Chemical Bonding PDF - 5. Chemical Bonding

5. Chemical Bonding 5.1 Introduction : Why are atoms held Octet rule : In 1916 Kossel and Lewis together in chemical compounds ? There must proposed an important theory for explaining be some force that holds them together. You the formation of chemical bond k...

5. Chemical Bonding 5.1 Introduction : Why are atoms held Octet rule : In 1916 Kossel and Lewis together in chemical compounds ? There must proposed an important theory for explaining be some force that holds them together. You the formation of chemical bond known as have already learnt in lower classes that the Electronic Theory of Valence. This theory is forces holding atoms together in a compound mainly based on octet rule developed by Lewis. are the chemical bonds. Octet rule is based on stability of noble gases How are chemical bonds formed between due to presence of eight electrons (ns2np6) in two atoms ? There are two ways of formation the valence shell. of chemcial bonds (i) by loss and gain of This rule states that during the formation electrons (ii) by sharing a pair of electrons of chemical bond, atom loses, gains or shares between the two atoms. In either process of electrons so that its outermost orbit (valence formation of chemical bond each atom attains shell) contains eight electrons. Therefore the a stable noble gas electronic configuration. atom attains the nearest inert gas electronic Which electrons are involved in the configuration. formation of chemical bonds ? The electrons The octet rule is found to be very useful present in the outermost shell of an atom are in explaining the normal valence of elements involved in the formation of a chemical bond. and in the study of the chemical combination 5.2 Kossel and Lewis approach to chemical of atoms leading to the formation of molecule. bonding : Number of attempts were made However it should be noted that octet rule is not to explain the formation of chemical bond in valid for H and Li atoms. These atoms tend to terms of electrons, but the first satisfactory have only two electrons in their valence shell explanation was given by W.Kossel and G.N. similar to that of Helium (1s2) which called Lewis independently. They gave a logical duplet. explanation of valence which was based on 5.2.1 Ionic bond the inertness of noble gases. On the basis of I. Formation of sodium chloride (NaCl) this they proposed a theory of valence known as Electronic theory of valence in 1916. The electronic configurations of Sodium and According to Lewis, the atom can be Chlorine are : pictured in terms of a positively charged Na (Z = 11) 1s22s22p63s1 or 2, 8, 1 'kernel' (the nucleus plus inner electrons) and Cl (Z = 17) 1s22s22p63s2 3p5 or 2, 8, 7 outer shell that can accommodate a maximum Internet my friend of eight electrons. This octet of electrons represents a stable electronic arrangement. Search more atoms which complete Lewis stated that each atom achieves their octet during chemical combinations. stable octet during the formation of a chemical Sodium has one electron in its valence bond. In case of sodium and chlorine this can shell. It has a tendency to lose one elctron to be achieved by transfer of one electron from acquire the configuration of the nearest nobel sodium to chlorine. Thus Na⊕ (2, 8) and Cl gas Ne (2, 8). Chlorine has seven elctrons in (2, 8, 8) ions are formed which held together. its valence shell. It has a tendency to gain one In case of other molecules like H2, F2, Cl2, electron and thereby acquire the configuration HCl etc. the bond is formed by the sharing of of the nearest nobel gas Ar (2, 8, 8). During the a pair of electrons between the atoms. In this combination of sodium and chlorine atoms, process each atom attains a stable outer octet the sodium atom transfers its valence electron of electrons. 55 to the chlorine atom, sodium atom changes Elements having low ionization enthalpy into Na⊕ ion while the chlorine atom changes can readily form ionic bond with elements into Cl ion. The two ions are held together having a high negative value of electron gain by strong electrostatic force of attraction. The enthalpy. Both these processes take place in formation of ionic bond between Na and Cl gaseous phase. All ionic compounds in the can be shown as follows. solid state have each cation surrounded by a Na + Cl Na⊕ + Cl specific number of anions and vice versa. 2,8,1 2,8,7 2,8 2,8,8 Na⊕ + Cl Na⊕Cl or NaCl Ionic bond II. Formation of calcium chloride (CaCl2) : The following representation shows the formation of compound calcium chloride from Cl the elements calcium and chlorine : Electronic configuration of Calcium :1s22s22p63s23p64s2 Na⊕ Cl :1s22s22p63s2 3p5 NaCl crystal lattice Cl + Ca + Cl Cl +Ca + Cl 2⊕ The arrangements of cations and anions 2,8,7 2,8,8,2 2,8,7 2,8,8 2,8,8 2,8,8 in a crystalline solid is ordered and they are Cl + Ca2⊕ + Cl CaCl2or Ca2⊕(Cl )2 held together by coulombic forces of attraction. 5.1.2 Ionic solids and Lattice Enthalpy : During their formation, these compounds Ionic solids are solids which contain crystallize from the gaseous state (MX(g) cations and anions held together by ionic MX(s)) to the solid state. The structure in bonds. Kossel treatment helps us to understand which they crystallize depends upon the size of the formation of ionic bonds between ions of the ions, their packing arrangement and other different elements. Formation of ions depends factors. The overall stability of the ionic solid on the ease with which an atom can lose or depends upon the interactions between all gain electrons. these ions and the energy released during the M(g) M⊕(g) + e formation of the crystal lattice. X(g) + e X-(g) electron gain enthalpy Let us consider the formation of NaCl ionic M⊕(g) + X (g) MX(g) MX(s) solid. Na(g) Na⊕(g) + e ∆iH = 495.8 kJmol-1 Do you know ? Cl(g) + e Cl (g) ∆egH - 348.7 kJmol-1 CsF is the most ionic compound. Na⊕(g) + Cl (s) NaCl (g) + 147.7 kJmol-1 Because Cs is the most electropositive while Conversion of NaCl(g) NaCl (s) F is the most elctronegative element. The is associated with release of energy which is electronegativity difference between them is -788 kJ mol-1. This released energy is much the largest. Hence ions are easily seperable, more than the absorbed energy. Thus stability the bond is weakest and the compound is of an ionic compound can be estimated by least stable ionic compound. knowing the amount of energy released Ionization is always an endothermic during lattice formation and not just by energy process while electron gain process can be associated with completion of octet around the exothermic or endothermic. Based on the ionic species in the gaseous state alone. ionisation enthalpy (∆iH) and electron gain Lattice Enthalpy : Lattice Enthalpy of an enthalpy, we can predict which elements can ionic solid is defined as the energy required form ionic compounds. to completely separate one mole of solid 56 ionic compound into the gaseous components. other at a certain internuclear distance they Lattice enthalpy of NaCl is -788 kJ mol-1 share their valence electrons. The shared which means that 788 kJ of energy is required pair of electrons belongs equally to both the to separate 1 mole of NaCl into one mole hydrogen atoms. The two atoms are said to of gasesous Na⊕(g) and Cl (g) to an infinite be linked by a single covalent bond and a distance. molecule H2 is formed. Table 5.1 : Lattice Enthalpy values of some H+H H : H or H − H ionic Compounds shared pair of electrons Compound Lattice enthalpy II. Formation of Cl2 : The Lewis-Langmuir kJmol-1 theory can explain the formation of chlorine LiCl 853 molecule, Cl2. The Cl atom with