1

CHEMICAL BOND

Why do atoms combine to form a molecule? Why does H atom combine with Cl atom toform HCl molecule? Why does Na atom combine with Cl atom to form NaCl? What is thenature of their linkage? All these questions will be answered in this chapter. When two atomsare tightly bound with each other and establish a contact, we say that a chemical bond hasbeen formed between them. Before this close contact, they were free and independent butafter the bonding between them, they are no longer independent and free. They are heldtogether by strong forces of attraction and it is not easy now to separate them from eachother. This is called a chemical bond.

Driving Force for Bond Formation:Every element has a tendency to acquire the stable electronic configuration of an inertgas(s2p6 called the octet or s2 called the duplet of He). This is because the electronicconfiguration of the inert gas atom is very stable. All other atoms want to make themselvesstable by attaining the nearest inert gas configuration. They do so in two different ways.

(i)By way of give and take of electrons between each other: This leads to theformation of ionic bond.

(ii)By way of mutual sharing of electrons: This leads to the formation of covalentbond.We shall discuss these two kinds of bonds in detail now.

IONIC OR ELECTROVALENT BOND:

The bond between a strong metal and a strong nonmetal is ionic or electrovalent in nature.In this case a give-and-take of electron takes place between them. The metal atom givesthe electron while the nonmetal atom takes. We know from periodic classification chapterthat metals have low ionisation energy and have strong tendency to lose electrons whilenonmetals have high electronegativity and have strong tendency to gain electrons. Thus whena metal atom sees a nonmetal atom, their mutual desire get chance to be satisfied. The metalatom gives its loosely bound valence electron and the nonmetal atom takes that electron toits valence shell and are thus converted to positive and negative ions respectively. Theoppositely charged ions are now held together tightly by electrostatic forces of attraction. Youknow from law of electrostatics that unlike charges (+ve and -ve) attract each other whilelike charges (+ve and +ve or -ve and -ve) repel each other. Due the strong attractionbetween the +ve and -ve ions formed during bond formation, in this case, they are linked orbonded with each other very firmly. This is called the ionic or electrovalent bond. Notethat by losing or gaining electron, each atom (now ion) acquires the stable inert gasconfigurations.

Group IA(alkali metals) and IIA(alkaline earth metals) possessing comparatively low ionisationenergy (IE) values are strong metals while group VIA(oxygen family) and VIIA(halogens)have high electron affinity(EA) and high electronegativity values and are strong non metals.Thus whenever any of the elements from strong metals(group IA or IIA) unites with any ofthe elements of strong non-metals(group VIA or VIIA) the bond between them becomesionic.

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LEWIS STRUCTURES OF IONIC BOND:

Ionic bond formation in NaClElectronic Configuration of Na(11): 1s2, 2s2, 2p6, 3s1

There is one electron (unpaired) in the valence shell of Na. This unpaired valence electronwhich is loosely bound will be donated by sodium completely to Cl. By that Na will beconverted to Na+. This is because Na atom has 11 protons(+11) and after losing one electronit will have 10 electrons(-10). So the net charge(+11-10) is +1. Let us see now what happensto Cl.

Electronic Configuration of Cl(17): 1s2, 2s2, 2p6, 3s2, 3p5.In its valence shell(3rd), there are three lone pairs and one unpaired electrons. This is knownafter making the box diagram of its 3s and 3p subshells as explained in the chapter, atomicstructure.

52 3p3s

For writing the Lewis structures for the ionic bond, you have to first find the number of lonepairs and unpaired electrons of each atom participating in the bond formation. The lone pairsare shown by two closely placed dots or symbols of a particular type while for unpairedelectron a single dot(or any other symbol) has to be placed separately, a little away fromthe lone pairs. Note that for better clarity, the symbols given for electrons should be differentfor different atoms. In the example below, the electron of Na is given the star(*) sign whilethat of Cl given dot sign. Then electron is transferred from the metal atom to nonmetal atomas shown below.

(ionic bond)[Ar][Ne]

1s22s22p63s61s22s22p61s22s22p63s51s22s22p63s1

Cl-

Na+

+

-Cl

+Na+ ClNa

The unpaired electron of Na atom is transferred to Cl atom, thereby each attains the neareststable inert gas configurations. Na after losing one electron becomes Na+ ion(Ne configuration)while Cl after gaining one electron becomes Cl- ion(argon configuration). The valence shellof Na(3s) has been removed now in Na+ while there are four lone pairs in Cl- ion. Thesetwo oppositely charged ions attract each other strongly and gives rise to ionic bond.

Ionic bond formation in CaOCa(20): 1s2, 2s2, 2p6, 3s2, 3p6, 4s2 .Hence Ca has one lone pair in the 4s orbital.O(8): 1s2, 2s2, 2p4. O has two lone pairs and two unpaired electrons in its valence

shell(2nd shell). The box diagram of the valence shell is shown below.

42 2p2s

2+

+1s22s22p63s2 1s22s22p63s4

Ca O

3p64s2

Ca + O2-

3

Calcium loses two electrons and these two electrons are gained by oxygen atom. The twounpaired electrons of O now get paired. O atom is thus converted to O2- ion by acceptingthe two additional electrons. Thus the two ion acquire stable inert gas configuration. Ca2+ hasAr(18) configuration while O2- has Ne(10)configuration. The two oppositely charged ions arebound together tightly by strong attractive forces. This is ionic bond.

SAQ 1:(i) Draw Lewis structures of the ionic bond formation in case of the following.

(a)KF (b)Na2O (c)MgBr

2

(ii) What kind of bond is formed between the following pairs and why?(i)K and O (ii)Mg and Cl

2(iii)Rb and S (iv)Cs and Br

(iii) What inert gas configurations do the following haveK+, S2-, Mg2+, F-, Rb+, Ca2+

Lowering of Energy during Bond Formation:Whenever a bond is formed between two atoms, may it be ionic or covalent bond, thereis a net decrease of energy of the system. This net decrease is called the bond energy.Because of the decrease of energy, the system becomes more stable and the bond betweenthe two atoms does not break of its own, unless and until the same amount of energy issupplied to the bonded atoms. During ionic bond formation, the main factor responsible forlowering the energy is called lattice energy.LATTICE ENERGY:The metal and nonmetal atoms first have to exist in the gaseous state before the exchangeof electron from metal to nonmetal can take place. Electron is transferred from the metal(say Na) to nonmetal (say Cl) to form gaseous Na+ and gaseous Cl - as per the followingequations.

Na(g) + Ionisation Energy ----------> Na+(g) + eCl(g) + e -----------------------> Cl - (g) + Electron Affinity

Note that metal atom(Na) will lose electron by absorbing some energy called the ionisationenergy(IE) while the nonmetal atom(Cl) will gain the electron lost by Na and will evolvesome amount of energy, called the electron affinity(refer the chapter, periodic classificationfor details). Now the turn comes for the formation of the ionic bond. Due to strong attractiveforces between the positive ions and negative ions, they come very close to each other andconsequently they no longer would remain in the gaseous state. They change their state fromgaseous to solid state. During this process of gas to solid conversion which takes place dueto strong attractive forces between the oppositely charged ions, a huge quantity of energyis released. This amount of energy released is called lattice energy. The major loss ofenergy during ionic bond formation is due to this lattice energy. In other words more thelattice energy, stronger is the ionic bond. The strength of the ionic bond is directly proportionalto the lattice energy.Lattice Energy: The amount of energy released when one mole of gaseous positive ionsreact with one mole of gaseous negative ions to form one mole of ionic solid is called latticeenergy.

Na+(g) + Cl -(g) ----------> [Na+][Cl-] (solid) + Lattice energy

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Note that during the formation of ionic bond, we came across with three steps.

(i) Formation of positive ion from metal: this involves absorption of energy(ionisationenergy)

(ii) Formation of negative ion from nonmetal: this involves evolution of energy(electronaffinity).

(iii) Formation of ionic solid from gaseous ions: this involves the evolution of hugeamount of energy(lattice energy).

So in one step(i), energy is used up and in steps(ii) and (iii) energy is released. The releaseof energy is more than the energy used up and hence some net quantity of energy is finallyreleased. This is the called the bond energy which contributes to the strength and stabilityof the ionic bond.It is to be remembered that if the lattice energy of an ionic compound is more, the ionic bondis stronger and vice versa. Lattice energy depends on the following two factors.(i) Size of the ion:If the size of the one or both the ions becomes greater, the ions go farther away from eachother and so the attraction between them decreases. This consequently reduces the latticeenergy. Take the case of NaCl and CsCl. Cl - is commonly present in both. But the cationsare different. Na+ ion has a much smaller size than Cs+. Therefore the force of attractionbetween Na+ and Cl- is much greater than between Cs+ and Cl-. So the lattice energy isgreater in NaCl. That is the reason why the ionic bond in NaCl is stronger than CsCl.Therefore again why NaCl melts at a very high temperature of nearly 8000C while CsClmelts at a low temperature of 1000C.(ii) Charge of the ions:If the magnitudes of charge in one or both the ions are more, then the forces of attractionbecome more and hence the lattice energy is more. Compare NaCl and MgO. The Mg2+ andO2- have greater charge than Na+ and Cl -.Therefore MgO has a much greater lattice energyand bond strength than NaCl. This is supported by the fact that while NaCl melts nearly at8000C, MgO melts at nearly 28000C.

SAQ 2: Na metal has an ionisation energy of 496 kjoule/mole. The electron affinity of Clis 348 kjoule/mole. The lattice energy for NaCl solid is 788 kjoule/mole. Calculate the netdecrease of energy during the formation of the ionic bond from one mole of Na(g) atomsand one mole Cl(g) atoms.

IONIC COMPOUNDS AND THEIR PROPERTIESThe compound in which the constituent units(ions) are held together by ionic bond is calledionic compound. These compounds are very typical and are different from other compounds.Let us know some of their important characteristics.1. State:All ionic compounds exist in solid states at room temperature. This is because the ionic bondis very strong and the ions stay very close to each other and so ionic compound remains assolid.2. Melting and Boiling Points:Ionic compounds melt and boil at very high temperatures. For example, NaCl which is anionic solid melts at about 8000C. Since ionic bond is very strong, a great amount of energy

5

is required to loosen the bonds between the oppositely charged ions, so that enough spaceis created for the free movement of ions and change into liquid state. Consequently themelting points of ionic solids are very high. For boiling the liquid (molten)ionic compound togaseous state we need still more energy. For this the ionic bonds have to be completelybroken which therefore needs much larger amount of energy. Thus the boiling point isenormously high.3. Solubility:Ionic compounds dissolve in water and have high degree of solubility in water. In fact,whether an inorganic compound is ionic or not, can be best known when you put a smallamount of the compound in water and shake. If it dissolves it is ionic, if not it is covalent.Take for instance, NaCl dissolves in water and hence it is ionic, while AgCl does not dissolvein water, it is not ionic. It is a covalent compound.To understand why water dissolves ionic compounds we have two explanations. Oneexplanation we take up now and the other, we shall take up a little later. Water is a stronglypolar compound (just wait for some time to know what is called polar compound) and hashigh dielectric constant. Dielectric constant is the property of the medium which controls theforce of attraction between two ions present inside the medium. The force of attractionbetween the oppositely charged ions is inversely proportional to the dielectric constant. This,you shall know more clearly in Physics in a law called Coulomb's law of forces in higherclasses. But this much you must know, at this stage, that more the dielectric constant, lessis the force of attraction between the oppositely charged ions. Since water has very highdielectric constant, the force of attraction between positive ion (say Na+) and negative ion(say Cl -) of the ionic compound put in water is largely reduced. As a result ions getseparated from each other and start moving freely and independent of each other in thesolvent (water). The ions mix uniformly throughout the solvent and forms a homogenoussolution. The other explanation shall be given after a few pages in this chapter.The ionic compounds are not soluble in non-polar organic solvents such as carbon tetrachloride,benzene etc. What is a non-polar solvent and why the ionic compounds are not soluble inthem, we also shall know all these a little later.3. Electrical Conductivity:Ionic compounds conduct electricity in the liquid (molten or fused)state or in aqueoussolution. They do not conduct electricity in the solid state.Let us know what these statements mean. If we make an electrical circuit in which a solidNaCl crystal (common salt crystal) is placed at some place in the path of current in thecircuit, will the current pass? Will an electric bulb placed in the circuit glow? The answeris NO. Solid NaCl is a poor conductor of electricity and electric current cannot pass throughsolid NaCl crystal. However, in stead of solid NaCl crystal if we take NaCl solution in wateror liquid NaCl (obtained by melting solid NaCl at about 8000C), then the current startsflowing through the circuit and the bulb glows. Thus we found that ionic compounds conductelectricity only in the liquid state and in aqueous solution.

Explanation:First of all let us know how electricity is conducted in a conductor. There are two ways forthe conduction of electricity. For metallic conductors such as Cu, Fe, Al etc. the electriccurrent is carried by free electrons. But for aqueous solution of an ionic compound which

6

contains free ions(both +ve and -ve), the current is carried inside the solution by these freeions. This is called an electrolytic conductor. For example, in NaCl solution, the currentinside the solution is carried by the ions. Positive ions move towards the cathode(-ve electrode)and negative ions move towards the anode(+ve electrode) and thus electricity is conductedthrough the solution. Same thing will happen if we take a liquid NaCl in stead of NaClsolution. In electrolytic conductor, the current is carried by the free ions.

One thing we must remember that in ionic compound, the positive and negative ions existeven in the solid state. It is wrong to write the formula of sodium chloride as NaCl. Thecorrect formula of sodium chloride should be Na+Cl -, because in all the states namely solid,liquid and in solution, sodium chloride contains Na+ and Cl- ions. We should never think thatsodium chloride undergoes ionisation when dissolved in water. Ions are present even beforedissolution i.e in the solid state. The ions in the solid state are however very tightly packedforming a crystal and so they are unable to move from their positions. They are rigid at theirpositions and cannot move from one place to another and thus cannot carry current. Howeverwhen the ionic solid is put in water, the ions get separated (for the reason already discussed)and freely moving ions are formed which carry the current. In liquid state also, the ions arefree (although not as free as in solution) and conduct electricity. In solid state the ions arenot free as they are tightly packed and do not conduct electricity.4. Crystal Structure:Ionic compounds are crystalline in nature. Have you seen a big crystal of common salt(NaCl)or any other crystalline salt such as copper sulphate crystal? Crystals have fixed geometricalshape. Amorphous substances are powder like(like talc-cum powder) in nature. If one seesto the internal structure of an ionic crystal one finds that hundreds and thousands of positiveand negative ions are arranged in a regular, orderly and systematic manner in the same shapeas the big crystal that we observe. If you break down a crystal to smaller and smaller unitsyou will arrive at the smallest unit which is called the unit cell of the crystal. This is the motifor the simplest pattern that is repeated again and again in all possible directions to get thebig crystal. This infinite arrangement of ions in the crystal in the systematic way is calleda crystal lattice. During the formation of this solid crystal lattice from the gaseous ions dueto strong forces of attraction, a large amount of energy is released which is called latticeenergy. A simple unit cell of a NaCl lattice is shown below. The big shaded balls stand forthe Cl - ions while the small black balls stand for the Na+ ions. We find that Cl - ions aresitting at the corners of an imaginary cube and also at the centres of each face. Since thereare six faces of a cube, we find in the unit cell, six Cl - ions are placed at the centres ofeach face. In between two Cl - ions, we find one Na+ ion. In other words Na+ ions are placedat the centre of each edge. There are 12 edges and hence we find 12 black balls(Na+) ateach edge centre. Besides these, there is one Na+ ion at the centre of the cube(called thebody centre) in between the Cl- ions present at any two opposite face centres. You imaginethat this unit cell(cube) that has been shown in the diagram below is repeated in all sides suchthat between any two successive Cl- ions you will find one Na+ ion and in between any twosuccessive Na+ ions you get one Cl- ion, we get an infinite network of ions arranged in a veryorderly manner which ultimately results in the big crystal. Hence the shape of the unit cellis same as the shape of the big crystal. Each Cl - ion (shaded ball) and each Na+ ion(blackball) shares more than one unit cell and this infinite network of ions is called the crystal

7

lattice. To put in more simple way, Na+ and Cl - ions are arranged in an alternate mannerin the cubical pattern as shown in the diagram below. Although in the diagram we find a lotof empty space in between the ions, in reality the ions are very closely packed and there isvery little space inside them. Due to close packing of ions the ionic compounds are very rigidand have high melting points.So ionic compound forms crystalline structure which is one of its special properties. Individuallyan ionic bond which represents the attractive forces between the positive and negative ionhas no specific direction and we say the ionic bond is non-directional. But collectively theions give a fixed geometry (shape) to the ionic crystal i.e the ionic solid. This is due torepeated arrangement of positive and negative ions in a regular and orderly manner followinga particular geometry while repeating. There are several types of crystal structures fordifferent ionic compounds such as cubic, hexagonal, rhombic, monoclinic, etc. But we shallnot discuss about them at this stage. Further details on crystal structures will not be includedin this book.

