{^.i...l11.,,||,,.l.it.,l{ ,t'1 r'":,Y'

Each Al atom contributes 1 valence electrons to its metallic bonding whereas each Na

atom contributes only { valence electron to its metallic bonding" Therefore, A/ has

n'nnil*, _:s:lii:"r,,. fol9r"tiuitvthan Na as

in A/than in Na"

ther€

JJC 2008 PageT of 42

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3.

3.1

IONIC BONDING

lonic Bondinq in lonic Compounds

Definition: lonic bond (which is formed by the transfer of electrons) is the strong electrostaticattraction between the _

Figure 3.1 Model of an ionic compound: Lattice of positive and negative ions

ooooooooooooooo

As shown by Figure 3.1, ionic bonding exists extensively throughout the entire structure.Thus, ionic compounds are said to have giant ionic structure.ln the giant ionic structure, cations and anions occupy fixed and alternate positions in the lattice structure.

Figure 3.2 lonic lattice of sodium chloride

Oppositely charged ions are formedfrom the transfer of electrons from themetalto the non-metal. Metals formpositive ions whereas non-metals formnegative ions to achieve a stable noblegas configuration.

CT

Na*

x The table below shows the chemicalformulae of ions|[ilt tv

of the first twenty elements.

VVI vil

xDot.and-crossdiagramS(a|soknownasLewisdiagrams1ofioniccompounds

- a simple representation of the electronic structure of a species by showing the valence electrons only

Sodium chloride. NaCi

[ -"].[,:]tJ-

[,i:,1 -[

ca J'.[,i]'l -

Calcium chloride,CaC/2

JJC2008 Page8ot42

Common mistakes: Remarks

[ - " -I *[ .. -l - Since Na atom has already given away its valence

I I lt" i | | :C/ :l electron to chtorine atom toior:m Na*, there shoutd notL xx J L o. J 5e any crosses around Na.

l

ifill .

Ifh I Nooutercircre shourd be drawn

L\_-iJ t\../J| 12* | o. I - Subscript'2'should not be used.

lc"l lt,^,'l Correctrepresentationis: [ .. l-f 12*f .o l-t--J [';:'Jz lac/:l lc"l llcrll.. J t I t .. J

ExefQlse 3: Draw the'dot-and-cross' diagram of

(a) Lithium hydride

i i I -- j

vv -i

I 3.2 Strenqth of lonic Bond

x Strength of ionic bond is indicated by the magnitude of lattice energy.

Lattice energy of an ionic compound is defined as the enthalpy change (AH) when one mole ofthe ionic compound is formed from its constituent gaseous ions under standard conditions.[Refer to Chemical Energetics Lecture Nofes/

(Lattice energy simply refers to the energy released when one mole of ionic compound is formed fromifs gaseous rbns.)

" e.g. tta*(g) + C/ -(g) -+ NaC(s) AH = - 77'l kJ mol-t = Lattice energy of Nac/

= 771 kJ of heat is released when 1 mole of NaC/ is formed.

+ 771kJ of heat is needed to break 1 mole of NaC/ into Na* and C/- gaseous ions.

Thus, the magnitude of tattice energy is an indication of the strength of ionic bond.


9*x9-It*r-

where g+ and q- are the charges of the cations and anions respectivery;and r+ and r- are the ionic radii of cations and anions respectivery.

Generally' {hs t 'r-or ' .r-''- the magnitude of lattice energy, the stronger the electrostatic attractionbetween the oppositely charged ions, the - : : ,,, . . , ine ionic UonO.

Two main factors affecting the magnitude of lattice energy (LE):

(a) Charge of ionThe more highry charged the ions, the greater the magnitude of LE.(Electrostatic attraction between higher charged anions and cations are stronger.)

e.g. Magnitude of lattice energy of MgO > MgC/2 > NaC/

(b) Size of ionFor ions with the same charge, the smaller the ions, the smallgr the interionic distance (r*+r_ ), thegreater the magnitude of LE' (Electrostatic attraction between small oppositely charged ions arestronger as smaller ions can approach closer to each other and they have higher charge density.)

e.g. Magnitude of lattice energy of NaF > NaC/ > NaBr > NaI

3.3

1. Volatitity

lonic compounds have r'-' ' | ' melting and boiling points because large amount of energy isrequired to overcome the strong electrostatic attraction between positive and negative ions"Melting and boiling points of ionic compounds increase with the strength of ionic bond.

E.g. metting point of MgO

MgF. 1261MgC/z 714CaF2 1423

2" ElectricalConductivity i

Poor electrical conductors in _ state because the ions in crystals are fixed at the lattice points.Good, electrical conductors in i'| anrt

,.., !.1' :'' , .. I

rx Magnitude of LE "c

Compare with Coulomb's Law:( where d = r* + 7- 1 11 orce

Table 3.1 Melting points of some ionic compounds in oC

fF-El fFBl fffi-f,] IEF Hc_tli;; ;;l li.lll'*'l lit $il lcarz Mzsl l:g ?g?Sl

r'1\f. i, !r t states because there are

Positive cations move towards the negative cathode; negative anions move towards the positive anode.

JJC 2008 Page 10 ot 42

3" Solubility

lonic compounds are generally soluble in water and insoluble in non-polar solvents (such as organicsolvents).

In dissolving ionic compounds in water, the strong electrostatic attraction between positive and negativeions may be overcome by the ^-''y'.""*l\ :';-,+"':r,

-,- .,released wh.en ions are'solvated by thepolarwater molecufes through the formation o1 :'7- ) ?'\ €''o'!{l'' +' t'.. : "

AHsoturion = - AHL"ti"""nurgy + AHnyor"rion

For example, when solid sodium chloride is added to water, an electrostatic attraction is established

between the ions (Na., C/-) and the polar water molecules. We call this attraction the ion-dipoleinteraction. This ion-dipole interaction releases sufficient energy (known as hydration energy) toovercome the ionic bonds between the positive and negative ions, resulting in the breakdown of the NaC/crystal lattice and the solvation of the ions.

Figure 3.3 How does an ionic compound dissolve in water?

I ,-## s,;i $ rr-n.^

-a *' *),,brI ";9 g,r ''5+

,,37rfrlv ,;d cr J. UI'.J ! ";t,l re

.:tlp

(a)lon-dipole interaction begins

to form: OF (of HzO) -

attratted to cations and H6*(of HzO) attracted to anions-

(b)Crystallattice begins

to break down.

(c)The ions are completelysolvated. lonic lattice has

broken down and ioniccompound has dissolved.

Figure 3.4 How does an ionic compound dissolve in a polar solvent?

lon-dipole interaction begins toform: the posilive end of polar

solvent molecules are attractedto the anions and the negativeend of the polar molecules areattracted to the cations. These

interactions release sufficienthydration energy to break the

ionic bonds.

JJC 2008 Page 1 1 of 42

rlons arecompletelysolvated bypolar solventmolecules.

Polar solventmolecules

Condition for solubilitv:

Since ions can form ion-dipole interaction with polar water molecules, ionic compounds are soluble in water ifthe energy released from the formation of ion-dipole interaction is sufficient to overcome the strong ionic

bonds.

With non-polar solvents, effective interaction between ions and the non-polar solvent molecules cannot be

established. Hence ionic compounds are insoluble in non-polar solvents.