electronic NaCl 788 configuration [Ne]3s23p5 is one electron BeF2 3020 short of Argon configuration. The formation CaCl2 2258 of Cl2 molecule can be understand in terms AlCl3 5492 of the sharing of a pair of electrons between For same anion and different cations : two chlorine atoms. Each chlorine atom 1. Cations having higher charge have large contributes one electron to the shared pair. In lattice energies than compounds having the process both the chlorine atoms attain the cations with lower charge. AlCl3 > CaCl2 > valence shell octet of the nearest noble gas NaCl (i.e. Argon) 2. As size of cation decrease, lattice energy Cl + Cl Cl Cl or Cl - Cl increases. LiF > NaF > KF. The dots represent the electrons. Such 5.2.2 Covalent bond : In 1919 Lewis structures are referred to as Lewis structures. suggested that there are atoms which attain The Lewis dot structure can be written for inert gas configuration (i.e. 1s2 or ns2np6 other molecules also in which the combining configuration) by sharing one or more electron atoms may be identical or different. Following pairs with similar or dissimilar atoms. Each are the important features of covalent bond. atom contributes one electron to the shared Each bond is formed as a result of sharing electron pair and has equal claim on the shared of electron pair between the two atoms. electron pair. Langmuir called the Lewis When a bond is formed, each combining electron pair bond a covalent bond. Thus the atom contributes one electron to the shared concept of covalent bond is known as Lewis pair. Langmuir concept. A shared pair of electron is The combining atoms attain the outer shell represented as a dash (−) and is responsible for nobel gas configuration as a result of the holding the two atoms together. sharing of electrons. A covalent bond may be defined as follows : Thus in H2O and CCl4 the formation of The attractive force which exists due covalent bonds can be represented as, to the mutual sharing of electrons between Cl the two atoms of similar electronegativity or having small difference in electronegativities H O H Cl C Cl is called a covalent bond. Cl H acquires duplet I. Formation of H2 molecule : The electronic O acquires octet All atoms acquire octet configuration of H atom is 1s1. It needs one H2O CCl4 more electron to complete its valence shell. When two hydrogen atoms approach each 57 III. Formation of Multiple bond : When 8. After writing the number of electrons as two atoms share one electron pair they are shared pairs forming single bonds, the said to be joined by a single covalent bond. remaining electron pairs are used either for When two combining atoms share two pairs multiple bonds or remain as lone pairs. of electrons, the covalent bond between them Table 5.2 includes Lewis representation of is called a double bond. For example a double some molecules. bond between two carbon atoms in ethylene Table 5.2 Lewis dot structures of some molecule. When two combining atoms share molecules/ions three electron pairs a triple bond is formed as Molecul/Ion Lewis Representation in the case of two nitrogen atoms in the N2 H2 H:H, H-H molecule [Fig 5.1(a)]. Some other examples O2 of multiple bonds are CO2 and C2H2 [Fig 5.1] O O O3 ⊕ N N O C O H C C H O N≡N O=C=O H-C≡C-H O O NF3 F (a) (b) (c) F N F Fig 5.1 Multiple bonding 5.2.3 Lewis structures (Lewis CO32 O 2 representations of simple molecules) : Lewis C dot structures show a picture of bonding in O O molecules and ions in terms of the shared HNO3 pairs of electrons and the octet rule. Although O O N O H such a picture does not explain completely the ⊕ bonding and behaviour of a molecule, it helps to understand the formation and properties of molecule. Problem 5.1 : To write Lewis structure of 5.3.1 Steps to write Lewis dot structures nitrite ion NO2. 1. Add the total number of valence electrons of Solution combining atoms in the molecule. Step I : Count the total number of valence 2. Write skeletal structure of the molecule electrons of nitrogen atom, oxygen atom and to show the atoms and number of valence one electron of additional negative charge. electrons forming the single bond between N(2s22p3), O (2s22p4) the atoms. 5 + (2×6) + 1 = 18 electrons 3. Add remaining electron pairs to complete Step II : The skeletal structure of NO2 is the octet of each atom. written as O N O 4. If octet is not complete form multiple bonds Step III : Draw a single bond i.e. one between the atoms such that octet of each shared electron pair between the nitrogen atom is complete. and each oxygen atoms completing the octet 5. In anions add one electron for each negative on oxygen atom. This does not complete the charge. octet of nitrogen. Hence, there is a multiple 6. In cations remove or subtract one electron bond between nitrogen and one of the oxygen from valence electrons for each positive atoms (a double bond). The remaining two charge. electrons constitute a lone pair on nitrogen. 7. In polyatomic atoms and ions, the least Following are Lewis dot structures of NO2. electronegative atom is the central atom for eg. 'S' is the central atom in SO42 , 'N' is the [ O N O] [O = N - O [ or [ O - N = O[ central atom in NO3. 58  Problem 5.2 : Write the Lewis structure of formal charge is as close to zero as possible. CO molecule. Formal charge is assigned to an atom based on Solution : electron dot structures of the molecule/ion. e.g. Step - I. Count number of electrons of O3, NH4⊕, [N = C =O] , [ S = C =N] carbon and oxygen atoms. The valence shell configuration of carbon and oxygen atoms are : 2s22p2 and 2s22p4 respectively. Formal charge on an atom in a Lewis structure The valence electrons available are of a polyatomic species can be determined 4 + 6 = 10 using the following expression. Step - II : The skeletal structure of CO is written as C O or C O [ Formal charge [[ on an atom in a Lewis structure = Total no. of valence electrons in free atom [ - Step - III : Draw a single bond (One shared electron pair) between C and O and complete the octet on O. The remaining two [ Total no. of non bonding or lone [ [ pairs of electrons - 1/2 Total no. of bonding or shared electrons [ electrons is a lone pair on C. or FC=VE − NE − (BE/2) The octet on carbon is not complete hence The structure having the lowest formal charge there is a multiple bond between C and has the lowest energy. O (a triple bond between C and O atom). 1. Let us consider ozone molecule O3 This satisfies the octet rule for carbon and Lewis structure of O3. oxygen atoms. 1 O C ≡ O or C O O O 2 3 Each H atom attains the configuration of Here three oxygen atoms are numbered as 1, helium (a duplet of electons) 2, 3. The formal charge on the central oxygen 5.4.1 Formal charge atom no.1 = 6-2 -1/2(6) = +1 The Lewis dot diagrams help us to get a Formal charge on the end oxygen atoms is picture of bonding in molecules which obey marked as 2 the octet rule.In case of polyatomic molecules = 6 - 4 - 1/2(4) = 0 double bonds or some times triple bonds are Formal charge on the end oxygen atoms present and can be represented by more than marked as 3 one Lewis structure. In the case of CO32 we = 6 - 6 - 1/2(2) = -1 can have three dot diagrams. Hence, O3 molecule can be represented along [ [ [ [ O O O C 2 or O C O O 2 or [ [ O C O O 2 with formal charge as follows : ⊕ O Double bonds can be present between O O Carbon and any one of the three oxygen The lowest energy structure can be atoms. Formal charges can help us in selected using formal charges from the number assingning bonds when several structures are of possible Lewis stuctures, for a given species. possible. Formal charge is the charge assigned 2. Let us take the example of