(One unit cell of NaCl crystal lattice)

-

+Na

Cl

SAQ 3:(i) Between NaCl and CsCl, which has a lower melting point and why?(ii) Between CaO and MgO, which has a stronger ionic bond and why?(iii) NaCl solid does not contain Na+ and Cl- ions but when it is dissolved in water

produces ions. Is it true? If not what is correct?(iv) Molten MgCl

2 conducts electricity because it has free electrons in it. Is it true? If

not why?(v) NaCl is soluble in benzene. State true or false.(vi) In the crystal lattice of NaCl, each Na+ ion is surrounded by how many Cl- ions and

each Cl- ion is surrounded by how many Na+ ions?(vii) Copper conducts electricity because of moving ions. Correct it if it is wrong.(viii) Why do the closed packed lattice of an ionic compound break when put in water

and dissolution occurs?(ix) Explain why Be being a group IIA element does not form ionic compounds. BeCl

2

is not ionic, it is covalent. Why?

8

COVALENT BOND

We learnt that the bond between a strong metal(like Na, K, Ca etc) and a strong non-metal(like F, Cl, O etc) is ionic in nature. In such compounds, the positive and negative ionsare arranged alternately and are packed closely to form the ionic solid. Now the questionarises, what kind of bond exists between atoms which are both nonmetals say between Hand H in H

2 molecule, between Cl and Cl in Cl

2 molecule, O and H in H

2O molecule, N and

H in NH3 molecule and so on. When both the atoms are nonmetals, no atom has the

tendency to lose electron. Rather both have the tendency to gain electrons. In that situation,the bond takes place by mutual sharing. Elements appearing in the right portion of theperiodic table which are mostly nonmetals and have high electronegativity form covalentbonds between themselves. Each atom contributes one electron to form a bond pair. In abond pair, if one electron spins clockwise, the other spins anticlockwise. The bondpair sits in between the two atoms. In ionic bond, this pair is present completely in thenonmetal atom making it a negative ion. But in covalent bond the electron pair called the bondpair is situated in between the two atoms and is shared by both the atoms. The two electronsare counted for the electronic configuration of both the atoms. Thus a bond is establishedbetween the two atoms on the basis of sharing. This is called a covalent bond.

H2 molecule:

covalent bondbond pair

HH+ HHHH or

Each H atom contributes one electron (the only electron it possesses) and the two electronsget paired and form the bond pair. This bond pair sits tightly in between the two H atomsand binds the two H atoms to form one H

2 molecule. This is a covalent bond. We represent

a covalent bond by a dash symbol in between two atoms.This bond pair is counted forthe electronic configuration of both the atoms. Thus after forming covalent bond theconfiguration of each H atom now becomes 1s2 i.e stable He configuration (duplet). Thuswe find that by way of covalent bond formation also, the atoms attain the stable configurationof the inert gas.Lewis Electron dot and Dash Structures:We can also write the formation of the covalent bond by showing the electrons as dots instead of the arrow as shown above. This is called Lewis electron-dot structure. When abond pair is replaced by a dash, it is called dash structure. The dot and dash structures inH

2 are as follows.

H H H H+ H Helectron dot structure dash structure

One thing we must bear in mind that for the formation of a covalent bond, each atom shouldpossess at least one unpaired electron. Each atom contributes this unpaired electron to formthe bond pair which is shared by both. If an atom has no unpaired electron, there are twopossibilities.

(i)It cannot form a bond.(ii)It breaks down a lone pair and gets two unpaired electrons and then forms twocovalent bonds.

9

So to conclude again we emphasize here that the atom should have unpaired electron forestablishing covalent bond with other atoms (either with identical atom or different atom).The number of unpaired electrons is called the valency or more correctly the covalencyof the element.Before we take up more examples on the Lewis structures of covalent molecules, first letus know how to find the number of lone pairs and unpaired electrons for common atoms.Do you remember what are lone pairs? The pairs of electrons present in the valenceshell(last shell) of an atom are called the lone pairs. So we have to write the electronicconfiguration and then box diagram of the valence shell and arrange electrons in themfollowing the Hund's rule.

H(1): 1s1:11s

Hydrogen has one unpaired electron.

Be(4): 1s2, 2s2:2s2 2p0

(ground state) one lone pair

Be has one lone pair and no unpaired electron in its ground state. Ground state means themost stable state in which the atom usually remains before bonding. Since Be has nounpaired electron, we may expect that it would not form any bond with any atom. But weknow that Be has a valency of 2 and therefore it has to use two unpaired electrons. Howcan we account for this? So the lone pair has to be broken. One of the electrons of 2s orbitaljumps to one vacant p orbital so as to form two unpaired electrons. This is called the excitedstate of Be atom. This is the higher energy state which the atom attains before bondformation.

11 2p2s(excited state)

So Be has now two unpaired electrons for bonding.

B(5): 1s2, 2s2, 2p1:

12p22s(ground state)

B has one lone pair and one unpaired electron. But B has a valency of 3. How to accountfor it? One elctron has to jump from the 2s orbital to a vacant p orbital, thus producing threeunpaired electrons.

1 2p22s(excited state)

So B has now three unpaired electrons for bonding.

C(6): 1s2, 2s2, 2p2:

10

2 2p22s(ground state)

C has one lone pair and two unpaired electrons in the ground state. But the valency of carbonis 4. So one electron from 2s orbital jumps to the vacant 2p orbital and thus produces fourunpaired electrons.

31 2p2s(excited state)

Thus C has four unpaired electrons for bonding.

N(7): 1s2, 2s2, 2p3

32 2p2sN has one lone pair and three unpaired electrons. Since valency of N is three, so the threeunpaired electrons solve the purpose. There is no necessity of breaking the lone pair as wewere doing before. In fact it is not possible in case of N. Why it is so, we shall know it later.

O(8): 1s2, 2s2, 2p4

42 2p2sThus O has two lone pairs and two unpaired electrons. The two unpaired electrons satisfythe valency of oxygen(2). Moreover breaking of lone pairs to get more unpaired electronsis not possible like N.

F(9): 1s2, 2s2, 2p5

52 2p2sF has thus three lone pairs and one unpaired electrons. Valency of F(1) is satisfied by oneunpaired electron. Breaking down of lone pairs is not possible like N and O.So let us make a summary table showing the number of lone pairs and unpaired electronsin the common elements which are used for bonding. The element belonging to the samegroup have same number valence electrons. Therefore they also have same number of lonepairs and unpaired electrons. For example, Al belongs to boron family and it has one lonepair and one unpaired electron in the ground state and three unpaired electrons in the excitedstate like B for bonding. Similarly Si has same configuration as C(group IVA), P same asN(group VA), S same as O(group VIA), Cl same as F(group VIIA)

TABLE TO BE USED ONLY FOR BONDINGName of the Number of Number ofElement lone pairs unpaired elctrons

Be 0 2B, Al 0 3

11

C, Si 0 4N, P 1 3O, S 2 2F, Cl, Br, I 3 1

Now let us return to Lewis structures. For that let us know the rules.

RULES:(for simple diatomic molecules)1. First write the symbols of the atoms which will form the covalent bond side by side.2. Place the lone pairs and unpaired electrons around the atom. The electrons areshown by dots or any other similar signs. The dots representing the lone pairs are closelyplaced whereas the dots for the unpaired electrons are kept separately and a little away fromthe lone pairs. Look to the example for forming Cl

2 molecule from two Cl atoms.

ClCl

You can clearly see the difference between lone pairs and unpaired electrons.3. Bring the atoms close to each other so that the two unpaired electrons of the twoatoms come close and form the bond pair. Give a oval shaped mark over the bond pair toshow its difference from the lone pairs. Thus we get the Lewis dot structure of Cl

2.

Lewis Electron dot structure:

ClCl

The bond pairs may be replaced by dash mark to get the Lewis dash structure.Lewis Dash Structure:

Cl Cl

Note that in both dot and dash structures, the lone pairs on each atom have to be shownas before. The readers here are made cautious not to forget placing the lone pairs in theLewis dot or dash structures. The bond pair is counted for the electronic configuration ofboth the atoms. Each Cl atom now acquires the stable electronic configuration of the inertgas(Ar=18).

TYPES OF COVALENT BOND ACCORDING TO THE NUMBER OF BONDPAIRS SHARED:There are three types of covalent bonds according to number of bond pairs shared betweentwo atoms.(i) Single Bond: If each atom contributes one unpaired electron to produce only onebond pair, a single bond results.

Example: H2 : dash structureelectron dot structure

HHHH

Cl2: Cl ClClCl or

HCl: ClCl orXH H

In H2 and Cl

2, only one bond pair is formed between the atoms, hence these are single bonds.

In HCl, each of H and Cl contributes one electron and one bond pair is formed between the

12

atoms and hence a single bond results. The electron of H has been shown by a differentsymbol(cross in stead of dot)merely to show that bond pair contains one electron from H andone electron from Cl. Although you can give a dot symbol for the electron of each atom, itlooks more convincing if you make different symbols for the electrons of atoms of differentelements. For same element(Cl

2, H

2) however you may show the electrons by the same

symbol(dot), but for dissimilar atoms, it is better to show the electrons by different symbols.But remember that this is not a binding rule.(ii) Double Bond: If two electron pairs are shared between two atoms by the contributionof two unpaired electron from each atom then we get a double bond.

XX X X orY YEach atom contributes two unpaired electrons. One unpaired electron of one atom(say X)gets paired with the one unpaired electron of the other atom(say Y) while the other unpairedelectron of first atom(X) gets paired with the other unpaired electron the second atom(Y)to produce two bond pairs. These two bond pairs are jointly shared by the two atoms. Thisis called a double bond. We shall see several examples of double bonds later.(iii) Triple Bond: If three electron pairs are shared between two atoms by thecontribution of three unpaired electrons from each atom, then a triple bond is formed. Takethe case of N

2 molecule.

Nitrogen atom has three unpaired electrons and one lone pair(see the table given before).So three unpaired electron of one N atom pair with three unpaired electrons of the other Natom to form three bond pairs. Thus a triple bond is formed.

XX XX

NNX NN orXX

Note that the electrons of the two N atoms have been shown in different symbols to clearlyindicate that each N atom contributes three unpaired electrons. You can also show theelectrons by the same symbol(dot), when you practise, as there is no strict rule for givingthe symbols for the electrons of atoms. Also see that each N atom carries a lone pair in theLewis structure.

LEWIS STRUCTURES OF SOME MORE COVALENT COMPOUNDS:RULES:When the molecule contains more than two atoms, say for example CH

4, H

2O, NH

3 etc. then

you have to first find out the central atom among them. For simple molecules like the aboveyou can make a correct guess. In CH

4, C is the central atom to which all the four H atoms

are attached. Similarly in H2O, O and in NH

3, N are the central atoms. But let us learn some

rules how to know the central atom.(i) H cannot be the central atom. The least electronegative (least nonmetallic) amongthe rest is to be selected as the central atom.In case of CH

4, H

2O and NH

3, there is no other choice than to make C, O and N as central

atoms respectively as H atom cannot be the central atom. But for molecule like HNO3,

which is the central atom? H cannot be the central atom, then between N and O which isless electronegative? Please refer the electronegativity table in the periodic classificationchapter for comparing the electronegativity values. Since N is less electronegative, it is takenas the central atom.(ii) Write the central atom first and place all other atoms in a symmetrical mannersurrounding the central atom. See the case of CH

4, H

2O and NH

3.

13

HH

H

NHH OH

H

H

H

C

(iii) Next, the unpaired electrons and lone pairs for each atom are placed in a verysymmetrical manner. Care should be taken to place the unpaired electron of one atomopposite to the unpaired electron of the other atom with which it will form covalent bond.Refer table given before to know the number of unpaired electrons and lone pairs availablein each of the common atoms for bonding. Then highlight the bond pairs formed by thepairing of unpaired electrons by giving a oval shaped mark for each bond. The lone pairs arekept as such without being used up. Thus the Lewis electron dot structure is complete. Youcan also draw the Lewis dash structures by replacing the dots of the bond pairs by dashes.Single dash(single bond) is used for a single bond pair, double dash(double bond) for the twobond pairs and triple dash(triple bond) is used for three bond pairs between the atoms. Forsometime you practise both dot and dash structures. After becoming familiar with them, youmay stop writing the dot structures and write only the dash structures which look more clean.

CH4: C has four unpaired electrons and no lone pair and H has one unpaired

electron.

(dash structure)(dot structure)

C

H

H

H

HX

X

X H

H

H

H

C

X

H2O: O has two lone pairs and two unpaired electrons. It is the unpaired electrons

which form the covalent bonds. Two unpaired electrons form two single bonds with two Hatoms as shown below.

or OH HX X HH O

You find that the two lone pairs of O atom have been shown in both the dot and dashstructures.

NH3: N atom has one lone pair and three unpaired electrons. These three unpaired

electrons form three single bonds with three H atoms as shown below.

orN

H

H

HXX

X

H

H

HN

CO2: C has four unpaired electrons and no lone pair in the excited state while each

O has two lone pairs and two unpaired electrons. The central atom is C as it is lesselectronegative than O. The two O atoms are to be connected with C on either side of it.The two unpaired electrons of each O has to match with two unpaired electrons of C bymeans of a double bond on either side of C.

CO OXX

XX

XX

X X

XX

X XXX

XX

XX

XX

OO Cor

14

C2H

4(Ethylene): C is the central atom and there are two carbon atoms which are to be

placed first. Then four H atoms are to be symmetrically distributed about the two carbonatoms. Obviously two H atoms are to be attached to each carbon atom. Each C has fourunpaired electrons while each H has one unpaired electron. Two of the unpaired electronsof each carbon will match with the unpaired electrons of two H atoms to form two C-Hsingle bonds on either side. Thus each carbon atom is left with two unpaired electrons. Theywill hand shake with each other with a double bond.

orH

H

H

HC CC

X X

XXC

H

H

H

H

C2H

2(Acetylene): In this case the two C atoms will have to use three unpaired

electrons each to form a triple bond between them.

