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Chemical Bonding and Molecular Structure

Atoms of different elements excepting noble gases do not have complete octet so they combine with other atoms to form chemical bond. The force which holds the atoms or ions together within the molecule is called a chemical bond and the process of their combination is called Chemical Bonding. It depends on the valency of atoms.

Normally only electrons in the outermost shell of an atom are involved in bond formation. Atoms may attain stable electronic configuration in three different ways by loosing or gaining electrons by sharing electrons.

Elements may be divided into three classes.

  • Electropositive elements, whose atoms give up one or more electrons easily, they have low ionization potentials.
  • Electronegative elements, which can gain electrons. They have higher value of electronegativity.
  • Elements which have little tendency to loose or gain electrons

Electronic theory of valency: This theory was proposed by Kossel and Lewis. The following are the important points of the theory —

  1. Electrons of outermost shell of atoms of an element take part in bond formation, i.e. when two atoms combine to form a bond; there is redistribution in the number of electrons in the outermost shell of both the atoms.

In transitional elements, electrons of (n–1) d sub-shell also participate in chemical bonding.

  1. Atoms of inert gas elements do not possess combining capacity because in their atoms outermost shell is saturated with electrons, i.e., they have stable octet of electrons in outermost shell (except helium which has a stable duplet of electrons).

When K-shell is the outermost shell of an atom, only two electrons are needed for stability.

The electronic configuration, ns2 np6 is associated with minimum energy and hence maximum  stability.

  1. The atoms which do not have stable electronic configuration in their outermost shell have a tendency to acquire their nearest inert gas like configuration. It is known as the octet rule. The rule is stated as, “It is tendency of every atom to complete its octet (or duplet) i.e. to acquire eight electrons in the outermost shell (or two electrons if only one shell is present) so that stability may be attained.”

This tendency of the atom to complete its octet (or duplet) is responsible for its chemical reactivity.

  1. An atom can acquire stability by loss, gain or sharing of electrons with other atoms.
  2. The number of electrons lost, gained or shared by one atom of an element is known as its valency.

Types of Chemical bonds: Depending upon the nature of atoms which combine to give molecules, the following types of bonds are formed:

  1. Electrovalent or Ionic Bond.
  2. Covalent Bond.
  3. Co-ordinate or Dative Bond.

Apart from these, some other types of bonds such as metallic bond and hydrogen bond etc. are also formed.

Electrovalent or Ionic Bond

lonic or Electrovalent bond is formed by complete transfer of one or more electrons from the valence shell of one atom to the valence shell of the other. The atom that loses electrons gets converted into a cation while the other which gains electrons gets converted into an anion.

\(Na.+\overset{{.\,\,.}}{\mathop{{.\,\underset{{.\,\,.}}{\mathop{{Cl}}}\,}}}\,:\to N{{a}^{+}}{{[:\underset{{.\,.}}{\overset{{.\,.\,}}{\mathop{{Cl}}}}\,:]}^{-}}\) or \(N{{a}^{+}}C{{l}^{-}}\)

The strong electrostatic force of attraction between these oppositely charged ions is called lonic or electrovalent bond and the number of electrons gained or lost by the atom  is called its electrovalency.

Favourable conditions for forming a stable lonic bond are:

(1) Low ionization energy of the atom forming the cation.

(2) High electron affinity of the atom forming the anion.

(3) High lattice energy of the crystal formed.

       Characteristics of ionic compounds are :

       (1) Each ion is surrounded by a uniformly distributed electric field i.e., each ion is non- directional. Therefore, ionic bond is also considered to be non directional. Hence ionic compounds do not show stereoisomerism.

       (2) These are highly soluble in solvent with high dielectric constant, such as water and other polar solvents, but insoluble in non- polar solvents like benzene, ether, etc. (like dissolves like).

When an ionic compound dissolves in water, the ions are hydrated and the energy evolved in the process is called hydration energy. Thus for an ionic solid to be soluble in water, the necessary condition is

\(\Delta {{H}_{{Hydration\,energy}}}>{{H}_{{Lattice\,energy}}}\)

Ionic compounds like \(CaC{{O}_{3}},C{{a}_{3}}{{(P{{O}_{4}})}_{2}},Ca{{F}_{2}},BaS{{O}_{4}},PbS{{O}_{4}},AgCl,\ AgBr,\ AgI\) etc., are insoluble in water due to their high lattice energies.

       (3) On account of strong electrostatic attraction between the ions in the crystal of an ionic compound, these have high melting and boiling points. Order of melting and boiling points of certain compounds are examplified below.

a) \(NaF>NaCl>NaBr>NaI\)                                                         b) \(MgO>CaO>BaO\)

(4) Ionic compounds conduct electricity in solution as well as in molten form (fused state) because ions become mobile. These free ions are able to move under the influence of an electric field and thus act as carriers of current. In solid state, they are bad conductors as the ions are tightly held.

