CHEMICAL BONDING - docshare04.docshare.tipsdocshare04.docshare.tips/files/20735/207353112.pdf ·...

CHEMICAL BONDING

In this chapter we shall be studying the following major articles.(1) Ionic Bond(2) Covalent Bond(3) Hydrogen Bond(4) Polarity of Molecules

So let us start with the study of ionic bond and properties of ionic compounds.Thus the complete study of ionic bond can be divided into two parts.

A. Conditions of formation of ionic bond.B. General properties of ionic compounds.

A. CONDITIONS OF FORMATION OF IONIC BOND.i. The difference of electronegativity between the combining elements must

be large. For example CCl4 is a covalent substance because the differenceof electronegativity between C (2.5) and Cl (3.0) is only 0.5. But NaCl is anionic compound because the difference (Na = 0.9) and (Cl = 3.0) is of 2.1units. The electronegativities are on Pauling's scale. Approximately if thedifference is more than 1.7 then ionic bond formation is expected. Thusone element is electropositive then the other the other one must beelectronegative.

ii. Ionisation energy of electropositive element must be low so that it caneasily lose the electron(s) to form the required cation.

iii. Electron affinity or quantitatively known as electron gain enthalpy ofelectronegative element must be numerically high so that theelectronegative element may easily gain the required number ofelectron(s) to form stable anion.

iv. Since noble gas configurations (eight electrons in the outermost shell,except He which is only two) are the most stable configurations, hence thetransfer of electron(s) from electropositive atom or group toelectronegative atom or group should be such that cations and anionsacquire stable configurations. For example in the formation of NaClsodium atom transfers its valence electron to chlorine atom. The cationsand anions both acquire stable configurations as shown below

Na+ = 1s2, 2s2p6

Cl– = 1s2, 2s2p6, 3s2p6

CHEMICAL BONDING

v. The cations and anions thus formed attract each other to form ioniccrystals. The energy released in this process is called Lattice Energy. Forthe formation of stable ionic compound, lattice energy must be high.

LATTICE ENERGY: It is defined as the amount of energy liberated (enthalpychange) when cations and anions in the gaseous state combine to form one moleof ionic crystals.For example:

Na g Cl g Na Cl s 1

L.E.H 777.8 KJ mol Some more examples of L.E. in KJ mol–1 are given below:LiF = 1033 KF = 813.4 CsF = 747.7LiCl = 845.2 KCl = 708.8 CsCl = 652.3LiBr = 797.9 KBr = 679.5 CsBr = 632.2LiI = 739.7 KI = 640.2 CsI = 601.2

NaF = 915 RbF = 778 BeF2 = 3505NaCl = 777.8 RbCl = 686.2 MgF2 = 2957NaBr = 739.3 RbBr = 659 CaF2 = 2630NaI = 692.0 RbI = 622.2 AlF3 = 5215

Calculation of Lattice EnergyIt is best calculated with the help of Born-Haber cycle using Hess Law. Forexample for NaCl.

fH

s 2 g s1Na Cl Na Cl2

=

sub.H H dissociation2

Na(g) Cl(g)

L.E.

I.E. e e E.A.

gNa gCl

rding to Hess Law

CHEMICAL BONDING

diss.f sub. I.E. EA L.E.

HH H H H H2

since f (NaCl) = – 411 KJ mol–1

sub. (Na) = + 107 KJ mol–1

I.E. (Na) = + 496 KJ mol–1

disso. (Cl2) = + 244 KJ mol–1

E.A. (Cl) = – 355 KJ mol–1

L.E.244411 107 496 355 H2

1

L.E. Na ClH 781 KJ mol

Another convenient but approximate method is by Kapustinskii Equation,according to this

L.E. = 1120 200 Z Z 34.51 KJ molr r

where ν = number of ions per molecule e.g.NaCl = 2, BaCl2 = 3r = inter ionic distance in the crystal= r+ + r– i.e. sum of ionic radii in terms of pmZ + & Z – are the valence charges of ions.Since for NaCl,

Na Clr r in NaCl crystal is 281 pm.

120 200 2 1 1 34.5L.E. 1281 281

1740 KJ molor 1

L.E.H 740 KJ mol

Factors affecting L.E.

I. Inter ionic distance 1r r ; L.E.

r r

II. ionic charge; more are the charges ion of the ion more is the L.E. e.g. forLiF it's value 1033 but for MgO it's is value is 3933 KJ mol–1.

CHEMICAL BONDING

III. Coordination number of ions in the ionic crystal; more are thecoordination numbers, more is L.E.

IV. There is one important generalization about L.E. specially applicable toAlkali metal halides (M+X–) that if

Mr is kept constant then as

Xr increases

L.E. decrease. Similarly ifXr is constant then as

Mr increases, L.E.

decrease. For exampleNaF, NaCl, NaBr, NaIL.E. (KJ mol–1) 915, 777.8, 739.3, 692

Mr = constant

Xr = increasing

L.E. is decreasing.On this basis we answer the following question–

Q.

2

2

RbI Cl

Rb ICl or

RbCl ICl

The answer is RbCl + ICl because L.E. of RbCl is more than that of RbI, hence RbClis more stable, so this is the product with ICl.

General characteristics of Ionic Compounds:(i) Physical state: Most of the ionic compounds are in solid crystalline state

at ordinary temperatures. Few ionic compounds are found in thesolution phase for example HCl (aq.) consists of H+ & Cl– in aqueousmedium. In solid crystalline state, the arrangement of ions is regular

depending on the radius ratiorr

as we have studied in solid state.

Radius of an ion depends on its effective nuclear charge (Zeffe.) i.e.

ion1r

Zeffe.

But Zeffe. = Z–Swhere Z = atomic number

CHEMICAL BONDING

and S = screening or shielding constant for the ion. The value of 'S' iscalculated theoretically with the help of some empirical rules given byslates. These rules give quite good results for ions of lower atomicnumber. This is calculated for all electrons except the one that is theoutermost hell electron.

The important rules to be used at present are:(1) All the other electrons in the (ns, np) group (leaving one) contribute to the

extent of 0.35 (except for is for which the value is 0.30).(2) (All the electrons in the (n–1) th shell contribute 0.85 each.(3) All the electrons in (n–2)th shell or lower contribute 1.0 each towards

screening constant (s)let us calculate 'S' for Na+

Na+ = 1s2, 2s2 p6

Leave one electron∴ S = 7 × 0.35 + 2 × 0.85 = 4.15Isoelectronic ions with same configurations are bound to have samevalue of 'S'.

Let us solve the following question–Q. In Na+F– crystal, the interionic distance is 2.31 Å. If 'S' for Na+ be equal to 4.15

calculate the radius of Na+ as well as that of F–.

Solution: ion1rZ S

ionCor rZ S

Where 'C' is simply proportionality constant

Na

Cr11 4.5

F

Cand r9 4.15

C Cor 2.31 given6.85 4.85

C 6.56

CHEMICAL BONDING

HenceNa

6.56r 0.9576 Å6.85

Fand r 1.3525 Å Ans.

(ii) Melting and boiling points:Since there are strong electrostatic attractions between the oppositelycharged ions in an ionic solid, hence their melting points and boilingpoints are much higher than most of the covalent compounds in whichmolecules are held by weak van der Waals attractions.For example melting point of NaCl is 1073 K while that of Urea is 406 K.

(iii) Ionic Reactions are faster than Molecular ReactionsFor example if we add some NaCl to an aqueous solution of AgNO3, awhite precipitate of AgCl is formed within no time.

3 aq. aq. s 3 aq.Ag NO Na Cl AgCl Na NO

white ppt.

On the other hand a reaction between H2 (g) and I2 (g) to form HI (g) iscomparatively very slow. Reactions of covalent compounds are calledmolecular reactions and these are slow because covalent bonds inreactants are broken and then new bonds for products are formed. Thisall takes time. White in case of ionic solutions ions are readily availableto react.

