Bond parameters
Bond Strength and Bond Length
The average distance between the centres of two nuclei of the bonded atoms measured in angstrongs. Bond energy or bond strength is the amount of energy required to break one mole of covalent bonds in gas phase. Greater the bond energy of a bond more strong the bond and more stable the bond. (Bond length µ1 / Bond strength)
Bond energy increases as the number of bonds increases between the atoms.
Bond Lengths \(C-C>C=C>C\equiv C\)
Bond Energies\(C\equiv C>C=C>C-C\)
Bond Lengths \(C-N>C=N>C\equiv N\)
Bond Energies\(C\equiv N>C=N>C-N\)
Bond energy increases as the contribution of p-orbitals increases in hybrid orbitals
sp < sp2 < sp3 (Bond energies)
sp3 < sp2 < sp (Bond Lengths)
Bond Angle
The angle between two bonds sharing a common atom is known as the bond angle. Bond angle depends on the geometry of the molecule. For example the bond angle in tetrahedral geometry is \({{109}^{0}}\) . 28¢ while bond angle in planar geometry is \({{120}^{0}}\). bond angle is also depends on the identity of atoms bonded, for example the geometry of \(\text{N}{{\text{H}}_{\text{3}}}\) and \(N{{F}_{3}}\) is same but bond angle of \(\text{N}{{\text{H}}_{\text{3}}}\) is greater than \(N{{F}_{3}}\) because F has higher electro negativity than H, the electron pair is attracted more towards F in \(N{{F}_{3}}\) i.e. the bond pairs of electrons are away from N or in other words distance between bond pairs ion is more. Hence repulsion between bond pairs in \(N{{F}_{3}}\) is less than \(\text{N}{{\text{H}}_{\text{3}}}\). Hence the lone pair repels the bond pairs of \(N{{F}_{3}}\) more than it does in \(\text{N}{{\text{H}}_{\text{3}}}\). As a result, the bond angle decreases to 102.40. Whereas in \(\text{N}{{\text{H}}_{\text{3}}}\) it decreases to 107.30 only.
Illustration : Explain why bond angle of PH3 is less than that of PF3.
Sol. \(\text{P}{{\text{H}}_{\text{3}}}\,\,\text{and}\,\,\text{P}{{\text{F}}_{\text{3}}}\) are pyramidal in shape with one lone pair on P. But \(\text{P}{{\text{F}}_{\text{3}}}\) has greater bond angle than \(P{{H}_{3}}\) (opposite to \(N{{H}_{3}}\) and \(N{{F}_{3}}\)). This is due to resonance in \(\text{P}{{\text{F}}_{\text{3}}}\),leading to partial double bond character as shown below
As a result repulsions between P – F bonds are large and hence the bond angle is large. There is no possibility of formation of double bonds in \(P{{H}_{3}}\)
Bond Order: It may be defined as the half the difference between the number of electrons present in the bonding molecular orbitals and the anti-bonding molecular orbitals i.e.
Bond order (B.O.) =\(\frac{No.of\,electrons\,in\,BMO-No.of\,electrons\,in\,ABMO}{2}\)
A positive bonding order suggests a stable molecule while a negative bond order or zero bond order suggest an unstable molecule.
Magnetic Behaviour: If all the molecular orbitals in species are spin paired, the substance is diamagnetic. However, if one or more molecular orbitals are singly occupied it is paramagnetic.
Illustration :
Arrange the species \({{O}_{2}},\,\,O_{2}^{-},\,\,O_{2}^{2-}\,\,and\,\,O_{2}^{+}\) in the decreasing order of bond order and stability and also indicate their magnetic properties.