(5) In solution, ionic compounds show ionic reactions which are quite fast and instantaneous. Energetics of formation of ionic bond are represented by the Born-Haber cycle which is as followers :

The heat of formation of an ionic solid is net resultant of the above changes.

\(\Delta {{H}_{f}}=\Delta {{H}_{{sub.}}}+\frac{1}{2}\Delta {{H}_{{diss}}}+I.E.-E.A.-U\)

The lattice energy is one of the deciding factors of ionic bond formation. It depends upon :

       (1) Size of the ions : Smaller the size of the ions greater is the attractive forces between the ions and higher is the lattice energy.

(2) Charge on the ions : Greater the charge on ions greater will be attractive forces between the ions and hence greater is the value of lattice energy

For example, lattice energy of bi-bivalent ions > bi-uni or uni-bivalent ions> uni-univalent ions. Order of lattice energy of certain compounds are :

(a) LiX > NaX > KX > RbX > CsX (where X = F, Cl, Br or I); (size of alkali metal increases)

(b) MgO > CaO > SrO > BaO (size of alkaline earth metal increases)

(c) \(MgC{{O}_{3}}>CaC{{O}_{3}}>SrC{{O}_{3}}>BaC{{O}_{3}}\)  (size of cation increases)

As lattice energy increases, melting and boiling points of ionic compounds increases. This is due to strong electrostatic force of attraction.

Fajan Rules:

In Ionic bond, some covalent character is introduced because of the tendency of the cation to polarise the anion. In fact cation attracts the electron cloud of the anion and pulls electron density between the two nuclei.

According to Fajan rules, the magnitude of covalent character in the ionic bond depends upon the extent of polarisation caused by cation. In general,

(1) Smaller the size of cation, large is its polarising power.

       (2) Among two cations of similar size, the polarising power of cation with noble gas configuration (ns2np6nd10) is larger than cation with noble gas configuration (ns2np6) e.g., Polarising power of Ag+ is more than K+

(3) Larger the anion, more will be its polarity.

Covalent Bond

Covalent bond is formed by mutual sharing of electrons so as to complete their octet or duplet in case of H, Li and Be. Depending upon whether one, two and three electrons are shared by each atom, single, double and triple bonds are respectively formed.  The number of electrons contributed by each atom for sharing is called its Covalency.

The formation of covalent bonds is best explained by a new approach according to which sharing of orbitals is possible only with overlapping. According to this concept, the formation of covalent bond involves the overlapping of half filled atomic orbitals of the atoms participating in bonding. The atomic orbitals undergoing overlapping must have electrons with opposite spin.

When two half-filled atomic orbitals overlap along their internuclear axis, the bond formed is called the sigma bond (or σ-bond) It may be formed by the overlap of two s-orbitals, two p-orbitals, one s– and one p-orbital.

Accordingly these bonds are designated as ss, pp and sp sigma bonds.  On the other hand, covalent bond formed by sideways or lateral overlap of p-orbitals is called pi-bond (or π–bond). It may be noted that

(1) All single bonds are σ–bonds.

(2) Multiple bonds contain one σ–bonds and the rest are πbonds.

(3) A πbond is never formed alone. First σ–bonds is formed and then the formation of the πbond takes place.

(4) A sigma bond is always stronger than pi-bond because the extent of overlapping of atomic orbitals along internuclear axis is greater than sideways overlapping.

The Lewis theory : The tendency of atoms to achieve eight electrons in their outermost shell is known  as lewis octet  rule.

Lewis symbol for the representative elements are given in the following table,

Polarity n a covalent bond: When two covalently bonded atoms have equal electronegativities, the bonded pair of electrons is equally shared between them, i.e., shared pair of electrons remains in the middle of the nuclei of bonded atoms. Such a bond is non-polar covalent bond. For example,

  •    Non-polar covalent bond is formed between identical non-metallic atoms or atoms having electronegativity difference equal to zero.
  •    Forces of attraction between non-polar molecules are weak, so their M.P and B.P. are low.

Polar covalent bond: When the two covalently bonded atoms are dissimilar, i.e., differ in their electronegativities, the bonded pair of electrons does not remain in between the bonded nuclei but is attracted towards the more electronegative atom resulting in the development of partial negative charge on this atom and equal but opposite (positive) charge on the other atom.

The magnitude of charge produced on the bonded atoms is proportional to their electronegativity difference. Such a covalent bond formed between atoms with a difference in their electronegativites is called polar covalent bond. For example.