4. Hardness

The crystals of ionic compounds are verypartially overcome to deform the crystal. However,cfeanfy using a sharp-edged razor).

When a force is applied along a particular plane, one layer of ions is displaced relative to the next.As a result, ions of the same charge meet and repel one another, shattering the crystal along the faultline.

Figure 3'5 Cleavage of ionic solids: a layer of ions (a) before and (b) after force is applied

Repulsion btwions of samecharge Lattice breaks

down along alayer of ions

because strong electrostatic attraction must bethe crystals are _ (i.e can be cleaved

Exercise 4:

(a) Both CaC/2 and CaF2 have structure.

F- is

-- than Cf, leading to ionic bonds in

CaF2. Larger amount of energy is required to overcome the stronger

For each pair of substances, state which one has a higher melting point and explain youranswer. (a) CaC/2 and CaF2 (b) NaCi and MgO

Therefore has a higher melting point than

(b) Both NaC/ and MgO have structure.

Mg2* and 02- are in size and of -charge than

Na" and cf respectively. Larger amount of energy is required to overcomethe stronger electrostatic attraction between chargedand _ ions in than the weaker electrostatic attractionbetween charged and _ ions in

Therefore has a higher melting point than

electrostatic attraction between _ andthe weaker electrostatic attraction between

ions in than

_ and ions in

To explain differences inm.p. or b.p. betweensubstances, alwaysinclude the following inyour answer.

(1) type of structure foreach substance.

(2) type of bondinqovercome duringmelting/ boiling andcOmpare their relativestrength.

(3) comPare the amountof energv neededand relate to m.p. or b.p.

Force applied n"r" I\u


4" COVALENT BONDING

Definition: Covalent bond is the strong electrostatic attraction between the

shared between them'

Cova|entbondisformedduetosharingofelectronsbetween2atoms.

Erectrons which are shared between the atoms are called bonding electrons. Two bonding electrons

constitute a bond pair. outer sheil erectrons that are not invorved in bonding are cailed rone electrons"

Two non-bonding electrons constitute one lone pair'xx a.

For examPle, I C, I C/ :xx ao

Around each C/ atom, no. of bond pair =no. of lone Pair =

1

3

x octet Rule: Non-metallic elements form covalent bonds by sharing electrons to achieve a stable noble

gas configuration.iil

r' rn morecures (a) to (c), the centrar atom has ress than g erectrons in the varence sheil after bonding. The

central atom in .these molecules are saio'io ue "ieitron-o"ficient

and thus have a tendency to accept

;;ilr;'ir"r "noinat

molecule by forming dative covalent bonds' (See Section 4'2)

rn morecures (d) and (e), the centrar atom has more than g erectrons in the varence sheil after bonding' This

is called expansion of octet configuration'


NitrogenNz

carbon dioxideCoz

WaterHzo

.XXH:OIH

XX

OxygenO2

ao xxo3Iooo xx

Octet rule is sometimes not obeyed' See the following tables for some examples'

(c) Aluminium chlorideA/C/3

XXX nrXXV,X

XX OX XX

IcIIA/: cilXX XX

@ Boron trifluoride(a) Beryllium chlorideBeCl2

XX XX

IcrI ae! cl{XX XX

aa

'tl :itoax"xaa-,Iiri.r'i.3

(O) Pnospnorus(V) chloride orphosPhorus Pentachloride

PC/s

.Oor.l oa-'O aoo

^to.li" | "'f"';L'j ".ti:onr o oo-o-'aaa

Chemicalformula

Dol-and-crossdiagram

- =-

-

-..-: ::::: _l -3

wilnmflflt,,tfllfllflt'

''{rlilltiilliii

x Expansion of octet configuration: llii:

The expansion of octet configuration is possible only if the central atom has u,,,,

to allow more than 8electrons in the outermost electron shell.

e.g. To form PC/5, P must have 5 unpaired electrons.

P: 1s2 2s2 2p6 3s2 3p3nrrn3s 3p

P*: [|-T-nPC4: TIIN

Only elements in Period 3 andhigher (Period 4, 5 ....) can expandoctet because octet is expandedthrough the use of d-orbitals. The firstset of d-orbitals is the 3d-orbitals.

One of the 3s electrons is promoted to an empty 3d-orbital which is energetically accessible. With 5

unpaired electrons, P can thus form 5 bonds and hence PCis is formed.

Questrbn: Does NC/s exist? N: 1s2 zsz 2po | | 3s

For N, the next available orbital for the promotion of electrons is 3s, which is from the next prin6ipalquantum shell. Promotion of electron to another principal quantum shell requires too much energy to be

feasible. Thus, N cannot expand its octet to form NC/s. NC/s does not exist!

Exercise 5:

Draw the'dot-and-cross' diagram of

(a) CCia (b) NF3 (c) CSz

a

4.2 Dative Covalent (Coordinate) Bond

Definition:Hffind is a *gffiruHhl in which

@

x ln forming dative covalent bond, one atom is the drrfrrHraa ' ^nd the other is the *ei#

For an atom to act as a-F4t must have

For an atom to act as anLt*it must have ffix Once formed, a coordinate bond has the same characteristics as a normal covalent bond. ln other

words,the only difference lies in the way the bonds are formed.

x The dative covalent bond can be represented asfi&r (A is the electron donor; B is the acceptor).

3d

JJC 2008 Page 14 ol 42

i,iI

lSome Examples of Da :

(a) HlEr-qlOLoa

!.o l- + H*H't t'H ,:"f'

(nLNHr*Ab

H._N-H +

I

H

ct c/tlA/_Ct + A/-C,tlct cl

Exercise 6:Use the concept of dative covalent bond,reactions.(a) NH3 + A/Ci3 + ?

ir i if

\,{ f\.-, A'. *i'i

It )

n

ld,\lcvc'7c'''r f't''

-_-_>

1 - -[,, N / :,t\4,F-B-F ---> i\-i'Y

!

-l L-

A dative covalent bond is formed so that H can achieve the stable noble qas {He) confiquration.

i-i i

I

F

A dative covalent bond is formed so that B can achieve the stable octet configuration.

, (c) AHrfrrof ,. r >a''1, /

\Li'

Jit'

! .j. . .

draw the structures of the products of the following

(b) BF + NaF-+ ?

IJrj n

,Jl 1',;,11- 1, -. ' 1

. :J

(d) Formation of #uring the reaction:ffil/i-J

H-N-H + H* rr _ f l-l-i, .' -/ ll - rvlrFl Eltl

7

r-* \"I

l'rt-! \- , -'I

tr

i.,i,rl

i nr,,.' a a:i

i"**-,i'.i''""1

, JJC 2008 Page '15 of 42I yc*vlX i-r, -

r,vaurf 1s hq w^V fvrlrlc' '

: 1i\,]{ i.nY rklL\r F t s

Note: HgO'can also be represented as:

,,{ .r [x'].

Itre pur,.1 r,r o r c\".

Bond energy of a covalent bond is the energy required to break one mole of "o,,r"l"nt

b*o"ln "gaseous molecule into constituent gaseous atoms [Refer to chemical Energetics LectureNotesJ

4.3 Strenqth of Covalent Bond

covalent bond strength is rerated to the bond energy. what is bond energy?