Sol. The molecular orbital configuration of \({{O}_{2}},\,\,O_{2}^{-},\,\,O_{2}^{2-}\,\,and\,\,O_{2}^{2+}\) are as follows:
O2 =\(\sigma\)1s2, \(\sigma\)*1s2, \(\sigma\)2s2, \(\sigma\)*2s2, \(\sigma\)2px2,\(\pi\)2py2, \(\pi\)2pz2, \(\pi\)*2py1 =\(\pi\)*2pz1
Bond order =\(\frac{10-6}{2}=2\), No. of unpaired electrons = 2
Thus paramagnetic
O2 –= \(\sigma\)1s2, \(\sigma\)*1s2, \(\sigma\)2s2, \(\sigma\)*2s2, \(\sigma\)2px2, \(\pi\)2py2, \(\pi\)2pz2, \(\pi\)*2py2 = \(\pi\)*2pz1
Bond order = \(\frac{10-7}{2}=1.5\), No. of unpaired electrons = 1
Thus paramagnetic
O22- = \(\sigma\)1s2, \(\sigma\)*1s2,\(\sigma\)2s2, \(\sigma\)*2s2, \(\sigma\)2px2,\(\pi \) 2py2,\(\pi \)2pz2, \(\pi \)*2py2 = \(\pi \)*2pz2
Bond order = \(\frac{10-8}{2}=2\), No. of unpaired electrons = 0
Thus paramagnetic
O2+ = \(\sigma\)1s2, \(\sigma\)*1s2, \(\sigma\)2s2,\(\sigma\) 2s2,\(\sigma\) 2px2, \(\pi \)2py2, \(\pi \)2pz2, \(\pi \)*2py1 = \(\pi \)*2pz0
Bond order = \(\frac{10-5}{2}=2.5\), No. of unpaired electrons = 1
Thus paramagnetic
Now as the bond order decreases in the order O2+>O2 >O2->O22-
So, same will be the stability order of the above species because stability is directionally proportional to bond order.
Note: Bond length is inversely proportional to bond order.
Illustration :
The Correct order of bond energies of the following is:
[A] N2>N2+ > N2– > N22– [B] N2+ > N2– >N22– > N2
[C] N2 > N2– = N2+ > N22– [D] N2– > N2 = N2+ >N22–
Solution: Bond order of \({{N}_{2}}=3,\,\,N_{2}^{+},\,\,N_{2}^{-}=2.5,\,\,N_{2}^{2-}=2\)
But N2– has more electrons in ABMO than \(N_{2}^{+}\) so correct order of bond energy will be
N2 > N2+ > N2- > N22-
Resonance
It is the description of the electronic structures of a molecule (or) an ion by means of several schemes of pairing of electrons, with the features of each scheme contributing in the discription. As a result of resonance the energy of a molecule is lowered and aquires minimum value. The energy of the true structure is less than any of the resonating structures(Resonance Hybrid). This lowering in energy is called resonance energy.
Resonance energy = Energy of most stable cannonical form - Experimental value of energy.
Conditions of Resonance
1. The relative positions of all the atoms in each of the resonating forms must be the same.
2. The number of unpaired and paired electrons in each of the resonating forms must be the same. So that a continuous change from one bond type to another bond type may occur.
3. Like charges should not reside on atoms close together in the contributing form, but unlike charges should not be greatly seperated.
4. Contributing structures should be of almost equal energy.
Ex:
The above are the resonating structures but can not explain paramagnetic character of
These structures explain paramagnetic character of .
Due to resonance double bond acquires single bond character, single bond acquires double bond character.
Greater the no. of resonating structures that can be written more stable the molecules.
Structures having lesser no. of bonds have higher energy.
Dipole Moment: It is a vector quantity and is defined as the product of the magnitude of charge on any of the atom and the distance between the atoms. It is represented by
Magnitude of dipole moment
\(\left| \mu \right|=q\left( \text{charge in}\,\text{esu} \right)\times r\left( \text{distance}\,\text{in}\,{{\text{A}}^{\text{0}}} \right)\)
The unit esu-cm or debye(D) \((D={{10}^{-18}}\,\,esu\,\,cm)\) is used in practice. In SI units charge q is measured in coulombs (C) and the distance, r in metre, m
1C = 2.998 × 109 esu and 1 m = 102 cm
1 C-m = 2.998 × 109 × 102 = 2.998 × 1011 (esu) cm
Therefore in SI system, the unit of dipole moment is coulomb metre
1 C-m =\(\frac{2.998\times {{10}^{11}}}{{{10}^{-18}}}=2.998\times {{10}^{29}}D\) or
\(1D=3.336\times {{10}^{-30}}\,\,C-m\)
\(\,\text{1D=3}\text{.336 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-30}\,}}\text{col-m}\)
Dipole moment is a vector quantity and is often indicated by an arrow parallel to the line joining the point of charge and pointing towards the negative end e.g..