CO2. to an atom in a molecule, assuming that all CO2 can be represented by the following electrons are shared equally between atoms, structures. regardless of their relative electronegativities. O =C=O O ≡C-O O -C≡ O While determing the best Lewis structure per 1 2 1 2 1 2 molecule the structure is chosen such that the A B C 59  Assigning the formal charges on the Problem 5.3 : carbon atom and the two oxygen atoms Find out the formal charges on S, C, N. numbered 1 and 2 (S = C = N) ; (S - C ≡ N) ; (S ≡ C-N) Structure A Solution : Number of electrons: 4 from carbon and 6 Step I from each Oxygen Write Lewis dot diagrams for the structure So total number of electrons=4+6+6=16 S=C=N S−C≡N Formal charge on C =4-0-1/2(8)=0 S≡C−N Formal charge on O-1 and O-2 =6-4-1/2(4)=0 A B C In this structure formal charge on all atoms is Step II zero. Assign formal charges Structure B. FC = V.E - N.E - 1/2 B.E. Formal charge on C =4-0-1/2(8)=0 Structure A : Formal charge on O -1 = 6 - 2 - 6/2 = 4 - 3= +1 Formal charge on S = 6 - 4 - 1/2(4) =0 Formal charge on O - 2 = 6 - 6 -2/2 = -1 Formal charge on C = 4 - 0 - 1/2 (8) = 0 Structure C. Formal charge on N = 5 - 4 - 1/2 (4) = -1 Formal charge on C =4-0-1/2(8)=0 Structure B : Formal charge on 1st O = 6 - 6-2/2 = -1 Formal charge on S = 6 - 6 - 1/2(2) = -1 Formal charge on 2nd O = 6-2-6/2=4-3=+1 Formal charge on C = 4 - 0 - 1/2 (8) = 0 We find that in structure A the formal charge Formal charge on N = 5 - 2 - 1/2 (6) = 0 on all atoms is 0 while in structures B and C Structure C : formal charge on Carbon is 0 while Oxygens Formal charge on S = 6 - 2 - 1/2(6) = +1 have formal charge -1 or +1. So the possible Formal charge on C = 4 - 0 - 1/2 (8) = 0 structure with the lowest energy will be Formal charge on N = 5 - 6 - 1/2 (2) = -2 Structure A. i. Incomplete octet ii. Expanded octet Use your brain power iii. Odd electrons Which atom in NH4⊕ will have formal i. Molecules with incomplete octet : charge +1? eg. BF3, BeCl2, LiCl In these covalent molecules the central Generally the lowest energy structure has atoms B, Be and Li have less than eight the smallest formal charges on the atoms. electrons in their valence shell but are stable. 5.4.2 Limitation of octet rule : Li in LiCl has only two electrons 1. Octet rule does not explain stability of some Be in BeCl2 has four electrons while molecules. B in BF3 has six electrons in the valence shell. The octet rule is based on the inert ii. Molecules with expanded octet : Some behaviour of noble gases which have their molecules like SF6, PCl5, H2SO4 have more octet complete i.e. have eight electrons in their than eight electrons around the central atom. valence shell. It is very useful to explain the F structures and stability of organic molecules. F F However there are many molecules whose existence cannot be explained by the octet S theory. The central atoms in these molecules F F does not have eight electrons in their valence F shell, and yet they are stable. eg. BeCl2, PF5 etc such molecules can be categorized as having SF6;12 electrons around sulfur 60  Cl 5.5 Valence Shell Electron Pair Repulsion Cl Theory (VSEPR) : Properties of substances are dependent on the shape of the molecules. P Cl Lewis concept is unable to explain the shapes Cl of molecules. Shapes of all molecules cannot Cl be described completely by any single theory. PCl5;10 electrons around Phosphorus One of the popular models used earlier to predict the shapes of covalent molecules is the valence shell electron pair repulsion theory proposed by Sidgwick and Powell. It is based on the basic idea that the electron pairs on the atoms shown in the Lewis diagram repel H2SO4; 12 electrons around sulfur each other. In the real molecule they arrange It must be remembered that sulfur also themselves in such a way that there is minimum forms many compounds in which octet rule repulsion between them. is obeyed. For example, sulfur dichloride the The arrangement of electrons is called as sulfur atom has eight electrons around it. electron pair geometry. These pairs may be shared in a covalent bond or they may be lone pairs. Rules of VSEPR : SCl2 8 electrons around sulfur 1. Electron pairs arrange themselves in such a way that repulsion between them is minimum. 2. The molecule acquires minimum energy and Use your brain power maximum stability. How many electrons will be around I in the 3. Lone pair of electrons also contribute in compound IF7 ? determining the shape of the molecule. 4. Repulsion of other electron pairs by the lone iii. Odd electron molecules pair (L.P) stronger than that of bonding pair Some molecules like NO (nitric oxide) and (B.P) NO2 (nitrogen dioxide) do not obey the octet Trend for repulsion between electron pair is rule. Both N and O atoms, have odd number L.P - L.P > L.P - B.P > B.P - B.P of electrons. ⊕ Lone pair -Lone pair repulsion is maximum N = O O=N−O because this electron pair is under the influence 2. The observed shape and geometry of a of only one nucleus while the bonded pair is molecule, cannot be explained, by the octet shared between two nuclei. rule. Thus the number of lone pair and bonded pair 3. Octet rule fails to explain the difference in of electrons decide the shape of the molecules. energies of molecules, though all the covalent Molecules having no lone pair of electrons bonds are formed in an identical manner that have a regular geometry. is by sharing a pair of electrons. The rule fails to explain the differene in reactivities of different molecules. Use your brain power Why is H2 stable even though it never satisfies the octet rule ? 61  Table 5.3 : Geometry of molecules (having no lone pair of electrons) Number of electron Arrangement of Molecular geomentry Examples pairs electron pairs 2 Linear Linear BeBr2, CO2, CO2 180º 3 Trigonal planar Trigonal planar BF3, BCl3, BH3 BCl3 120º 4 Tetrahedral Tetrahedral CH4, NH4+ SiCl4 5 Trigonal bipyramidal Trigonal bipyramidal PCl5, SbF5, A5F5 6 octahedral octahedral SF6, TeF6, SeF6, Depending on the number of lone pair and bonded pairs of electrons the molecules can be represented as AB2E, AB3E, AB2E2, AB4D, AB3E2, AB5E, AB4E2 where A is the central atom, B - bonded atom E - lone pair of electrons. examples of the above type of molecules are given in table 5.7. Table 5.4 : Geometry of some molecules (having one or more long pairs of electrons) Molecule type No. of lone No. of bonding Arrangement of shape examples pairs pairs bonded electron pairs AB2E 1 2 Bent SO2, O3 SO2 119.3º 62  AB3E 1 3 Trigonal NH3, PCl3 pyramidal 106.7º AB2E2 2 2 Bent H2O, OF2, H2S, SCl2, etc. 104.5º AB4E 1 4 See saw SF4 AB3E2 2 3 T-shape ClF3, BrF3,ICl3, etc 86.2º AB5E 1 5 square BrF5, IF5 pyramid AB4E2 2 4 square planar XeF4 The VSEPR theory is therefore able to The lone-pair-bond pair repulsions are predict and also explain the geometry of large stronger and the bonded pairs are pushed inside number of compounds, particularly of p-block thus reducing the bond angle to 1070181. and elements. shape of the molecules becomes pyramidal. Let us explain the bond angles in NH3 2. Water molecule H2O : The central atom and H2O. oxygen has six electrons in its valence shell. 1. Ammonia NH3 : Expected geometry is On bond formation with two hydrogen atoms tetrahedral and bond angle 1090 28'. Central there are 8 electrons in the valence shell of nitrogen atom has in all 8 electrons in its oxygen. Out of these two pairs are bonded valence shell, out of which 6 are involved pairs and two are lone pairs. in forming, three N-H covalent bonds Due to lone pair - lone pair repulsion the lone the remaining pair forms the lone pair. pairs are pushed towards the bond pairs and There are two types of repulsions between the bond pair- Lone pair repulsions becomes electron pairs. stronger thereby reducing the HOH bond i. Lone-pair-bond pair angle from the tetrahedral one to 1040351 and ii. Bond pair - bond pair the geometry of the molecule is angular. 