(or, bond energy is the energy released when a covalent bond is formed between 2 atomsr, fh;;;;;;6The greater the bond energy, the stronger the covalent bond. i J - . . , :

x Factors affecting covalent bond strength:

(a) Number of bonding electrons (bond order)Bond strength: Triple bond > Double Bond > Single Bond

e.g. Bond energy: C=C

(b) Bond length (which in turn depends on size of atom)For a fixed number of bonding electrons between atoms, when thesize of the two atoms increases, bond length andthe covalent bond strength

e.g. Bond energy of Ct--St > Br-Br > I-I

Table 4.1 Bond length and bond energy

Bond Length: thedistance betweenthe nuclei of 2atoms joined bychemicalbond.

Bond Bond length / nm Bond enerqy / t<.1 moF-ct-ctBr-Br

l-l

0.1990.2280.266

244193'|51

Bond tsond lenqth / nm Bond energy / kJ mol']c-cC=CC=C

0.1540.1330.120

3506't0840

5"

x

(c) Bond polarity [Refer to section 7.3.1: potaisation of covatent Bond]

Polarisation of covalent bond such as 6*H--ct'- leads to some ionic character. This ioniccharacter gives rise to-additional attraction to strengthen the bond.

COVALENT COMPOUNDS

There are two types of covalent compounds which possess different physical properties due to thedifference in their structures.

Most iovalent compounds have simple covalent (or molecular) structure.some covalent compounds such as diamond, graphite, silicon and silicon dioxide (i.e sand, quartz) havegiant covalent (or molecular) structure. -'.'-- \.'-

3The following diagram shows the difference between these two structures.

Simple covalent structureExample: carbon dioxide

O:6-O o-c-o; "' \

--- intermolecular forces

o-g-O --z/-/'o:c=o

\, covalent bond

Giant covalent structureExample: diamond

I

:l'---'Fl*l rro

r represent a carbon atom iEvery atom is bonded to others by stronqcovalent bonds.

+ Covalent bonds exist extensively throughoutthe whole structure, thus compound is sjiO tohave qiant covalent structure.


J

GIANT COVALENT STRUCTURE6"

6.1

x Giant covalent structure consists of covalently bonded atoms at the lattice points that can be extendedinfinitely in three dimensions to form a giant lattice.

Atoms in giant covalent structures are held by strong covalent bonds. Therefore, giant covalentstructures

(a) have v"' "'''' melting and boiling points because large amount of energy is required toovercome the strong covalent bonds between atoms.

(b) are Dol'r electrical conductors (except graphite and silicon) because there is no mobileelectron (charged particle) as allvalence electrons are localised in the covalent bonds btw atoms.

(c) are n I olvrrz{ q. in all solvents because large amount of energy is required to overcome thestrong covalent bonds between atoms.

(except graphite) because breaking or deforming the structurerequires breaking the strong covalent bonds between atoms.

6.2 Diamond and Graphite

x Bonding and Structure' Diamond and graphite have giant covalent structure.

In diamond, each C atom uses all its 4 valence electrons to form 4 single covalent bonds with 4 other Catoms in a 'r:'lj- " , !' ' i arrangement. (See Figure 6.1)ln graphite, C atoms are 'rlr'i]

"',"1. -'' in flat, parallel layers. Each C atom forms 3

single covalent bonds with 3 other C atoms. The fourth unpaired valence electron of each C atom isdelocalised along the layers. The parallel layers are held by weak van der Waals, forces.

Csc i 'r1i \-r ); i

Poorelectricalconduction

Figure 6.1 Structure of Diamond ( . = carbon atoms)

Electrical Conductivity

Figure 6.2 Structure of Graphite ( . = carbon atoms)

Diamond is a poor electricar conductor because there is r. ) i,- (' 1as all valenceelectrons of carbon atoms are tocalised in the covalent bonds between C atoms.

; ;;; ;;" *i,r" a ;; ;1,;;;; ",.Figure 6.2) becausethere are ]-;': ''i '' I - -'1 ,' which are free to move r.1 1 .o, -,r the layers.

'What is the difference between these 2 terms: delocatised electrons and mobile electrons?i.D.elocalised electrons: they are n,ol-fixed at a position but are free to move to a certain extent- ]

' -iMoDf le efectrons: they are able to move freelv in anv direction . I , "*

(d)

l

x Hardness

Diamond is v-.lY 7' because breakingstrong covalent bonds between C atoms.

Graphite is 'l"i- because the layers are held by van der waats' forces which are inr\ ' ' enoughto allow the layers to slide over each other. This makes graphite suitable for use as a l rbricant.

or deforming the structure requires br.eaking of the


L.-= )

6.3 Silicon and Silicon Dioxide

r{'

Silicoir has a giant covalent structure similar to that ofdiamond. Each Si atom forms 4 single covalent bonds with

4 other Si atoms in a tetrahedral arrangement.

Silicon is a semi-conductor with a high melting point of14100C.

SiO2 (s/rca, crystalline form in quaftz, impure form in sand)

has a giant covalent structure.

Each Si atom forms 4 single covalent bonds with 4 oxygenatoms in a tetrahedral arrangement. Each O atom forms 2single covalent bonds with 2 Siatoms.

Silicon dioxide has a high melting point of 1600oC.

and boiling points of compounds with simple covalent structureincrease with strength of intermolecular forces.

Note that the covalent bonds between atoms within themolecule are not brsken during melting/boiling.

Figure 6.3 Structure of Silicon (left) and

Silicon dioxide (right)

F Metallic bond, ionic bond and

covalent bond ate strong

bonds. Thus metals, ionic cpds

and cpds with giant covalentstructure have high mP and bP.

) lntermolecular forces are weakforces. Thus cpds with simPle

covalent structure have low mP

and bp.

( . silicon atoms)

/\| . Silicon atoms I

I o oxvoen atom IJ

7"

x

x

SIMPLE MOLECULAR STRUCTURE

Most covalent compounds have simple molecular structure (also known as simple covalent structure)"

In solid state, simple molecular structure consists of simple discrete molecules at the lattice points.

WITHTN the molecule, atoms are held by :!'1'.) r \ ! I :i\ 1" t ''r : "

BETWEEN molecules, molecules are held by ' ;' '*-.' i ' -

7.1 Phvsical Properties of Govalent compounds with Simple Covalent Structure

1. Volatility

Compounds with simple covalent structure have

melting and boiling points (high volatility) because little

energy is required to overcome the weakbetween the molecules. Melting

2. ElectricalConductivity

Compou"nds with simple covalent structure are poor electrical conductors in any state because there isNO i- .''.;,' f ., li ':, . t " ":' l'.'

(Exception: For polar molecules which react with water to form ions, the resulting aqueous solutionconducts electricity because of the presence of mobile ions. E.g. When hydrogen chloride gas is added

to water, H+ and Cf ions are formed. Thus, the resulting hydrochloric acid can conduct electricity dueto the presence of mobile H' and Cf ions.)

Solubility j t

Non-polar molecules are generally more soluble in non-polar solvents such as hexane, benzene

and methylbenzene. Polar molecules are generally more soluble in polar solvents such as water"

Hardness

Compounds with simple covalent structure are soft because breaking or deforming the structure only

requires the breaking of the weak intermolecular forces between molecules.

a

3"

4"


7.2

x

x

Shapes of Molecules

Can be predicted using the Valence Shell Electron Pair Repulsion (VSEPR) theory.