% Ionic character of a covalent bond = \(\frac{\exp erimental\,dipole\,moment}{theoretical\,dipole\,moment\,assu\min g\,100%\,ionic\,character}\times 100\)
Illustration:
The dipole moment of KCl is 3.336 × 10–29 Cm. The interatomic distance K+ and Cl– ion in KCl is 260 pm. Calculate the dipole moments of KCl, if there were opposite charges of the fundamental unit located at each nucleus. Calculate the percentage ionic character of KCl.
Sol. From the given data
q = 1.602 × 10–19C
r = 260 pm = 260 × 10–12 m = 2.6 × 10–10 m
Magnitude of dipole moment for 100% ionic character
|\(\mu \)| = qr = (1.602 × 10–19) (2.6 × 10–10) = 4.165 × 10–29 Cm
Actual dipole moment = 3.336 × 10–29 Cm
\(\therefore\)% of ionic bond =\(\frac{3.336\times {{10}^{-29}}}{4.165\times {{10}^{-29}}}\times 100=80.1\%\)
The bond is 80.1% ionic.
In general a polar bond is established between two atoms of different radii and different electronegativities while positive centres (nuclei) of different magnitudes combine to share an electron pair. Greater the values of the dipole moment, greater is the polarity of the bond.
The following points may be borne in mind regarding dipole moments :
i) In case a molecule contains two or more polar bonds, its dipole moment is obtained by the vectorial addition of the dipole moments of the constituent bonds.
ii) A symmetrical molecule is non-polar even though it contains polar bonds.
For example, carbon dioxide, methane and carbon tetrachloride, being symmetrical molecules, have zero dipole moments.
Dipole moment of methyl chloride is a vectorial addition of dipole moments of three C – H bonds and one C – Cl bond.
Dipole moment gives valuable information about the structure of molecules. For example, carbon dioxide is assigned a linear structure since its dipole moment is zero.
We have seen that in a polar covalent bond between two atoms (say A and B), there is a partial separation of charge. This bond is, therefore, said to have a partial ionic character. Greater the difference of electronegativity between A and B, greater is the degree of ionic character (or polarity measured by dipole moment of AB) of the bond. Pauling gave a fairly accurate rule by which the nature of the bond can be predicted. According to this rule, “If the difference on the electronegativity scale between the two atoms is 1.9, the bond is 50% ionic in character. When the difference is greater than 1.9, the bond is correspondingly more ionic”. For example, when the electro negativity difference is 0.8, 1.2, 2.2 and 2.6, the corresponding partial ionic character is 12%, 25%, 61% and 74% respectively.
Illustration:
The resultant dipole moment of water is 1.85D ignoring the effects of lone pair. Calculate,the bond moment of each OH bond {given that bond angle in H2O = 1040, cos 1040 = –0.25}
Sol. R2 = P2 + Q2 + 2PQ cos
(1.85)2 = x2 + x2 + 2x2\((-\frac{1}{4})\)
(1.85)2 = 2x2 –\(-\frac{{{x}^{2}}}{2}\Rightarrow \frac{3{{x}^{2}}}{2}\) \(\therefore\)x = 1.51D
Bond Length,Bond Angle
Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule. Bond lengths are measured by spectroscopic, X-ray diffraction and electron-diffraction techniques about which you will learn in higher classes. Each atom of the bonded pair contributes to the bond length (Fig. 4.1). In the case of a covalent bond, the contribution from each atom is called the covalent radius of that atom.
The covalent radius is measured approximately as the radius of an atom’s core which is in contact with the core of an adjacent atom in a bonded situation. The covalent radius is half of the distance between two similar atoms joined by a
Fig. 4.1 The bond length in a covalent molecule AB.R = rA + rB (R is the bond length and rA and rB are the covalent radii of atoms A and B respectively)
covalent bond in the same molecule. The van der Waals radius represents the overall size of the atom which includes its valence shell in a nonbonded situation. Further,the van der Waals radius is half of the distance between two similar atoms in separate molecules in a solid. Covalent and van der Waals radii of chlorine are depicted in Fig.4.2.