63  Lone pair v. If an atom possesses more than one on nitrogen unpaired electrons, then it can form more + than one bond. So number of bonds formed will be equal to the number of half-filled 1s orbitals of H sp3 hybrid orbital of N NH3 orbitals in the valence shell i.e. number of Lone pair in an unpaired electrons. sp3 hybrid orbital on oxygen vi. The distance at which the attractive and + O-H s bond between an repulsive forces balance each other is the oxygen sp3 hybrid and an equilibrium distance between the nuclei H Is orbital 1s orbital of H sp3 hybrid orbital of O H 2O of the bonded atoms. At this distance Fig 5.1 : Formation and orbital pictures of NH3 the total energy of the bonded atoms is and H2O molecules minimum and stability is maximum. Advanced theories of Bonding : vii. Electrons which are paired in the The Kossel and Lewis approach valence shell cannot participate in bond to chemical bonding is the first step in formation. However in an atom if there understanding the nature of chemical bond. is one or more vacant orbital present then More advanced theories of bonding were put these electrons can unpair and participate forth to account for the newly discovered in bond formation provided the energies properties of compounds in the light of quantum of the filled and vacant orbitals differ mechanical theory of atomic structures. Two slightly from each other. important approaches regarding nature of viii. During bond formation the 's' orbital which chemical bond are valence bond theory and is spherical can overlap in any direction. molecular orbital theory. The 'p' orbitals can overlap only in the x, y 5.4 Valence Bond Theory : or z directions. [similarly 'd' and 'f' orbitals In order to explain the covalent bonding, are oriented in certain directions in space Heitler and London developed the valence and overlap only in these direction]. Thus bond theory on the basis of wave mechanics. the covalent bond is directional in nature. This theory was further extended by Pauling 5.4.2 Interacting forces during covalent bond and Slater. formation : By now we have understood that 5.4.1 Postulates of Valence Bond Theory : a covalent bond is formed by the overlap of i. A covalent bond is formed when the two half filled atomic orbitals and the bonded half-filled valence orbital of one atom atoms are stable than the free atoms and the overlaps with a half filled valence orbital energy of the bonded atoms is less than that of of another atom. free atoms. So lowering of energy takes during ii. The electrons in the half-filled valence bond formation. How does this happen ? orbitals must have opposite spins. This happens due to interactive forces iii. During bond formation the half-filled which develop between the nuclei of the two orbitals overlap and the opposite spins atoms and also their electrons. These forces of the electrons get neutralized. The may be attractive and repulsive between increased electron density decreases the nuclei of A and electrons of B and those nuclear repulsion and energy is released arising from attraction between nuclei of atom during overlapping of the orbitals. A and electrons of B and the repulsion between iv. Greater the extent of overlap stronger electrons. is the bond formed, however complete The balance between attractive and overlap of orbitals does not take place due repulsive forces decide whether the bond will to internuclear repulsions. be formed or not. 64  When the attractive forces are stronger and stability decreases. (See Fig. 5.2 potential than the repulsive forces overlap takes place energy diagram) between the two half filled orbitals a bond is If atoms containing electrons with formed and energy of the system is lowered. parallel spins are brought close to each other, This lowering of energy during bond the potential energy of the system increases formation is depicted in the potential energy and bond formation does not take place. diagram. To understand this let us consider 5.4.3 Overlap of atomic orbitals : Formation the formation of H2 molecule from atoms of a bond has been explained on the basis of of hydrogen each containing one unpaired overlap of atomic orbital having same energy electrons. When the two atoms are for away and symetry. In the preceding section, we have from each other there are no interactions seen that the strength of the bond depends on between them. The energy of the system is the the extent of overlap of the orbitals. Greater sum of the potential energies of the two atoms the overlap stronger is the bond. which is arbitarily taken as zero. The orbitals holding the electrons vary in shape, energy and symetry. So the extent of overlap depends on the shape and size of the orbital On the basis of the above considerations we have 2 types of bonds. i. sigma bond (σ) ii. pi bond (π) i. Sigma Bond : When the overlap of the bonding orbitals is along the internuclear axis it is called as sigma overlap or sigma bond. Fig. 5.2 : Potential energy diagram for The σ bond is formed by the overlap of formation hydrogen molecule following orbitals. Repulsive forces be stabilize the a. Two 's' orbitals system with increase energy of the system b. One 's' and one pz orbital while attractive forces decrease the energy. c. Two 'p' orbitals Experimetally it has been found that during a. s-s overlap : eg. H2 formation of hydrogen molecule the magnitude The 1s1 orbitals of two hydrogen atoms of the newly developed attractive forces overlap along the internuclear axis to form a contribute more than the newly developed σ bond between the atoms in H2 molecule. repulsive forces. As a result the potential electronic configuration of H : 1s1 energy of the system begins to decrease. As the atoms come closer to one another the energy of the system decreases. The overlap increases only upto a certain distance + between the two nuclei, where the attractive s s and repulsive forces balance each other and the system attains minimum energy (see Fig H H 5.3). At this stage a bond is formed between the two atoms of hydrogen. If the two atoms + + + + + are further pushed closer to each other the repulsive forces become more predominant 1s orbitals of H H2 molecule and the energy of the system starts increasing 65 b. p-p overlap trivalency, carbon shows tetravalency in spite This type of overlap takes place when two of their electronic configuration e.g. BeH2, p orbitals from different atoms overlap along BF3, CH4, CCl4 etc. the internuclear axis eg. F2 molecule. Be : 1s2, 2s2 B : 1s2, 2s2, 2p1 C : 1s2, 2s2, 2p1x, 2p1 + In order to explain the observed valency p p in these and such other compounds a concept Electronic configuration of fluorine 1s2, of hybridization was put forward. 