VSEPR theory states that(1) electron pairs (i.e bond pairs and lone pairs) of srtral -)

atom arrange themselves as fat as,possible in space

to minimise mutual repulsion (see Table 7.1),

(2) lonepair-l'one pair repulsion > lone pair-bond pairrepulsion > band pair.bond pair repulsion, \

(3) multipte bond (i.e. double bond or triple bond) is 1

treated as a single electron pair. I 1

Determines the arranqement ofelectron pairs around centralatom.

Determ ines the @tion oflone_.rejtas well as bond anqle.

I

(r if-rl ' r ioi; .r-t ^

y;t -i:stf )

Table 7.1 How electr:qn paiii will anange around ihe eailiilltfito minimise their mutual repulsions (VSEPR):;G*=

7

t

r rr:':- i i

?.r t

?electron pairs . $-blectron pairs 4\lectron pairs

i,!-.'- ':i.,".., "r' r:.' 1: "':' ''

. .^,!''Arrangement' \r J; rr 1{

Dr0

r,,1200

.\.

Arrangement: ',{,"l^." Arranoement' rt-r

lectron parrst

-@etectron pairs

Arrangemeng TP i 6o!'lAL P1b';l; :)'!.', i t/;' t,_

I_L_

I

I

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Eoro-OvvoCLgoo-cF

x

,

)

ln an ammonia molecule, there "r" a electron pairs on the central N atom. The electron pairs are in'a

tetrfdned,fal arrangement. Since tnere are 3 bond pairs anO \ lone pair on N, ammonia is

with bond angle l0*o because the \-$A&ltslY: l!- repulsion is

greaier than the L$/d\r'v*b0,trlhd\v repulsion.

In a water molecule, there are + electron pairs on the central O atom. The electron pairs are in atQ\\AWdYaf arrangement. Sinc,e th.erg.are L bond pairs and L .lone pairs on o, water is

crr4^r4// Jo.J ""gli U*C4"fu"."u"" n" tolnrePttrz-'tqa-etailrz repulsion is greater

than the repulsion, which is in turn greater than therepulsion.

Exercise'8: By considering the no. of lone and bonding pairs of electrons, predict and draw the shape of the

following molecules or ions.$(a) QSz a B

4'e ' 'ial

o''(

-I.

6Gl-o ,t e

x1 r--) o

.t\r[, iE,

j

*-\

E:f

!*-\ .tt!

I

I

1

Ii:

I

:

(b) 'tFs

-lt1

(c) H.Q*

thl '\\:) -'

sb\l!\r: /t''\.

.l' : 'i:t' "i

,tfl

i*i r ;:-':.... ...r_".- i_li-i

l*t

Reminder:

For electron pairsarranged in trigonalbipyramidal manner,the lone pairs arealways placed at theequatorial positionto minimise repulsion.

(d) IJ-t-

r--{lill-1

-t^l,r-, \ l

:-L' -* "' .; ',\ -'','i'',.,j| ! \t l'

\.\1\",ir1t

I

"1t

iinf,a" .


"*t-,,uf !1.{:,,',C,tt .

7.3 @Pure covalent and ionic bonds are extreme cases. In reality, most compounds have bonding which isi

7.3.1 lonic Characterin Covalent Bond - Mx|Gi$€h{aGl9rinacova|entbondistheresultofthebetweenthetwo

atoms involved in the bonding.

"iE{'@o

aJ

Between two atomswitfrci*iler.(*rue)d-tnmget&uity, theelectron cloud [email protected] results in theformation of a mrffi

lf the elalmgtetiulty of the two atoms is#!da#elent, ldr*don occurs as[hs el*r'an Glgul will be ehi&edmore towardsthe@Hence, the more electronegative atom.acqu ires aSrcharge (pc**'.rccnergc) wh ilethe less electronegative atom acquires a.A*charge (p.*C*n h.f3e). + dipde F:r,"t4Such a covalent bond would thus have someiricrlaracleJ and is called a pufnntunt,hnrdpor in shbrt, aprhhtd.

lf the da**qnll*ty ofthe two atoms iscq;.diffffirt? the deltrum]would ber;rptuifitnnfsmed tr the moreelrtmgufi+aqatem.qlgLtt'W"J+ivQ > crThis results in theformation of cation andanion. Hence, an*rni*-bilftls formed.

reg .t,C

xUsua|ly,ifthecovalentbondisformedbetweentwn"dirnrre&-etoms,the@Exception: C{4*,S,s4,$H bonds because the electronegativity of 6; H, F and S are simi{ar.

x Presence of ionic character in covalent bond affects> hoh^4o.\ c.ovoJc-qfi1""r,1n7,

(1) the sgtbof the covalent bond (see Section 4.3, page 16), and eX-{+adtrD, sl+e-,r^q{nenV,A&l -

(2) the neta*'of in'!ffrnslcsntarJa-ee: bcbr r{Fs-mcffrles (see section g).

Bond Bond energy / kJ moFr Bond Bond enerqv / kJ mol-'H_HP-PP-H

436208322

H-Hct-ctH-C/

436244431

Example: J91 P2 O2(b)

The P-H bond energy is the mean (average) of the H-H and P-P values. Explain why the H-C/ bondenergy is not the mean of the H-H and C/-C/ values.

Because the electronegativities of H and P are similar, P-H bond formed will behdn-Po\oY . As such, the p-H bond energy is the average of tJre H-H.and p-Pvalues because all three bonds share the same characteristics of being non-polar. lncomparison, H-C/ bond is Sqt (due to different electronegativities of H and C/)while H-H and C/-C/ bonds are non-polar. The presence of slight ionic character in H-C/bond - stvtv''qtngn I the bond, giving a bond energy higher than the mean of theH-H and CI-CI values.

tEat t'o"l I Large extent of polarisation can result in weakening of the covalent bond.

JJC 2008 Page25 of 42

a

7.3.2

x

anion towards itself. The cation is said to be polarising the anion.

between the two ions, and gets shared by them.

the electron cloud is being shared between Af* and O2-.

x This concept explains why beryllium chloride and aluminium chloride are essentially covalent as C/-

*g5,r#:Ti":3r:t,:l: "uoj".ted to sreater polarisation. Note that Beo and A/zor are essentiallv ionic due

x The degree of covalent character in an ionic bond depends on the extent to which polarisation occurs,which is in turn determined by:

(1) the polarising power of cation hoW paqgpa c-m 7A?AW LÂrdr\t cAvAck d?Cations with

-

charge density (i.e. small size and hiqh charge) can polarise anions to " S

greater extent.

E.g. charge density of cations: Na*

Degree of covalent character: NaC/ <

(2) the ease of polarising the anions (polarisability of anions)

anions are more easily polarised (more polarisable) by cations.

E.g sizeof anions: ;nc*<cta'thal. F- < Cf < Br- < I-Degree of covalent character: A/F3 < A/C/3 < A/Brs < AlIs

If,ts ,s why AICQ is covalent while AIFa is ionic ( F - is less polansahle ).

Polar Molecules

The presence of polar bond(s) in a molecule does not necessarily mean that the molecule is polar.

E.g. 1: CO2 molecule is non-polar even though it contains polar C=O bonds.