  •    For polar covalent bond electronegativity difference of bonded atoms 0.
  •    Polarity of covalent bond Electronegativity difference of bonded atoms.
  •    Forces of attraction between polar molecules are stronger dipole-dipole attractions.
  •    A diatomic molecule with a polar covalent bonds is called a dipole, for example,

  • The relative order of electronegativities of few of the common elements is—


Dipole Moment (µ):  The property which measures the extent of polarity in a bond is called dipole moment (µ). Molecules like  A(δ+)-B(δ-) having two polar ends are known as dipole and possess dipole moment, which is defined as, “the product of magnitude of the partial charge + or δ-) developed on any of the covalently bonded atoms and the distance between these two atoms, i.e., the bond length”.  Thus:

Dipole moment (µ)= Magnitude of charge (e) x distance (d).

      The dipole moment of any molecule may be determined experimentally. Its value provides a measure of the polar character of the molecule.

  •    µ = 0, the molecule, is non-polar.
  •   µ≠0 , the molecule is polar.

Unit of dipole moment: As e is of the order of 10–10 e.s.u. and d is of the order of 10–8 cm, µ is of the order 10–18 e.s.u. cm. 10–18 e.s.u. cm is called 1 Debye. The unit of dipole moment is Debye (D).

      1 Debye = 1 x 10–18 e.s.u. cm.

  •    In C.G.S. system, separation of 1 electronic charge (4.8 x 10–10 e.s.u) with 1Å bond length will give a        dipole moment of 4.8 x 10–18 e.s.u. cm (4.8 Debye).
  •    In M.K.S. system, for a transfer of electronic charge at distance of‘d’ meter, the dipole moment is  given by,

µ = 1.6 x 10–19 coulomb x d meter

= 1.6 x 10–19 x d coulomb meter.

Thus, the dipole moment for hundred percent ionic character in a binary compound A+B is 1.6 x10–19 x d C m where d is the internuclear distance.

  •    In M.K.S. system, the dipole moment of a binary molecule having electronic charge and internuclear  distance 10–10 meter is:

µ = 1.6 x 10–19 coulomb x 10–10 meter

= 1.6 x 10–29 coulomb meter

On equating the values of dipole moment in both the units, we get 4.8 Debye = 1.6 × 10–29 coulomb meter.

1 Debye           = \(\frac{{1.6\times {{{10}}^{{-29}}}}}{{4.8}}\)

= 0.33 × 10–29 coulomb meter

  • Dipole moment is indicated by an arrow. The tail of the arrow is on the +ve end and arrow head is on   the –ve end of the molecule, for example, H(δ+)—Cl(δ-)
  • Dipole moment is a vector quantity, i.e., it depends upon both magnitude as well as direction. Thus the dipole moment of a molecule containing two or more polar covalent bonds is given by the vector addition of the dipole moments of all the constituent bonds. Thus, the resultant of the dipole moments of all the bonds will give the overall polarity (dipole moment) of the molecule.
  • If the resultant is zero, µ will be zero and the molecule will be non-polar. If the resultant is not zero,  the molecule will be polar. For example, for the following general triatomic molecule.

The dipole moment is given by: \(\mu =\sqrt{{(\mu _{1}^{2}+\mu _{2}^{2}+2{{\mu }_{1}}{{\mu }_{2}}\cos \theta )}}\)

As the value of bond angle, θ , decreases, the value of dipole moment increases.

Question.       A molecule having the formula XY has dipole moment 1.2 Debye and bond length is 1Å. What fraction of electronic charge exists on each atom?

Solution:        Given is dipole moment, µ = 1.2 Debye = 1.2 x 10–18 e.s.u. cm.

Bond length, d = 1Å = 10–8

hence,          µ = e x d

[‘e’ is the charge on each atom]

∴ 1.2 x 10–18 = e x 10–8

\(e=\frac{{1.2\times {{{10}}^{{-8}}}}}{{{{{10}}^{{-8}}}}}=1.2\times {{10}^{{-10}}}e.s.u\)

Fraction of electronic charge

\(=\frac{{1.2\times {{{10}}^{{-10}}}e.s.u}}{{4.8\times {{{10}}^{{-10}}}e.s.u(\text{charge on an }{{\text{e}}^{\text{–}}})}}=\frac{1}{4}\)

Question.       Dipole moment of H2O molecule is 1.85 Debye. Calculate the bond moment of O—H bond. Bond angle is 105°.

Solution:        H2O molecule has a bent shape as shown below.

Let ‘µ’ be the bond moment of O—H bond. Since ‘µ’ is a vector quantity, so the net dipole moment of H2O molecule is given by:

Question.       Assuming 100% ionic character, calculate the dipole moment of NaCl. The radii of Na+ and Cl ion are 0.102 and 0.181 nm respectively.

Solution:        Given rNa+ = 0.102 nm

rCl = 0.181 nm

Therefore, The internuclear distance or bond distance

d=rNa++rCl       = 0.102 +0.181 nm

= 0.283 nm

= 0.283 x 10–9 meter

= 0.283 x 10–7 cm.