(iv) Electrical conduction:Ionic solids cannot conduct electricity through them due to fixedpositions of ions in them. But in molten state or in solution phase theyare very good conductor of electricity due to the movement of ions. Butcovalent solids are mostly non-conductors because neither free ions norfree electrons are available in them for conduction of current. Someexceptions like graphite are of course available which conduct electricitydue to mobile electrons in that.

(v) Isomorphous nature:Those substances which have similar type of crystal structures are calledisomorphous. Many ionic compounds show isomorphism for exampleFeSO4.7H2O and ZnSO4.7H2O generally isomorphous ionic solids have

CHEMICAL BONDING

same formula type, similar relative size of ions and similar polarizingproperties. That’s why BaSO4 and KMnO4 are isomorphous but NaClO3 isnot with NaNO3.It is because 3ClO is pyramidal but 3NO is planar.

Simple ionic compounds like Na Cl (not the ionic-coordinationcompounds) do not show Isomerism. It is a characteristic of manycovalent compounds.

(vi) Hard and Brittle:Ionic solids, due to strong electrostatic attractions between oppositelycharged ions in them, are hard. But they are brittle also because due tohammering (some force being applied) a layer of ions in the crystalmoves on the other layer such that similarly charged ions come closer toeach other, where due to repulsion between them the crystal breaksdown. Thus ionic crystals can be brought into the powered form.Remember metals are also hard but they are ductile i.e. converted intosheets.

(vii) Solubility :Ionic compounds prefer a polar solvent for dissolution to a non-polarsolvent. That’s why water is he best solvent for them. However theirsolubility in a solvent depends on:

1st. Dielectric constant of the solvent: It is defined as the efficiency of asolvent to break the electrostatic attractions between oppositelycharged ions of a solute when placed in it. More is the dielectric const. ofa solvent better it is a solvent for ionic compounds Dielectric constant forH2O is about 80, C2H5OH has value of 27 and C6H6 only about 6 units.

2nd. Heat of salvation of ions:Once a solute dissociates into its ions in a solvent, then these ions areimmediately solvated molecules. Solvation of ions is always anexothermic process. If solvent is water it is called hydration. Smaller isthe size of the ion, more is the heat of salvation or hydration.That’s why the heat of hydration of Mg2+ is 1908 KJ mol–1 for Na+ it is 406KJ mol–1 and for Cs+ it is 255 KJ mol– because as regards size Mg2+ < Na+ <Cs+. Among the alkali and alkaline earth metal ions, the largest heat ofhydration is for Be2+, it being the smallest in the entire s-block (exceptfree H+).

CHEMICAL BONDING

Therefore as a general rule we can say that an ionic compound is likely tobe soluble in water of numerically.

heat of hydration of ions > lattice energy.The dissolution of an ionic compounds may be exothermic orendothermic, it all depends on the values of ∆Hhydration and ∆HL.E., as isclear from the following table.

Tablesoln. of some ionic compounds in water

S.No. Compound ∆HL.E. (KJmol–1)

∆Hhyd. (KJmol–1)

∆H(soln.) – ∆Hhyd. – ∆HL.E.

(KJ mol–1)1. NaCl –777 –773 +42. LiF –1033 –1005 +283. KF –813 –819 –64. RbF –778 –792 –14

Thus we see that dissolution of NaCl and LiF are endothermic but dissolution ofKF and RbF are exothermic.But an ultimate explanation for the solubility of an ionic compound can beprovided by thermodynamics where we have read that for the occurence of aspontaneous process ∆G must be negative.

soln. soln.G H T S

and soln. hyd. L.E.H H H

[Put values with sign)As regards ∆S(solution):Remember crystal breaking into ions and mixing of ions in water (solvent) areboth ∆S > 0 processes. But ordering (hydration) of water molecules around theions is ∆S < 0 process.Smaller is the ion with more charge, better is the ordering of solvent moleculesaround it and hence more negative is –1 charge overall∆S(soln.) is > 0.Thus for Na+Cl–, ∆H(soln.) = + 4 KJ mol–1

and ∆Ssoln. > 0.Also T is bound to be (+) ve∴ T∆S > ∆H

CHEMICAL BONDING

hence ∆G for dissolution of NaCl will be (–)ve. That’s why NaCl is readily solublein water.Some important trends of solubility of ionic compounds in water. (Veryimportant from practical point of view.)(i) Many salts of ammonium and alkali metal ions are soluble in water

except LiF, Li3PO4, Li2C2O4, Li2CO3.(ii) All nitrates and almost many perchlorates and acetates are soluble in

water except, the CsCl4 and acetates of Ag(I) and Fe(III)(iii) Many halides are water soluble except, chlorides, bromides and iodides

of Ag(I), Hg(I), Pb(II) and Cu(I). Remember AgF is soluble in water butfluorides of Ca(II), Sr(II) and Ba(II) are insoluble.

(iv) Many sulphates are soluble except sulphates of Ba(II), Ag(I), Sr(II), Pb(II)and Hg(I).

(v) NH4OH, alkalimetal hydroxides, Ba(OH)2, Sr(OH)2 are soluble in water.Ca(OH)2 is slightly soluble. Other metal hydroxides are insoluble.

(vi) Ammonium and alkali metal carbonates are soluble.(vii) (NH4)2S and alkali metal sulphides are soluble.(viii) (NH4)3PO4 and alkali metal phosphates are soluble.(ix) Some solubility (g/L) order are given below:

LiF < LiCl < LiBr < LiILiF < NaF < KF < CsFAgI < AgCr < AgCl < AgFBe(OH)2 < Mg(OH)2 < Ca(OH)2 < Sr(OH)2 < Ba(OH)2

CsClO4 < RbClO4 < KClO4 < NaClO4 < LiClO4

BaSO4 < SrSO4 < CaSO4 < MgSO4 < BeSO4

CHEMICAL BONDING

Part IICOVALENT BONDWe have already noted that if the difference of electronegativity betweencombining elements is low then covalent bond is formed. But combiningelements both must be electronegative not electropositive like metals whichform metallic bonds the nature of covalent bond is not so simple as that of ionicbond. Hence in this part we wish to study, nature of covalent bond and shapes ofmolecules or molecular ions formed by covalent bonds. The following foursystematic attempts have been made to do this job. These are known as conceptsof covalent bonds, because it is impossible to view inside a molecule during itsformation that how the bonds are being formed. Of course the final outcome ofthe formation of bonds is a molecule or molecular ion which can be analysed forits shape and geometry with the help of many physical methods includingspectroscopy. Therefore the success of a concept lies in the fact that how much itcan explain the observed physical as well chemical properties of the covalentcompounds?The four concepts which we must study are:(i) Lewis concept – concept of electron sharing(ii) Valence – shell electron pair repulsion (VSEPR) model(iii) Valence Bond Theory (VBT)(iv) Molecular Orbital Theory (MOT)Let us begin with Lewis concept.(i) Lewis Concept:

The first systematic approach for explaining the nature of covalent bondwas made by G.N. Lewis. According to this concept, a covalent bondbetween two atoms similar or dissimilar is formed by sharing of twoelectrons each providing one electron for example:H2 : H : H or H—HFurther sharing of electrons must be such that each atom acquires anoctet of electrons around it. This is also known as octet rule. The basis ofoctet rule is the stable configuration of Noble Gase.Thus

: F : : F : : F : F : or F—F

CH4 (Methane)

CHEMICAL BONDING

Diagram hereOctet around carbon is complete 'H' atoms can acquire only Heconfiguration.But many exceptions of this octet rule were discovered for example.

BF3 (Boron trifluoride):

Diagram hereThough octet around is 'F' is complete but not around the control atom'B'.

NH3 (Ammonia):

diagram here.Here octet around 'N' is complete but not & by four bond pairs. Thereare three bond pairs and a lone pair.