5- 5+ b-o-c-o++_+Rsason:6- and 5+ charges of any polar bond produce a dipole

(separation of charge, ie presence of 5+ and 5- charge)

which has a dipole moment p. Dipole moment can be

understood as the magnitude of the dipole.

In CO2, the dipole moments cancel out each other because the dipoles, with equal magnitude, are

acting in opposing directions. Thus, CO2 molecule hasand hence is

E.g. 2: BF3 molecule is non-polar even though it contains polar B-F bonds.

6- F\\6+ 6_B_F

To determine overall dipolemoment, treat each bonddipole as a vector, and sum upall the vectors.

lonic bond withcovalent character

Mg'*

MgC/2

< A/3*

< A/C/3

7.4

x--3

A non-polar molecule is onewhere overall dipole moment ofthe molecule is zero. (Eitherbecause there is no polar bond orthe dipole moments of the polarbonds cancel out each other.)

ur/

The dipole momentscancel out each other.


I

-

Does the molecule have polar bonds?

---lr x Tqche@

-

>J

Motecule is non-polar.

Some other examples:

(a) PC/F

PCG is a non-polar molecule

because the dipole moments

cancel out each other.

(b) sFq

SF+ is a polar molecule since ithas an overall dipole moment.

(c) SFe

SFe is a non-polar molecule

because the dipole moments

cancel out each other.

(d) rFs

IF5 is a polar molecule since ithas an overalldipole moment.

(e) XeF4

XeFr is a non-polar moleculebecause the dipole momentscancel out each other.

Note: lf the central atom has no lone pair of electrons and is bonded to only one type of atom (e.9. CO2,

BFs, PC/s and SF5), the molecule is non-polar because the dipole moments would cancel out one

another.


Exercise 9: Determine whether the following molecules are polar or non-polar.

(a) HzO

(b) cc/4

(c) CFC/3

(d)

Dipole moment of C-F bond is greater than

between C and F is bigger than that between

and hence it is a polar molecule.

IC/3

TRY THIS: J97P3Q3

The diagram shows a liquid flowing from a burette and a charged rodbeing brought near the flow.

Which liquid would be deflected as shown?

A bromine

B cyclohexane

C hexachloroethane

D trichloromethane

that of C-C/ bond because the electronegativity difference

C and C/. Thus CFC/3 would have an overall dipole moment

-ffi --nieoatiueucr'giig9{

rort


8"

x

INTERMOLECULAR FORCES (IMF)

Recall: Simple Molecular Structure

WITHIN the molecule, atoms are held by strong covalent bonds.

BETWEEN molecules, molecules are held by WEAK intermolecular forces.

Three types of intermolecular forces:

(1) Induced dipole-induced dipole attraction (ID-ID) wQql'( -{ +

(2) Permanent dipole-permanent dipole attraction (PD-PD) i*\,.$: $it"",'(3) Hydrogen bonding ':Jvo$Qt1\'if,

The strength of the intermolecular forces between the molecules affects the

(a) Melting and Boiling Points

When compounds with simple covalent structure are heated to melt or boil, it is the

which are overcome. See Figure 8.1. (Covalent bonds are not broken!!)

The stronger the lMF, the ht the m.p. and b.p. of the covalent compound because more

energy is required to overcome the stronger lMF.

.btf tf

Figure 8.1. Boiling of water. Note that the covalent bonds between atoms are not broken.

Only the weak intermolecular forces are overcome.

(b) Solubility of the covalent compound (?)polar covalent compounds are more soluble in qltleov^s sofuents. Non-polar molecules are soluble in

solvents. Molecules which can form hydrogen bonding with water are expected to be

more soluble in

-.8.'! Permanent Dipole-Permanent Dipole Attraction (PD-PD)

tB A polar molecule has a permanent dipole moment.

xThee|ectrostaticattractionbetween3rr|*wlusisknownasthen.'#JiFi-.lErdiaD (o r i nteraction).

permanent dipole - permanent dipole attraction

Strength of IMF(comparing moleculeswith similar M):ID-ID < PD-PD <

hydrogen bonding

tfI

C .bI

e

,/\6+,/\.G,/ \j\

6+

JJC 2008 Page29 ot 42

tP

aaI

a

:::=.-:a:

t.a\ t."\- .t\^_ ^

*/T= (;;/z=* ) ,f23.:e-permanent dipole-permanent dipole attraction

x As the magnitude of the dipole moment increases, strength of permanent dipole-permanent dipoleattraction 'tf'}{n'Qo| (molecules which are more polar are held by stronger PD-PD).

8.2 Induced Dipole-lnduced Dipole Attraction (ID-ID)

|rWareheldtogetherbyinduceddipo|e.induceddipo|eattraction.x How does induced dipole-induced dipole attraction arise?

Non-polarmoleculese.g. Br2,CO2, He

ID-ID attraction

I0-/

Qiu-lnduced dipole

(a* )

temporarydipole

(a) Distribution of electron cloudshould be even for a non-polarmolecule.

(b) However, at a particular instant,uneven distribution of electroncloud may occur due to thecontinuous motion ofelectrons in the molecule.This results in the formation of atemporary dipole.

G

{.

o'temporarydipole

(c) The temporary dipole inducesanother dipole in a neighbouringmolecule, resulting in theformation of an induced dipole.The attraction between them iscalled induced dipole-induceddipole attraction.

ID-ID attraction and PD-PD attraction are commonly known as van der Waals' forces.

Factors affecting the strength and extensiveness of van der Waals' (VDW) forces(a) Strength of VDW forces depends on the number of electrons in the molecule.

When M, of the molecule increases, the no. of etectrons in the molecule lr\A$+ . As theelectron cloud gets larger, it becomes more polarisable and the strength of VDW forces betweenmolecules rnenl)t95

(b) Extensiveness of VDW forces depends on the surface area of the molecule.(For molecules with the same no. of electrons) When sgrface area of molecule increases,VDW forces between molecules become tl^s\ 1€v as there is greater surfacearea of contact between molecules.

JJC 2008 Page 3O of 42

-I

Exercise 10: Describe and explain the variation in the boiling points of noble gases.

. All noble gases have *qufu structurd with

between their respective

rnCr<r'tSQ-lAs the no. of electronsvV\) tsrbt tvu,v?qV

down Group 0, the strength ofdown the group.

. Thus, down the group an increasing amount of energy is needed to overcome thev\lw ftvr4 between atoms and hence, the boiling points of noble

gases rvr rrv0A )&

x For organic compounds, branching of the carbon chain causes the molecules to be more spherical, hence

reducing their surface area and the extensiveness of VDW forces between them

= branched organic compounds have a lower boiling point than their straight-chain isomer.

@l:Exp|ainthedifferencesinboi|ingpointsofthefo||owingisomers

-x-t

H-C-llI

l*-g-flIlrc-HI

x-c-rtIlFe{IX.

pentaneb.p.36'C

ttafi-c-HIH-f-x xtlg-g+C-lltl

n-*i-rr H

Iil'.

' ? methylbutane. b"p.28'C

'r|'t"F9-"

*trl'e-c -c -c

*ltrtl

' H#-n H

Ilt'

2"2dimethylpropana

8.3

x

2-methylbutane molecule has a

b.p. 1S'C

surface atea than pentane molecule due to

of carbon chain. As a result, less energy is required to overcome the

between 2-methylbutane molecules. Hence 2-methylbutane has alower boiling point than pentane.