For 100% ionic character, e, charge on any atom

= 4.8 x 10–10 e.s.u.

Dipole moment of Na+Cl = e x d

=4.8 x 10-10 esu x 0.283 x10-7cm

= 13.584 Debye.

Question.       The dipole moment of KCl is 3.336 x 10–29 coulomb meters, which indicates that it is a highly polar molecule. The inter-atomic distance between K+ and Cl in this molecule is  2.6 x 10–10 meter. Calculate the dipole moment of KCl molecule if there were opposite charges of one fundamental unit located at each nucleus. Calculate the percentage ionic character of KCl.

Solution:        Given is the observed dipole moment of KCl = 3.336 x 10–29 coulomb meter

The interatomic distance,

D = 2.6 x 10–10 meter

Fundamental unit charge = e, electronic charge = 1.6 x 10–19 coulomb

Dipole moment of KCl for 100%  ionic character = e x d

= Electronic charge × interatomic distance

= 1.6 x 10–19 x 2.6 x 10–10 coulomb meter

= 4.16 x 10–29 coulomb meter

                        % ionic character of KCl molecule

= \(\frac{{\text{Observed dipole moment }\times \text{ 100}}}{{\text{Dipole moment for 100 }\!\!%\!\!\text{ ionic character}}}\)

= \(\frac{{3.336\times {{{10}}^{{-29}}}\times 100}}{{4.16\times {{{10}}^{{-29}}}}}\) =  80.19


  • This concept was put forward by Heitler and London and modified by Pauling and Slater. According to this concept, “the formation of a covalent bond between two atoms results by coupling of electrons with opposite spins belonging to outermost shell orbitals to the two atoms.”
  • Thus, the presence of one or more unpaired electrons in the Valence shell of an atom is responsible for chemical bonding. Half filled atomic orbitals having electrons spinning in opposite directions overlap with each other to form a covalent bond. The overlapping results in the coupling (pairing) of electrons and thus leads to lowering of potential energy of the system.
  • Noble gases do not take part in chemical bonding because they do not have unpaired electron(s) in their outermost shell. The combining capacity or covalency of an atom is equal to the number of unpaired electrons present in its ground or excited state. Important points of orbital theory of covalency are:
  •    Every atom has a tendency to get its unpaired electrons paired up with opposite spinning electrons of  other atom. It results in lowering of energy of the system which is stabilized and hence atoms are  chemically bonded.
  •    Atomic orbitals of the two atoms having unpaired electrons overlap with each other. Thus, overlapping of atomic orbitals of the outer shell of two atoms gives rise to a covalent linkage.
  •    Greater the extent of overlapping of the atomic orbitals, the stronger is the bond formed between the  atoms.


The tetravalency shown by carbon is actually due to excited state of carbon which is responsible for carbon bonding capacity.

If the bond formed is by overlapping then all the bonds will not be equivalent so a new concept known as hybridization is introduced which can explain the equivalent character of bonds.

s and p orbital belonging to the same atom having slightly different energies mix together to produce same number of new set of orbital called as hybrid orbital and the phenomenon is called as hybridization.

Important characteristics of hybridization

(i)   The number of hybridized orbital is equal to number of orbitals that get hybridized.

(ii) The hybrid orbitals are always equivalent in energy and shape.

(iii) The hybrid orbitals form more stable bond than the pure atom orbital.

(iv) The hybrid orbitals are directed in space in same preferred direction to have some stable arrangement and giving suitable geometry to the molecule.

Depending upon the different combination of s and p orbitals, these types of hybridization are known.

(i)   sp3 hybridization: In this case, one s and three p orbitals hybridise to form four sp3 hybrid orbitals. These four sp3 hybrid orbitals are oriented in a tetrahedral arrangement.

For example in methane CH4

(ii)  sp2 hybridization: In this case one s and two p orbitals mix together to form three sp2 hybrid orbitals and are oriented in a trigonal planar geometry.

The remaining p orbital if required form side ways overlapping with the other unhybridized p orbital of other C atom and leads to formation of p bond as in H2C = CH2

(iii)       sp hybridization: In this case, one s and one p orbital mix together to form two sp hybrid orbitals and are oriented in a linear shape

The remaining two unhybridised p orbitals overlap with another unhybridised p orbital leading to the formation of triple bond as in HC º CH.

Shape                                                              Hybridisation

Linear                                                               sp

Trigonal planar                                                  sp2

Tetrahedral                                                       sp3

Trigonal bipyramidal                                          sp3d

Octahedral                                                        sp3d2

Pentagonal bipyramidal                                     sp3d3

Question.   Which of the following molecule has trigonal planer geometry?