2s2, 2px2, py2, pz1, The 2pz orbitals of the fluorine It was suggested that one elctron in '2s' atoms overlaps along internuclear axis to form orbital is promoted to the empty '2p' orbital. p-p σ overlap. Thus in the excited state Be, B and C have two, F2 molecule :1s2, 2s2, 2px2, py2, pz1 three and four half filled orbitals, respectively. c. s-p σ bond Electronic configurations in excited state : In this type of overlap one half filled 1s 2s 2p 's' orbital of one atom and one half filled 'p' Be orbital of another orbital overlap along the B internuclear axis. eg. HF molecule C In the excited state Be, B and C have 2, + 3 and 4 half filled orbitals. So Be, B and C s p can form 2, 3 and 4 bonds respectively. This Electronic configuration : concept helps to understand how Be forms 2 H 1s1 ; F : 1s2, 2s2, 2px2, py2, pz1 bonds whereas B and C form 3 and 4 bonds, respectively but it cannot explain how all bonds 1s1 orbital of hydrogen and 2pz1 of fluorine have same bond length and bond strength. overlap to form s - p σ overlap. For example, in BeF2 berylium will use ii. p-p overlap/ π overlap/π bond : one s and one half-filled p orbitals to overlap When two half filled orbitals of two with two half filled 'p' orbitals on fluorine, atoms overlap side ways (laterally) it is so in the molecule there will be one s-p bond called π overlap and it is perpendicular to the and one p-p bond. which will not be of equal interuclear axis. strength, but actually both Be-F bonds are of the same strength.Similar situation is seen in BF3, BH3, CH4, CCl4. In CH4 all bonds are of equal strength although the overlaps are between s, px, py, pz orbitals of carbon and 's' orbital of hydrogen, experimentally all C-H p- orbital p- orbital π - overlap bond lengths bond strengths and bond angles 5.4.4 Hybridization : The valence bond theory are found to be identical. explained well the formation of covalent bond This can be explained using another by the overlap of orbitals in case of simple concept. ''Hybridiztion'' in the valence molecules like H2, F2, H-F etc. Accordingly bond theory. This concept helps to explain the maximum number of covalent bonds the observed structural properties of many which an atom can form equals the number of molecules. unpaired electrons present in its valence shell. Hybridization refers to mixing of valence orbitals of same atom and recasting them But the theory does not explain how berylium into equal number of new equivalent orbitals- forms two covalent bonds or how boron shows Hybrid orbitals. 66 Steps considered in Hybridization vi. A hybrid orbital has two lobes on the two i. Formation of excited state sides of the nucleous. One lobe is large ii. Mixing and Recasting of orbitals and the other small. i. Formation of the excited state : The paired vii. Covalent bonds formed by hybrid orbitals electrons in the ground state are uncoupled are stronger than those formed by pure and one electron is promoted to the a vacant orbitals, because the hybrid orbital has orbial having slightly higher energy. Now total electron density concentrated on the side number of half filled orbitals is equal to the with a larger lobe and the other is small valency of the element in the stable compound. allowing greater overlap of the orbitals. e.g. in BeF2, valency of Be is two. In the excited state one electron from 2s orbital is uncoupled 5.4.5 Types of Hybridization and Geometry and promoted to 2p orbital. of Molecules : Different types of hybrid 2s 2p orbitals are obtained from the atomic orbitals that participate in hybridization. Ground state s and p orbitals can hybridize to form the Excited state following hybrid orbitals ii. Mixing and Recasting : In this step the two i. sp3 's' and 'p' orbitals having slightly different ii. sp2 energies mix with each other. Redistribution iii. sp of electron density and energy takes place and i. sp3 Hybridization : In this type one 's' and two new orbitals having exactly same shape three 'p' orbitals having comparable energy and energy are formed. mix and recast to form four sp3 hybrid orbitals. These new orbitals arrange themselves in space It should be remembered that 's' orbital is in such a way that there is minimum repulsion spherically symmetrical while the px, py, pz, and maximum sepration between them. orbitals have two lobes and are directed along So during formation of sp hybrid orbitals as in x, y and z axes, respectively. Be the two sp hybrid orbitals are 1800. The four sp3 hybrid orbitals formed are Conditions for hybridization : equivalent in energy. and shape. They have 1. Orbitals belonging to the same atom can one large lobe and one small lobe. They are participate in hybridization. at an angle of 109028 with each other in space 2. Orbitals having nearly same energy can and point towards the corners of a tetrahedron undergo hybridization, so 2s and 2p orbitals CH4, NH3, H2O are examples where the orbitals undergo hybridization but 3s and 2p orbitals on central atom undergo sp3 hybridization. do not. Characteristic features of hybrid orbitals : i. Number of hybrid orbitals formed is exactly the same as the participating atomic orbitals. ii. They have same energy and shape. iii. Hybrid orbitals are oriented in space in such a way that there is minimum repulsion and thus are directional in nature. Fig 5.3 : Formation of sp3 hybrid orbitals iv. The hybrid orbitals are different in shape Formation of methane (CH4) molecule : from the participating atomic orbitals, but Ground state electronic configuration of they bear the characteristics of the atomic Carbon is 1s2, 2s2, 2px1, py1, pz0. In order to orbitals from which they are derived. v. Each hybrid orbitals can hold two form four equivalent bonds with hydrogen the electrons with opposite spins. 2s and 2p orbitals undergo hybridization. 67 Electronic 1s 2s 2p Two sp2 hybrid orbitals overlap axially configuration of two 's' orbitals of hydrogen to form sp2-s σ carbon bond. The unhydrized 'p' orbitals on the two Ground state carbon atoms overlap laterally to form a lateral π overlap. Thus the C2H4 molecule has Excited state four sp2-s σ bonds. One sp2-sp2 σ bond one p-p sp3 Hybrid orbitals π bond. (four sp3 hybrid orbitals.) Electronic configuration of carbon One electron from the 2s orbital of 1s 2s 2p Carbon atom is excited to the 2pz orbital. Then the four orbitals 2s, px, py, pz mix and recast to Ground state form four new sp3 hybrid orbitals having same Excited state shape and equal energy. They are maximum sp2 Hybridized apart and have tetrahedral geometry. Each sp2 hybrid orbitals pure 'p'orbital hybrid orbital contains one unpaired electron. Each of these sp3 hybrid orbitals with one sp2 orbital PZ PZ electron overlap axially with the 1s orbital of H H sp2 sp2 C C sp2 sp2 hydrogen atom to form one C-H sigma bond. H H sp2 C C sp2 Thus in CH4 molecule we have four C-H bonds formed by the sp3-s overlap. 1s 2p orbital orbital σ bond 109.5º Formation of Boron trifluoride (BF3) + molecule : 1. Need of hybridisation : Observed valency 1s orbitals of H sp3 hybrid orbital of C CH4 of boron in BF3 is three and its geometry is triangular planar which can be explained on ii. sp2 Hybridization : This hybridization the basis of sp2 hybridisation. involves the mixing of one s and two p orbitals 2, sp2 hybridisation of Boron atom : In BF3 to give three sp2 hybrid orbitals of same energy molecule central boron atom undergoes and shape. The three orbitals are maximum sp2 hybridisation. The ground state apart and oriented at an angle of 1200 and electronic configuration of Boron (z = 5) is are in one plane. The third p orbital does 1s22s22px12py12pz0 not participate in hybridization and remains at right angles to the plane of the sp2 hybrid orbitals. BF3, C2H4 molecules are examples of 1s2 2s2 2px12py02pz0 sp2 hybridization. To explain valency of boron in BF3 one electron from 2s orbital of boron atom is uncoupled and promoted to vacant 2py orbital. Thus excited state electronic configuration of boron becomes s22s22px12py12pz1 Fig 5.4 : Formation of sp2 hybrid orbitals Formation of C2H4 molecule : This molecule 1s2 2s 2px12py02pz0 contains two carbon atoms each bound to The three orbitals i.e. 2s, 2px of and 2py of boron undergoes sp2 hybridisation to form two hydrogen. Each carbon atom undergoes three sp2 hybrid orbitals of equivalent energy sp2 hybridization. One 's' orbital and two 'p' which are oriented along the three corners of orbitals on carbon hybridize to form three sp2 an equilateral triangle making an angle of hybrid orbitals of equal energy and symetry. 