Fig. 4.2 Covalent and van der Waals radii in a chlorine molecule. The inner circles correspond to the size of the chlorine atom (rvdw and rc are van der Waals and covalent radii respectively).
Some typical average bond lengths for single, double and triple bonds are shown in Table 4.2. Bond lengths for some common molecules are given in Table 4.3.The covalent radii of some common elements are listed in Table 4.4.
Bond Length, Both Angle
It is defined as the angle between the orbitals containing bonding electron pairs around the central atom in a molecule/complex ion. Bond angle is expressed in degree which can be experimentally determined by spectroscopic methods. It gives some idea regarding the distribution of orbitals around the central atom in a molecule/complex ion and hence it helps us in determining its shape. For example H–O–H bond angle in water can be represented as under :
Bond Enthalpy
It is defined as the amount of energy required to break one mole of bonds of a particular type between two atoms in a gaseous state.The unit of bond enthalpy is kJ mol–1. For example, the H – H bond enthalpy in hydrogen molecule is 435.8 kJ mol–1.
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamisamaaBa % aaleaacaaIYaaabeaakmaabmaabaGaam4zaaGaayjkaiaawMcaaiab % gkziUkaadIeadaqadaqaaiaadEgaaiaawIcacaGLPaaacqGHRaWkca % WGibWaaeWaaeaacaWGNbaacaGLOaGaayzkaaGaai4oaiabgs5aenaa % BaaaleaacaWGHbaabeaakiaadIeadaahaaWcbeqaaiaadAfaaaGccq % GH9aqpcaaI0aGaaG4maiaaiwdacaGGUaGaaGioaiaadUgacaWGkbGa % amyBaiaad+gacaWGSbWaaWbaaSqabeaacqGHsislcaaIXaaaaaaa!53BD! {H_2}\left( g \right) \to H\left( g \right) + H\left( g \right);{\Delta _a}{H^V} = 435.8kJmo{l^{ - 1}}\)
Similarly the bond enthalpy for molecules containing multiple bonds, for example O2 and N2 will be as under :
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGpb % WaaSbaaSqaaiaaikdaaeqaaOWaaeWaaeaacaWGpbGaeyypa0Jaam4t % aaGaayjkaiaawMcaamaabmaabaGaam4zaaGaayjkaiaawMcaaiabgk % ziUkaad+eadaqadaqaaiaadEgaaiaawIcacaGLPaaacqGHRaWkcaWG % pbWaaeWaaeaacaWGNbaacaGLOaGaayzkaaGaai4oaiabgs5aenaaBa % aaleaacaWGHbaabeaakiaadIeadaahaaWcbeqaaiaadAfaaaGccqGH % 9aqpcaaI0aGaaGyoaiaaiIdacaWGRbGaamOsaiaad2gacaWGVbGaam % iBamaaCaaaleqabaGaeyOeI0IaaGymaaaaaOqaaiaad6eadaWgaaWc % baGaaGOmaaqabaGcdaqadaqaaiaad6eacqGHHjIUcaWGobaacaGLOa % GaayzkaaWaaeWaaeaacaWGNbaacaGLOaGaayzkaaGaeyOKH4QaamOt % amaabmaabaGaam4zaaGaayjkaiaawMcaaiabgUcaRiaad6eadaqada % qaaiaadEgaaiaawIcacaGLPaaacaGG7aGaeyiLdq0aaSbaaSqaaiaa % dggaaeqaaOGaamisamaaCaaaleqabaGaamOvaaaakiabg2da9iaaiM % dacaaI0aGaaGOnaiaac6cacaaIWaGaam4AaiaadQeacaWGTbGaam4B % aiaadYgadaahaaWcbeqaaiabgkHiTiaaigdaaaaaaaa!797F! \begin{array}{l} {O_2}\left( {O = O} \right)\left( g \right) \to O\left( g \right) + O\left( g \right);{\Delta _a}{H^V} = 498kJmo{l^{ - 1}}\\ {N_2}\left( {N \equiv N} \right)\left( g \right) \to N\left( g \right) + N\left( g \right);{\Delta _a}{H^V} = 946.0kJmo{l^{ - 1}} \end{array}\)
It is important that larger the bond dissociation enthalpy, stronger will be the bond in the molecule.For a heteronuclear diatomic molecules like HCl, we have
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamisaiaado % eacaWGSbWaaeWaaeaacaWGNbWaaeWaaeaacaWGNbaacaGLOaGaayzk % aaaacaGLOaGaayzkaaGaeyOKH4QaamisamaabmaabaGaam4zaaGaay % jkaiaawMcaaiabgUcaRiaadoeacaWGSbWaaeWaaeaacaWGNbaacaGL % OaGaayzkaaGaai4oaiabgs5aenaaBaaaleaacaWGHbaabeaakiaadI % eacaaI0aGaaG4maiaaigdacaGGUaGaaGimaiaadUgacaWGkbGaamyB % aiaad+gacaWGSbWaaWbaaSqabeaacqGHsislcaaIXaaaaaaa!55C1! HCl\left( {g\left( g \right)} \right) \to H\left( g \right) + Cl\left( g \right);{\Delta _a}H431.0kJmo{l^{ - 1}}\)
In case of polyatomic molecules, the measurement of bond strength is more complicated. For example in case of H2O molecule, the enthalpy needed to break the two O – H bonds is not the same.