                        (A) CO2                                                (B) PCl5

                        (C) BF3                                                 (D) H2O

Solution.    BF3 has trigonal planer geometry (sp2 – hybridized Boron).

Hence (A) is correct.

Rule for determination of total number of hybrid orbitals
  • Detect the central atom along with the peripheral atoms.
  • Count the valence electrons of the central atom and the peripheral atoms.
  • Divide the above value by 8. Then the quotient gives the number of s bonds and the remainder gives the non-bonded electrons. So number of lone pair = (Non bonded electrons)/2
  • The number of s bonds and the lone pair gives the total number of hybrid orbitals.

An example will make this method clear

SF4       Central atom S,   Peripheral atom F

∴ total number of valence electrons = 6+(4 x7) = 34

Now 34/8= Number of hybrid orbitals = 4s bonds + 1 lone pair)

So, five hybrid orbitals are necessary and hybridization mode is sp3d and it is trigonal bipyramidal (TBP).

Note: Whenever there are lone pairs in TBP geometry they should be placed in equatorial position so that repulsion is minimum.

  1. NCl3    Total valence electrons = 26

Requirement  = 3 s bonds + 1 lone pair                                   

Hybridization = sp3

Shape  = pyramidal

        2. BBr3        Total valence electron  = 24

Requirement  = 3s bonds                                                                   

Hybridization = sp2

Shape  = planar trigonal

  1. SiCl4    Total valence electrons = 32

Requirement  = 4s bonds                                                                   

Hybridization =  sp3

Shape  = Tetrahedral

     4. ClF3     Total valence electrons : 28

Requirement : 3 s bonds + 2 lone pairs                                         

Hybridization : sp3d

Shape : T – shaped

5. XeF4     Total valence electrons : 36

Requirement:4s bonds+ 2 lone pairs                                     

Hybridisation : sp3d

Shape :  Square planar

Now three arrangements are possible out of which A and B are same. A and B can be inter converted by simple rotation of molecule.

The basic difference of (B) and (C) is that in (B) the lone pair is present in the anti position which minimizes the repulsion which is not possible in structure (C) where the lone pairs are adjacent.

So in a octahedral structure the lone pairs must be placed at the anti positions to minimize repulsion. So both structure (A) and (B) are correct.

            6. NO2     Total valence electron: 18

Requirement : 2s bonds + 1 lone pair                                       

Hybridisation: sp2

Shape: angular

7.CO2      Total valence electrons : 16

Requirement: 2s bonds                                                                    O = C = O

Hybridisation: sp

Shape: linear

8. BF4      Total  valence electrons = 32

Requirement= 4 s bonds                                                       

Hybridisation: sp3

Shape: Tetrahedral

9. XeO3    Total valence electrons : 26

Requirement: 3 s bonds + 1 lone  pair                               

Hybridization: sp3

Shape: Pyramidal


Molecules exist in a variety of shapes. A number of physical and chemical properties of molecules arise from and are affected by their shapes. For example, the angular shape of the water molecules explains its many characteristic properties while a linear shape does not.

The determination of the molecular geometry and the development of theories for explaining the preferred geometrical shapes of molecules is an integral part of chemical bonding. The VSEPR theory (model) is a simple treatment for understanding the shapes of molecules.

Strictly speaking VSEPR theory is not a model of chemical bonding. It provides a simple recipe for predicting the shapes of molecules. It is infact an extension of the Lewis interpretation of bonding and is quite successful in predicting the shapes of simple polyatomic molecules.

The basic assumptions of the VSEPR theory are that:

Pairs of electrons in the valence shell of a central atom repel each other

  1. These pairs of electrons tend to occupy position in space that minimize repulsions and thus maximize distance between them.
  2. The valence shell is taken as a sphere with the electron pairs localizing on the spherical surface at maximum distance from one another.
  3. A multiple bonds are treated as a single super pair.
  4. Where two or more resonance structures can depict a molecule, the VSEPR model is applicable to any such structures

For the prediction of geometrical shapes of molecules with the help of VSEPR model, it is convenient to divide molecules into two categories

Regular Geometry

Molecules in which the central atom has no lone pairs

Irregular Geometry

Molecules in which the central atom has one or more lone pairs, the lone pair of electrons in molecules occupy more space as compared to the bonding pair electrons. This causes greater repulsion between lone pairs of electrons as compared to the bond pairs repulsions. The descending order of repulsion

(lp – lp)       >    (lp – bp)            > (bp – bp)

where lp-Lone pair; bp-bond pair

Question:       Interpret the non-linear shape of H2S molecule and non-planar shape of PCl3 using valence shell electron pair repulsion (VSEPR) theory.

Solution:        In H2S, two bonded pairs and two lone pairs are present, i.e., sulphur is in sp3 hybridized state. The angle is less than 109°28´ as contraction occurs due to presence of lone pairs. Thus, H2S has V-shaped structure. In PCl3, three bonded pairs and one long pair are present, i.e., phosphorus is also in sp3 hybridized state but it has pyramidal structure.