PCl5 (Phosphorus pentachloride):

diagram hereThere is expansion of octet around 'P'. It has a surrounding of 10electrons.Similarly expansions of octet are found in molecules like SF6 and IF7.Well, all the structures given above and similar are called Lewisstructures or electron dot structures. These are very helpful specially inOrganic Chemistry.However, the main objections about this concept are that it does notexplain the nature of sharing between two similarly charged particlesand moreover does not given any clue regarding the geometry or shapeof the molecule. It dos not say any thing about the relative strengths ofdifferent covalent bonds.

(ii) Valence-shell Electron Pair Repulsion (VSEPR) model:This model was given by Gillespie and Nyholm. This is a simple andreliable method for predicting the shape of molecules and polyatomicions. It is based on Lewis dot structures of molecules or ions and takesinto account the bond pairs (shared electron pairs) and lone pairspresent in them.

CHEMICAL BONDING

The central idea of this model is that a molecule or polyatomic ionwould be stable only when the electron pairs are so oriented thatrepulsions between them is minimum.Further, since a bond pair of electrons is concentrated in the bondingregion between the two atoms by the strong attractive forces of twopositive nuclei of the bonded atoms, hence it is relatively compact. Onthe other hand for a lone pair present on an atom, there is only onenucleus attracting this pair, hence lone pair or pairs are less compactthan bond pairs. Therefore the relative strengths of electron-pairrepulsions are(lone pair – lone pair) > (lone pair – bond pair) > (bond pair – bond pair)orL–L > L–B > B–BBefore we proceed for predicting shapes of molecules on the basis ofVSPER model, we must clearly make a distinction between the followingtwo terms:

(i) Electron pair geometry or simply geometry:It includes the spatial positions of all the bond pairs and lone pairs ofelectrons around a central atom of the molecule or ion, whose structurewe wish to discuss.

(ii) Molecular geometry or shape:It means arrangement of its atoms in space. Thus the positions occupiedby lone pairs are not specified when we describe the shape of amolecule.For example for Ammonia (NH3) the geometry is Tetrahedral but shape isPyramidal.

Geometrics and shapes of different types of molecules or polyatomic ions:For this purpose let us suppose 'A' is a central atom and 'B' are atoms bondedwith it. 'L' represents a lone pair on the central atom. So following cases mayarise.(i) AB2 :

The central atom ‘A’ is bonded with two 'B' atoms and 'A' does not haveany lone pair on it. It means there are only two bond pairs in themolecule; the repulsion between them can be minimum only when theyare maximum far apart. Such a situation can be explained only by lineargeometry or shape of AB2. i.e.

B x—• A •—x Blinear geometrylinear shape

The bond angle which is defined as the angle between the two adjacentbond pairs is 180°, presently do not consider multiple bonds, we shallhave their study as σ and π bonds later on.AB3 (Geometry: Triangular planar or trigonal planar)Now there can arise following two situations.

(i) AB3 (ii) AB2Lexample: BF3 SnCl2

For BF3 For SnCl2geometry and shape will geometry will be triangularbe same as shown below: planar but shape is called V-

shape as shown below: A| 120⁰ B B

The bond angle is expectedto be less than 120° due to

B B

B

A

Triangular Planar unequal repulsions.(L–B > B–B)

AB4 (Tetrahedral Geometry)Following four cases are possible.Formula: AB4 AB3L AB2L2 ABL3

Shape: Tetrahedral Pyramidal V-shape Linear

diagram hereRegular tetrahedron, bond angle 109.5°Compression by one lone pair∴ bond angle is lessCompression, by two lone pairs bond angle is further less than 109.5°.AB5 Trigonal bipyramidal geometry or TBP)Following four cases are possible.Formula: AB5 AB4L AB3L2 AB2L3

Shape: TBP Seesaw T-shape LinearExample: PF5 SF4 ClF3 XeF2 (Draw images exactly as given in hard copy data)

Now before we draw their figures it is necessary to look at the geometry oftrigonal or triangular bipyramid.As shown below there are three corners of a equilateral triangle at the centre ofwhich the atom 'A' is situated. The three 'B' atoms are occupying the threecorners of this triangle. The remaining two 'B' atom are above and below thecentre.

diagram here.Thus there are two types of positions in th above geometry. The corners of thetriangle are called Equitorial positions. While the positions above and below thecentre are called Axial positions. then in order to have minimum repulsion in themolecule, lone pair or pairs (maximum three) must occupy the equatorialpositions. Further if all the 'B' atoms are not same then more electronegativeatom/s will occupy the axial positions (This will be explained with the help ofhybridization of orbitals in VBT later on). Finally it is the nature of TBP that axialbonds are longer than the equatorial bonds.So now we are in a position to draw molecular shapes for all the four cases of AB5

types. They are as shown in the following figure:

diagram here.Fig. Molecular shape for AB5, AB4L, AB3L2 and AB2L3

It may be further noted that bond angles in all cases except AB5 will deviate fromtheir standard values due to the presence of cone or more lone pairs. That’s why,

the equatorial bonds angle in SF4 is 103° FClF is about 87° in ClF3. Ofcourse nosuch problem will arise in XeF2 due to the symmetry.

AB6 [Octahedral geometry]Here following three cases are possible.Formula : AB6 AB5L AB4L2

Shape : Octahedral Square pyramidal Square planarExample : SF6 : BrF5 XeF4

Now it is to be noted that unlike triangular bipyramid, an octahedron has nodistinct axial and equatorial positions. Here all the six positions are equivalent.The six 'B' atoms bonded to central atom 'A' can be placed on three axes at rightangles to one another (the x, y and z axes) and are equidistant from the centralatom.Now in case of AB5L, it makes no difference, the lone pair (L) may occupy any ofthe six positions. But in AB4L2, the two lone pairs must be at maximum distancefrom each other so as to have minimum repulsion. This is best achieved byplacing the lone pairs above and below the square plane which contains 'A' at thecentre and four 'B' atoms at its four corners.Therefore the figures for all the three cases can be drawn as given below:

diagram here.Fig. : Molecular shape for AB6, AB5L and AB4L2.

AB7 (Pentagonal bipyramidal geometry)So far we have only two types of cases for thisFormula : AB7 AB6LShape : pentagonal bi- distorted octahedralpyramidalExample : IF7 XeF6

A pentagonal bipyramid will have a regular pentagonal in a plane such that atom'A' is at its centre and five 'B' atoms are occupying its five corners. The remainingtwo 'B' atoms are above and below the centre of the pentagon. Thus in AB7 typeof molecule there are two types of bond angles, 72° in the plane or say betweentwo equatorial bonds and 90° between an axial and a equatorial bond. Obviouslythere are two types of bond lengths also.But the structure of AB6L cannot be predicted correctly by VSEPR model. Nodefinite location of lone pair can be predicted due to over-crowding of 6 'B'atoms around the central atom 'A'.However the structure of an example of the type :XeF6 is said to be distortedoctahedral as shown below:

diagram here.pentagonal bipyramidal structure of IF7

Distorted octahedral structure of : XeF6.(iii) Valence Bond Theory (model) or VBTThis theory was mainly developed by Pauling. According to this theory:(a) A covalent bond between two atoms is formed by maximum partial

overlapping of two orbitals, each atom providing one orbital containingone electron. The overlapping means to form a region between thenuclei of the two atoms in which the probability of finding sharedelectrons (bond pair) is maximum.

(b) More is the extent of overlapping, stronger is the bond.(c) If two atomic orbitals overlap along the molecular axis (the line joining

the nuclei of two bonded atoms) the bond formed is called σ and if theyoverlape perpendicular to the molecule axis, the bond formed is calledπ.

(d) σ bond is stronger than π.(e) If there is overlapping between a filled and a vacant orbital then a

coordinate bond is for.