1200. 68  Thus boron in hybridised state has electronic The remaining two unhybridized p following configuration. orbitals overlap laterally to form two p-p π bonds. So there are three bonds between the 1s2 three hybrid orbitals two carbon atoms : one C-C σ bond (sp-sp) 3. orbital overlap : Each sp2 hybrid orbital overlap, two C-C π bonds (p-p) overlap and of boron atom having unpaired electron fourth sp-s σ bond between C and H satisfy the overlaps axially with half filled 2pz orbital fourth valency of carbon. of fluorine atom containing electron with Electronic configuration of carbon opposite spin to form three B-F sigma bond 1s 2s 2p by sp2-p type of overlap. Ground state 4. Bond angle : Each F-B-F bond angle in BF3 Excited state molecule is 1200. 5. Geometry : The geometry of BF3 molecule is sp Hybridized trigonal planar. sp hybrid orbitals pure 'p'orbitals σ bond σ bond σ bond one π bond Second π bond 5.4.6 Importance and limitation of valence BF3 bond theory Importance of valence bond theory (V.B) iii. sp hybridization : In this type one 's' and V.B. theory introduced five new concepts in one 'p' orbital undergo mixing and recasting to chemical bonding. form two sp hybrid orbitals of same energy i. Delocalization of electron over the two and shape. The two hybrid orbitals are placed nuclei. at an angle of 1800. Other two p orbitals do ii. shielding effect of electrons. not participate in hybridization and are at right iii. covalent character of bond. angles to the hybrid orbitals. For example : iv. partial ionic character of a covalent bond. BeCl2 and acetylene molecule HC ≡ CH, v.The concept of resonance and connection In cross-section between resonance energy and molecular stability. 5.4.7 Limitations of valence bond theory Schematic representation of hybrids shown together i. V.B. Theory explains only the formation Fig 5.5 : formation of sp hybrid orbitals of covalent bond in which a shared pair of Formation of acetylene molecule : This electrons comes from two bonding atoms. molecule contains two carbon and two However, it offers no explanation for the hydrogen atoms. Each carbon atom undergoes formation of a co-ordinate covalent bond in sp hybridization. One s and one p orbitals which both the electrons are contributed by mix and recast to give two sp hybrid orbitals one of the bonded atoms. arranged at 1800 to other. ii. Oxygen molecule is expected to be Out of the two sp hybrid orbitals of carbon dimagnetic according to this theory. The atom one overlaps axially with 's' orbital of two atoms in oxygen molecule should have hydrogen while the other sp hybrid orbital completely filled electronic shells which give overlaps with sp hybrid orbital of other carbon no unpaired electrons to the molecule making atom to form the sp-sp σ bond. The C-H σ it diamagnetic. However, experimentally the bond is formed by sp-s overlap. molecule is found to be para-magnetic having 69 two unpaired electrons. Thus, this theory fails It has been found that the MO theory to explain paramagnetism of oxygen molecule. gives more accurate description of electronic structure of molecules. Problem 5.4 The concept of an orbital is introduced by Explain the formation of BeCl2 quantum mechanics. The quantum mechanical Electronic configuration of berylium wave equation is a differential equation 1s 2s 2p and it solution is called wave function. The Ground state square of the wavefunction gives a measure of probability of finding an electron within a Excited state cetain region of space of an atom. It is nothing sp Hybridized but an atomic orbital. sp hybrid orbitals pure 'p' MO theory does not concentrate on orbitals individual atoms. It considers the molecule as a whole rather than an atom for the bonding. Formation of BeCl2 molecule. Accordingly a molecular orbital MOT is the This molecule has one Be atom and two property of a molecule similar to what an chlorine atoms. Electronic configuration of atomic orbital is to an atom. Hence, molecular Be is 1s2, 2s2, 2pz0. The 2s and 2pz orbitals orbital can be depicted through a square of undergo sp hybridization to form two sp wavefunction that gives the probability of hybrid orbitals oriented at 1800 with each finding an electron within a certain region other. 2pz orbitals of two chlorine atoms of space in a molecule. Like atomic orbitals, overlap with the sp hybrid orbitals to form molecular orbitals have energy levels and two sp-p σ bond. definite shapes. They also contain maximium two electrons with opposite spins. 5.5.1 Formation of molecular orbitals : According to the MO theory the formation of molecular orbitals from atomic orbitals is expressed in terms of Linear Combination of Atomic Orbitals (LCAO). Formation of molecular orbitals can be understood by considering the interference of the electron waves of combining atoms. Interference iii. Valence bond theory does not explain the of electron waves can be constructives or bonding in electron deficient molecules like destructive i.e. the waves can reinforc each B2H6 in which the central atom possesses less other or cancel each other. So we can say that. number of electrons than required for an octet Two atomic orbitals combine in two ways of electron. to form molecular orbitals. 1. By addition 5.5 Molecular orbital theory : You are of their wave functions. 2. By subtraction of familiar with the valence bond theory which their wave functions. Addition of the atomic describes the formation of covalent bonds orbtials wave functions results in formation of by overlap of half filled atomic orbitals. a molecular orbital which is lower in energy This theory is successful to give satisfactory than atomic orbitals and is termed as Bonding electronic description for a large number of Molecular Orbital (BMO). Subtraction of the molecules. In some cases it gives to incorrect atomic orbitals results in the formation of a electronic description. Therefore another molecular orbital which is higher in energy bonding description called Molecular Orbital than the atomic orbitals and is termed as Theory (MO) has been introduced. Antibonding Molecular Orbital (AMO). 70  In bonding molecular orbital the large the overlap, greater is the electron density electron density is observed between the nuclei between the nuclei and so stronger is the of the bonding atoms than the individual bond formed. atomic orbitals. On the other hand in the 5.5.3 Types of molecular orbitals : In antibonding orbital the electron density is diatomic molecules, molecular orbitals formed nearly zero between the nuclei. by combination of atomic orbitals are of two So placing an electron in bonding orbital types (i) σ (ii) π. leads to formation of a covalent bond. While According to this nomenclature a σ placing electron in antibonding orbital makes designates a molecular orbital which is the bond unstable. symetrical around the bond axis and π designates a molecular orbitals those are unsymetrical. This is clear if we consider a linear combination of i. two 's' orbitals ii. two p orbitals. Antibonding MO, σ* 1s Node Substract (1s-1s) Energy of Isolated H atoms Isolated H atoms Add (1s+1s) Fig. 5.5 : formation of MOs 5.5.2 Conditions for the combination of Bonding MO, σ* 1s Atomic Orbitals : Atomic orbitals can be combined linearly which give molecular Fig 5.6 : Linear combination of two s orbitals orbitals only if following conditions are i. The s orbitals are spherically symmetric fulfilled. along x, y and z axes, combination of