HC XX CH H CC H

(iv) Octet Rule:You already know that by sharing electrons and by forming covalent bonds each atom attainsthe stable electronic configuration of the nearest inert gas. Carbon(6) in the above example,attains the configuration of Ne(10) and H attains the configuration of He(2). If you considerthe valence shell only(which is drawn in the Lewis structure while the inner electrons arenot shown), each atom except H will attain the stable octet(s2p6) configuration of an inertgas. H will attain the stable duplet(s2) configuration. So if you count the dots enjoyed by eachatom(other than H), you should find the total electrons inclusive of the lone pairs will be equalto eight in the Lewis structures. H atom will enjoy two electrons. In the above example, eachcarbon is enjoying 8 electrons(six from the triple bond and two from the one C-H singlebond).H is enjoying two electrons. In all the previous examples that we have seen so far e.gCH

4, H

2O, NH

3, CO

2, C

2H

4, all the atoms other than H namely C, O, N enjoy eight

electrons(octet) each. This is called the octet rule. During bond formation, the central atomand other atoms excepting H will have eight electrons in the valence shell(Lewis structure).As a matter of first choice, the central or any other atom(except H) cannot violet the octetrule. However if it is impossible for the central atom to attain octet configuration, it will violetthe octet rule. In such cases the atom can enjoy more or less than eight electrons. We shalllook to those cases a little later.

TYPES OF BONDS ON THE BASIS OF ELECTRON SHARING:On the basis of the nature of sharing of electrons by the atoms, there are three kinds ofcovalent bonds.(i) Non-polar covalent bond (ii)Polar covalent bond (iii)Coordinate covalent bondor Dative bond

Non-Polar Covalent Bond:When the two atoms are identical, the bond is called non polar covalent bond.

Ex: H H , ClCl , XX

X NN X

X

15

This is a case of equal contribution and equal sharing. The three bond pairs are equallyshared by the two identical atoms. We shall know more about the non-polar covalent bondafter we know about polar covalent bond.Polar Covalent Bond:When the two atoms are different, the bond is called polar covalent bond.Take the case of HCl.

H X Cl+ -

Each atom contributes one electron to form the bond pair. But the bond pair is not equallyshared by the two dissimilar atoms. You know that Cl is more electronegative than H(Cl=3and H=2.1). So Cl atom will attract the bond pair a little more than the H atom. Thus thebond pair is slightly tilted or shifted to the side of the Cl. Note that electron pair does notentirely go to Cl. In that case, it would form H+ ion and Cl- ion and make the ionic bondinstead of covalent bond. The electron pair will remain in between the two atoms and willbe shared by the two atoms but the sharing will be unequal. The electron pair instead ofsitting exactly at the mid point between the two atoms, sits a little closer to Cl atom i.e alittle away from H atom. Therefore Cl atom will gain a little additional share over H'selectron while H atom loses a little share over its own electron. This results in the developmentof a fractional positive charge(+δ) on H atom and a fractional negative charge(-δ) on theCl atom. In general, the less electronegative element(like H) will have +δ charge and moreelectronegative element(like Cl) will have -δ charge over them. Had the charges been +1and -1 in stead of fractional(+δ and -δ), the bond would have been ionic. But in this casethe bond is covalent but having polar or a little ionic character. This separation of chargeswithin a neutral molecule i.e development of partial ionic character in the covalent moleculeis called polarity and the bond is called a polar covalent bond. This is a case of equalcontribution but unequal sharing of electrons.In H

2O molecule, the two O-H bonds are polar covalent bonds. There is +δ charge on the

less electronegative H atom and -δ charge on more electronegative O atom.

+-+OH HXX

In NH3 also, the three N-H bonds are polar, with N having -δ charge and H having +δ

charge. So in general, the bond between any two dissimilar atoms is polar in nature with +δremaining on the less electronegative atom and -δ remaining on the more electronegativeatom.(iii) Coordinate Covalent Bond(Dative Bond):If out of the two atoms, one atom contributes two electrons (electron pair) and the otheratom does not contribute any electron, then the bond formed between them will be calledcoordinate covalent or dative bond. It is a case of one-sided contribution but both-sidedsharing. Dative means donated i.e out of the two electrons contributed by one atom, as if,one electron is completely donated to the other atom because it has no electron to contribute.See this example.

(i) H3O+ (hydronium ion):

16

X

XH

H O H+

+

HOH

HX

X

H

O HH

+

When water molecule reacts with hydrogen ion(H+), we get hydronium or hydroxoniumion(H

3O+). This is formed when one lone pair of O atom in water forms a coordinate

covalent bond with H+ ion. H+ has no electron to contribute. For forming a bond with Oatom, it has to seek the donation of electron by O atom which is called the donor atom.H atom will be called the acceptor atom. Out of the two lone pairs present on O atom, ituses one lone pair and forms a new bond with H. This lone pair is converted to the bondpair. As if out of the two electrons, it completely donates one electron to H+ ion thus thepositive charge on H is neutralised. By losing one electron completely, O now carries the +vecharge which is customarily shown above the bracket symbol. Thus a dative bond is formedbetween O and H. A dative bond is shown by an arrow mark as shown in the figurefrom the donor atom to the acceptor atom.

(ii) NH4+ ion:

NH3 reacts with H+ ion to form ammonium ion(NH

4+) by forming a dative bond between

N and H+. The lone pair of N takes part in the formation of this dative bond.

or NH

H

HH ++

HXX

X

H

H

HN+

HNH

H

H

X

XX

SAQ 4: Indicate which of the following forms non-polar and which polar covalentbonds HF, O

2, Br

2, HBr, I

2, N

2 , H

2S

LEWIS STRUCTURES OF SOME MORE COVALENT MOLECULES:

We know that while drawing the Lewis structures, we should see that the central atom andall other atoms excepting H contain 8 electrons(octet) in the valence shell. For that purposeswe have to make a dative bond in place of double bond or vice versa whenever situationdemands. Note that a dative bond is considered to be a single bond having two electrons forsharing. Look to the Lewis structure of SO

2 and SO

3.

SO2:

Both O and S belong to the same family. Each of them(see table) has two lone pairs andtwo unpaird electrons. Since S is less electronegative, it is the central atom. The O atomshave to be placed on either side of S. Then the lone pairs and unpaired electrons aredistributed on each atom. First of all, we have to establish a double bond between the S andO oxygen atom by employing two unpaired electrons of each. Now we do not have anyunpaired electron left on S to form covalent bond with the other oxygen. We have now twopossibilities.

17

X S OO XXX

XX

X X

XX

X X XX

XX

XX

XX

O OSor (i) (wrong)

X S OO X XX

XX

X X

XX

X X XX

XX

XX

XX

O OSor XX (ii) (correct)

(i) One of the lone pairs of S has to be broken and make two additional unpairedelectrons and form a double bond with the other O atom using its two unpaired electrons.(ii) The other O atom has to make pairing of the two unpaired electrons and form thethird lone pair, thus giving a room free for forming a dative bond with S. S uses one of itslone pairs and forms a dative bond with O atom which now has three lone pairs. Note thatO has four orbitals (one s and three p)in its valence shell and hence it has to make one orbitalfree for accommodating the lone pair of S in forming the dative bond. Therefore it has tomake pairing of its two unpaired electrons to form the third lone pair.Out of the two possibilities let us find which one will be correct..

Look to the S atom in the 1st case. Count the electrons it is sharing. It is sharing 10 electrons(four bond pairs and one lone pair). This violates the octet rule which says that usually thecentral atom should share a total of eight electrons in the valence shell. In the secondpossibility, a dative bond has been made in place of a double bond. In this case the centralS atom shares eight electrons (three single bonds and one lone pair). Thus the second Lewisstructure is correct.

SO3:

In SO3, there are three O atoms which surround the central S atom. There will be one more

dative bond as compared to SO2 which will be formed with the help of the remaining lone

pair of S.

X S OO X XX

XX

X X

XX

X X XX

XX

XX

XX

O OSorXX

OXX

XX

XX

OXX

XX

XX

If we make two double bonds in place of two dative bonds, S would share 12 electrons. Inthis structure S shares 8 elctrons. Note that wherever there is dative bond with O, thatO atom has three lone pairs.

LEWIS STRUCTURES OF MOLECULES CONTAINING MORE THAN THREEATOMS:Rule: H atom cannot be directly attached to the central atom.Take the examples of the common mineral acids: H

2SO

4, H

3PO

4, HNO

3

H2SO

4:

S is the central atom in this case. First, we have to keep four O atoms distributed aroundS. Then keep two H atoms symmetrically adjacent to two O atoms on either side of S butaway from it. This is because, H atom cannot be directly attached to the central atomaccording to rule. So that they have to be attached to O atoms. Then place the lone pairsand unpaired electrons on each atom. S and O each contain two lone pairs and two unpaired

18

electrons while H has one unpaired electron only. After placing the electrons, make planningto establish bonds between the atoms. Towards the side where H atom has been placed, Shas to establish single bond with O by using one unpaired electron. So S uses two unpairedelectrons on either side of it to form two single bonds with the two O atoms to which Hatoms are to be connected. Each such O atom uses one unpaired electron for S and one forthe H atom on the other side to form single bonds with S and H respectively. Now we haveonly two lone pairs on S atom. If we break the lone pairs and make double bonds with theremaining O atoms, we will violate the octet rule. S will share 12 electrons. In such case,we must think of dative bonds. S uses two lone pairs separately to establish two dative bondswith the two O atoms. Thus S shares now eight electrons. The O atoms which are bondedto S atom by dative bonds have now three lone pairs each.

or S OO

O

O

HHXX

XX

X X

XXXX

XX

XX

XX

XX

XX

XXXX

XX

X XH H

O

O

O OSXX

X X

X X

X X

XX

X X

XX

X X

H3PO4:P is the central atom. Four O atoms are placed symmetrically surrounding P. Three H atomsare placed beside the three O atoms. One O atoms is left without H. P has one lone pairand three unpaired electrons like nitrogen atom. O has two lone pairs and two unpairedelectrons. Place the electrons and plan for establishing bonds. P has to use the unpairedelectrons in the direction of H atom to make three single bonds with the three O atoms.These three O atoms will make single bonds with P as well as with H atoms by using twoof unpaired electrons each. P has now one lone pair which it will use for making dative bondwith the remaining O atom. It cannot make double bond by breaking the lone pair becauseby that it will violate the octet rule.

XX

XXXX

XXXX

XX

XX

XX

OO

O

O

HH XX

XX X

X

XXXX

XX

XX

XX

XX

XX

X

XX

XX

H H

O

O

O Oor

P

X

H

P

H

HNO3:

N is the central atom. Three O atoms are to be placed first surrounding N. Then place oneH atom beside any one O atom. After placing the valence electrons (N has one lone pairand three unpaired electrons while for O and H you know), first make a double bondbetween N and one free O(not connected with H) by using two unpaired electrons of each.N has one more unpaired electron which is to be used to make a single bond with O in thedirection of H. That O atom will have to form single bond with N as well as with H on eitherside of it by using two of its unpaired electrons. Now we have only one lone pair on N whichis to be used to make a dative bond with the remaining O. If you make a double bond, theoctet rule will be violated.

19

XXX

X

X orN O H

O

O XX

XX

X X

X

XX XX

XX

XX

O

O

HON

XX X

X

X X

XX

XX

XX

LEWIS STRUCTURES OF NEGATIVE IONSFor writing the Lewis structures of negative ions like SO

42-, NO

3-, CO

32- etc. we shall have

to first write the Lewis structures of the corresponding acids and then remove (wipe out)the H atoms from that structure. Note that while rubbing of the H atoms, leave the electronof H atom on the O atom, so that a new lone pair is formed. The presence of additionalelectron coming from H atom forms a negative charge on O atom.

CO3

2-:First let us draw Lewis structure of H

2CO

3. C is the central atom and has four unpaired

electrons for bonding. First we have to make a double bond between C and O by using twounpaired electrons each. Then we make single bonds between C and the remaining two Oatoms (to be connected with H atoms) by using the remaining unpaired electrons of C andone unpaired electrons of each of the two O atoms. These two O atoms also make singlebonds with H atoms by using each of its remaining unpaired electrons with unpaired electronsof H atoms. Note that we cannot make a dative bond in place of a double bond in betweenC atom and O atom. That will violate the octet rule as in that case C will share only sixelectrons.

XXX

X

XX orO H

O

O

X X

XX

XXX

XX

XX

O HO

XX X

X

XX

X X

XX

XXC CXH

O

H

Now the two H atoms have to be removed with the creation of a new lone pair andconsequently -1 charge on each oxygen. In stead of writing -1 on each O, we shall writethe total charge of 2- above the bracket as shown below.

XXX

X

XX orO

O

O

X XXX

X X

X

XX

XX

O O

XX X

X

XX

X X

XX

XXC CX

O

XX

2- 2-

Note that the electron of H indicated by a symbol of triangle(D) is present on each O atom.This is responsible for the negative charge of -2 for the two additional electrons left by Hatoms.Similarly we can write the structure of SO

42- and NO

3- ion first writing the structures of

H2SO

4 and HNO

3 respectively and then rubbing off the H atoms as explained before. Refer

the structures of these two acids given earlier. The structures of their ions would be asfollows. We give only the dash structures.

SO4

2-:

2-

XX S OO

O

O

XX

XX

XX

XXXX

XX

XX

X X

XX

X X NO3

- :

-

X

XX

XX

XX

O

O

ON

XX X

X

XX

XX

Likewise the reader is supposed to practise writing Lewis structures of some more ions.

20

SAQ 5:(i) Write the Lewis structures (dot or dash) of the following:

PCl3, HClO

2, HClO

3, ClO

4-, NO

2-

(ii) Why do N, O and F atoms cannot break their lone pairs and create additionalunpaired electrons?

LEWIS STRUCTURES OF MOLECULES VIOLATING THE OCTET RULE:There are many molecules in which the central atom cannot, by any means, share eightelectrons. It shares either less than eight electrons, in such case we call it to be an electrondeficient molecule or in other cases it shares more than eight electrons. See the followingcases.

BeCl2, BCl

3, PCl

5, SF

6

BeCl2:

Be has two electrons in the valence shell for bonding (see table). It forms two single bondswith two chlorine atoms. Each Cl atom uses its one unpaired electron for the purpose. ThusBe shares only four electrons. This is a violation of octet rule. Note that when there is noother way, octet rule is violated. So you should not remain under the impression that octetrule can never be violated.

or Be ClClCl ClBe

BCl3:

B has three electrons in the valence shell for bonding (see table). Each Cl has one unpairedelectron for bonding. So there will be three single bonds and B will share only six electrons.This is violation of octet rule.

ClCl

or B ClClCl ClB

PCl5:

P has one lone pair and three unpaired electrons in its valence shell. But it needs fiveunpaired electrons to make five single bonds with five Cl atoms. So it has to break the lonepair and produce two more unpaired electrons. Now it has five unpaired electrons to makefive single bonds. In PCl

5, P thus shares 10 electrons in stead of 8. This is again a violation

of octet rule.