2,2-dimethylpropane has the lowest boiling point because it's molecule has the smallest surface area, and

hence, themolecules.

VDW forces exist between the 2,2-dimethylpropane

Hvdrooen Bondinq ,. s\rt\r;1 ir-\ -('t \ b\;v{l\L 'i

When a H atom is {;s!!y bonded 1s s F$hty'^'^.trrnr! (such asf;€*llfa *fdhCeis produced in which the H atom acquires 6+ charge and the electronegative atom X acquires 6- charge.

x This particularly strong permanent dipole-permanent dipole a molecule and the

l^^^ f^;- r v6'^r "nolher molecule is known as E4[ggtlbtlding:

Example 1: Hydrogen bonding between H2O molecute{-ft$L\lliifiGft1w" ,"6" 7 ,rti\rr.rirZ* rzrt \

./ .-\ i xc.,H H-'- - - - {o^

! !-! tl,-d\ 1r'

Example 2: Hydrogen bonding between NHe molecules i\M,'vruNv\Ji H Hlr

*,\$i/'"Y\ ,- 1,. / A

5+ n iilcl\l\4"'-i^ [] Hw)\' {' JJc 2oo8 Page 31 of 42

criteria for existence of hydrogen bond between molecules(a) a H atom bondedrulltEtly to a @(F, o or N only)(b) at least a pairof lc*r*5cn the highly@

@ing on physicat properties(a) Boiling point (and melting point) of HF, H2O and

the presence of ltvrlr,.gr i.rtun"ri.", vr-*e\,-e,l befwNH_3 are unexpectedly l,r\tlu due torn their molecules.

For moleculeswith similar Mr,

strength of IMF:H-bonding >

PD-PD attraction> lD-lD attraction

(c) compounds capable . of _forming hydrog.en_ bonding tend to be s.o\ulble with

E.g. NH3, carboxyric acids (RCooH;, atcorrJs (RoH) "ni "riner.--

s5:

Example 1: NH3 molecules form hydrogen bonding with water molecules. C t,43 t:lv,trtr? ir,rvv"-r{- e-H \ r-

H

-tt' Y-,,/^

H

ti.{rtr-i-e..^!c*crl j'"Example 2: Alcohol (RoH) molecules form hydrogen bonding with water molecules.

p

d

{c

- Ht)-{ ^ H\ .t*- ")r.. \\ ô': *{ * n{1rl\ /<^/ /''u

\/H

rI'

(b) So|idiceisl€[5..dense.thanliquidwaterbecausethetweenH2omolecu|esresu|lsinicehaving.anflfFErlg|^+ll'Êachoatominso|idiceiSffiottltuttt*nd!h"''^g r'brrC '+^ +.-^ ^*'-^r ll"t"tt'O ln arlla|16l manner. tv rU\-lv\€ €{e{prrti

1

In water

R\ .+t\vv:/

Ox/f'

H

- JtH/\,C*./ \ ln*,qx -:/n^ \

H

JJC 2008 Page32of 42

Xtsxot,/\HR

J+

=\aui \ur'"1crt r.:\€r_.J;i'. €fiu,. {.1'*L -l!t:.\*p*.h:.,r,,\r,1 ,-t- r'r,r t..'{ h]< icil,11v* 1,'1r_ p1{4r

ffi**6-o6-*

where X, Y F, O or N

Numler of,piriod' Nualrrolpsr-id<|.

(d) M, doubles when cr-fln3s C*

hydrogen.bonding

o

o-nitrophenol

monomer

hydrogen bonding

(

re\S \t[\1 t; .

h,r'J, w{ta'J, r'" gv

J..Jf tL'r,fE\ -i r-i:--"{\6-{,,

i'r:.-,J t --i\ t., i; i .\.j-rfl,:,lf--

#of the dimer will

be €fi: that of

JFt€iffiFtI.

oi*u,i,"tio ;ffiilffi;if*i#*'Xt*"t'",the carboxylic acid molecules willform "

y^'*wfl€* . l^V '*L !!^"*1-- t t't '' I

( *] -1',;'l-1 : [,]'

As rt Y'.)rc lrr;r^''1. :' \'f, tH

ln some compounds, functionalgroups are so oriented that hydrogen bonding can take place between

them**.'|etoformintramo|ecutarhydrogenbonding.Thiswouldhaveanimpactontheir mp and bP. For examPle,

intramolecular\of hydrogen bond

aa -/i \--/ \

.zH6+

V'I

t( )lYo\

oA\

Exercisel2: Explain why the boiling point of water is higher than that of (a) HzS and (b) NHg.

(a) Both comPounds have i.O\Y-trn\r-. . A larger amount of energY is

needed to overcome the jt!*s" between tlz,D ,molecules

than the l^..'€, it-x,- u- between

f-r , i rr"f "*rl"r.

Therefore, b.p. of water is higher than that of HzS. r.''\ i f \ u: ie. '- t$rz' f' r ';-'

Both COmpOundS 6"u" livufl € rxrtt f t ur uV strUCture with .^^''{ ''rfu' tCl'+ t+;l' lit - r:i ..

between their respective molecules. A larger amount of energy is needed to overcome tne

f-r{',r',tr,:.r r"\.Tlr-rl(,v l-:rr',',1 between HzO molecules than that between

NHg mo'lecules as the O-H bond is r,v.i.N.(,, polar than the N-H bond. Therefore, b.p. of water is

higher than that of NHs.

.o/

c,,

\6+ a-r/3\

F

F

H

p-nitrophenol

The -OH and -NOz grouPs are toofar away to form intramolecularhydrogen bonding.

Hence, p-nitrophenol will have a*5ftbo (than o-nitroPhenol) dueto#@ct&:.

After forming intramolecular hydrogenbonding, the H atom is no longer available forintermolecular hydrogen bonding with othero-nitrophenol molecules.

Hence, o-nitrophenol will have(than p-nitrophenol) due to lcer*nic

(b)


a

9. GOVALENT BONDING: INVOLVEMENT OF ORBITALS

9.1 Overlap of Atomic Orbitals

x ffitflEnd is formed due to the sharing of electrons between 2 atoms. Since electrons are housedorbitals, formation of covalent bond involve%

i_.\ r,,l't,'.1

i--

.-

in

x The electron pair in the overlapping orbitals is shared by both atoms i.e. bonding electrons.

x The the

9.2 Tvpes of Govalent Bonds

xThereare,whichareformedbydifferentwaysofover|appingtheorbita|s.i. @formedbyhurlrmd6oforbitals)

Head-on overlap (or head to head overlap) results in the formation of sigma bond (o bond).

E.s. (i) H2: +H,,^[.1\-/

1s orbital>4\1--t v - t z+F

2p-orbital

)

(ii) F2:

t-t

i

,l'i" rr'

H

t>'i l .,; l1^r'

1s orbital Head-on overlap of 1s and 2porbitals

H-H

2p orbital

tl ffi)(formedby ffkp of orbitals)

Sometimes, after a sigma bond is formed between the two atoms, the p orbitals ofthe 2 atoms, that are parallel to each other, also overlap sideways (side-to-side

overlap). This type of overlap results in the formation of a pi bond (n bond).