On the basis of above points, e can explain the formation of molecules like H2,HCl(g) and Cl2 as given below:H2 (S–S overlapping) : as shown in the figure below:

diagram here.HCl gas molecule (s-p overlapping)H = 1s' Cl = 1s2, 2s2p6, 3s2p5

Z = 1 Z = 17

Thus we see that H-atom has 1s orbital available for overlapping and Cl-atom hasone of the 3p orbitals (containing one unpaired electron) available foroverlapping. The '3p' orbital may be 3px or 3py or 3pz.If we consider X-axis to be the molecular axis then the orbital or Cl available foroverlapping or bonding will be called a 3px orbital. This overlapping is shown inthe figure below:

diagram here.Cl2 molecule (p-p overlapping) : It is shown in the figure given below :

diagram here.NOTE:(i) In all the above three cases we are getting one single bond along the

molecular axis. Hence each bond is abetween two atoms in always a

(ii) The overlapping between two atomic orbitals will take place only whenthe sign of their wave functions are same. e.g. following overlapping arenot permitted.

diagram here.Net overlape = Zero Not allowed.

(iii) Provided the principal quantum numbers of s & p orbitals are same, theextent of overlapping has the order.p p p s s s

But this concept of overlapping of pure orbitals cannot explain the moleculargeometry, bond strengths etc. of all the molecules.For example let us consider the case of BeF2 molecule. According to VSEPR modelwe have seen that it must be a linear molecule. Let us now develop its structurewith the help of orbital overlapping.Be = 1s2, 2s2p0

Z = 4

(ground state)Since Be atom has to form two covalent bonds with two fluorine atoms, hence itmust have two unpaired electrons. For this it is supposed that Be atom is firstexcited by promotion of a electron from 2s to 2p subshell. The energy needed forthis excitation will be compensated by the energy released due to formation ofbond. We will have to do such excitations at many places in future examples butone thing in certain that excitation of electron/s with change in principalquantum number (n) is not permitted because it requires much more energywhich cannot be compensated by bond energy.Thus,Be = 1s 2s 2p

Z = 4

(excited state)and F = 1s2, 2s2 p5

Z = 9

So each 'F' atom has a '2p' orbital say 2px (x-axis being molecular axis) whichcontains unpaired electron.Hence BeF2 can be viewed as, Be forming two covalent bonds around it linearlyby overlapping of 2s orbital of Be with a 2p orbital of F on one side and 2p orbitalof Be with a 2p orbital of another 'F' on the other side, s shown in the figurebelow:F (2p—2s) Be (2p—2p) FSince p-p overlape is more than s-p overlape hence in BeF2 molecule the two Be-Fbonds must be of different strengths. Is it so really? The answer is no.

Then, how to make the two bonds of equal strength? The only solution is that Bemust provide similar orbitals on both sides of it along the molecular axis tooverlape equally on both sides with 2p orbitals of two 'F' atoms.To solve such problems, Pauling proposed the concept of hybridization of atomicorbitals.According to that, in the case of BeF2, if 2s and 2p orbitals of Be combine to formtwo new orbitals each possessing 50%, 50% 5 & p character and be linearlyarranged about the Be atom to overlape with 2p orbital of F atoms, then both thebonds will be of equal strength. Such a mixing of a s-orbital with a p-orbital of thesame atom is called sp hybridization.So let us now first understand the main points of hybridization of orbitals.Hybridization of atomic orbitals:(i) It is a hypothetical concept.(ii) Only those orbitals of an atom whose energies are comparable take part

in this concept.(iii) It is a concept of mixing of orbitals, hence it is immaterial whether

orbitals are empty or possess single electron or even pair.(iv) The number of hybrid orbitals produced will be equal to the total

number of orbitals taking part in hybridization.(v) The hybrid orbitals produced as result of some type of hybridization will

be of same energy (degenerate).(vi) The electrons possessed by parent orbitals will be filled in he newly

formed hybrid orbitals according to Hund's rule.(vii) All hybrid orbitals have directional character i.e. these will be directed in

the space about the atom depending on the type of hybridization.Since hybrid orbitals can overlape (due to specific direction) to a largerextent than the pure orbitals, hence bonds formed by hybrid orbitals arestronger and directional. Following table gives the geometries ofdifferent types of hybridization.

Table (Geometry of different hybrid orbitals)

diagram here.

sp Linear

sp2 Planar

sp3 Tetrahedral

sp3d Trigonal bipyramidal

sp3d2 Octahedral

NOTE:(i) Besides sp3d2, sp3d3 hybridization is also possible; the seven hybrid

orbitals thus produced are directed towards the seven corners of apentagonal bipyramid from its centre.

(ii) In case of hybridizations involving’d’ orbitals i.e. sp3d, sp3d2 and sp3d3,can also be presented as dsp3, d+2sp3 and d3sp3 respectively for simplecovalent compounds. But in case of metal complexes (coordinationcompounds) these will have a difference that in former outer ‘d’ orbital/sand in later inner’d’ orbital/s will be used.

(iii) Though in most of the cases, π-bonds are formed by pure 'p' or’d’orbitals (s-orbitals do not form π-bonds) but in some rare cases, hybridorbitals can also form π-bonds.

(iv) If we can draw Lewis dot structure for a molecular or polyatomic ion,then the type of hybridization, the central atom will show can be decidedby Steric Number Rule which was given by Gillespic. According to thisrule:Steric Number of an atom = number of atoms immediately bondedwith that atom + number of lone pairs of electrons on that atom.(This rule does not work for odd electron molecules and free radicals)

Thus steric number of Nitrogen in NH3 according to Lewis dot structure is

diagram here.= 3 + 1 = 4For 'S' in SO2

i.e. = 2 + 1 = 3For 'Xe' in XeF2

i.e. = 2 + 3 = 5and so on.

Steric numbers are generally calculated for the central atom of the moleculebecause that decides the geometry of the molecule.Steric number is related to hybridization in the following manner.Steric Number hybridization2 sp3 sp2

4 sp3

5 sp3d6 sp3d2

7 sp3d3

Now we shall discuss the individual steric number and the correspondinghybridization.sp hybridization (steric No. = 2)Geometry = Linear; shape = Linearstandard Bond angel = 180°Some examples are given below with their structures:Example StructureBeF2 F—Be—FCO2 O = C = O

CO O C

C2H2 H—C —HCH2 (Triplet carbine) H—C—HN2 : N

3N (azide ion)

— —

sp:N: N N:

2NO (nitronium ion) O N: O

Miscellaneous examples:SiH3HCO (silyl isocyanate)CH3CN ( ethane nitrile or methyl cynide)sp2 hybridization (steric No. = 3)In this, as discussed earlier in VSEPR model following two types of cases areavailable.(i) AB3 (ii) AB2Lgeometry: Triangular planer Triangular planarshape: Triangular planer V-shape

ideal bond angle 120° bond angle < 120°Some example of AB3 type in which 'A' os central atom which is sp2 hybridisedare given below:C2H4 H H

C = CH Hsp2 sp2

C4H6 each (c) is sp2

23CO O

||O– O–

Graphite an allotrope of carbon inwhich each (c) is sp2

(detailed study in IV groupof p-block)

BF3 F|BF F

SO3 O||SO O

3NO ON—O–

O(oxy acids of Nitrogen)

Some examples of AB2L type are : (all with V-shape): CH2 (singlet carbine) CSO2 S

O OSnCl2 Sn

Cl ClNOCl N

Cl O

O3 OO O

2NO N—O–

OSome miscellaneous examples are:B3N3H6 (inorganic benzene or borazine)(SiH3)3N (trisilylamine) N3H (hydroazoic acid)CH3CONH2 (acetamide)H3BO3 (boric acid) FNO3 (fluorine nitrate)CH3NCO (methyl isocyanate) ClO2 (Chlorine dioxide)sp3 hybridization (steric No. = 4)Geometry: Tetrahedral; standard bond angle 109.5°Shapes: AB4 (Tetrahedral) e.g. CH4