ClClClCl

PP

ClCl

orClClCl Cl

SF6:

S has two lone pairs and two unpaired electrons in its valence shell like O. In SF6, S needs

six unpaired electrons to make six single bonds with six F atoms. So it has to break both ofits lone pairs and produce four more unpaired electrons. Thus there are six unpaired electronsfor the purpose.(Note that the oval shaped markes are not given in the Lewis dot structures afterBCl

3)

21

FFF F

FF

F

F

FF

F

FSS or

SAQ 6:(a) Draw the Lewis dash structures of the following. N

2O

3, N

2O

5

(b) Why it is possible for P and S atom to break their lone pair and produce addtionalunpaired electrons?

(c) How is the energy of the system lowered during covalent bond formation?

PROPERTIES OF COVALENT SUBSTANCESSubstance which contains covalent bonds is called a covalent substance. This may be acompound or an element. Covalent substances exhibit different properties as compared toionic compounds. These are discussed below.

1. Physical State:Covalent compounds are usually gases, liquids or soft and low melting solids at roomtemperature. Say for example, at room temperature CO

2, CO, NH

3, H

2, N

2, O

2, PH

3, HCl,

HBr, HI, SO2, SO

3, H

2S, C

2H

4, C

2H

2 etc are all gases while Br

2, CCl

4, H

2O, benzene(C

6H

6),

chloroform(CHCl3) etc. are liquids and I

2, P

4, S

8 etc. are soft solids.

2. Melting and Boiling Points:The covalent substances have low melting and boiling points. For example the boiling pointof NH

3 is -330C and its melting point is still less. The same is the case with other gases. The

melting and boiling points of covalent liquids and solids are also low. Take for example,sulphur (S8) melts at 1130C while an ionic compound like NaCl melts nearly at 8000C.Because their melting and boiling points are low, covalent substances remain mostly as gasesor liquids. The reason for this will discussed later.3. Electrical Conductivity:Covalent substances do not conduct electricity in any state. However some covalent compoundsdo conduct electricity in solution as they undergo ionisation when dissolved in water. Forexample, HCl(gas) is a covalent compound, but when HCl is dissolved in water, it undergoesionisation to give H+ and Cl- ions. Due to presence of free ions in solution, it conducts electriccurrent. All mineral acids like H

2SO

4, HNO

3, H

3PO

4 etc. are covalent compounds and they

do not possess free ions in pure state. But when they dissolve in water, undergo ionisationproducing ions. That is why they conduct electricity. Let us know now how the covalentcompounds like HCl, H

2SO

4 etc. undergo ionisation in water.

Ionisation of HCl(g) in water:

(hydrated positve and negative ions)Ionization finishedIonisation begins

+

-+H

HO

-

++

HH

O

-

+

+

HH

O-

+

+

H

HO

OH

H

+

+-

ClO

H

H

+

-+

OHH +

-

+

OH

H

+

-+

O

H

H

+

-

+OH

H+

-

+

H

-

-

+

+

+

+

+-

+

H

H

H

HH

HO

O

O+

+++

-

-

+

-

+

HH

H

H

HH

O

O

O-+

H Cl

22

HCl is a covalent molecule and has polar covalent bond. H is the +ve pole and carries +δcharge while Cl is the negative pole and carries -δ charge. H

2O is also a strongly polar

compound which is acting as a solvent. H2O has two O-H bond dipoles. O is the negative

pole having -δ charge and H is the positive pole carrying +δ charge. When H-Cl moleculeenters in water, large number of water molecules start surrounding each H-Cl molecule insuch a manner that the -ve poles of H

2O molecules face the +ve pole of HCl and the +ve

poles of H2O molecules face the -ve pole of the same HCl molecule. Water is the solvent

and is present in large amount. Large number of H2O molecules face the two poles of each

HCl molecule as explained and as shown in the above diagram. Due to strong attractionbetween water molecules and each pole of HCl, the two atoms H and Cl are pulled apartfrom each other with the bonding pair of electrons shifting to the more electronegative atomchlorine. Ultimately the water molecules succeed in breaking the bond between H and Cl andbringing about ionization. The ions H+ and Cl - do not remain in the naked state. They arejailed inside the solvent cage. Water molecules surround the +ve ion(H+) with their negativepoles facing the +ve ion and also water molecules surround the -ve ion(Cl -) with theirpositive poles facing the -ve ion. These are called the hydrated ions(hydrated H+ ion andhydrated Cl- ion) as shown in the above diagram. We represent this ionisation as follows.

HCl(g) + aq. -----------> H+(aq) + Cl-(aq)The process of surrounding of the water molecules to an ion is called hydration of ion.During hydration of ions, some amount of energy is liberated. This is called hydrationenergy. We shall study more about this in higher classes. These hydrated ions are responsibleto conduct electricity in aqueous solution. Ionisation of polar covalent molecules like HCl,HBr, H

2SO

4, HNO

3 etc. in water is due to hydration of ions. The strongly polar solvent

molecules are responsible to break the covalent bond and produce hydrated ions.

SAQ 7: In the light of hydration of ions, explain why ionic compounds are soluble in water.

(3) Solubility:Covalent compounds which are non-polar or weakly polar in nature are not soluble in stronglypolar solvent like water. They are soluble in non-polar solvents like carbon tetrachloride(CCl

4),

benzene(C6H

6) or weakly polar solvents like chroroform(CHCl

3), acetone etc. Take for

example, I2 is a non-polar covalent solid and is very poorly soluble in water, but highly soluble

in non-polar solvents like CCl4. H

2, N

2, O

2, Cl

2 which are non-polar gases are very poorly

soluble in water. CCl4 which is a non-polar compound (a liquid) does not mix with water.

Benzene which also a non-polar liquid does not mix with water. However the covalentcompounds which are polar in nature like HCl, HBr, NH

3, SO

2, glucose(C

6H

12O

6), ethyl

alcohol(C2H

5OH) etc. are soluble in water. This is because water is also a polar compound.

Some of the polar compounds like HCl, HBr etc.dissolve due to ionisation (accompanied byhydration), some other covalent molecules like glucose, ethyl alcohol dissolve due to someother factor. This, we shall discuss a little later. So now we know that not only ioniccompounds are soluble in water but also polar covalent compounds are soluble in water. Asa matter of fact, polar covalent compounds can be considered as partially ionic and behavesimilar to ionic compounds so far as solubility in water is concerned. All these observationshave resulted a famous slogan in the principle of solubility:

Like dissolves like.

23

This means that if the solvent is polar(H2O) then it will dissolve polar compounds and ionic

compounds.(Note that ionic compounds are 100% polar while polar covalent compounds arepartially ionic and so ionic character is common in both). If the solvent is non-polar (CCl4type), then it will dissolve non-polar compounds like H

2, N

2, I

2, P

4 etc. We shall discuss about

its reason a little later.

SAQ 8:(i) NH

3 is a covalent molecule but aqueous solution of NH

3 conducts electricity. Explain.

(ii) CO2 is weakly soluble in water while HCl is highly soluble in water. Why?

(iii) Indicate which among the following are weakly soluble and which highly soluble inwater. I

2, HI, SO

2, H

2, H

2SO

4, CCl

4, N

2.

(4) Shape:Covalent molecule has a fixed shape. This is because covalent bond is directional in nature.This is unlike an ionic bond which is non-directional. There is a fixed bond angle betweentwo covalent bonds within a molecule. For example, the two O-H bonds in H

2O make an

angle of 104.50. This makes the shape of H2O V-shaped(or bent). Any two N-H bonds in

NH3 make an angle of 1070. Thus the shape of NH

3 is pyramidal. N is sitting at the apex

of an imaginary pyramid and the three H atoms are sitting at the three corners of thetriangular base. The two C=O bonds in CO

2 make an angle of 1800.Thus the shape of CO

2

is linear. Thus all the covalent molecules have fixed shapes and fixed bond angles. This hasbeen revealed by actual experimental techniques. We shall see the details on the head, shapeof covalent molecules in our next section.(5) Isomerism:The covalent compounds exhibit a phenomenon called isomerism. This we shall discuss in ourorganic chemistry section.

SHAPES OF COVALENT MOLECULES:

VSEPR(VALENCE SHELL ELECTRON PAIR REPULSION) THEORY:

It has been already said that a covalent molecule has a fixed shape or molecular geometry.VSEPR theory gives simple rules for determining the shapes of covalent molecules. This ismerely a shape predicting tool.This theory was first published in 1940 by N.J. Sidgwick and H.M. Powell. The theoryindicates that the positions of bonds about an atom are determined by the number ofelectron pairs associated with the central atom. The geometry of the molecule is knownby knowing the geometry of the central atom. In 1957, R.J. Gillespie and R.S. Nyholmextended this theory to its final form. Take the case of BeCl

2.

BeCl2:

Cl Be ClX X or Cl Be ClThere are two pairs of electrons associated with the central atom. These are the two bondpairs as you see in the structure above. Electron pairs, both being negative in charge, repeleach other. As a result the bond pairs stay as much away from each other as possible. If

24

they come closer then there will be unfavourable repulsion between them and this arrangementis unstable. They attain most stable arrangement when the bond pairs exert minimum repulsionbetween them. In the above example, the two bond pairs of electrons will remain at an angleof 1800 apart. In this arrangement, the repulsion is minimum. It is the most stable arrangement.If the angle between the bonds is changed from this value, they come closer and repulsionbetween them would be more. This will be an unstable arrangement. Therefore the geometryof the bonds has to be linear (angle 1800) and since Cl atoms are attached to these bonds,the geometry of the molecule has also to be linear. The shape of BeCl

2 is truly linear having

bond angle 1800. Take the case of BCl3 molecule.

BCl3:

X X

ClCl

Cl

BX

or ClCl

Cl

B

In BCl3 there are three pairs of electrons associated with the central atom, B. These are

bond pairs of electrons. To minimise the repulsion between them they should remain fartheraway from each other. This is achieved when they remain in one plane and the anglebetween them is 1200. This is called trigonal planar geometry, because, if we join theterminal of these bonds by imaginary lines, we get an equilateral triangle. Since Cl atoms areattached to B by these bonds, the shape of BCl

3 is also trigonal planar with a bond angle

of 1200. Note that the broken lines in the Lewis dash structure are merely the outlines ofa equilateral triangle and not the actual bonds. The actual bonds are shown by solid lines.CH

4:

In methane there are four bond pairs which try to remain away from each other to minimizethe repulsion between them. They do so by remaining at an angle of 109028' away from eachother. This is called a tetrahedral geometry. The central atom(C) is situated at the centreof the tetrahedron and the four bonds are directed towards the four corners of the tetrahedron.The broken lines show the outline of a tetrahedron while the solid lines show the actualbonds. The geometry of bonds is tetrahedral, so also the shape of the molecule, CH

4.

H

H

H

H

C

PCl5:

In PCl5, there are five bond pairs which try to remain away from each other to minimize

repulsion between them. They do so by assuming trigonal bipyramidal geometry. Three bondsremain in trigonal planar geometry i.e directed to the three corners of an equilateral trianglelike BCl

3, the remaining two bonds lie perpendicular to the plane carrying the three bonds.

One of these bonds lie above the plane and the other below the plane. If the terminal of thesefive bonds are joined by imaginary lines, we find two pyramids, one above and one belowthe plane. So it is called trigonal bipyramidal geometry.

25

Cl

P

Cl

Cl

Cl

Cl

Since the five bonds make the trigonal bipyramidal geometry, the molecule, PCl5 also has the

same shape. The geometry of the molecule is same as the geometry of the bondsdistributed about the central atom. There are three bond angles in this geometry. Thebonds lying on the plane has the angle of 1200 between them; each of the planar Cl andperpendicular Cl make an angle of 900 while the perpendicular Cl atoms make an angle of1800 between them.SF

6:

In SF6, there are six bond pairs which try to remain away from each other to minimize

repulsion between them. They do so by assuming the octahedral (square bipyramidal)geometry. The four bonds lie in one plane directed towards the four corners of a square andthe remaining two bonds lie perpendicular to the square plane, one lying above the plane andone below the plane. When the terminal of the bonds are joined by imaginary outlines weget two square pyramids lying opposite to each other. This is called the square bipyramidalor more popularly the octahedral geometry. Note that the bonds are shown in solid lines whilethe outlines of the geometry are shown by broken lines.

F

F

FF

F

F

S

LONE PAIRS AND MOLECULAR GEOMETRY:

The lone pairs of electron associated with the central atom behave in the same manneras the bond pairs in deciding the geometry of the molecule. The lone pair repels a bondpair and also another lone pair in the same manner a bond pair repels another bond pair.(i) Trigonal Planar Geometry:Supposing central atom has two bond pairs and one lone pairs. The three electron pairs willrepel each other and be stable at the maximum separation between them. The result is thatthey assume trigonal planar geometry as explained before.

lone pairB

A

X (V-shaped geometry)

Out of the three pairs, two are bond pairs, the other is a lone pair. The geometry of all the

26

electron pairs(taking both bond pair and lone pair into account) is trigonal planar. This iscalled electronic geometry. But the actual shape of the molecule does not take the lone pairinto account. To present in a simpler way, when photograph of the molecule is taken, the lonepairs present on the central atom do not appear in the photograph. What shape we actuallyobserve is electronic geometry minus the lone pairs. This is called the molecular geometry(actual shape). In the above case, the lone pair is not observed, the resultant geometry ofthe molecule becomes V- shaped or angular. The bond angle for a trigonal planar geometryi.e 1200. Here we shall learn another finer aspect of VSEPR theory.

The repulsion between a lone pair and a bond pair is little more than the repulsionbetween a bond pair and another bond pair. In the above example, there are two bondpair-lone pair repulsion and one bond pair-bond pair repulsion. Since the former is little morethan the latter, the angle between two bond pairs is slightly reduced. But the angle remainsclose to the original value and the reduction of the angle is very small. So remember thatwhenever there is one or more lone pairs of electrons on the central atom the bondangle is slightly reduced from the expected value. We shall see several examples later.

(ii) Tetrahedral Geometry:Case-I: If there are three bond pairs and one lone pair i.e four pairs of electrons

associated with the central atom, then they will result tetrahedral geometry as explainedbefore. The electronic geometry is thus tetrahedral. But if you don't look at the lone pair, howdoes the molecule look like? The ultimate geometry (molecular geometry) becomes pyramidal.The central atom sits at the apex of the pyramid and the other three atoms at the threecorners of the triangular base.

(lone pair)

AB

C

X(pyramidal geometry)

The broken lines connecting A, B and C give the outlines of the triangular base of thepyramid. The solid lines give the actual bonds. The bond angle is close to tetrahedral angleof 1090 28'. There is a slight reduction of bond angle from this value due to presence of onelone pair. You know that lone pair-bond pair repulsion is greater than bond pair-bond pairrepulsion. That is why a slight reduction in angle occurs.

Case-II: If there are two bond pairs and two lone pairs, the electronic geometry asusual is tetrahedral, out of which two sites are taken by lone pairs and the other two sitesby bond pairs. If we disregard the lone pair sites and only see the bond pairs, the moleculargeometry is V-shaped. But the bond angle is close to 1090 28'(tetrahedral angle). But dueto presence of two lone pairs, there will be a little reduction of bond angle from this value.The reduction in this case is little more than that when there was one lone pair. We shalltake some examples of each type later.