Question:       Using VESEPR, draw the molecular structures of OSF4 and XeF4 indicating the position of lone pairs and hybridization of central atoms.


  1. Bond Length: The distance between the nuclei of two atoms bonded together is termed as bond length or bond distance. It is expressed in angstrom (Å) units or picometer (pm).


  •       Bond length in ionic compound =  rc+ + ra
  •       Similarly, in a covalent compound, bond length is obtained by adding up the covalent (atomic) radii of two bonded atoms.

Bond length in covalent compound (AB) = rA +rB

  •       The factors such as resonance, electronegativity, hybridization, steric effects, etc., which affect the radii of atoms, also apply to bond lengths.

Important features

(i)   The bond length of the homonuclear diatomic molecules are twice the covalent radii.

(ii)  The lengths of double bonds are less than the lengths of single bonds between the same two       atoms, and triple bonds are even shorter than double bonds.

Single bond > Double bond > Triple bond (decreasing bond length)

(iii) Bond length decreases with increase in s-character since s-orbital is smaller than a
p – orbital.

\(\displaystyle s{{p}^{3}}C-H=1.112{{A}^{\circ }};\)                          \(\displaystyle s{{p}^{2}}C-H=1.103{{A}^{\circ }};\)              \(\displaystyle spC-H=1.08{{A}^{\circ }};\)

(25% s-character as in alkanes) (33.3% s-character as in alkenes) (50% s-character as in alkynes)

(iv) Bond length of polar bond is smaller than the theoretical non-polar bond length.

2.   Bond Energy or Bond Strength: Bond energy or bond strength is defined as the amount of energy required to break a bond in molecule.

Important features

(i)      The magnitude of the bond energy depends on the type of bonding. Most of the covalent bonds have energy between 50 to 100 kcal (200-400 kJ). Strength of sigma bond is more than that of a  π-bond.

(ii)     A double bond in a diatomic molecules has a higher bond energy than a single bond and a triple bond has a higher bond energy than a double bond between the same atoms.

\(\displaystyle C\equiv C>C=C>C-C\) (decreasing bond length)

(iii)     The magnitude of the bond energy depends on the size of the atoms forming the bond, i.e. bond length. Shorter the bond length, higher is the bond energy.

(iv)    Resonance in the molecule affects the bond energy.

(v)     The bond energy decreases with increase in number of lone pairs on the bonded atom. This is due to electrostatic repulsion of lone pairs of electrons of the two bonded atoms.

(vi)    Homolytic and heterolytic fission involve different amounts of energies. Generally the values are low for homolytic fission of the bond in comparison to heterolytic fission.

(vii)    Bond energy decreases down the group in case of similar molecules.

(viii)   Bond energy increase in the following order:

s < p < sp < sp² < sp³

C-C                    > N-N                         > O-O

(no lone pair)      (one lone pair)          (two lone pair)

  1. Bond angles: Angle between two adjacent bonds at an atom in a molecule made up of three or more atoms is known as the bond angle. Bond angles mainly depend on the following three factors:

(i)   Hybridization: Bond angle depends on the state of hybridization of the central atom

Hybridization      sp3                     sp2                     sp

Bond angle        109o28’             120o                   180o

           Example            CH4                    BCl3                  BeCl2

Generally s- character increase in the hybrid bond, the bond angle increases.


(ii) Lone pair repulsion: Bond angle is affected by the presence of lone pair of electrons at the central atom. A lone pair of electrons at the central atom always tries to repel the shared pair (bonded pair) of electrons. Due to this, the bonds are displaced slightly inside resulting in a decrease of bond angle.

(iii) Electronegativity: If the electronegativity of the central atom decreases, bond angle decreases.



An atom of hydrogen linked covalently to a strongly electronegative atom can establish an extra weak attachment to another electronegative atom in the same or different molecules. This attachment is called a hydrogen bond. To distinguish from a normal covalent bond, a hydrogen bond is represented by a broken line eg X – HY where X & Y are two electronegative atoms. The strength of hydrogen bond is quite low about 2-10 kcal mol–1 or 8.4–42 kJ mol–1 as compared to a covalent bond strength 50–100 kcal mol–1 or 209 –419 kJ mol–1

Conditions for Hydrogen Bonding:

  • Hydrogen should be linked to a highly electronegative element.
  • The size of the electronegative element must be small.

These two criteria are fulfilled by F, O, and N in the periodic table. Greater the electronegativity and smaller the size, the stronger is the hydrogen bond which is evident from the relative order of energies of hydrogen bonds.