AB3L (Pyramidal) e.g. NH3

AB2L2 (V-shape) e.g. H2OABL3 (linear) e.g. ClO–

Examples:Central atom (sp3) ExamplesBe [BeF4]2–

B 4BF

C alkanes, : CH3, diamond, CCl4Si SIO2, SiCl4, silicates like

[Si3O9]6– etc.N 3 4 3 2 2 2

hydrazineNH , NH , NF , NH , NH —NH

NH2OH,P 4 4 6 4 10 4P , P O , P O , PCl , oxyacids

of phosphorus, POCl3O H2O, H2O2, O2F2

S S8, SCl2, H3S+, (CH3)2SO, SOCl2,SO2Cl2, oxyacids of sulphur

Cl oxyacids of chlorine (in general(halogens) halogen), Cl2O, OF2

Xe XeO3, XeO4

Miscellaneous examples:

Borax anion [B4O5(OH)4]2–

B2H6 (diborane)sp3d hybridization (steric No. = 5)Geometry: triangular bipyramidalstandard bond angles 120° and 90°Shapes: AB5 (Trinagular bipyramidal) e.g. PF5, PAB4L (seesaw) e.g. SF4, XeO2F2

AB3L2 (T-shape) e.g. ClF3, 3XeF

AB2L3 (Linear) e.g. XeF2, 3 2I , ICl

As we have already read in the study of VSEPR model that in TBP structure thereare two types of positions, equatorial (E) and axial (a). The lone pair/s alwaysoccupy (E) and more electronegative atom/s prefer axial positions.Thus the structures of some of the examples are:PCl5 (g):

diagram heremeans covalent bond direction of lone pair.PCl3F2 :

diagram hereSF4 :

diagram hereShape: seesaw, decrease in the equatorial bond angle is due to presence of lone?????.ClF3 :

diagram here.Shape: T-shape.XeF2 :

diagram here.Shape: Linear.

In all examples of sp3d (or dsp3) hybridization the five orbitals of the central atomi.e. s + p + p + p + d combine to form five new hybrid orbitals which are directtowards the five corners of a TBP.sp3d hybridization involves 2z

d orbital.

Thus it can be considered as made up of two parts:

23

z x yz

Equitorialaxial

sp d p d s p p

According to Bent Rule, more negative substitutes prefer hybrid orbitals havingless 'S' character because more is the 'S' character more is the electronegativecharacter. Hence more electronegative atoms prefer axial positions to equatorialposition.However if more negative atoms are there i.e. more than two then the third willnaturally occupy one of the 'E' positions. e.g. PCl2F3 will be represented as shownbelow:

diagram here.Well! as regards the positions of lone pairs, we know they occupy 'E' positionsbut not axial, why?Let us consider the case of ClF3.We can draw following three structures for it.

diagram here.We know (c) is considered to be the correct structure. But why this is so? Theanswer can be provided if we consider the quantum of repulsions in eachstructure. Thus,In structure (A):Repulsions are due to:L–L (One) at 180°L–B (Six) at 90°B–B (Three) at 120°In structure (B):L–L (One) at 90°L–B (Three) at 90°and (Two) at 120°B–B (Two) at 90°and (One) at 120°

In structure (C):L–L (One) at 120°L–B (Four) at 90°and (Two) at 120°B–B (Two) at 90°Now since L–L repulsions are most forceful and less is the angle more is therepulsion. Hence the structure (B) can be quickly eliminated because it has one L–L repulsion at 90°.Now the choice lies between (A) & (C). Both these have one L–L repulsions at180° and 120° respectively. But then (C) has only 4 L–B repulsions as compared to(A) which has 6 L–B repulsions. Obviously the structure (C) must representminimum repulsion and hence maximum stability.This explains, why lone pairs do not occupy axial positions in sp3d hybridization.sp3d2 hybridization (Steric No. = 6)Geometry: OctahedralStandard bond angle: 90°Shape: AB6 (Octahedral)

e.g.

2 3 46 6 6 6 6

perxenate ionSF , PF , SiF , AlF , XeO

AB5L (square pyramidal)e.g. BrF5, XeOF4

AB4L2 (square planar)e.g. XeF4, [ICl4]–

sp3d2 or (d2sp3) hybridization involves

2 2 2x y z x y zs p p p d d

orbitals of the central atom.

Since an octahedron has all its six corners equivalent hence presence of conelone pair may be anywhere but when there are two lone pairs then they must beat maximum distance.With these things in mind, let us drain the structures of some of the exampleslisted above.SF6 :

diagram here.Shape: Octahedral.BrF5 :

diagram here.Shape: Square pyramidal.XeF4 :(Xenone tetrafluoride)

diagram here.Shape: Square planar.sp3d3 hybridization (Steric No. = 7)Geometry: Pentagonal bipyramidalStandard bond angles: 72° and 92°Shapes: AB7 (pentagonal bipyramidal) e.g. IF7

AB6L (distorted octahedral) e.g. XeF6

sp3d3 or d3sp3 hybridization involves

x y z xy yz zxs p p p d d d orbitals of the central atom.

Structure of IF7 (iodine heptafluoride) is as shown below:

diagram here.Structure of XeF6 (Xenone hexafluoride):It has been very difficult to determine the structure of XeF6 (g). It is supposedthat the lone pair can acquire any position hence it looks like a distortedoctahedron. Another suggestion is that Xe atom is situated at the centre of anoctahedron and six 'F' atoms occupy its six corners as usual. The lone pair in oneof the sp3d3 hybrid orbital of Xe is directed towards one of the faces of thisoctahedron as shown below. So it is as distorted octahedron.

diagram here.distorted octahedron

Otherwise XeF6 solid is found to be an ionic compound consisting of [XeF5]+ unitsinterlinked through F–. The structure of [XeF5]+ cation will be square pyramidaldue to sp3d2 hybridisation.NOTE:Now we shall try to answer some important questions related to some p-blockcompounds. The answers will be based on the structure of compounds. We shall

not repeat these explanations later on when we study p-block compounds in thenext chapter.

1 • Chemistry 19 September

Some Important Questions and their Answers:Q.1. Azide ion 3N is linear but hydroazoic acid (N3H) is a bent molecule, why?

Ans. This can be explained by drawing their structures and noting thehybridization of central atom. Thus,In 3N , the central 'N' is sp hybridized hence it is linear.sp2 sp sp2

Bh = 1.16 Å, Bh = 1.16 Å(Bond length)But in N3H, one of the terminal 'N' is attached to 'H' and that (N) is sp2

hybridized, as shown in the figure:H 112°

N N N

sp sp2

sp2

1.24 Å 1.13 Å

Q.2. Compounds SiH3NCO and CH3NCO appear to be similar but their structuresare not same, why?Ans. The structure of methylisocyanate CH3NCO isHH—C—NH sp3 sp2 CspO

Since steric number of 'N' is three hence it is sp2 hybridized, so CNC anglecannot be 180°. 'N' atom has a lone pair on it. Hybridizations of other has alone pair on it. Hybridizations of other atoms are also shown in the abovestructure.Now let us draw the structure of SiH3NCO (silylisocyanate). If we draw it at parwith CH3NCO then it must beHH—Si—N = C = OH


But lone pair on 'N' won't allow it to be linear the fact is that, it is linear and 'N'atom does not possess lone pair. How this is possible? This is possible bydonation of lone pair by 'N' to Si, which has all its '3d' orbitals vacant. This typeof donation which is some times also known as back bonding, will form a -bond of coordinate nature and is called d -phybridization state with a filled 'p' orbital as shown below:N =Z = 7

s p p psp hybridizationused for pπ – pπ bond with carbon atomused for pπ – dπbond or SiTherefore the correct presentation of SiH3NCO will beH dπ – pπ

H—Si—N = C = OH sp3 sp spHence the part of the molecule from Si to O is linear. This type of d – pbonding is not possible in CH3NCO because carbon atom does not have anyvacant orbital and moreover it cannot have 'd' orbital in it.Q.3. Which is a better electron pair donor (Lewis base) and why?CH3NH2 or CH3CNAns. Let us compare their structures:H HH—C—NH Hsp3; ('s' is 25%)HH—C—CHsp; ('s' is 50%)Remember more is the’s’ character in the hybridization more is theelectronegativity of that atom and less will be the tendency to donate theelectrons. Hence CH3NH2 is a better donor.