(lone pair)

(lone pair)

AB

X(V-shaped geometry)

27

Similarly the presence of one or more lone pairs in trigonal bipyramidal or octahedral geometrychange the ultimate molecular geometry. The following table gives a list of different moleculargeometry (actual shapes) for each type of electronic geometry.

DOUBLE AND TRIPLE BONDS AND MOLECULAR GEOMETRY:For predicting molecular geometry, we consider a double bond or a triple bond as one pairof electron. This is because a double or a triple bond take up the same amount of spaceas a single bond in describing a geometry. So remember a very important rule in theVSEPR theory that a double bond or a triple bond should be thought as a single bond forthe purpose of predicting electronic geometry and then molecular geometry. Take the caseof CO

2.

XX

XX

XX

XX

OO CThe central atom is actually associated with four pairs of electrons (two from each doublebond). But for the purpose of knowing its shape, we shall consider each double bond asone electron pair. Because a double bond is consuming the same space as a single bond.So there are two electron pairs associated with the central atom so far as geometry isconcerned and the electronic geometry therefore is linear and molecular geometry is alsolinear as explained before in case of BeCl

2. We introduce a new term called the Steric

Number(SN), at this point, to denote the number of electron pairs which are relevant fordetermining the shape.

STERIC NUMBER(SN):Steric Number(SN) is the number of electron pairs associated with central atom whichactually determine the shape. For finding SN of the central atom, the single bond(s), doublebond(s), triple bond and lone pair(s) associated with it are counted. Double and triplebonds are counted as one each for finding the SN. Take the following examples.

(i) OH H : SN = 4 = 2 (for two single bonds) +2 (for two lone pairs)

(ii)N

H

H

H

: SN =4 = 3 (for three single bonds) + 1 (for one lone pair)

(iii) CO OXX

X X

X X

X X:SN = 2 (each double bond is taken as 1)

(iv)XX

XX

XX

XX

O OSXX : SN = 3 = 1(for one double bond) + 1(for one dative bond) + 1(for

one lone pair)Likewise you can find the steric number (SN) of the central atom after you have correctlydrawn the Lewis structure of the molecule.

STERIC NUMBER, ELECTRONIC AND MOLECULAR GEOMETRY:We have already seen that if there are two electron pairs, according to VSEPR theory thepairs will stay 1800 apart and the resulting geometry is linear. Now we shall say the samething in a different way. We shall use the term SN(steric number) in place of electron pairswhich determines the shape. For SN=3, we have seen that the electronic geometry is trigonalplanar and if there is no lone pair, the molecular geometry (actual shape) is same aselectronic geometry. If there is one lone pair then the molecular geometry is reduced to V-

28

shape. The following table gives a list of different steric numbers, their electronic geometryand possible molecular geometry. The table is highly useful in predicting the shapes ofcovalent molecules.

PREDICTING SHAPE OF COVALENT MOLECULES (VSEPR THEORY)

SN Electronic Geometry Molecular Geometry Bond Angle(0)

2 linear linear 180

X A X B

3 trigonal planar (i)trigonal planar(no lone pair) 120

X XA B

C

≈120

(ii) V-shaped(angular) (A little less) (one lone pair)

BA

X

4. tetrahedral (i)tetrahedral(no lone pair)

X X

A

B C

D

109028'

(ii)pyramidal(one lone pair)

X

D

CB ≈109028 '

(a little less)(iii)V-shaped(three lone pairs)

X

D

B ≈109028'

(a little less)5 trigonal bipyramidal (i)trigonal bipyramidal 120(base)

(no lone pair) 90 and 180

X

E

D

C

B

A X

29

(ii)see-saw shaped ≈120, 90, 180(one lone pair)

X

E

D

C

B

(iii)T-shaped ≈90, 180(two lone pairs)

C

D

E

X

(iv) linear 180(three lone pairs)

D

E

X

6 octahedral (i)octahedral 90, 180(no lone pair)

X

F

E

D

XC

BA

(ii)square pyramidal ≈90(one lone pair)

E

D

XC

BA

(iii)square planar 90(two lone pairs)

D

XC

BA

7. pentagonal bipyramidal (i)pentagonal bipyramidal 72,90,180(no lone pair)

(ii) distorted octahedral(no lone pair)

30

PREDICTION OF SHAPES USING VSEPR THEORY:For predicting shapes of covalent molecules and ions, first draw the Lewis structure correctly.Then count the steric number(SN) of the central atom as explained before. Finally look tothe above table to predict the shape. Note that the shape of the species is the moleculargeometry given in the third column of the table(not the electronic geometry). Let us applyto some simple covalent species.

1. CH4: C

H

H

H

H SN = 4, and there is no lone pair on the

central atom. Hence Molecular geometry: TETRAHEDRAL and Bond Angle = 1090 28'

H

H

H

H

C

2. NH3:

NH

H

H

SN: 3+1 = 4, Electronic Geometry:

tetrahedraland there is one lone pair, so Molecular geometry(actual shape): PYRAMIDAL

H

H

H

N

Bond angle ≈ 1090 28' Actual: 1070.

3. H2O: OH H SN = 2+2=4, Electronic Geometry :

tetrahedraland there are two lone pairs, so Molecular Geometry(actual shape): V-shaped.

H

H

O

Bond angle ≈ 1090 28' Actual: 104.50

4. SO2:

XX

XX

XX

XX

O OS XX SN = 1+1+1=3, Electronic Geometry: trigonal planar

and there is one lone pair, so the Molecular Geometry: V-shaped.

OS

OBond angle ≈ 1200 Actual: 119.50

5. BCl3:

Cl

B ClClSN = 3, Electronic Geometry: trigonal planar:

Since there is no lone pair, the Molecular Geometry is same: trigonal planar

BClCl

ClBond angle : 1200

6. BeCl2: Be ClCl SN = 2, Electronic geometry: linear

Molecular Geometry: linear and bond angle = 1800.

31

7. PCl5:

ClCl

P

ClClCl SN: 5, Electronic geometry: trigonal bipyramidal

There is no lone pair on the central atom, Molecular Geometry= trigonal bipyramidal.

PCl

Cl

Cl

Cl

Cl

Bond angle: 1200 among the triangular base,

900 between a bond in the triangular base and upper or lower bond. 1800 is the angle betweenthe upper and lower bonds.

8. SF6: S

F

F

F

FFF

SN: 6, Electronic Geometry: Octahedral.

Since there is no lone pair on central atom, the Molecular geometry is same(octahedral).

S

F

F

FF

FF

Bond angle: 900 and 1800.

SAQ 9:(i) Among Electronic and Molecular geometry, which gives the actual shape of themolecule?(ii) The central atom of certain covalent molecule is associated with three electron pairsout of which two are bond pairs and one is a lone pair. Predict what will be its actual shape?What would be the expected bond angle?(iii) Although central atoms in each of CH

4, NH

3 and H

2O has four electron pairs(steric

number is 4), the bond angles are 109028' 1070 and 104.50 respectively. Explain.(iv) What is the steric number of C in CO

2 molecule. Predict its shape and bond angle.

(v) Explain why the bond angle in SO2 is close to 1200 while that in H

2O is close to 1090

28' although both are V-shaped(angular) molecules.(vi) Explain why NH

3 is pyramidal while BCl

3 is trigonal planar.

(vii) Draw the Lewis structures of the following. Also predict the shape and expectedbond angle on the basis of VSEPR theory.

HCN, CCl4, CHCl

3, C

2H

4, C

2H

2, H

2SO

4, HNO

3, CO

32-, SO

32-, H

3PO

4, HNO

2, HClO,

HClO2, HClO

3, HClO

4, ClF

3, SnCl

2, XeF

2, XeF

4.

(viii) What are the shapes of all diatomic molecules like CO, NO etc. Do you applyVSEPR theory to them?(ix) What are paramagnetic covalent molecules? Name a few. State why the Lewisstructure does not correctly explain their properties.

32

SAQ 10: Indicate which of the following compounds are ionic and covalent in nature.Indicate which contain both types of bonds.

NaBr, CO2, NH

3, KF, CH

4, CaO, SO

2, AgCl, HCl(g), HCl(aq), Br

2, RbF

2,

BeCl2, NH

4Cl, H

2SO

4(l), H

2SO

4(aq), AlCl

3(anhydrous), MgCl

2, SO

3, PCl

5, NaNO

3

POLARITY OF MOLECULESWe have seen before that the covalent bond between two dissimilar atoms is polar in nature,because the bond pair of electron is unequally shared by them. The atom which is moreelectronegative acquires a fractional -ve charge(-δ) and the less electronegative atom acquiresfractional +ve(+δ) charge. This charge separation is called polarity. The polarity is measuredquantitatively by a term called dipole moment(μ). More the polarity or ionic character of acovalent bond, more is the dipole moment.

-+

H Cl

The dipole moment is just like an electric field which is directed from the positive pole tonegative pole as shown by an arrow mark in the figure. This dipole moment is directlyproportional to distance between two poles(bond length) and also to the magnitude of fractionalcharge present at each pole(δ).

μ = q X dwhere μ = dipole moment, q = magnitude of factional charge on each pole;d = distance between the two opposite poles = bond length

Unit of dipole moment :CGS: Debye(D) ; 1D = 10-18 esu.cm (where the unit of charge is esu, i.e electrostaic unitand unit of distacne is cm.)1 unit of electrical charge = 4.8 X 10-10 esuMKS : C.m (unit of charge is Coulomb); One unit of charge = 1.602 X 10-19 coulDipole moment as a vector quantity : Dipole moment is a vector quantity and its directionin chemical convention is from positve pole to negative pole.The dipole moment(μ) varies from polar bond to polar bond. The dipole moment of H-Cl bondis different from dipole moment of H-F bond. Could you guess which has a greater dipolemoment? More is the electronegativity difference, more is the charge separation i.e morepolar character and hence more is the dipole moment. F is more electronegative than Cl, soH-F bond is more polar than H-Cl bond.SAQ 11: Arrange the following in the order of decreasing polarity of the bond.

HF, HCl, HBr, HI

Diatomic molecules such as HCl, HBr, HI, HF etc.contain only one bond each. Since thisbond is polar, molecule is said to be a polar molecule. But if a molecule has more thanone dipoles, the molecule as a whole may or may not polar. Look to the example of CO

2.

We know from VSEPR theory(discussed before) that CO2 molecule is linear in shape. The

C atom and the two O atoms lie in one straight line.

- -+O C O

We already know that dipole moment has a direction which is from +ve pole to -ve pole. Any

33

such parameter which has both a magnitude and direction is called a vector quantity. CO2

molecule has two dipole moments (which are two vectors) which are equal in magnitude butopposite in direction. So they cancel each other and the net dipole moment becomes zero.That is why CO

2 is a non-polar molecule although there are two C=O bond dipoles. We shall

learn more about vectors and dipole moments in higher classes.Unlike CO

2, sulphur dioxide molecule is not non-polar. SO

2 molecule is V-shaped(we already

know its shape from VSEPR theory). Note also that each lone pair orbital also consists adipole whose direction is away from central atom. The lone pair dipole moment vector is notshown in the diagram below.

-X

X XX

XX

OX

XX XO

S -

+

The bond dipoles make an angle of 1200 between them and they cannot cancel each othercompletely. There remains a net(resultant) dipole moment and so the molecule is polar.So a molecule which has a net dipole moment is said to be a polar molecule.Take the case of CCl

4. The shape of carbon tetrachloride is tetrahedral. Carbon atom sits

at the centre and the four Cl atoms sit at the four corners of the tetrahedron.-

-

+

--

Cl

Cl

Cl

Cl

C

There are four C-Cl dipoles and the direction of each dipole moment is from +ve pole(C)to -ve pole(Cl). Since all are C-Cl bonds, the magnitudes of their dipole moments are same.These four dipole moments go away from each other from a common point in asymmetrical manner and hence cancel out together and the net dipole moment becomeszero. Hence CCl

4 is a non-polar molecule, although it has four dipoles. The details about the

calculations on dipole moments (vectors) are beyond the scope of this book.

CONCLUSION:1. Remember one important thing that if a molecule has a symmetrical shape andstructrue (i.e all the dipoles have same magnitue), then the dipole moments cancel outtogether and the molecule becomes non-polar.2. If a molecule has one or two lone pairs and it is not symmetrical structurally,then it is undoubtedly polar.BCl

3 has trigonal planar shape while BeCl

2 has linear shape. All these are symmetrical

shapes. Hence they are also non-polar molecules like CCl4. Similarly PCl

5 is trigonal

bipyramidal in shape and SF6 has octahedral shape which are also symmetrical. They are

also non-polar. The individual dipoles cancel out in all these cases.Examples of non-polar molecules: H

2, N

2, Cl

2, Cl

2, Br

2, SnCl

4, CO

2, CS

2, CH

4(methane),

CCl4, BCl

3, BeCl

2, PCl

5, SF

6, C

6H

6(benzene), C

2H

4(ethylene), C

2H

2(acetylene) etc.The dipole

moment of these molecules is zero.

Examples of strongly polar compounds: H2O, NH

3, HF, CH

3Cl(methyl chloride), HCl

etc.

34

Examples of weakly polar compounds: CO, N2O, PH

3, HI, acetone etc.

The dipole moments of a few common compounds are given in the following table forreference and comparison.

Dipole Moment (m) of Some covalent Molecules( in Debye unit)Covalent dipole Covalent dipolemolecule moment(D) Molecule moment(D)HF 1.75 HBr 0.78H

2O 1.85 HI 0.38

NH3

1.46 H2S 1.1

NF3

0.24 SO2

1.63CH

3Cl 1.86 PH

30.55

HCl 1.03 CH3Br 1.8

HCN 2.9 CH3I 1.35

N2O 0.17 CH

3NH

21.24

CH3COOH 1.74

SAQ 12.(i) Arrange the following in the order of increasing polarity (use the table).

(a)HI, HBr, HCl, HF (b)HF, H2O, NF

3 and NH

3,

(c) CO2, SO

2, H

2S, CH

3Br

(ii) Indicate which are polar and which are non-polar among the following.CCl

4, CO

2, HCl, CHCl

3, BCl

3, BeCl

2, H

2O, NH

3

(iii) Explain why NH3 is strongly polar(μ=1.46D) while NF

3 is weakly polar(m=0.24D).

(iv) Between H2O and H

2S, which is less polar and why?

MODERN THEORY OF COVALENCY:Modern theory of covalency describes the formation of a covalent bond in terms of overlapof atomic orbitals. A covalent bond is formed by the overlapping of atomic orbitals.Theatomic orbital(s,p etc.) of one atom containing an unpaired electron overlaps with atomicorbital(s,p etc.)of another atom containing an unpaired electron to form a covalent bond. Theunpaired electrons of both the atoms get paired and this pair(bond pair) is localised at thecommon space of overlap. Take for example, in H

2 molecule, the spherically symmetrical 1s

orbital of one H atom containing one unpaired electron overlaps with the 1s orbital of theother H atom containing one unpaired electron and the two electrons get paired and remainwithin the common space of the overlap. From the diagram below you find the shaded portionis the common space of overlap between two spherically symmetrical orbitals. The bond pairremains within that common space. So the condition for an orbital for making overlap withanother orbital is that it should contain one unpaired electron.

TYPES OF ORBITAL OVERLAPS

35

1. s-s overlap:When s orbital of one atom overlaps with s- orbital of the other atom, we call it a s-s overlap.The overlap in H

2 molecule shown above is an example of s-s overlap.