- ,j trlJJC 2008 Page 34 of 42

$[. '{

'-ie.y\'1[, | /

\iiA: ')""

'-:v'\'{ !-u,v'€ 'i' -r"-' '

]fo- '- i -i

* '- l'i

5ii

fi

o.r,,molecule

The 'ro 3n rrbilale ricrtF l-re' ^^ to.form

-$ond. The twcflpTuf*c|e overlap

sideways to form ar5.r#Hence the tnnx|gs,elms are boundtogether by a iltbtFbuftl (comprise of **-.551lrt anrt Arrbrd\

(i i) N itrogen 1'l s22s22p,1 2prt 2p.' )

Nz:molecule

The two-form a -rb.rd The ftrnFr€|filake

Similarly, the two rffr*te.overlapJialrcys to form another#Hence the two nitrogen atoms are boundtogether by a triple bond (comprise of 1 obondand2nbonds).

lr i-;v*'\

/.

R

The extent of overlap of orbitals in a n bond is t€ JJ than that in a o bond. Thus a n bond 1. u'i'tt -'than a o bond.

1n sll"ngttinla hnsJs, ruf the bonds is artl{r+C and thest are-*.nt

E.g.

Each N-H bond is a o bond

\"fi,--?FVfi" oc=t

C=O bond is made up or ^-t r bsnd and orrrbend

1 s22s

u1S

22pu3s23ps

rurtmm runmilTl

-

N:N N=N triple bond consistof emrorb*d and ttnd:ndo

tfff*r of Atomic Orbitals

ln 414gugi6$ng6g{Jles (e.g Oz, HF, N2, etc), a#end is formed by erndrnffV'€ftn*@|HlHffl'

CantheSameconceptofatomicorbita|overlapbeapp|iedtffiszConsider*elrwhichisa-culewithtwoequalBe_C/bondlengthandstrength.Whichatomicorbitals of Be could be used to overlap with the 3p orbital of C/ in forming the Be-C/ bond?

r(** B€ *-- C-f.

9.3

x

a

Be: 1s22sz

Be:

(gffitf state)

Be*:'(r5r{ette)

u ul-T-t--l1s 2s 2p

M nnr-n1s €.'/ zP

mtl ,\ 1

sp hybrid orbitals ,e+*f*- a.l

CI:3p3s2s 2p

l'.,-:/.1-,1 :,, ..11

I t, .t.

-t

i.( i .,' .,:,liTo form two covalent bonds with Cl, Be needs to have two half-filled valence orbitals. This can be achieved

by the promotion of a 2s electron to an empty 2p orbital. Note that while 2p orbital is directional (pointing in x,

y ot z- axis), 2s orbital is non-directional (spherical). That means one Be-C/ bond will point at a definite

direction (say z-axis) while the other bond points at some undefined direction. This contradicts the linear

shape observed for BeC/z molecule.

Toresolvesuchdiscrepancies,roi9proposed.Thistheorysuggeststhatbeforec5*efiqrfa5i13b occurs,.l#lrof the ddrfrl arecl-to produce a new set of lfrrfilrElf

Note: No. of hybrid orbitals generated = no. of pure atomic orbitals used in hybridisation


x lQpesoffrryenHisat$oa:

(a) €p.@r{d4tmtioD'.trg.and *|lor[ital undergo hybridisation, formingr;yhisit spstl€sto each other.

@io

3F,6,,, *'$,.*#$ $riapq, exanrptr,,..'_!- r\t r-, - 3e- Cl

4oOnes One p Trrs ttr cibital:

(b) .ge&{ry*#t$#*rn: flises*.r6 .f,m*tr{trtiitals undergo hybridisation,nyn**"e**ta*s, which are4*&\o each other.

sro.{ffi--& ASna.*-.. ri*o.p i--flftrF


form ing thr.nqrlfri#efit .sp2

c/I

B

ct/ \cl' n'4fi)1"

(c) qA"tq*!a#i6{a: m*and thprcrbifr*ls undergo hybridisition, forming fxeqd*n6.ep3hytriAerhitals, w h ich . a rsq€€€$ 1e each oth e r.

:-+: tQQ* ,fl: i Y""' .

spr etJCC* M 'r' -$r

toc @ "7\ n;f-H

O.trs t, *ri*d,p:: f*ur,srt',offiais, (meth-ane) i

1'r'r,1,",t-\ n.p-rr-.: ,ntttLlr C(i''L* '''r,f Ltt'i,,ifit t)

x How to determine the state of hybridisation for the central atom?

1. Draw dot-and-cross diagram.

2" Determine the number of electron pairs (bond pairs and lone pairs) surrounding the central atorh.(multip|ebondisregardedasasinglee|ectronpair)

Match the number of electron pairs to the number of hybrid orbitals needed. L

Number of epairs aroundcentralatom

No. of hybridorbitalsneeded

State ofhybridisation forthe centralatom

Arrangement of hybrid orbitals Anglebetween

hybrid

orbitals

Example

2 2 sp Linear 1900 BeCl2

3 3 sp2 TrigonalPlanar 1200 BC/3

4 4 sp3 Tetrahedral 109.50 CH+

5 c sp3d Trigonal Bipyramidal 1200, goo PC/5

6 6 sptd2 Octahedral 900 SFe

----7 EXeIgjSe_I3: Determine the hybrid state of allcarbon atoms in the following molecules"

(a) CO2

/sPo-c-o

(c) CH2=CH2

Hr. ,f ,v- '.f'\^=,( |

y'Y-'\HH

9.4 Hvbridisation in Methane. Ethane. Ethene. Ethvne and Benzene

Note: Only the orbitalsof the central atomundergo hybridisation.

(d) cH3cooH

H-

vtA

'te

Molecule Shape HybridState of G

Hybridisation and Bonding

Methane.CHa

;fcar*e*al .#, Observations / Facts:

. tetrahedral molecule with a bond angle of 109.5o

. four C-H bonds, identical in length and strength

'@. m(s#&dc) 1s

€s: mkrtits[sKq 1s

u2s

n2s

ffi

2p

The four atomic orbitals (one 2s and three 2p) are mixed to give '1.new, identic"l 5$r" hybrid orbitals, each containing one electron.

4 sp3 hybrid orbitals of C atom qffl**..dsn with 1s orb1al of H,

forming#

tetrahedral

H| {€eEl/:H<C\----__\.' H

H

\\$rr"' signifies bonds into the plane of the paper

> sigriifies bonds out of tbe plane of the paper

signifies bonds on the plane of the paper


.-:

Molecule Shape HybridState of C


EtlfdfieCzH.

.lFctmhskak

withrespect toGhGcalem,,*

"d'"Similarly,

Each of the two carbon atoms has 4 sp3 hybrid orbitals.

\_s,*u./,-Y'

1 ep1.&ybrid.,ff,b.ltal of C atom qrm@,,&ead-on with the c*? ayurioorbital cjf anotherGetom, forming *,@.,eM.*The rest of the $*ry4**erffib.@f,,ffi,#l,C atom o*psaadon with

1Sffi,s*r**,,forming a totalof$G** Meb in an ethane molecule.

EffileQsf+

-@TlEnar with

respect toeafr?qelffib

(moleculeis planar,

flat)

.ocp? Observations / Facts:

. trigonal planar molecule with a bond angle of 120o

o four C-H bonds, identical in length and strength, and a C=C bond

C: EM(ground state) 1s

c*r U(excited state) 1s

2s 2p

n3 equivalent spz 2p orbital (unhybridised)hybrid orbitals

Three atomic orbitals (one 2s and two 2p) are mixed to give J new,

identical Q:r hybrid orbitals, each containing one electron.