Table 4.2 Average Bond Lengths for Some Single, Double and Triple Bonds
Table 4.3 Bond Lengths in Some Common Molecules
Table 4.4 Covalent Radii, *rcov/(pm)
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGib % WaaSbaaSqaaiaaikdaaeqaaOGaam4tamaabmaabaGaam4zaaGaayjk % aiaawMcaaiabgkziUkaadIeadaqadaqaaiaadEgaaiaawIcacaGLPa % aacqGHRaWkcaWGpbGaamisamaabmaabaGaam4zaaGaayjkaiaawMca % aiaacUdacqGHuoardaWgaaWcbaGaamyyaaqabaGccaWGibWaa0baaS % qaaiaaigdaaeaacaWGwbaaaOGaeyypa0JaaGynaiaaicdacaaIYaGa % am4AaiaadQeacaWGTbGaam4BaiaadYgadaahaaWcbeqaaiabgkHiTi % aaigdaaaaakeaacaWGpbGaamisamaabmaabaGaam4zaaGaayjkaiaa % wMcaaiabgkziUkaadIeadaqadaqaaiaadEgaaiaawIcacaGLPaaacq % GHRaWkcaWGpbWaaeWaaeaacaWGNbaacaGLOaGaayzkaaGaai4oaiab % gs5aenaaBaaaleaacaWGHbaabeaakiaadIeadaqhaaWcbaGaaGOmaa % qaaiaadAfaaaGccqGH9aqpcaaI0aGaaGOmaiaaiEdacaWGRbGaamOs % aiaad2gacaWGVbGaamiBamaaCaaaleqabaGaeyOeI0IaaGymaaaaaa % aa!71B1! \begin{array}{l} {H_2}O\left( g \right) \to H\left( g \right) + OH\left( g \right);{\Delta _a}H_1^V = 502kJmo{l^{ - 1}}\\ OH\left( g \right) \to H\left( g \right) + O\left( g \right);{\Delta _a}H_2^V = 427kJmo{l^{ - 1}} \end{array}\)
The difference in the \(\Delta\)HV value shows the second O – H bond undergoes some change because of changed chemical environment. This is the reason for some difference in energy of the same O – H bond in different molecules like C2H5OH (ethanol) and water. Therefore in polyatomic molecules the term mean or average bond enthalpy is used. It is obtained by dividing total bond dissociation enthalpy by the number of bonds broken as explained below in case of water molecule,
Average bond enthalpy =\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aI1aGaaGimaiaaikdacqGHRaWkcaaI0aGaaGOmaiaaiEdaaeaacaaI % Yaaaaaaa!3C14! \frac{{502 + 427}}{2}\)
= 464.5 kJ mol–1.
Bond Order
In the Lewis description of covalent bond, the Bond Order is given by the number of bonds between the two atoms in a molecule.The bond order, for example in H2 (with a single shared electron pair), in O2 (with two shared electron pairs) and in N2(with three shared electron pairs) is 1,2,3 respectively. Similarly in CO (three shared electron pairs between C and O) the bond order is 3.For N2,bond order is 3 and its \(\Delta\)HV is 946 kJ mol–1; being one of the highest for a diatomic molecule.