2. s-p overlap:When spherically symmetrical s orbital of one atom overlaps with dumbbell shaped p orbitalof the other atom we call it a s-p overlap. In HCl molecule, H contains the unpaired electronin 1s orbital while Cl contains the unpaired electron in the 3p orbital(write the electronicconfiguration and make box diagram of the valence shell to confirm this). When 1s orbitalof H overlaps with 3p orbital of Cl, we get a covalent bond. The electron pair remains withinthe common space of overlap.

3. p-p overlap:When p-orbital of one atom containing unpaired electron overlaps with the p orbital ofanother atom containing unpaired electron, we call it a p-p overlap. Note that both the orbitalsare dumbbell shaped. In Cl

2 molecule, the unpaired electron of each is present in 3p orbital

and the overlap of the two p orbitals gives rise to covalent bonding.

TYPES OF COVALENT BOND ON THE BASIS OF OVERLAP1. Sigma(σσσσσ) bond:When the overlap of the orbitals takes place in a head-on or direct manner, it forms a sigmabond. In all the three examples cited above, namely s-s, s-p and p-p overlaps, the orbitalsoverlap in a head-on or direct manner(just like a head-on bus-truck accident). This type ofoverlapping is very strong and therefore sigma bond is very strong.2. Pi(πππππ) bond:When the atomic orbitals overlap in a side-wise or lateral manner, a pi(π) bond is formed.Take the case of N

2 molecule which has a triple bond between them. Refer its Lewis

structure for the purpose.

z

y

zy

x

36

Look to the electronic configuration of the valence shell of N atom. It is 2s22p3. Accordingto Hund's rule we have three unpaired electrons in each of p

x , p

y and p

z orbitals which are

lying mutually perpendicular to each other. If px orbital of one N atom overlaps directly with

px orbital of the other N atom, both lying along x-axis, then a sigma(σ) bond will be formed.

At that time, the py orbitals of the two atoms which are lying parallel to each other cannot

go for the direct(head-on) overlap, rather they would go for a side-wise overlap to form api(π) bond. Similarly at the same time the two parallel p

z orbitals will overlap in a sidewise

manner to give another pi bond. Note that a p- orbital has two lobes, a head and a tail.The side wise overlap of two p-orbitals will involve a partial overlap of the head portion anda partial overlap of the tail portion as shown below. These two partial overlaps give rise toone pi bond.

zz pp

y

y

pp

In the N2 molecule shown in the first diagram, however, actual sidewise overlapping of the

py-p

y and p

z-p

z has not been shown to avoid clumsiness in the diagram. If the p

y and p

z

orbitals are brought very near as shown in the second diagram, then the diagram will becomevery much unclean. Therefore the overlapping have been shown by two connections for eachpi bond i.e connection of head region and tail region for each pi bond. Thus in N

2 molecule

there is one sigma bond and two pi bonds. Since pi bond is formed by the sidewise overlapping,it is not as effective as head-on overlapping(like a sidewise accident of truck and bus).Hence pi(πππππ) bond is weaker than a sigma bond.We shall not discuss more about the modern theory of covalency now and reserve theremaining part for the higher classes.

HYBRIDISATION OF ATOMIC ORBITALSValence bond theory explained before could explain satisfactorily the formation of

covalent molecules by the overlapping of orbitals. However it failed to explain the shapes andbond angles in a large number of covalent molecules and ions. VBT could not explain whyBeCl

2 is linear, BCl

3 is trigonal planar, CH

4 is tetrahedral, PCl

5 is trigonal bipyramidal and

the shapes of many other molecules and ions could not be explained by VBT. In order toexplain clearly the shapes and bond angles of all covalent molecules and ions Pauling andSlater introduced the the hypothetical concept of hybridisation in VBT. According to this

37

theory, the central atom in a molecule or ion does not often use pure s, p, d orbitals foroverlapping with orbitas of other atoms. In stead it uses hybrid or mixed type of orbitalshaving character of more than one type of pure orbitals. For example, the hybrid orbitalshaving the charcter of both s- and p- orbitals are formed by the hybridisation or mixing ofpure s and p orbitals. Note that during bond formation, the pure orbitals mix up to form hybridorbitals each of which possess the characters of the pure orbitals used for hybridisation.So hybridisation is the process of intermixing of pure atomic orbitals and subsequentredistribution to form a set of equivalent orbitals having equal energy which describea regular geometry in space.

The following table gives the hybridisation of central atom depending on its Steric Number(SN).Steric Number Hybridisation2 sp3 sp2

4 sp3

5 sp3d6 sp3d2

7. sp3d3

sp3 Hybridisation:

C(6) : s p3

︵excited state ︶

px

py

pz

s

+

sp3 hybrid orbitals

In this case one spherically symmetrical s-orbital hybridises with 3 dumb-bell shaped p-orbitals and each hybrid orbital possesses 25% s-character and rest p-character. 4 sigmabonds are formed by the sp3-s overlaps. The shape is tetrahedral

Orbital Diagram of Methane:

C

H

H

HH

109028'

38

Orbital Diagram of NH3:

SN of N = 4, hence hybridisation is sp3

N(7) : 2s22p3

s p3

lone pair

H

HN

H

NH H

H

Three N-H sigma bonds are formed by sp3-s overlaps. One hybrid orbital contains a lonepair. Bond angle is 1070, a bit less than tetrahendral angole due to presence of a lonepair(lone pair-bond pair repulsion being greater than bond pair-bond pair repulsion)Orbital Diagram of H

2O:

O(8): 2s22p4 s p3

lone pairs

H

HO

OH

H

In H2O, there are two O-H sigma bonds and two hybrid orbitals contains lone pairs. The

bond angle in this case is 104.50, little more less than tetrahedral due to presence of two lonepairs.

sp2 Hybridisation:Orbital Diagram of BCl

3 :

B(5) : 2s22p1 s p2 (excited state)

+px

py

sp2 hybrid orbitalss

39

In this case one s-orbital hybridises with two p-orbitals to form 3-hybrid orbitals directedtowards the corners of an equilateral triangle(trigonal planar geometry). Each hybrid orbitalhas 33.3% s-character.

Cl

ClCl

B B

Cl

Cl Cl

1200

There are three sp2-p overlaps to result three B-Cl sigma bonds. The shape is trigonal planar.

Orbital Diagram of Ethylene(C2H

2) :

s px pzpy (excited state of carbon)

C CH

H

H

H

Here there are one C-C sigma bond due to sp2-sp2 overlap, four C-H sigma bonds due tosp2-s overlaps and one pi bond due to the side wise overap of two unhybridised parallel p

z

orbitals.

sp Hybridisation:Orbital Diagram of BeCl

2 :

2s 2p1 1

sp hybridisation

(excited state of Be)

40

+

s-orbital p-orbital two sp hybrid orbitals

In this case one s- and one p-orbital hybridises to give two hybrid orbitals lying in one straightline. Each hybrid orbital has 50% s-character.

Be ClClBe ClCl

In BeCl2 there are two Be-Cl sigma bonds due to sp-p overlaps.

Oribtal Diagram of Acetylene(C2H

2):

s px py pz

(excited state of carbon)

C C HH

In acetylene, there is one C-C bond due to sp-sp overlap, two C-H sigma bonds due to sp-s overlaps and two pi bonds due to p

y-p

y sidewise overlap and p

z-p

z sidewise overlap. It is

a linear molecule.

In the same manner, when steric number is 5, 6 and 7 the hybridisations of central atomswill be respectively sp3d, sp3d2 and sp3d3. We have already studied the shapes of manymolecules and ions having these SN values.

INTERMOLECULAR FORCES

In ionic compounds we found that the positive and negative ions are arranged in an alternatemanner in a regular order in all directions through infinite distance. This is an infinite networkstructure and no single and isolated ionic molecule (say NaCl) exists. But if we consider thecovalent substances the situation is different. Let us take the case of CCl

4 which is a liquid

at room temperature. The bonds within each CCl4 molecule is covalent. There are four

covalent (single) bonds in CCl4 and it has tetrahedral shape. But in a bottle containing a few

millilitres of carbon tetrachloride, there are thousands and thousands of molecules. What kind

41

of relationship exists between these molecules? Are the molecules entirely free and have noattraction between themselves? No doubt, they are independent of each other and existseparately and there is no real chemical bond (ionic or covalent) between the moleculesunlike ionic compounds in which ions are interconnected in an infinite network. Between thecovalent molecules, however, there exists a very weak attraction and molecules are associatedwith each other in a very temporary manner. In some compounds this cohesive associationis extremely weak and in some cases the associations are relatively stronger. But in no casethese attractions are comparable with real chemical bonds such as ionic or covalent bond.These are called the intermolecular forces of attractions which exist only in covalent substancesas the covalent substances exist in molecular form. Depending on the nature of associations,the intermolecular forces are of two types.

(i)Hydrogen Bonding (ii)Vander Waals' Forces.

HYDROGEN BONDING:This is not true bonding and is wrong to call it a bonding. The reader should not considerthis as a real bonding. This is a form of weak intermolecular force of attraction betweencovalent molecules, although it is strongest among all intermolecular forces of attractionwhose energy range lies between 1 to 10 KCal/mole.Conditions:The compound in order to exhibit this type of intermolecular forces should satisfy the followingconditions.(i) The compound should have H atom and one of the most electronegative and smallsize atoms such as F, O or N.(ii) H should be directly attached (bonded) to one of the most electronegative atomscited above(F, O or N).Water, ammonia and hydrogen fluoride satisfy the above requirements and hence are associatedby means of hydrogen bonding.

DONOR-ACCEPTOR CONCEPT IN H-BONDING:In the light of Bronsted-Lowry concept, the H atom of the molecule which provides

the +ve pole is called the donor atom while the electronegative atom(F, O or N) providingthe lone pair(-ve pole) is called the acceptor atom in the H-bonding. In water, H is the donoratom and O is the acceptor atom. Similarly in HF, H is the donor and F the acceptor atoms.

F......H > O.....H > N......H∼10 ∼7 ∼2 kcal/mole

F.....H strength is highest and that of N....H being lowest.Hydrogen bonding in Water(H

2O):

In a sample of water, H2O molecules are strongly associated by H-bonding interactions.

We know that H2O molecule is strongly polar in nature and contains two O-H polar covalent

bonds, in which there is +δ on H and -δ on O atoms. There is attraction between +ve pole(H)of one water molecule and -ve pole(O) of another neighbouring H

2O molecule. This dipole-

dipole attraction(H-bonding) is continued throughout the bulk of the water. O has two lonepairs and has two covalently bonded H atoms. Notice that each water molecule can potentiallyform four hydrogen bonds with surrounding water molecules. Both the two positve poles(Hatoms) and -ve poles(lone pairs) are used up in H-bonding interactions with other molecules.This is why H

2O is a perfect example of H-bonding

42

OH H

OH

HOH

H

+

--

+

HO

H

HO H

All the H2O molecules present in a sample are associated in a gigantic(polymeric)

network by H-bonding i.e +ve and -ve poles of adjacent molecules remain in the attractiveinteractions. The hydrogen bonding interactions are shown by broken lines between oppositepoles of neighbouring molecules. Each H

2O molecule is surrounded by 4 other water molecules

with which it makes H-bonds in a tetrahedral geometry. This is called coordination number(CN)of water which is 4 usually. This number changes with the change of temperature andpressure.The length of hydrogen bonds(.......) depends on bond strength, temperature, and pressure. Thebond strength itself is dependent on temperature, pressure, bond angle, and environment (usuallycharacterized by local dielectric constant i.e polarity of the solvent). The typical length of ahydrogen bond in water is 1.97 Å (197 pm). The O-H covalent bond length is fixed andis much smaller(96 pm) than the variable length of H-bond.Hydrogen bonding in Hydrofluoric acid(HF):

H-F bond is more polar than O-H bond. Hence the H-bonding interaction in HF isstronger than that in H

2O. But the H-bonding in HF is not ideal as that found in H

2O. Each

F has three lone pairs and only one covalently bonded H. So two lone pairs remain idlewithout being engaged in H-bonding. So coordination number of HF is 2 instead of 4 that wefound for H

2O. So the number of H-bonds in HF is half as that in H

2O, although individually

the H-bond is stronger in HF. The H-bonded structure of HF is shown below.

FH

FH

FH

FH95pm 155pm

1160

nThe H-F-H angle is 1160 while F-H-F angle is 1800. In solid state, the value of n is very large.But in liquid state n=5 or 6. Usually HF remains as a hexamer in the liquid state (HF)

6.

Hydrogen bonding in Ammonia(NH3):

Ammonia contains three N-H bonds which are polar in nature. N has only one lonepair and is covalently bonded with three H atoms. Hence the H-bonding in NH

3 is also not

ideal like H2O. NH

3 suffers from double disadvantage of having least strength of H-bonding

and lower degree of H-bonding. Recent research has revealed that NH3 does not show

detectable H-bonding with itself becausue NH3 can act as acceptor but not as donor.

However NH3 forms H-bonding with H

2O which acts as donor for H-bonding.

43

N HH

H H HHN

H HHN N H

HH

+- - - -+ + +

Consequences of Hydrogen Bonding:

(i) Melting and Boiling Points: Physical State:Compounds which exhibit hydrogen bonding have higher boiling points than those which donot exhibit hydrogen bonding. This is because we have to supply large amount of energy tobreak this innumerable hydrogen bonds. It must be noted that individually the hydrogenbonding interaction is weak as it is not a true bonding, but collectively in presence ofthousands and thousands of hydrogen bonding interactions, their cumulative (joint) effectbecomes very strong. So the boiling points of such compounds are relatively higher thancompounds which do not exhibit hydrogen bonding.

(a)H2O boils at 1000C while H

2S boils at -670C. Therefore water at room temperature

is a liquid while H2S is a gas. There is H-bonding interaction present in water but not in H

2S.

H2S, no doubt, exhibits intermolecular interaction (about which we shall study soon) but the

interactions are much weaker than the H-bonding interactions present in water.(b)HF has the boiling point of +190C while HCl has the boiling point of -850C. In cold

seasons, HF remains as a liquid while HCl is always a gas. The high boiling point of HF isdue to the presence of H-bonding in HF.

(c)NH3 has the b.p of -330C while PH

3 has a b.p of -880C. This is due to the

presence of H bonding in NH3.

(ii) Solubility in Water:Compounds which exhibit H-bonding are soluble in water. This is because water also exhibitsH-bonding. If both solute and solvent exhibit H-bonding, then the solute molecules exchangewith solvent molecules and H-bonding interactions are also exchanged. This results in mixingof the solute with the solvent. Just like you make friendship with people of your nature andtemperament and not with people who have completely different nature. In the same waythe nature of intermolecular association within the solute molecules should be same as thatwithin solvent molecules. Then only the exchange between them is possible and dissolutionis possible.

(a) Glucose and sucrose (table sugar) are covalent compounds, but are highly solublein water, whereas CCl

4 is also a covalent liquid but it is not miscible with water. This is

because glucose and sucrose themselves exhibit H-bonding interactions within theirmolecules(they possess a lot of -OH bonds in their structures) and hence are soluble in waterwhich also exhibit H-bonding. However CCl

4 does not exhibit H-bonding and so it cannot

dissolve in water.(b)Ethyl alcohol(C

2H

5OH) which is used as liquor is miscible with water in all

proportions although it is a covalent compound. It is because ethyl alcohol exhibits H-bondinglike water, so they are miscible.SAQ 13:(i) H-bonding term is a misnomer (a wrongly coined term) as it does not represent a

44

real chemical bond. State FALSE or TRUE.(ii) State which of the compounds exhibit hydrogen bonding.