Types of Hydrogen Bonding:

  • Intermolecular hydrogen bonding: This type of bonding takes place between two molecules of the same or different types. For example,

Intermolecular hydrogen bonding leads to molecular association in liquids like water etc. Thus in water only a few percent of the water molecules appear not to be hydrogen bonded even at 90°C.

Breaking of those hydrogen bonds throughout the entire liquid requires appreciable heat energy. This is indicated in the relatively higher boiling points of hydrogen bonded liquids.

Crystalline hydrogen fluoride consists of the polymer (HF)n. This has a zig-zag chain structure involving

Intramolecular hydrogen bonding: This type of bonding occurs between atoms of the same molecule present on different sites. Intramolecular hydrogen bonding gives rise to a closed ring structure for which the term chelation is sometimes used. Examples are
o-nitrophenol, salicylaldehyde.

Effect of Hydrogen Bonding

Hydrogen bonding has got a very pronounced effects on certain properties of the molecules. They have got effects on

  • State of the substance
  • Solubility of the substance
  • Boiling point
  • Acidity of different isomers

Question.          H2O is a liquid at ordinary temperature while H2S is a gas although both O and S belong to the same group of the periodic table.

Solution:         H2O is capable of forming intermolecular hydrogen bonds. This is possible due to high electronegativity and small size of oxygen. Due to intermolecular H-bonding, molecular association takes place. As a result the effective molecular weight increases and hence the boiling point increases. So H2O is a liquid. But in H2S no hydrogen bonding is possible due to large size and less electronegativity of S. So it’s boiling point is equal to that of an isolated H2S molecule and therefore it is a gas.

Question.       Ethyl alcohol (C2H5OH) has got a higher boiling point than dimethyl ether (CH3-O-CH3) although the molecular weight of both are same.

Solution:                     Though ethyl alcohol and dimethyl ether have the same molecular weight but in ethyl alcohol the hydrogen of the O-H groups forms intermolecular hydrogen bonding with the OH group in another molecule. But in case of ether the hydrogen is linked to C is not so electronegative to encourage the hydrogen to from hydrogen bonding.

Due to intermolecular H-bonding, ethyl alcohol remains in the associated form and therefore boils at a higher temperature compared to dimethyl ether


We have enough reasons to believe that a net attractive force operates between molecules of a gas. Though weak in nature, this force is ultimately responsible for liquifaction and solidification of gases. But we cannot explain the nature of this force from the ideas of ionic and covalent bond developed so far, particularly when we think of saturated molecules like H2, CH4, He etc. The existence of intermolecular attraction in gases was first recognised by Vanderwaal’s and accordingly intermolecular forces have been termed as Vanderwaal’s forces. It has been established that such forces are also present in the solid and liquid states of many substances. Collectively they have also been termed London forces since their nature was first explained by London using wave mechanics.

Nature of Vanderwaal’s Forces:

The Vanderwaal’s forces are very weak in comparison to other chemical forces. In solid NH3 it amount to about 39 KJ mol–1 (bond energy N-H bond = 389 KJ mol–1). The forces are non directional. The strength of Vanderwaal’s force increases as the size of the units linked increases. When other factors (like H-bonding is absent), this can be appreciated by comparison of the melting or boiling points of similar compounds in a group.

Origin of Intermolecular Forces:

Intermolecular forces may have a wide variety of origin.

  • Dipole-dipole interaction: This force would exist in any molecule having a permanent dipole e.g. HF, HCl, H2O etc.
  • Ion-dipole interaction: These interactions are operative in solvation and dissolution of ionic compounds in polar solvents.
  • Induced dipole interaction: These generate from the polarisation of a neutral molecule by a charge or ion.
  • Instantaneous dipole-induced dipole interaction: In non polar molecules dipoles may generate due to temporary fluctuations in electron density. These transient dipole can now induce dipole in neighbouring molecules producing a weak temporary interaction.


Metals are characterised by bright, lustre, high electrical and thermal conductivity, malleability, ductility and high tensile strength. A metallic crystal consists of very large number of atoms arranged in a regular pattern. Different model have been proposed to explain the nature of metallic bonding two most important modules are as follows

The electron sea Model

In this model a metal is assumed to consist of a lattice of positive ion (or kernels) immersed in a sea of mobile valence electrons, which move freely within the boundaries of a crystal. A positive kernel consists of the nucleus of the atom together with its core on a kernel is, therefore, equal in magnitude to the total valence electronic charge per atom. The free electrons shield the positively charged ion cores from mutual electrostatic repulsive forces which they would otherwise exert upon one another. In a way these free electrons act as ‘glue’ to hold the ion cores together.

The forces that hold the atoms together in a metal as a result of the attraction between positive ions and surrounding freely mobile electrons are known as metallic bonds.