Isoelectronic molecules and ions have identical bond orders; for example, F2 and O22– have bond order 1. N2, CO and NO+ have bond order 3.
A general correlation useful for understanding the stablities of molecules is that: with increase in bond order, bond enthalpy increases and bond length decreases.
Resonance Structures
It is often observed that a single Lewis structure is inadequate for the representation of a molecule in conformity with its experimentally determined parameters. For example, the ozone, O3 molecule can be equally represented by the structures I and II shown below :
Fig. 4.3 Resonance in the O3 molecule (structures I and II represent the two canonical forms while the structure III is the resonance hybrid)
In both structures we have a O–O single bond and a O=O double bond. The normal O–O and O=O bond lengths are 148 pm and 121 pm respectively. Experimentally determined oxygen-oxygen bond lengths in the O3 molecule are same (128 pm). Thus the oxygen oxygen bonds in the O3 molecule are intermediate between a double and a single bond. Obviously, this cannot be represented by either of the two Lewis structures shown above.
The concept of resonance was introduced to deal with the type of difficulty experienced in the depiction of accurate structures of molecules like O3. According to the concept of resonance, whenever a single Lewis structure cannot describe a molecule accurately, a number of structures with similar energy, positions of nuclei,bonding and non-bonding pairs of electrons are taken as the canonical structures of the hybrid which describes the molecule accurately.Thus for O3, the two structures shown above constitute the canonical structures or resonance structures and their hybrid i.e., the III structure represents the structure of O3 more accurately.This is also called resonance hybrid. Resonance is represented by a double headed arrow.
Some of the other examples of resonance structures are provided by the carbonate ion and the carbon dioxide molecule.
PROBLEM 3
Explain the structure of CO32– ion in terms of resonance.
SOLUTION
The single Lewis structure based on the presence of two single bonds and one double bond between carbon and oxygen atoms is inadequate to represent the molecule accurately as it represents unequal bonds.According to the experimental findings, all carbon to oxygen bonds in CO32– are equivalent.Therefore the carbonate ion is best described as a resonance hybrid of the canonical forms I, II, and III shown below.
Fig.4.4 Resonance in CO32–, I, II and III represent the three canonical forms
PROBLEM 4
Explain the structure of CO2 molecule.
SOLUTION
The experimentally determined carbon to oxygen bond length in CO2 is 115 pm.The lengths of a normal carbon to oxygen double bond (C=O) and carbon to oxygen triple bond (C\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyyyIOlaaa!37BF! \equiv \)O) are 121 pm and 110 pm respectively. The carbon-oxygen bond lengths in CO2 (115 pm) lie between the values for C\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyyyIOlaaa!37BF! \equiv \)O and C\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyyyIOlaaa!37BF! \equiv \)O. Obviously, a single Lewis structure cannot depict this position and it becomes necessary to write more than one Lewis structures and to consider that the structure of CO2 is best described as a hybrid of the canonical or resonance forms I, II
and III.
Fig. 4.5 Resonance in CO2 molecule, I, II and III represent the three canonical forms.
In general, it may be stated that
1. Resonance stabilizes the molecule as the energy of the resonance hybrid is less than the energy of any single cannonical structure; and,
2. Resonance averages the bond characteristics as a whole.Thus the energy of the O3 resonance hybrid is lower than either of the two cannonical froms I and II (Fig 4.3).
Many misconceptions are associated with resonance and the same need to be dispelled. You should remember that :
1. The cannonical forms have no real existence.
2. The molecule does not exist for a certain fraction of time in one cannonical form and for other fractions of time in other cannonical forms.
3. There is no such equilibrium between the cannonical forms as we have between tautomeric forms (keto and enol) in tautomerism.
4. The molecule as such has a single structure which is the resonance hybrid of the cannonical forms and which cannot as such be depicted by a single Lewis structure.