HCl, NH3, PH

3, HF, C

2H

5OH, glucose, H

2O, H

2S

(iii) Explain why glucose being a covalent compound is highly soluble in water.(iv) NH

3 is strongly soluble in water while CO

2 is not. Explain.

(v) Explain why HF is a weak acid while HCl, HBr and HI are strong acids.(vi) Explain why HF has a lower boiling point than water although the strength of H-bondis greater in HF.(iii) Density of ice and water :

We already know that due to directional nature of H-bond and the ideal case of H-bond in H

2O, each H

2O molecule is H-bonded with 4 other neighbouring H

2O molecules

describing a tetrahedral geometry. Two of the H atoms(+ve poles) and the two lonepairs(negative poles) are the four sites of coordination with water molecules. In solid state(iceat 00C), the coordination number of water molecule is exactly 4. The tetrahedral arrangementof the covalently bonded and H-bonded structures of H

2O molecules produce a large number

of adjacent hexagonal holes in the solid state. One such hole is shown below.

OH

H

OH

H

OH H

OH

H

OH

H

OHH

--

-

-

-

-

+ +

+

+

+

+

Thus volume increases in the solid state for which the density of ice at 00C is less(0.9175g/cc). When ice melts at 00C, the density increases to 0.99987 g/cc. This is because, meltingbrings about breaking of some H-bonds(tetrahedral structures) and partial filling of the hexagonalholes with H

2O molecules. The coordination number increases from 4 to nearabout 5. Thus

volume decreases and density increases. As water is further heated, more and more H-bondsare broken and the coordination number goes on increasing maximum upto 6 at 40C at whichthe hexagonal holes are completely filled up and the volume becomes minimum and hencedensity of water becomes maximum(1.000 g/cc). Further increase in temperature beyond40C brings about further breaking of H-bonds and thermal expansion occurs for whichvolume again increases and thus density again decreases with temperature. At 100C thedensity of water is 0.99975 g/cc and it further decreases with increase in temperature.(iv) Acid strength :

HF is weaker acid than other hydrohalic acids. This because of two reasons(i) Due to strong H-bonding network, the dissociation of HF is weak.(ii) Due to high H-F bond energy, the dissociation of HF is weak.

(v) Viscosity : Greater the degree of H-bonding greater is the viscosity of the liquid. Forexample water has a greater viscosity than ethyl alcohol(C

2H

5OH) as the degree of H-

bonding in the former is greater as one +ve pole is absent in ethyl alcohol.(N.B : Readers should note that many other application of H-bonding in organic chemistryhas not been discussed here. You will get in higher standard books)Types of H-bonding:

45

(a) Intermolecular H-bonding :(b) Intramolecular H-bonding.

Whatever discussion we had made on H-bonding is on intermolecular H-bonding, i.e the H-bonding interactions within neighbouring molecules of the same substance. In intramolecularH-bonding, there is H-bonding within each molecule.

CH

CC

C

X

In this case both +ve pole(H) and the negative pole(F,

O, N) are present within the same molecule. Many organic compounds show this type of H-bonding. The consequences are exactly opposite to intermolecular H-bonding. The compoundsshowing intramolecular H-bonding have lower boiling point and less soluble in water thanthose showing the other type. The detailed discussion on this is beyond the scope of thisbook.

VAN DER WAALS FORCES:Covalent compounds which do not exhibit H-bonding interactions, are associated with stillweaker intermolecular forces of attraction. These very weak intermolecular attractions arecommonly called Van der Waals forces. This is divided into two types.(a) Dipole-dipole Interaction (Keesom forces)This exists in polar compounds which do not exhibit H-bonding. This is analogous toH-bonding interaction but is weaker than it. For example, HCl, H

2S, PH

3 etc. which are

analogous to HF, H2O and NH

3 respectively in structure are associated in an analogous

manner as described in H-bonding. But the attractions are much weaker and are not calledH-bonding. They are simply called dipole-dipole attraction.

H Cl H Cl H Cl H Cl H Cl H Cl+ - - - - - -+ + + + +

Due to smaller size of F, O and N(so that the molecules can come very close to each other)as well as their higher electronegativity(so that the magnitude of +δ and -δ charges aremore) the H-bonding interactions in HF, H

2O and NH

3 are much stronger than normal dipole-

dipole attractions present in HCl, H2S, PH

3 etc.. These normal dipole-dipole interactions

belong to the Van der Waals forces.(b) London Forces( Induced dipole-induced dipole interaction):These are the weakest among all intermolecular forces. This type of interaction exists in non-polar molecules like I

2, H

2, N

2, benzene, Br

2, CCl

4 etc. Since non-polar molecules do not

carry any fractional charges(+δ and -δ) on their atoms, we may think that there cannot beany attraction between the molecules. This is not so. Although these are non-polar, theyacquire small polarity on temporarily basis. This temporary polarity constantly fluctuates inits direction. Due to rapid rotation, vibration, translation(movement) of molecules(particularlyin the gaseous state), the electron cloud of the non-polar molecules are temporarily shiftedto one side. As a result a +δ charge develops on the side in which electron density is lessand -δ charge develops on the side in which the electron density is more. This is called a

46

temporary dipole. These temporary dipoles continuously change their directions .

The attractions between the opposite poles of the neighbouring temporary dipoles are veryweak and are called London forces which are named after its discoverer London. Note thatsome authors address the intermolecular forces among nonpolar compounds as Van derWaals forces. This is not untrue. London forces are merely one kind of forces under Vander Waals category. Dipole-dipole attractions are relatively stronger while London forces areweaker. But both belong to Van der Waals type of forces.IMPORTANT NOTE : In dipole-dipole attraction, the major contribution(85%) comes fromLondon forces and rest 15% from the actual dipole-dipole interactions. Hence London formcesare the most dominant among all Van der Waals forces.Note that Van der Waals forces increase with increase in size of the molecules.Example:(i) F

2, Cl

2, Br

2, I

2: If we consider the halogens which are covalent substances, we find

that the melting points increases smoothly from F2 to I

2. While F

2 and Cl

2 are gases and

Br2 is a liquid and I

2 is a solid at room temperature. This is because the size of halogen

increases from F to I.(ii) The boiling point increases smoothly in the order: H

2S< H

2Se < H

2Te. This is

because size increases from S to Te in group VIA and so the Van der Waals forcesincreases and hence the boiling point increases. Note that H

2O has not been included in the

comparison with the other hydrides of the oxygen family members as shown above. Thisis because, H

2O exhibits H-bonding while others exhibit Van der Waals forces. In HX, the

H-bonding increases in the order HCl < HBr < HI and their boiling points also follow thesame order. If we consider only the polarity, the order would have been the opposite. Butin Van der Waals forces, size the the biggest reason. HF has the highest boiling point amongall HX because it exhibits H-bonding.

SAQ 14:(i) Why do covalent substances have lower melting and boiling points compared toionic compounds?(ii) Why is I

2 highly soluble in benzene but very poorly soluble in water?

(iii) Why sugar does not dissolve in benzene or CCl4.

(iv) What kind of intermolecular forces exist in the following:(H-bonding or Vanderwaals forces)I

2, HCl, HF, H

2S, PH

3, H

2O, CCl

4, H

2, NH

3,

H2Se, O

2, benzene, C

2H

5OH, glucose

(v) Indicate which will have dipole-dipole interactions and which will have London

47

forces. Which is weaker interaction? What are both associations commonlycalled?

HCl, I2, H

2S, N

2, O

2, CCl

4, CO

2, PH

3

(vi) Arrange in the order of increasing boiling point among : PH3, AsH

3, SbH

3 and BiH

3.

METALLIC BONDING:This the third type of bonding which is comparable with ioinic and covalent bonding.

The bonding or interaction present between the atoms of a metal is called metallic bonding.The valence electrons of metal atoms are are loosely bound to their nuclei and are

delocolised from their assigned orbitals. In other words, valence electrons are not associatedwith any specific atom, but wander freely among adjacent metal atoms. These electronsconstantly interchange between different atoms. The delocalisaton of valence electrons ispossible due to overlapping of the valence shell atomic orbitals of neighbouring atoms. Forexample, in Na metal, the 3s orbital of each Na atom overlaps with 3s orbitals of eightneibouring Na atoms(since Na has a bcc lattice- refer solid state for the details). Thusinfinite number of overlappings of atomic orbitals occur in the bulk of the metal lattice to forminfinite number of molecular orbitals having very close energy values. Thus the valenceelectrons are free to move about all the molecular orbitals extending the whole metallicstructure.

The phase of the metal consisting of the delocalised valence electrons is the mobilephase called the 'electron sea' or 'electron gas'. Each metal atom releases its valenceelectron(s) to the electron sea or gas, but its core electrons( for example 2s2 2p6 for Na)remain static around its nucleus. Thus metal atoms devoid of their valence electrons remainas positve ions or cations. These are called positive cores or kernels which constitute thestatic phase. The 'electron sea' consitituing the delocalised electrons is free to flow aroundthe regular arrangement of metal cations, but cannot escape completely without getting extraenergy from somewhere else. But the positive ions(cores) remain fixed at their positions aslong as any external stress is not applied.

Metallic bonding is the electrostatic force of attraction between the static phase(positiveions or atomic cores) and the mobile phase(electron gas or sea). This strong force binds allthe atoms together to form a gigantic network. Thus metal consists of a lattice of positveions(not neutral atoms as we generally believe) immersed or suspended in a sea of electrons(freeelectrons).These free electrons move in a raìdom manner in all possible directions.

These free or delocalised electrons present in the metals are responsible to explainall the properties of metals such as

48

(a) electrical (b) thermal conductivity(c) malleability and ductility (d) metallic lustre(e) high melting and boiling point (f) strength(g) hardness

(a) Electrical conductivity : When a PD is applied on a metal, the free electrons whichare moving in a random manner now drift in a particular direction and conduct electricity. Theelectrical conductivity decreases with increase in temperature as the electrons collide morefrequently with vibrating atoms.(b) Thermal conductivity : The free electrons enhance the thermal conductivity ofmetals. Thermal conduction, in contrast to electrical conduction, consists of electrons movingin random directions and generally increases at higher temperatures as more frequent collisionshelp propagate thermal vibrations.(c) Malleability and Ductility: Since the binding forces in metals are non-directional,metals can be reshaped easily and do not fracture when stretched or hammered. Thus metalsare not brittle and can be shaped according to our requirement. When external stress isapplied(hammered into thinner sheet with an impact stress or drawn into wire by appling atensile stress) the postive ions are deformed easily in the electron sea while the electron seais redistributed surrounding the new positions of the cations so as to change the shape of themetal. In other words, the delocalised electrons in the ‘sea’ of electrons in the metallic bond,enable the metal atoms to roll over each other when a stress is applied. Metallic bonding doesnot oppose deformation because a change in the positions of the metal nuclei does notincrease the energy of the structure. The electron sea does not allow the positive ions tocome closer for unfavourable repulsion. The fluid nature of the electron sea is responsiblefor the moldability of metals.SAQ 15: Explain why ionic solids are brittle although ionic bond is non-directional likemetallic bonding.(d) Lustre: Metals exhibit shiny lustre. This is due to reflection of light by the free electronsin all possible directions more correctly the scattering of light by the free electrons. While mostmetals have silvery white and greyish white lustre when freshly cut or polished, a few aredifferently colourd. For example, gold is yellow, copper is red while cesium is blue which dependon the the colour the electron sea reflects.

(e) High Metling and Boiling points:Metals have in general high metling and boiling points like ionic soilds. When a metal is

melted, the metallic bonding is weakened and for that a large quantity of energy is required. Inthe molten state, the metalling bonding still exists but in a weakened state. But when we boil amolten metal, the metalling bonding is completely broken. The network structure of the metallicbond is totally dismantled and individual atoms with their valence electrons are segregated. Toachieve that still greater amount of energy is required. That is why boiling point of a metal ismuch above its metling point i.e the gap between the two points is large. This gap is very smallin nonmetals as for them the weak Van der Waals forces are weakended.

So stronger the metallic bond the greater is the melting point. In fact the strength ofmetallic bond in metals is best know by comparing their boiling points, not metling points.Because at boiling points, the metallic bond is completelely destroyed. Heat of sublimation orheat of vaporisation of metals are estimated to know their relative bond strengths.

49

(f) Strength : The metals have high tensile strength due to high metallic bond strength. Impactstrength is also high due redistribution of atomic cores in the mobile electron sea.(g) Hardness: Usually metals are hard due to greater strength of metallic bonding. Hardnessincreases with increase in the strength of metallic bonding. Group 1 metals are soft because oflow metallic bond strength. Among them Li is the hardest and Cs is the softest.

Relative strength of bonds:IONIC BOND ≈ COVALENT BOND > METALLIC BOND > H-BONDING > VAN

DER WAALS FORCESIt is a common belief that ionic bond is stronger than covalent bond. But actually the

two bond strengths are comparable which usually varies between 250 - 400 kJ on an averagescale. This notion arises from the fact that ionic compounds have much greater melting pointsthan the covalent compounds. When an ionic solid is melted, the ionic bonds in the infinitenetwork are weakened, but when a covalent substance is melted the covalent bonds are notweakened or destroyed, in stead the weak Van der Waals forces between the molecules areweakened. That is why the notion creeps in our minds that ionic bonds are stronger than covalentbonds. In fact, the reverse statement will not be untrue. Take for instance Li+F–, a strongly ionic

compound has the bond energy of 765 kJ/mole, while a C N covalent bond strength is 891 kJ/

mole.Li+ F– (s) + 765 kJ/mole ------> Li(g) + F(g)

C N + 891 kJ/mole ------> C(g) + N(g)

So there are many covalents bonds which are stronger than many ionic bonds. Metallic bond isweaker than these two bonds in a general scale. Look to the heat of vaporisation of the followingmetals in kJ/mole: Hg(61), Mg(148), Ag(285), Au(336), Fe(416), V(514), W(849).Thus we findthe metallic bond strength for most transition metals are even greater than many covalent andionic bonds. Therefore these three bonds namely covalent, ionic and metallic are included undertrue chemical bonding of comparable bond energy. H-bonding and Van der Waals forces are notconsidered as true chemical bonding as there energy values are much less i.e 8.4 - 42 kJ/molefor the former and less than 4.2 kJ/mole for the latter.

50

RESPONSE TO SAQs(Chemical Bond)

SAQ 1:(i) (a) K(19): Valence shell: 4s1(one unpaired electron like Na)

F(9): Valence shell : 2s2, 2p5(three lone pairs and one unpaired electron likeCl)

-+ FK F+ K

K+ has Ar configuration(3s23p6) and F- has Ne configuration(2s22p6)(b) Na(11): 3s1(one unpaired electron);

O(8): 2s2, 2p4(two lone pairs and two unpaired electron)

+

+++

2-

Na O NaNa O Na

O gains two electrons from two Na atoms and become O2- while each Naloses one electron and becomes Na+. Both have stable Ne configuration.