Q.4. The two π-bonds in the molecule of CO2 are coplanar or not. Explain.Ans. No, they are not coplanar but they are in perpendicular planes, because ifx-axis is the molecular axis then one -bond is formed by py – py overlape andthe other by pz – pz.Q.5. In BF3 molecule, the B—F bond length is found to be less than expectedsingle bond length. Why?Ans. Since 'B' in BF3 is sp2 hybridized hence it possesses a vacant 'p' orbital,which can form back bonding (pπ – pπ) with filled 'p' orbital of F.This is possible partly with each 'F' atom. It is called delocalized bonding.Hence each B—F bond develops parial double bond character which reducesthe bond length as shown in the figure below :F|BF Gp – p delocalised bond.Q.6. (SiH3)3N is planar but (CH3)3N is not, why?Ans. The structure of (CH3)3N (trimethylamine) is as given below:H3C CH3

N sp3 (pyramidal shape)|CH3

But in (SiH3)3N (trisilylamine), 'N' atom does not possess the lone pair, rather itforms delocalized p – d bonding with vacant 'd' orbitals of 'Si' atoms asshown below:SiH3

NH3Si SiH3

sp2

….means p – d bond (delocalized)(All the three Si atoms and N atom are in the same plane)But this is not possible in (CH3)3N, as (C) atoms do not have vacant orbitals.Q.7. CO2 is linear but SO2 is bent, why?Ans. It is clear from the hybridization of central atom.O = C = Osp


linearSO ObentQ.8. Why the 'N' atom in CH3ONH2 or in general RCONH2 (alkanamide) is said tbe in sp2 hybridized state?Ans. At first glance the structure of ethanamide CH3CONH2 would beO HCH3—C—NHTherefore 'N' atom must be in sp3 hybrid state. But RCONH2 exhibit followingproperties:(i) There is not free rotation possible about C—N bond.(ii) C—N bond length is 1.32 Å, while the standard C—N single bond length is1.47 Å.these observation can be explained if there is double bond between C and N in—CONH2 group. Infact RCONH2 is a resonance hybrid of following twostructures:O H O HC—NR H R H

sp2

Q.9. In 23CO ion, all the C—O bond lengths are same or different? Explain.

Ans. These are same, each of 1.29 Å. Infact standard C—O single bond length is1.43 Å and C = O is 1.22 Å. This suggests that 2

3CO must be a resonance hybridof the following structures.O– O O–

O = CO O O O OQ.10. Which of the following is correct with respect to CO bond length.(A) COF2 > CH3COCH3

(B) COF2 < CH3COCH3

(C) COF2 = CH3COCH3

(D) COF2 does not exist hence comparison cannot be done.


Ans. The correct answer is (A). Because in COF2 partial double bond characterin both C—F bonds due to delocalized overlapping between a filled 'p' orbitalof F and vacant 'p' orbital of C, weakens the C—O bond and hence its lengthincreases. This is also known as Synergy Effect. While this is not possible inCH3

C = OCH3

Q.11. Which is are true about Fluorine nitrate FNO3?(A) Oxidation state of 'N' in it is +5(B) It contains F—N bond(C) It contains O—O bond(D) It contains a peroxide group.Ans. The correct answer is (A) only. Its strcture isO OF NO(sp2)Q.12. Which is a stronger Lewis base NH3 or PH3 (Phosphene) and why?Ans. In NH3, 'N' atom is sp3 hybridized and the lone pair is present in a sp3

hybrid orbital

N =

Z = 7 s p p psp3 hybridization

all sp3 and degenerate

in sp3 (directed) ['S' characterNH HHbond angleOn the other hand the bond angle in PH3 is about 93.5° which suggests that 'P'is not hybridized he.Hence lone pair on 'P' in PH3 is present in pure 'S' orbital which is compact dueto its spherical nature.


Hence PH3 cannot donate its lone pair so easily as NH23 can do through its sp3

hybrid orbital.This is ome times known as Drago's rule according which of central atom is inIII period or below in periodic table and electronegativity of atoms attached tothat is < 2.5 then the lone pair will remain inactive in 'S' orbital.Q.13. The observed N—F bond length in NF3 is found to be more than expectedsingle bond length. Why?Ans. This is due to repulsion between lone pair on N and lone pairs on F atoms.NF FFrepulsionQ.14. In which O—O bond length is more and why H2O2 or O2F2 (dioxygenfiluoride)Ans. Hydrogen peroxide (H2O2) and dioxygen difluoride (O2F2) have similarbook like structures.H FO Omore repulsionO OH FBut bond polarities are in opposite directions. Thus electron pair density on thetwo oxygen atoms are more in H2O2 hence repulsion between them increasesO—O bond length in H2O2.Q.15. The structure of Caro's acid H2SO5 (peroxy monosulfuric acid) is as givenbelow:OH—O—S—O—O—H

In the first step dissociation this acid which H ( liberated as H+?Ans. The correct answer is 'stable due to charge delocalization (resonance).Q.16. Which of the following groups contains all its members, paramagnetic?(A) 2 3NO, NO , O , CO

(B) 2 3 2 2O , O , ClO , NO


(C) 2 3 3 3SO , SO , O , N

(D) 2 2 2CO , NOCl, Cl O, XeFAns. The correct answer is (B)O2 molecule is paramagnetic with 2 unpaired electrons.This cannot be explained by VBT. It will be discussed in the study of MOT.Others can be identified by their structures.

3O (ozonide ion) OO sp3 Onumber of unpaired electron/s= 1ClO2 (chlorine dioxide) ClO Osp2

= 1NO2 (nitrogen dioxide monomer)NO O= 1Q.17. How many σ and π bonds are there in a molecule of 1, 3, 5-Pentanetricarbonitrile.Ans. Draw its structure and you will find that there are 19Q.18. Which of the following molecule/ion will have square-pyramidal shape?(A) [IOCl4]– (b) [ICl2]+

(C) FClO3 (D) XeO2F2

Ans. The correct answer is (A), whose structure is given below:OCl ClICl Clsp3d2 hybridization.Q.19. Which of the following reactions is/are almost impossible to occur as perbalanced equations given below:(A) 3

3 3 2 3 3H BO 3H O 3H O BO (B) 3 3CF I OH aq. CF OH I


(C) 4 2 2 3CCl 3H O H CO 4HCl

(D) 3 3 3 33 3SiH N BF H Si N BF

Ans. The correct answers is (A), (B), (C) and (D).Infact none of the above reaction can take place as given above becauseH3BO3 (aq.) is a monobasic acid, the bond polarities in CF3I does not allow I– tobe released. CCl4 does not have vacant orbital with (C) so as to interact withH2O.In (SiH3)3N, 'N' does not have lone pair due to p – d bonding.