two '1s' i. The combining atomic orbitals must have orbitals centred on two nuclei of two atoms, comparable energies. led to two σ molecular orbitals which are So 1s orbitals of one atom can combine symetrical along the bond axis. One of which with 1s orbital of another atom but not with is σ bonding and other σ* antibonding (Fig. 7) 2s orbital, because energy of 2s orbital is ii. If we consider 'z' to be internuclear axis much higher than that of 1s orbital. then linear combination of pz orbitals from ii. The combining atomic orbitals must have two atoms can form σ 2pz bonding σ*(2pz) the same symetry along the molecular molecular orbitals. axis. Conventially z axis is taken as the internuclear axis. So even if atomic orbitals + have same energy but their symetry is not same they cannot combine. For example, 2s πp* orbital of an atom can combine only with 2pz orbital of another atom, and not with 2px or 2py orbital of that atom because the + symmetries are not same. pz is symetrical along z axis while px is symetrical along x πp Fig 5.7 : Formation of axis. π and π* molecular orbitals iii. The combining atomic orbitals must overlap to the maximum extent. Greater 71  The px, py orbitals are not symetrical along c. Bond order of molecule : The bond order the bond axis, they have a positive lobe above of the molecule can be calculated from the axis and negative lobe below the axis. Hence the number of electrons in bonding Nb and linear combination of such orbitals leads to the antibonding MOs (Na). formation of molecular orbitals with positive N - Na and negative lobes above and below the bond Bond order = b 2 axis. These are designed as π bonding and π 5.5.5 Key ideas of MO theory : antibonding orbitals. The electron density i. MOs in molecules are similar to AOs of in such π orbitals is concentrated above and atoms. Molecular orbital describes region below the bond axis. The π molecular orbitals of space in the molecule representing the has a node between the nuclei (fig. 5.9) probability of an electron. 5.5.4 Energy levels and electronic ii. MOs are formed by combining AOs of configuration : We have seen earlier that on different atoms. The number of MOs combination of two 1s orbitals; two molecular formed is equal to the number of AOs orbitals σ 1s and σ 1s are formed. Similarly combined. two 2s orbitals yield σ* 2s and σ 2s molecular iii. Atomic orbitals of comparable energies orbitals. and proper symetry combine to form The three 2p orbitals on one atom combine molecular orbitals. with three 2p orbitals on another atom forming iv MOs those are lower in energy than the six molecular orbitals, designated as σ 2pz, starting AOs are bonding (σ) MOs and πpx, πpy and σ *2pz, π*px, π*py those higher in energy are antibonding (σ) The molecular orbitals in homonuclear MOs. diatomic molecules have been determined v. The electrons are filled in MOs begining experimentally. with the lowest energy. For diatomic molecules of second row elements vi. Only two electrons occupy each molecular except O2 and F2 the rank order of energies is orbital and they have opposite spins that σ 1s < σ* 1s < σ 2s < σ* 2s < π2px = π2py > < is, their spins are paired. σ 2pz < (π*2py = π*2px) < σ* 2pz vii. The bond order of the molecule can be For O2 and F2 increasing order of energies was calculated from the number of bonding found to be : and antibonding electrons. σ 1s < σ* 1s < σ 2s < σ* 2s < σ2pz < (π2px = π2py) < π*2px < π*2py) < σ* 2pz 5.5.6 MO description of simple diatomic The sequence of filling the molecular orbitals Molecules give electronic configuration of molecules. The 1. Hydrogen molecule.-H2 electronic configuration of molecules provides the following information. a. Stability of molecules : If the number of electrons in bonding MOs is greater than the number in antibonding MOs the molecule is stable. b. Magnetic nature of molecules : If all MOs in a molecule are completely filled with two electrons each, the molecule is diamagnetic (i.e. repelled) by magnetic field. However, if at least one MO is half filled with one electron, the molecule is paramagnetic Fig. 5.8 : MO diagram for H2 molecule (i. e. attracted by magnetic field). 72  Hydrogen molecule is formed by the 3. N2 molecule : 7 linear combination of two Hydrogen atoms, each having one electron in its 1s orbital. Linear combination of two 1s orbitals gives two molecular orbitals σ1s and σ*1s. The two electrons from the hydrogen atoms occupy the σ1s molecular orbital and σ*1s remains vacant. Electronic configuration of Hydrogen molecule is written as σ1s2 Bond order = (bonding electron − antibonding electrons) ÷ 2 For hydrogen Bond order=(2-0)÷2=1 So in H2 molecule there exists one covalent bond between the two hydrogen atoms. The bond length is 74 pm and the bond Fig. 5.10 : MO diagram for N2 dissociation energy is 438 kJ mol. As there are N : 1s2 2s2 2p3 no unpaired electrons present the H2 molecule Electronic configuration of N2 molecule (14 is diamagnetic. electrons) is 2. Lithium molecule. Li2 : (σ1s)2 (σ*1s)2 (σ2s)2 (σ* 2s)2 (πx)2 (πy)2(σ 2pz)2 10 - 4 Bond order of N2 molecule = =3 2 N2 molecule is diamagnetic. 4. O2 molecule : Fig. 5.9 : MO diagram for Li2 Each Lithium atom has 3 electrons with electronic configuration 1s2, 2s1, so Li2 molecule will have 6 electrons. Linear combination of atomic orbitals gives four molecular orbitals σ1s, σ*1s, σ2s, σ*2s Electronic configuration of Li2 molecule will Fig. 5.11 : MO diagram for O2 be (σ1s)2, (σ*1s)2, (σ2s)2 8 O : 1s 2s2 2p4 2 Bond order=(4-2)÷2=1, The electronic configuraton of O2 molecule This means in Li2 molecule there is one bond (16 electrons) is between the two Lithium atoms. Such Li2 molecules are found in the vapour state. As (σ1s)2 (σ*1s)2 (σ 2s)2 (σ*2s)2 (πx)2(πy)2 (σz)2(π*x)2 there are no unpaired electrons the molecule is (π*y)2 diamagnetic. 10 - 6 Bond order of O2 molecule = =2 2 73 O2 molecule is paramagnetic due to presence Bond angle can be determined of 2 unpaired electrons in the π* orbitals. experimentally using spectroscopic techniques. 5. F2 molecule : Value of bond angle gives an idea about the arrangement of orbitals around the central atom and the shape of the molecule. Table 5.5 bond angles of some molecules Molecule Bond of angle 1 H2O 1040281 2 NH3 107 3 BF3 120 5.6.2 Bond Enthalpy : The amount of energy required to break one mole of bond of one type, present between two atoms in the F F-F F gaseous state is termed as Bond Enthalpy. For Atom Molecular Atom Configuration Configuration Configuration diatomic molecules the dissociation energy Bond Order = 1 is the same as bond enthalpy. Bond enthalpy Fig. 5.12 : MO diagram for F2 for H2 molecule is 435.8 kJ mol-1. The bond 9 F : 1s2 2s2 2p5 enthalpy is a measure of strength of the bond The electronic configuration of F2 molecule between two atoms and can be measured (18 electrons) is (σ 1s)2 (σ*1s)2 (σ 2s)2 (σ*2s)2 experimentally. N-N bond in N2 is stronger (πx)2(πy)2 (σz)2(π*x)2 (π*y)2 than the O-O bond in O2. Larger is the bond 10 - 8 dissociation energy stronger is the bond in Bond order of F2 molecule = =1 the molecule. For heteronuclear diatomic 2 F2 molecule is diamagnetic molecule HCl the bond enthalpy was found to be 431.0 kJ mol-1. In case of polyatomic molecules the bond Can you tell? enthalpy and bond dissociation energy are not Why He2 molecule is not stable ? identical. Bond enthalpy is the average of the Draw MO diagram for it sum of successive bond dissociation energies. For example dissociation of water. 