(b) Mg(12): 3s2(one lone pair);Br(35): 4s24p5 (three lone pairs and one unpaired electron like F and Cl)

++

-2+-

Br Mg BrBr Mg Br

Each Br gains one electron and become Br-(Kr configuration) and Mg losestwo electrons and becomes Mg2+(Ne configuration).(ii) Ionic bond is formed between the atoms of each pair given in the question. This is

because the metal atom belongs to group-IA or IIA(strong) and the non-metal atombelongs to VIA and VIIA.

(iii) K+, S2-, Ca2+ :(18) - [Ar]; Mg2+, F -:(10): [Ne]; Rb+:(36): [Kr]SAQ 2: Ionisation energy(IE) of Na = + 496 kjoule/mole (energy absorbed)

Electron affinity(EA) of Cl = -348 kjoule/mole (energy evolved, so -ve sign)Lattice Energy(LE) of Na+ and Cl- = -788 kjoule/mole (energy evolved, so

-ve)Net Energy evolved = Net decrease in energy = +496 -348 - 788 = - 640 kjoules/

mole.SAQ 3:(i) CsCl has lower melting point due to weak ionic bond. This is because the lattice

energy in CsCl is lower.(ii) MgO has stronger ionic bond. This is because its lattice energy is greater due to the

smaller size of Mg2+.(iii) False. Na+Cl- solid also contains Na+ and Cl- ions.(iv) False: conducts electricity because of free ions.(v) False(vi) six and six(vii) Copper conducts electricity because of free electrons.(viii) The force of attraction between the +ve and -ve ions is largely reduced due to high

51

dielectric constant of water as it is a highly polar solvent.(ix) Be has very high ionisation energy for which it cannot lose electrons easily. Socannot form Be2+.SAQ 4: Non-polar: O

2, Br

2, I

2 and N

2Polar: HF, HBr, H

2S

SAQ 5:

(i) PCl3 :

PCl

Cl

ClHClO

2 :

Cl O HO

HClO3:

Cl O HO

O

ClO4

- :Cl OO

O

O

-

For ClO4

-, first you have to write the structure of HClO4 and then rub off H atom.

NO2

- :XX

XX

O ONXX

XX

X

-(first HNO

2 structure to be written)

(ii) N, O and F belong to the 2nd period of periodic table and their valence shells have2s and 2p subshells. There does not exist any 2d subshell in the second shell. If they breakone of their lone pairs, the unpaired electron has to be sent to a vacand d- orbital. Since thereis no d-orbital in case of these elements, breaking of lone pair is not possible.

SAQ 6: (a) (i) O N O N O (ii)

O

O N O N O

O

(b) P and S belong to 3rd period of the periodic table. Their valence shell consists of 3s,3p subshells. 3d subshell also exists there as they are third period elements, but it is vacant.After breaking the lone pair, one unpaired electron can be sent to one vacant d- orbital.Hence P and S can break their lone pairs.(c) When two atoms are brought closer to each other in order to form a covalent bond,four forces come into play. The attractive forces between nucleus of one atom and theelectron of the other atom. The repulsive forces between the electrons and between thenuclei of the two atoms. The net result of all the four forces is attraction. As the twoatoms come closer and closer, the net attractive force increases. This results in the releaseof more and more amount of energy. At the closest approach(the distance between the atomsis called the bond length) the release of energy becomes the maximum. The two atomsremain at that position and a stable covalent bond is formed between them.SAQ 7:The difference between ionic compound like NaCl and covalent compound like HCl is thatin NaCl positive and negative ions exist even in the solid state in the crystal lattice, while inHCl, ions do not exist in any state. Only when HCl is put in water, it undergoes ionisationto produce hydrated H+ and Cl - ions. When ionic solid like NaCl is put in water. The watermolecules surround in the same manner as we found in case of HCl. Negative poles of H

2O

molecules face the positive ion(Na+) while the positive poles of H2O face the negative ion

52

(Cl-). Due to strong attraction, the ions are pulled apart from each other and completelycaptured inside the solvent cage. Thus the ions are not only separated but also hydrated inthe same manner as HCl. In the case of HCl, the covalent bond breaks to form ions whichwere not present before. In case of NaCl, the ionic bond is loosened to form free movinghydrated ions. The mechanism of hydration is similar in both the cases.SAQ 8:(i) NH

3 reacts with H

2O to form NH

4OH which is a weak electrolyte. It

dissociates(although weakly) into NH4

+ and OH- which conduct electricity.(ii) CO

2 is a nonpolar compound while H

2O is a polar solvent. So CO

2 does not dissolve

in water. HCl is a polar covalent compound, so it is soluble in water. Like dissolveslike.

(iii) Weakly soluble: I2, H

2, CCl

4 and N

2; Highly soluble: HI, SO

2, H

2SO

4.

SAQ 9:(i) Molecular geometry gives the actual shape of the molecule.(ii) It will V-shaped(see table) and the bond angle is close to 1200 as the electronicgeometry is trigonal planar. The actual bond angle will be slightly less than that because ofgreater repulsion between lone pair and bond pair(VSEPR theory) than bond pair and bondpair.(iii) Refer the Lewis structure of these three molecules. All of them have tetrahedralelectronic geometry and are expected to have bond angle of 109028'. However there is slightreduction bond angle as we go from CH

4 to NH

3 and a little more reduction in the angle as

we move to H2O. In CH

4 there is no lone pair in the central atom, in NH

3 there is one lone

pair on nitrogen and in H2O, there are two lone pairs on oxygen atom. According to VSEPR

theory, lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion. Thepresence of one lone pair supresses the bond angle from 1090 28' to 1070 in NH

3 and two

lone pairs suppresses the angle little more to 104.50 in H2O.

(iv) Refer the Lewis structure of CO2. The SN of C is 2 and its shape is linear and bond

angle 1800.(v) Refer the Lewis structures of SO

2 and H

2O. The electronic geometry of SO

2 is

trigonal planar(Steric number is 3) and hence its bond angle is close to 1200. However theelectronic geometry of H

2O is tetrahendral(steric number=4) and its bond angle is close to

109028'.(vi) Refer the Lewis structures of NH

3 and BCl

3 given before. The steric number(SN)

in NH3 is 4 and its electronic geometry is tetrahedral. Since there is one lone pair in the

central atom, its molecular geometry becomes pyramidal. However the steric number(SN)in BCl

3, is 3 and its electronic geometry is trigonal planar. Since there is no lone pair, its

actual shape(molecular geometry) is also trigonal planar.(vii) In all the bits of this question, the Lewis structures have been given in dash formand the following abbreviations have been used. While drawing the Lewis structure no carehas been taken to make the structure resemble the true shape of molecule or ion. The stericnumber(SN) has been found for the central atom in each case.

Steric Number of the central atom =SN, Electronic Geometry = EGMolecular Geometry(actual shape) = MG, Bond Angle= BATrigonal planar: TP, Tetrahedral: THTrigonal bipyramidal: TB Octahedral: OH

(a) H C N SN=1+1=2, EG = Linear, MG=linear, BA=1800

53

(b) C Cl

Cl

Cl

ClSN = 4, EG= TH, MG=TH B A = 1 0 9 0

28'

(c) C H

Cl

Cl

Cl, SN = 4, EG= TH, MG=TH BA=1090 28'

(d) C CH

H

H

HIn this case there are two central atoms(C) and we have to

consider the geometry at each carbon separately and so SN is counted separately.SN(at each C)=2+1=3, EG=TP, MG= Planar

(note that TP geometry at each carbon and hence shape of molecule becomes planar. Wedon't write TP because there are four H atoms and two carbon atoms and we thus cannotwrite trigonal planar. Hence the shape is said to be planar(not trigonal planar).

(e) C C HH . In this case, like C2H

4, there are two central atoms(C) and so SN

is counted separately.SN(at each C)= 1+1=2, EG=linear, MG=linear

(since the shape is linear at each carbon atom, the molecule as whole is linear).

(f) S

O

O

O HOH , SN=4, EG=TH, MG=TH, BA=1090 28'

(g)N

O

O O H, SN=3, EG=TP, MG=TP, BA=1200

(h)

2-

XX C

O

O O SN= 3, EG=TP, MG=TP, BA=1200

(i)

2-

XX SO O

O

SN = 4, EG=TH(one lone pair), MG=PYRAMIDAl,

BA ≈ 1090 28' (A little less than it)

54

(j)P

O

O HOH

O

H

SN=4, EG=TH, MG=TH, BA= 1090 28'

(k) O N O H SN=3, EG=TP(one lone pair), MG= V-shaped, BA≈1200

(l) H O Cl SN(O)= 4, EG=TH(two lone pairs), MG=V-shaped,

BA<1090 28'

(m) OH O Cl SN= 4, EG=TH(two lone pairs), MG=V-shaped, BA<109028'

(n)OH O Cl

O

SN= 4, EG=TH(one lone pair), MG= pyramidal,

BA≈109028'

(o)OH O Cl

O

O

SN=4, EG=TH, MG=TH, BA=109028'

(p)Cl F

F

FSN=5, EG=TB(two lone pairs), MG=T-shaped,

BA=90 and 1800

(q) Cl Sn Cl SN=3, EG=TP(one lone pair), MG=V-shaped, BA≈1200

(r) Xe ClCl SN=5, EG=TB(three lone pairs), MG=linear, BA=1800

(s) Xe ClCl

Cl

ClSN=6, EG=OH(two lone pairs), MG=Square planar, BA=900

(viii) All diatomic molecules like CO and NO are linear and so VSEPR theory is notapplied to them.(ix) The molecule which contains one or more unpaired electrons after bond formationbetween the atoms are called paramagnetic molecules. The common examples are O

2, NO

and NO2. Lewis structure does not correctly explain their paramagnetic properties. Look to

the Lewis structure of O2 molecule.

O O In this structure, we do not find an unpaired electron. The two unpaired

electrons of each O atom have been used for double bond formation and each O atomcontains two lone pairs. So where is the unpaired electron? Actually, O

2 molecule has two

55

unpaired electrons. That is why, the example of O2 has not been given anywhere in the text.

SAQ 10: IONIC: NaBr, KF, MgCl2, CaO, HCl(aq), RbF

2

COVALENT: CO2, NH

3, CH

4, SO

2, AgCl, BeCl

2, H

2SO

4(l),

AlCl3(anhydrous), SO

3, PCl

5, HCl(g), Br

2

BOTH IONIC and COVALENT: NH4Cl(NH

4+ and Cl-) H

2SO

4aq. (H+ and

SO42-), NaNO

3(Na+ and NO

3-). Within NH

4+, SO

42-

and NO3-, there are covalent bonds)

SAQ 11: Since the electronegativity decreases from F to I, the polarity also decreaesin the same order: HF>HCl>HBr>HISAQ 12:(i) (a) HI<HBr<HCl<HF (b)NH

3<HF<H

2O

As the electronegativity increases the polarity increases.(c)CO

2 (0 D)< H

2S(1.1D) < SO

2(1.63D)< CH

3Br (1.8D)

CO2 is non-polar(μ=0D) while others are polar. Refer the table to get the order.

(ii) Non-polar: CCl4, CO

2, BCl

3, BeCl

2; Polar: HCl, CHCl

3, H

2O, NH

3

(iii) NF

FF

+-

--

- -

-+H

NH

H+

+

+

You know that the shape of NH3 is pyramidal and there is a lone pair on N atom.

The shape of NF3 is same as that of NH

3. Remember that a lone pair also gives rise to a

polarity and hence dipole moment. The direction of its dipole moment is always from the Nnucleus to the outer part as shown by the upward arrow mark. In NH

3, the three N-H dipoles

point upwards(from +ve H pole to -ve N pole). The lone pair dipole points upwards. Thevector sum of all these dipole moments is large and hence NH

3 is strongly polar. In NF

3, the

three N-F dipoles point downards(from the positve N pole to -ve F pole) while the lone pairdipole points always upwards. The vector sum of these four dipole moments is small andhence NF

3 is weakly polar. The details of vector calculation is beyond our scope. You are

supposed to get only the concepts now.(iv) Since S is less electronegative than O, the polarity and hence dipole moment in H

2S

is less than H2O.

SAQ 13:(i) TRUE(ii) NH

3, HF, C

2H

5OH, glucose, H

2O exhibit Hydrogen bonding.

(iii) Glucose exhibits hydrogen bonding and so also H2O. So glucose is souble in water.

(iv) NH3 is highly soluble in water because both NH

3 and H

2O exhibit hydrogen bonding.

CO2 is a nonpolar molecule and hence is feebly(poorly) soluble in water.

(v) Due to H-bonding in HF, the molecules are associated strongly with each other andhave little tendency to undergo ionisation. In other acids, H-bonding is absent. Theother reason for the weakness of HF is the high bond strength of H-F bond comparedto H-Cl, H-Br and H-I bonds.

(vi) HF has lower number of H-bonds i.e its degree of H-bonding is nearly half as thatin H

2O. Hence its boiling point is less.

SAQ 14:(i) While melting or boiling a covalent substance(say for example we melt I

2 solid and

56

then boil the I2 liquid), the covalent bonds present in individual molecules are not broken.

You know that the covalent molecules are associated by intermolecular attractions namelyH-bonding or Vanderwaals forces. Individually the intermolecular force of attraction is weakwhich is much weaker than the covalent bonds.While melting or boiling a covalent substance,these intermolecular forces(H-bonding or Vanderwaals forces) are weakened or broken.The covalent bonds present in the individual molecules are not broken. Less energy isrequired to break these intermolecular forces, hence their melting and boiling points arelower. In ionic comounds, on the other hand, we have to weaken or break the muchstronger ionic bonds. That is why the melting and boiling points of ionic compounds aremuch higher.(ii) I

2 is a non-polar covalent compound and exhibits Vanderwaals forces among their

molecules. Water is a strongly polar solvent and exhibt much stronger H-bonding interactionwithin their molecules. Since the nature of their intermolecular attractive forces are different,they would not like each other and hence won't mix with each other. On the other handI

2 is readily soluble in benzene or CCl

4, because benzene and CCl

4 are non-polar in nature

and exhibit Vanderwaals forces within them. So nature of solute and solvent are similiar andso the solute molecules can easily exchange with solvent molecules. Therefore I

2 is highly

soluble in them.(iii) Sugar(sucrose) has a large number of -O-H groups and exhibit H-bondinginteractions. Benzene is a non-polar solvent and exhibit Van der waals forces among theirmolecules. The nature of solute and solvent are different. So sucrose is not soluble inbenzene. Now you understand the cause behind the slogan, "LIKE DISSOLVES LIKE".(iv) Hydrogen Bonding: HF, , H

2O, , NH

3, C

2H

5OH, glucose

Vanderwaals forces: I2, HCl, H

2S, PH

3,CCl

4, H

2,H

2Se,O

2, benzene

(v) Dipole-dipole interactions: HCl, H2S, PH

3 because they are polar molecules.

London Forces: I2, N

2, O

2, CCl

4, CO

2 because they are non-polar

molecules.London forces are weaker than dipole-dipole interactions. They are together knownby a common term Van der waals forces.

(vi) PH3 < AsH

3 < SbH

3 < BiH

3 (As size increases V.W forces increases)

Note that the position of NH3 is third as it exhibits poorest kind of H-bonding.SAQ 15 :• Deforming an ionic solid pushes ions of the same charge closer together• The repulsion from these like charges opposes any deformation• This explains why ionic solids are brittle - they will resist deformation strongly up to a point, but beyond that point, they fail.

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