Polarity of Bonds
The existence of a hundred percent ionic or covalent bond represents an ideal situation. In reality no bond or a compound is either completely covalent or ionic. Even in case of covalent bond between two hydrogen atoms, there is some ionic character.When covalent bond is formed between two similar atoms, for example in H2, O2, Cl2 ,N2 or F2, the shared pair of electrons is equally attracted by the two atoms. As a result electron pair is situated exactly between the two identical nuclei.The bond so formed is called nonpolar covalent bond. Contrary to this in case of a heteronuclear molecule like HF, the shared electron pair between the two atoms gets displaced more towards fluorine since the electronegativity of fluorine (Unit 3) is far greater than that of hydrogen. The resultant covalent bond is a polar covalent bond.
As a result of polarisation, the molecule possesses the dipole moment (depicted below) which can be defined as the product of the magnitude of the charge and the distance between the centres of positive and negative charge.It is usually designated by a Greek letter ‘\(\mu\)’. Mathematically, it is expressed as follows :
Dipole moment (\(\mu\)) = charge (Q) × distance ofseparation (r) Dipole moment is usually expressed in
Debye units (D). The conversion factor is
1 D = 3.33564 × 10–30 C m
where C is coulomb and m is meter.
Further dipole moment is a vector quantity and by convention it is depicted by a small arrow with tail on the negative centre and head pointing towards the positive centre. But in chemistry presence of dipole moment is represented by the crossed arrow put on Lewis structure of the molecule. The cross is on positive end and arrow head is on negative end. For example the dipole moment of HF may be represented as :
This arrow symbolises the direction of the shift of electron density in the molecule. Note that the direction of crossed arrow is opposite to the conventional direction of dipole moment vector.
In case of polyatomic molecules the dipole moment not only depend upon the individual dipole moments of bonds known as bond dipoles but also on the spatial arrangement of various bonds in the molecule. In such case, the dipole moment of a molecule is the vector sum of the dipole moments of various bonds. For example in H2O molecule, which has a bent structure, the two O–H bonds are oriented at an angle of 104.50. Net dipole moment of 6.17 × 10–30 C m (1D = 3.33564 × 10–30 C m) is the resultant of the dipole moments of two O–H bonds.
Net Dipole moment,\(\mu\)= 1.85 D
= 1.85 × 3.33564 × 10–30 C m = 6.17 ×10–30 C m
The dipole moment in case of BeF2 is zero. This is because the two equal bond dipoles point in opposite directions and cancel the effect of each other.
In tetra-atomic molecule, for example in BF3, the dipole moment is zero although the B – F bonds are oriented at an angle of 120o to one another, the three bond moments give a net sum of zero as the resultant of any two is equal and opposite to the third.
BF3 molecule;representation of (a) bond dipoles and (b) total dipole moment
Let us study an interesting case of NH3 and NF3 molecule.Both the molecules have pyramidal shape with a lone pair of electrons on nitrogen atom. Although fluorine is more electronegative than nitrogen, the resultant dipole moment of NH3 ( 4.90 × 10-30 C m) is greater than that of NF3 (0.8 × 10-30 C m). This is because, in case of NH3 the orbital dipole due to lone pair is in the same direction as the resultant dipole moment of the N – H bonds,whereas in NF3 the orbital dipole is in the direction opposite to the resultant dipole moment of the three N–F bonds.The orbital dipole because of lone pair decreases the effect of the resultant N – F bond moments, which results in the low dipole moment of NH3 as represented below:
Dipole moments of some molecules are shown in Table 4.5.
Just as all the covalent bonds have some partial ionic character, the ionic bonds also have partial covalent character. The partial covalent character of ionic bonds was discussed by Fajans in terms of the following rules:
1. The smaller the size of the cation and the larger the size of the anion, the greater the covalent character of an ionic bond.
2. The greater the charge on the cation, the greater the covalent character of the ionic bond.
3. For cations of the same size and charge, the one, with electronic configuration (n-1)dnnso, typical of transition metals, is more polarising than the one with a noble gas configuration, ns2 np6, typical of alkali and alkaline earth metal cations.
The cation polarises the anion, pulling the electronic charge toward itself and thereby increasing the electronic charge between the two. This is precisely what happens in a covalent bond, i.e., buildup of electron charge density between the nuclei. The polarising power of the cation, the polarisability of the anion and the extent of distortion (polarisation) of anion are the factors, which determine the per cent covalent character of the ionic bond.