5.6 Parameters of covalent bond : A covalent H2O(g) H(g) + OH(g) ∆H1 = 502 kJ mol-1 bond is characterised by different parameters. OH (g) H (g) + O (g) ∆H2 = 427 kJ mol-1 These parameters help to understand how Average bond enthalpy of O-H bond in strong is the bond between the two atoms, H2O : what is the distance between them and what is 502 + 427 ∆aH = = 464.5 kJmol-1. the shape of the molecule. 2 5.6.1 Bond angle : The electrons which In both the above equations the bond participate in bond formation are present between O and H is broken but the amount of in orbitals. The angle between the orbitals energy required to break the bond is different, holding the bonding electrons is called the i.e. enthalpies of two O-H bonds in water are bond angle. different. This difference arises due to the fact that cleavage of the two O-H bonds in water takes place in two steps. In the first step one O-H bond breaks leaving behind OH radical. Now Bond angle the electronic enviornment around oxygen to 74 which hydrogen is attached is different than Table 5.7 Average bond lengths for some single, that around oxygen in H2O molecule and double and triple bonds this causes a change in the successive bond Type Covalent Type of Covalent dissociation energy. of bond bond bond Same difference is observed in enthalpy bond length length values of O-H bond in C2H5OH. Oxygen here (pm) (pm) is attached to C2H5 group therefore hydrogen of O-H 96 H2(H-H) 74 O-H is in different environment than hydrogen C-H 107 F2(F-F) 144 of H-O-H. N-O 136 Cl2(Cl-Cl) 199 In the same way, the bond enthalpy value C-O 143 Br2(Br-Br) 228 of any covalent bond is slightly different for C-N 143 I2(I-I) 267 each bond of that kind in a given molecule and C-C 154 N2(N≡N) 109 also different molecules. The average values of bond enthalpy, ∆aH , are determined from C=O 121 O2(O=O) 121 the experimentally measured values of large N=O 122 HF (H-F) 92 number of compounds containing a particular C=C 133 HCl 127 bond. Average bond enthalpy data are given (H-Cl) in Table 5.4. In general stronger bond implies C=N 138 HBr 141 larger bond enthalpy. (H -Br) C≡N 116 HI (H-I) 160 Do you know ? C≡C 120 Each atom of the bonded pair contributes Among diatomic molecules the bond to the bond length. Bond length depends upon order and bond enthalpy of N2 is highest. the size of atoms and multiplicity of bonds. Bond order = 3, Bond enthalpy = 946 kJ mol-1 It increases with increase in size of atom Table 5.6 Bond enthalpies and decreases with increase in multiplicity of Bond ∆aH / kJ mol-1 bond. It is generally measured in picometre 0 C-H 400 - 415 (pm) or in Angstrom unit ( A ) − > = > ≡ >, so N-H 390 C - C single bond is longer and C ≡ C triple O-H 460-464 bond is shorter. Some typical average bond lengths of C-C 345 C - C single double and triple bond and others C- N 290 -315 are shown in table 5.4. C-O 355 - 380 5.6.4 Bond Order : According to the Lewis C - Cl 330 theory the bond order is equal to the number C - Br 275 of bonds between the two atoms in a molecule. O-O 175 - 184 The bond order in H2, O2 and N2 is 1, 2 and 3 C=C 610 - 630 respectively. Isoelectronic molecules and ions C≡C 835 have identical bond order. For example N2, CO C=O 724 - 757 and NO+ each have bond order 3. F2 whereas C≡N 854 O22- has bond order 1. Stabilities of molecules can be determined by knowing the bond 5.6.3 Bond length : Bond length implies the order. As bond order increases, bond enthalpy equilibrium distance between the nuclei of two increases and bond length decreases. covalently bonded atoms in a molecule. Bond lengths are measured by X-ray and Electron diffraction techniques. 75 5.6.5 Polarity of a Covalent Bond : Covalent 1 D = 3.33564 × 10-30 Cm bonds are formed between two atoms of the C : coulomb ; m : meter same or different elements. Thus covalent bond Dipole moment being a vector quantity is is formed between atoms of some elements represented by a small arrow with the tail on of H-H, F-F, Cl-Cl etc. The shared pair of the positive centre and head pointing towards electrons is attracted equally to both atoms the negative centre. and is situated midway between two atoms. δ+ H F-δ (µ =1.91 D). The crossed arrow Such covalent bond is termed as Nonpolar ( ) above the Lewis structural indicates Covalent bond. the direction of the shift of electron density H : H H-H towards the more electronegative atom. electron pair at the centre Non polar Dipole moments of polyatomic molecules covalent bond : Each polar bond in a polyatomic molecule When a covalent bond is formed between two atoms of different elements and have has its own dipole. The resultant dipole of different electronegativities the shared electron the molecule is decided by (i) shape of the pair does not remain at the centre. The electron molecule that is the spatial arrangement of pair is pulled towards the more electronegative bonds (ii) contribution of the individual dipoles atom resulting in the separation of charges. This and those of the lone pair of electrons, if any give rise to as Dipole. The more electronegative The dipole moment of polyatomic molecule atom acquires a partial -ve charge and the is the vector sum of the dipole moments of other atom gets a partial +ve charge. Such a various bonds and lone pairs. bond is called as polar covalent bond. The Consider BeF2 and BF3. examples of polar molecules include. HF, HCl BeF2 is a linear molecule and the dipoles are in etc. opposite direction and are of equal strengths, H : F δ+H−Fδ- H-F so net dipole moment of BeF2 = 0 Polar covalent bond Fluorine is more electronegative than F Be F + =0 Hydrogen therefore the shared electron pair is Bond dipoles net dipole in BeF2 = 0 more towards fluorine and the atoms acquire BF3 is angular partial +ve and -ve charges, respectively. In BF3 bond angle is 1200 and molecule is Polarity of the covalent bond increases as the symetrical. The three B-F bonds are oriented difference in the electronegatively between the at 1200 with each other and sum of any two is bonded atoms increases. When the difference equal and opposite to the third therefore sum in elctronegativities of combining atom is of three B-F dipoles = 0. about 1.7 ionic percentage in the covalent bond is 50%. F F B Can you tell? + =0 F net dipole in BF3 = 0 Which molecules are polar ? Bond dipoles H-I, H-O-H, H-Br, Br2, N2, I2, NH3 In case of angular molecules both lone 5.7 Dipole moment pairs and electonegativity difference contribute Definition : Dipole moment (µ) is the product to dipole moment. of the magnitude of charge and distance Lone pairs and dipole moment : In some between the centres of +ve and -ve charges. molecules the central atom has unshared or µ=Q×r lone pairs of electrons. These lone pairs also Q : charge ; r : distance of separation. contribute to overall dipole of the molecule. unit of dipole moment is Debye (D) Nitrogen in NH3 and Oxygen in H2O posses 76 lone pairs, these reinforce the dipoles due to In CHCl3 the dipoles are not equal and N-H and O-H bonds. Both these molecules are do not cancel hence CHCl3 is polar with a non highly polar. zero dipole moment. δ- δ- CHCl3 H C Nδ- Oδ- Cl Cl H δ+ Hδ+ Cl Hδ+ Hδ+ Hδ+ µ = 1.04 µ= 1.48 D µ= 1.85 D Nitrogen has only one lone pair while Dipole moments of some molecules are oxygen has two lone pairs which reflects in the shown in table 5.8. higher dipole moment of water. Table 5.8 dipole moments and geometry of Consider NH3 and NF3 some molecules Both have pyramidal shape with a lone Types of Example Dipole Geometry pair of electrons on nitrogen atom. Here molecule moment hydrogen is less electronegative while, µ (D) fluorine is more electronegative than nitrogen. Molecule HF 1.91 linear The resultant dipole moment of NH3 is 4.90 × AB 10-30 Cm while that of NF3 is 0.80 × 10-30 Cm. This difference is because in case of NH3 the HCl 1.03