Thus, each C atom has one unhybridised 2p orbital and three equivalent

sp'hybrid orbitals.

', o,,,,. %,....,rrr\\ H

^/@),

H\ /rH/t'o'l^-l^ \1v-lr, l

H/ ffi,YHf ry?*hyb*iC-'esital of C atom eiruh€E*#ead€n with the'se&'+r$l*orbital of another C atom, forming 3 @@u*brend:"r

The rest of the.s*?*ybrridbditets of each C atom exsls&cadeon with

desb*tsrkef,lJeforming a totalof 44*H,*&efigein an ethene molecule.

The renh$riCe*,"gpr'*€itfta+*tfl€"ttom mdrpc',git*G{ffiy with the.JJllbldlEidi€n4l''-8*:',,sbi&d " of another &*rl€H, forming a @rbend"(electron clouds lying above and below the plane).

n


\


Observations / Facts:

. linear molecule with a bond angle of 180'

r two C-H bonds, identical in length and strength, and a C:C bond

C:

(ground state)

C":

(excited state)

u1s

M1s

utflTl2s 2p

NMTIIq/-J2Pmm

2 equivalent sP two 2p orbitalshybrid orbitalq

(unhybridised)

Two atomic orbitals (one 2s and one 2p) are mixed to give 2' new,

identical &, tFnyOtid orbitals, each containing one electron'

Thus, each c atom has @ (2p' and 2pr) and

2 equivalen! sP hYbrid orbitals.

Planes at righl angle

H-C- c-H€0.'120 nm

. 4|!!*|.Iid orbital of c atom .tlrltprh.-Lon with theaphtbddnrlilelof another€atern, forming a€.Cclerd-

o The remaining ee*bliarc$ital ofrssglF€ atom cnrlaFr&erton with

{S,ltlitrtcfH, formingl€ullalggg$ in an ethyne molecule'

. SilrrliteLntrD atom dtEfElEriir!f;lv with ?trt?urbil* of..anotheF9

. atom. The same appties to ne pair otinq9dda..dsf,r.rbitals, giving

tvvo&C*&9ldgwhich are pltpailludB{ to each other'

@n

c-c

Etryp$#z


Molecule Shape HybridState of C


BenzeneCeHe

Symbolfor

benzene:

1-\\z/

trigonalplanar withrespect towch Catom

(moleculeis planar,

flat)

sp2H

Each C atom uses two of its sp2 hybrid orbitals to overlap head-on withthe spz hybrid orbital of 2 adjacent C atoms, forming a total of 6 C-C obonds

Each C atom uses its remaining sp? hybrid orbita,l to overlap hea$-gnwith 1s orbital of H, forming a total of 6 C-H o bonds.

d H

Each C atom has an unhybridised 2p orbital. These 6 orbitals(perpendicular to the planar arrangement of C atoms) overlap side-wavtogether, forming electron clouds above and below the plane of the

ring. These electron clouds contain a total of six delocalised n

electrons.

1O RECYCLING

10.1 Whv the Need for Recvclinq?

" . Metals are produced commercially from ores.. Common ores are oxides, sulphides, halides, silicates, carbonates & sulphates.. These are finite and non-renewable.r Difficult to predict and estimate how long the reserves will last.r Efforts must be made to conserve resources via successful recycling which is cost-effective.

10.2 lmportance of Recyclins

. Saye money.The high cost of extracting aluminium by electrolysis can make recycling worthwhile.

. Saye energy and fuels: Large amount of electricity is needed to extract aluminium and coke to extractiron.

. Saye environment Dig up less metal ore, mine less coaland drill less oil.

. Save the reserues for future.

. Solye the problem of waste disposa/: Recycling metal or plastic articles prevents them from becominglitter.

t However, recycling may not be always worthwhile because of the cost of sorting, collecting and

processing th6 waite material. The higher the value of the material, the more economical it is to recycle.

Thus, almost all gold is recycled, but only 40% of aluminium is recycled.

* Recycling plastics presents a particular problem because of the difficulty in identifying the type of polymer

used. lt ii easy to separate iron from copper in recycling but PVC and polyethene are much more difficult

to distinguish bnd separate. One way to get round this is to indicate the type of polymer by means of aspecialsymbol.

JJC 2008 Page 4O ol 42

Metals lonic compounds Govalent Compounds

Type of structure Giant metallic Giant ionic Giant covalent

(or giant molecular)

Simple covalent(or simple molecular)

Bonding Strongelectrostaticattraction betweenmetal cations andthe surroundingmobile electrons

Strong electrostaticattraction betweencations and anions

Strong covalentbonds betweenatoms

Strong covalent bondshold atoms togetherwithin the separatemolecules.

Separate moleculesare held by weakintermolecular forces.

Properties(i) Melting point,

boiling point

(ii) Hardness

(iii) ElectricalConductivity

(iv) Solubility

High mp and bp

Hard, yetmalleable

Conduct in thesolid or liquid(molten) state

lnsoluble in polarand non-polar"solvents, butsoluble in liquidmetals (eg.forming alloys)

High mp and bp

Hard and brittle

Does not conductin the solid state;

Conduct in theliquid or aqueousstate.

Soluble in polarsolvents (eg. HzO);

lnsoluble innon-polar solvents(es. CCL)

High mp and bp

Hard(except graphite)

Does not conduct inall states

Graphite is theexception

lnsoluble in allsolvents

Low mp and bp

Soft

Does not conduct inall states

Less solublefinsolublein polar solvents;

(More) soluble innon-polar solvents

Comparing thefour crasses of substances

2. Solubility in Water

A compound with simple covalent structure is expected to dissolve in water if

(a) it is capable of forming hydrogen bonding with water molecules; or/and

(b) it is able to form ions which can be solvated by water molecules through the formation

of ion-diPole interaction.

An ionic compound will dissolve in water if the hydration energy released from the iondipoleinteraction is zumcient,to overcome the strong ionic bonds in the ionic compound"

Exercise 14: Explain the following observations in terms of the structure and bonding of the substances'

(a) SiOz is a solid whereas CO2 is a gas at room temperature.

iOl Hyd?ogen chloride has a higher b;iling point than fluorine even though their M are similar.

(a) SiO2 has g ,,i('iir,\ "-i {':'*---o.'t*itructure. COz has a -iivr,'l'\r I \ilr- / 'lqq-:ctrUCture.

Larger amount of energy is needed to overcoms tlg f ii'l':, I " ''i:i'- L' t ' i betweenr..1 .d.\, i L^!.,^^^

in SiOz than the -'"q"r'i 'r' 4ir.'t '-:;i{:'f\Ii "rtl'''') betweent^\

and exist as a gas at room temperature.

Both HC/ and F2 have -f ii''"br'': i'*;{ i{'''"'-i r'v structure. HC/ is [- but Fz isI

\ ,l { 1. r. , . Larger amount of energy is needed to, overcpme the

_ --fTlt , . . . than the ,l\. ' "1! lt '':"-

. Therefore, HC/ has a higher b.p. than F2 even though their M. are similar'

(b)

JJC 2008 Page 41 ol 42

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JJC 2008 H2 Chem - Chemical Bonding