Comprehension (I)Paragraph for Questions No. 20, 21, 22.According to VBT the percentage of 'S' character in sp, sp2 and sp3

hybridizatinos is 50, 33.3 and 25 respectively. But this not true completely forall species. Depending upon the bond angle (s–p hybrid orbital can be calculated with the help of following formula:

s p 1coss 1 p

Q.20. In which the s% is maximum. Givencos 109.5° = – 0.3338, cos 107° = – 0.29(A) N—H bond in NH3 (B) P—H bond in PH3

(C) P—F bond in PF3 (D) All have equal s%Ans. The correct answer can be found out if we know the bond angles whichare

3, 3 and 3

3 :s0.29

s 1

s 0.224 or 22.4%

Further as 'The bond angle in PH3 and PF3 both are less than NH3 hence s% will be less.Hence correct answer is (A).On PH2 it is -orbital.Q.21. On the basis of your answer of Q.No. 20, which of the following iscorrect?(A) PH3 is stronger Lewis base than NH3


(B) NH3 is stronger Lewis base than PH3

(C) They are equally strong Lewis bases(D) P—H bond is more polar than N—H.Ans. The correct answer is (B) (refer to Q.No. 12)Q.22. s% in Phosphorus-halogen bond will be in the order.(A) PF3 < PCl3 < PBr3 < PI3

(B) PF3 > PCl3 > PBr3 > PI3

(C) PF3 = PCl3 = PBr3 = PI3

(D) None of theseAns. The correct answer is (A), because in the series PF3, PCl3, PBr3, PI3 thebond angle increases. And as 'bond angles are:PF3 PCl3 PBr3 PI3

Q.23. In which of the following sets, each member shows 'sp' hybridization ofall its 'c' atoms?(A) CaC2 CH2 (Triplet), (CN)–

(B) C6H6, CH3, CH2 (singlet)(C) CO2, C2H2, HCN(D) C3O2, CO3

2–, C2H4

Ans. The correct answers are (A) and (C).Q.24. Which of the following is non-linear.(A) 3I

(B) 3N

(C) XeF2 (D) SO2

Ans. The correct answer is (D), SO2 is a bent molecule with the followingstructures. Sulphur is sp2.S S

O O O O

Comprehension (II) [A Quiz]An element 'J' belongs to 's'-block and another 'Q' belongs to 'p'-block of theperiodic table such that the sum of their atomic numbers in equal to theatomic number of an element 'X' which belongs to '3d' series of the periodictable. In the ground state electronic configuration of 'X', the total number of


electrons for which sum of quantum numbers i.e. (n + l) equal to 5.0, is eight.The element 'J' forms a series of three compounds (R, S, T) with oxygen suchthat oxidation number of oxygen increases from Relement 'Q' forms two compounds (G, H) with oxygen in which oxidationnumber of oxygen decreases from G to H. The ionic compound JQ is such thatits aqueous solution does not give any precipitate with AgNO3 solution.Now answer the following questions:Q.25. Identify the elements J and Q and the compound JQ.Q.26. Write the formula of the compounds R, S, T, G and H.Q.27. What is atomic number of element X?Solution: The element 'J' can 'K' (Potassiumj) and 'Q' can be F (Fluorine). Sothat sum of their atomic numbers become 19 + 9 = 28. Also compound JQbecomes KF which does not give any precipitate with AgNO3 (aq.) because AgFis water soluble. the element X must be Ni (Z = 28) so thatNi = 1s2, 2s2p6, 3s2p6d8, 4s2

Z = 28

For these eight electrons n = l = 3 + 1 = 5No other electron has n + l = 5Thus the compounds R, S & T becomeK2O K2O2 KO2

Potassium : oxide peroxide superoxideO.No. of oxygen –2 –1 –0.5The compounds G & H will be OF2 and O2F2 respectively. Oxidation number ofO in OF2 is +2 and that in O2F2 is +1.

(iv) Molecular Orbital Theory (MOT)This theory was given by Mulliken. One important limitation of VBT was that iscould not explain the paramagnetic nature of O2. MOT is superior to that in thisrespect, l it explains the paramagnetic nature of O2.MOT is based on an important principle known as linear combination of atomicorbitals i.e. LCAO.The important points or principles of MOT are as follows:


(i) When two atoms combine to form a molecule then their orbitalsundergo linear combination. This combination will be most effectivewhen the combining orbitals are of similar energy. That’s why in theformation of a homonuclear diatomic molecule the '1s' orbital of oneatom will combine with '1s' orbital of another atom and so on. Thuswhen two atomic orbitals combine they produce two new molecularorbitals (m.o.).

(ii) Since an atomic orbital is represented bylinear combination of two atomic orbitals means a linear combinationof two atomic orbitals means a linear combination of twoOr more qualitatively we assume that two electron waves interferewith one another. Since interference can be constructive ordestructive so linear combination of their wave function can beaddition or subtraction.Therefore if we suppose a and b be two wave functions then, theymay combine to give two new wave functions as ( a + b) and ( a ~ b)( a + b) will represent bonding molecular orbital. It will be of lowerenergy than the average every of a & b.But ( a ~ b) will represent antibonding molecular orbital and itsenergy will be higher than the average of atomic orbitals. Infact thisupward energy displacement is slightly greater than the downwarddisplacement. The antibonding m.o. is identified by putting an asterisk(*) on its symbol as we are going see soon after.A bonding molecular orbital is such that electron density increases inthe internuclear region hence the repulsion between the two nuclei isminimum. that's why it is of lower energy and more stability. Itcontributes towards bonding.But there is a node at the centre of an antibonding nolecular orbitalwhich reduces the electron density in the internuclear region hencethe electron repulsion between the two nuclei is maximum. That’swhy its energy is higher and stability is less.In other words it works against bonding, hence the name.

A pictorial presentation of linear combination of 's' and 'p' atomic orbitals isshown below:

diagram here.


Fig. : Bonding and antibonding bonding molecular orbitals obtained from themixing of s and p-orbitals. Note that x-axis is taken as molecular axis hence 2px

+ 2px mixing is forming y + 2py and 2pz + 2pz (which isnot shown above) will mix similarly to form orbitals.(iii) A molecular orbital may be further gerade (means symmetrical) or

ungerade means anti-symmetrical. Ifmoving equal distance from the centre of internuclear region then it isungerade otherwise gerade. For example consider.

1s or in general any

diagram here.nodal plane

We find that on moving equal distances from the centre of nodal planein the opposite direction if m.o. becomes positive on the left ten itbecome negative on the right or vice verse. Hencem.o. In this manner we can decide their nature. The results will be:

x geradex ungerade

y or z ungeradey or z gerade

NOTE:The principal quantum number may be any other high number butshape and characgters will remain same. Mind it we have selected x-axis as molecular axis, that’s why py and pz are forming -m.o.

(iv) Each molecular orbital irrespective of its nature will behave like anatomic orbital i.e. it cannot accommodate more than two electrons init and that with antiparallel spin as demanded by Pauli exclusionprinciple.

(v) For a diatomic molecule in this manner a set of various molecularorbitals is obtained which are arranged in the increasing order ofenergy. Then the total number of electrons of both the atoms are fedinto these molecular orbitals from the molecular orbital of lowestenergy level to higher and higher till all the electrons are exhausted.During this course Hund's Rule is obyed wherever needed.


(vi) Finally the bond order for the molecule is calculated by the formula:

Bond order = b aN N2

Where Nb and Na means total number of bonding and antibondingelectrons in that molecule. The bond order may come out to be zero,an integer or even fractional but never negative.

So now we can draw molecular orbital energy level diagrams for varioushomonuclear diatomic molecules.

diagram here.or the molecular orbital configurationof hydrogen molecule can be written as :H2 = 2, 0

From this we derive that

(i) Bond order in H2 =2 0 1.0

2

(ii) H2 molecule is diamagnetic as there is no unpaired electron with themolecule.

diagram here.2 2B.O. Zero

2

It means He2 molecule has no net stability. Indeed it has not been possible toform He2 molecule.

diagram here.2 2 2 0

2Li 1s , *1s , 2s , *2s 4 2B.O. 1.0

2

Li2 molecule is diamagnetic. Practically Li2 molecules have been observed in thegas phase.Now we need not to draw a complete diagram for then next molecule Be2. Itwill be of the same nature as given above for Li2. But total number of electronsbeing eight [configuration of Be2 will be


2 2 2 22Be 1s , *1s , 2s , *2s

4 4B.O. Zero2

2 is not possible or not stable at all, like He2.So far we have drawn configurations for molecules H2, He2, Li2 and Be2. But ifwe are asked to draw configurations of their cations, then how to draw themthe answer is very simple, we have to first draw the configuration of parentmolecule and then remove the required number of electron or electrons fromthe highest occupied m.o. downwards. For anions add e upwards.Therefore:

1 02

1 0H 1s , *1s ; B.O. 0.5,2

paramagnetic

2 12

2 1H 1s , *1s ; B.O. 0.5,2

paramagnetic

2 12

2 1He 1s , *1s ; B.O. 0.5,2

paramagnetic

Infact all the above species have been found to exist under some specialcircumstances. Since less is the bond order less must be the bond energy andmore bond length. The observed facts are in accordance with it e.g.Species Bond Energy (KJ mol–) Bond length (pm)H2 432 74

2H 255 106

2He 230 108

From these data, it is also clear that relatively 2H is more stable than 2He .

Now although in the sequence we must draw configurations for B2, C2, N2, butdue to some scientific reason we wish to deal with O2, F2 and Ne2 before B2, C2

and N2. We shall certainly learn in brief the reason also.So let us draw configuration for O2.

diagram here.NOTE:X-axis is taken as molecular axis.


2 can be written asO2 = 2, 2, 2, x

2,

2 2 1 1y z y z2p , 2p , *2p , *2p , *2

At this stage we can introduce two new terms which are:'HOMO' means highest occupied molecular orbital 'LUMO' means lowestunoccupied molecular orbital.Therefore in O2 molecule 'HOMO' level has two diagenerate orbitals which are

y and z. And the LUMO level is x. Infact these terms are applicablefor all electronic configurations in terms of molecular orbitals. You will realizetheir significance soon.Well, so for O2 we find that Bond order is10 6 2.0

2

And a very important achievement of MOT is that we find two unpairedelectrons in its configuration hence it is paramagnetic.Now let us draw configurations of some important anions and cations of O2.

2 2 2 2 22 xO 1s , *1s , 2s , *2s , 2p ,

2 2 1 0y z y z2p , 2p , *2p , *2p , *2

10 5B.O. 2.52

and it is paramagnetic with one unpaired electron.2 2 2 2 2

2 xO 1s , *1s , 2s , *2s , 2p ,

2 2 0 0y z y z2p , 2p *2p , *2p

, , *2HOMO LUMO

10 4 3.0, Diamagnetic2

2 2 2 2 22 xO 1s , *1s , 2s , *2s , 2p ,

2 2 2 1y z y z z2p , 2p , *2p , 2p , *2p

(Superoxide ion)10 7B.O. 1.5,

2

Paramagnetic (one unpaired electric)


2 2 2 2 2 22 xO 1s , 1s , 2s , *2s , 2p ,

2 2 2 2 0y z y z x2p , 2p , *2p , *2p , *2p

(Peroxide ion)10 8B.O. 1.0, Diamagnetic

2

All the above results for O2 and its ions can be presented in the form of a tablegiven below:TableSpecies 2

2O 12O

2O 12O 2

2O

B.O. 1.0 1.5 2 2.5 3.0Number ofunpaired electrons 0 1 2 1 0

2F : Since atomic number of Fluorine is 9 hence F2 molecule will have 18

electrons.Therefore electronic configuration of F2 will be like O2 as given below:F2 = 2 2 2 2 2

x1s , *1s , 2s , *2s , 2p ,

2 2y z2 2

y z

*2p , *2p *2p2p , 2p , ,HOMO LUMO

10 8B.O. 1.0, Diamagnetic.2

Similarly for Ne2 (Total 20 electrons) we have:Ne2 = 2 2 2 2 2

x1s , 1s , 2s , *2s , 2p ,

2 2 2 2 2y z y z x2p , 2p , *2p , *2p , *2p

10 10B.O.2

= zero, which means Ne2 molecule is not possible. Infact

all noble gases are monoatomic, all the and attempts to form their moleculeshave not been successful.

Let us now answer a important question:


Q.1. Geneerally paramagnetic species are colored and diamagnetic colorless,then why O2 gas is colorless and F2(g) is very light yellow-green gas. Infact allthe halogen molecules are diamagnetic but none of the halogens is colorless,why?Answer: Answer of such a question is based on the difference of energybetween HOMO and LUMO level. Since this difference in O2 molecule is largeenough so that electronic excitation by partial absorption of light cannot takeplace in the visible range of the spectrum. No doubt unpaired electrons areeasily excited but the minimum energy needed for this is equal to energydifference between HOMO and LUMO.Therefore O2 does not show absorption in he visible part of the spectrum,rather it shows this excitation in the U.V. region, where we cannot see henceoxygen remains colorless.Remember, when any species absorbs a particular color (wavelengths) then itscomplimentary color is exhibited. For example of a specie is absorbs violetradiations ( – 435 nm) and it appears yellow green because this is thecomplimentary color of violet.Since the energy gap between HOMO and LUMO for F2 is less than that for O2,hence F shows absorption in violet part of the visible spectrum and it appearslight yellow green.Q.2. Whose ionization energy (I.E.) will be more that of oxygen atom oroxygen molecule and why?Ans. We know that I.E. is the energy needed to remove the most looselybonded electron in a gaseous atom molecule in ground state, to infinity. So letus compare the energy levels of most loosely bonded electron in these twospecies. Look at the vertical energy level diagram drawn for O2 earlier. We seethat the most loosely bonded electron must be in HOMO, which are twounpaired electrons so that any one can be removed to infinity. But the HOMOlevel consists of antibondingthan the 2p level of oxygen atom. Hence it is easier to remove a electron fromO2 than from O.Therefore I.E. of oxygen atom > I.E. of O2. Similarly you can decide for N2 andN.Energy level diagrams for B2, C2 and N2 :


We had not considered these three molecules in the sequence i.e. just afterBe2. Let us see why we did it? Suppose we draw the configuration of B2 on thepattern of O2, then it will be:

2 2 2 2 2 0 02 x y zB 1s , *1s , 2s , *2s , 2p , 2p , 2p

[Total 10 electrons]Z 5

Well, if this configuration is correct then B2 molecule must be a diamagneticspecies, while actually it is a paramagnetic molecule having paramagneticmoment equivalent to 2 unpaired electrons.Hence above configuration must not be correct. There must be some changein the energy order of m.o.s.Yes, there is a change and that is, that x m.o. occupies a place above ( y

and z).the reason behind this change is that energy gap between x m.o. inB2, C2 and N2 molecules is so small that they can merge with each other to formtwo new m.o. This is a different type of hybridization, applicable to m.os. Inthis type of mixing the energy gap between the two m.o. increases. Thus wecan say that now xs m.o. with the increased energy gapbetween them. x now occupies a place below the original level of

xs occupies a place higher than the original x, but this new level is nowabove the level of y and z. Similar interaction takes place betweenand x so that x goes down and xs goes up but this does not bringany change with respect to other molecular orbitals. For the sake ofconvenience we shall retain the original symbols o molecular orbitals and notthe hybrid symbols.therefore the increasing energy level order for B2, C2 and N2 will be:

y z1s, *1s, 2s, *2s, 2p , 2p

x y z2p , *2p , *2p , *2

Hence the correct configuration of B2 will be:

1 1 0y z2 2 2 2 x

2

2p , 2p 2pB 1s , *1s , 2s , *2s , ,HOMO LUMO

6 4B.O. 1.0,2

Paramagnetic with 2-unpaired electrons.

CHEMICAL BONDING - docshare04.docshare.tipsdocshare04.docshare.tips/files/20735/207353112.pdf ·...

Documents

Transcript of CHEMICAL BONDING - docshare04.docshare.tipsdocshare04.docshare.tips/files/20735/207353112.pdf ·...