Through the electron sea predated quantum mechanics it still satisfactorily explains certain properties of the metals. The electrical and thermal conductivity of metals for example, can be explained by the presence of mobile electrons in metals. On applying an electron field, these mobile electrons conduct electricity throughout the metals from one end to other. Similarly, if one part of metal is heated, the mobile electrons in the part of the metals acquire a large amount of kinetic energy. Being free and mobile, these electrons move rapidly throughout the metal and conduct heat to the other part of the metal.

On the whole this model is not satisfactory.

Question.         Sodium metal conducts electricity because it

                        (A) is a soft metal                                

                        (B) contains only one valence electron

                        (C) has mobile electron                        

                        (D) reacts with water to form H2 gas

Solution:         (C)


In Molecular Orbital Theory (MOT) the atoms in a molecule are supposed to loose their individual control over the electrons. The nuclei of the bonded atoms are considered to be present at equilibrium inter-nuclear positions. The orbitals where the probability of finding the electrons is maximum are multicentred orbitals called molecular orbitals extending over two or more nuclei.

In MOT the atomic orbitals loose their identity and the total number of electrons present are placed in Mo’s according to increasing energy sequence (Auf Bau Principle) with due reference to Pauli’s Exclusion Principle and Hund’s Rule of Maximum Multiplicity.

When a pair of atomic orbitals combine they give rise to a pair of molecular orbitals, the bonding and the anti-bonding. The number of molecular orbitals produced must always be equal to the number of atomic orbitals involved. Electron density is increased for the bonding MO’s in the inter-nuclear region but decreased for the anti-bonding MO’s, Shielding of the nuclei by increased electron density in bonding MO’s reduces inter nuclei repulsion and thus stabilizes the molecule whereas lower electron density even as compared to the individual atom in anti-bonding MO’s increases the repulsion and destabilizes the system.

In denotation of MO’s, s indicates head on overlap and  p represents side ways overlap of orbitals. In simple homonuclear diatomic molecules the order of MO’s based on increasing energy is

This order is true except B2, C2 & N2. If the molecule contains unpaired electrons in MO’s it will be paramagnetic but if all the electrons are paired up then the molecule will be diamagnetic.

Bond order = \(\displaystyle \frac{{no.\,\,of\,\,{{e}^{-}}s\,\,occupying\,\,bonding\,\,MO’s-no.\,\,of\,\,{{e}^{-}}s\,occupying\,antibonding\,\,MO’s\,}}{2}\)

Application of MOT to homonuclear diatomic molecules.

H2              molecule                 :     Total no. of electrons = 2

Arrangement          :      \(\displaystyle \sigma _{{1s}}^{{\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}2\stackrel{\scriptscriptstyle\to}{\leftarrow}}}\)

Bond order             :     ½ (2 – 0) = 1

H2+          molecule                 :     Total no. of electrons = 1

Arrangement         :      \(\displaystyle \sigma _{{1s}}^{{\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}1}}\)

Bond order                               :           ½ (1 – 0) = 1/2

He2             molecule                :     Total no. of electrons = 4

Arrangement          :    \(\displaystyle \sigma _{{1s}}^{{\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}2}}\) \(\displaystyle \sigma _{{1s}}^{{*\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}2\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}}}\)

Bond order             :     ½ (2 – 2) = 0

Thus, He2 molecule does not exist.

Question.         Compare the bond energies of O2, &

Solution:                     Higher the bond order greater will be the bond energy

Now formation of means to remove an electron from anti-bonding one, which means increase in B.O.

B.O. of  = ½ (6-1) = 2.5

means introduction of an e in the anti-bonding thereby reducing the bond   order.

Bond order of  = ½ (6 – 3) = 1.5

So bond energy of > O2 >O2

Question.         Give MO configuration and bond orders of H2, H2, He2 and He2. Which species among the above are expected to have same stabilities?

Solution:         H2  = s1s2                                                        

                                        Bond order = 1

H2 = s1s2s*1s1

Bond order = 0.5

He2 = s1s2s1s*2                        

                                        Bond order = 0

He2 = s1s2s*1s2s2s1

Bond order = (3-2)/2 = 0.5

H2 and He2 are expected to have same stabilities

M.O. of Some Diatomic Heteronuclei Molecules

The molecular orbitals of heteronuclei diatomic molecules should differ from those of homonuclei species because of unequal contribution from the participating atomic orbitals. Let’s take the example of CO.

The M.O. energy level diagram for CO should be similar to that of the isoelectronic molecule N2. But C & O differ much in electronegativity and so will their corresponding atomic orbitals. But the actual MO for this species is very much complicated since it involves a hybridisation approach between the orbital of oxygen and carbon.

HCl Molecule: Combination between the hydrogen 1s A.O’s. and the chlorine 1s, 2s, 2p & 3s orbitals can be ruled out because their energies are too low. The combination of H 1s1 and 3px1 gives both bonding and anti-bonding orbitals, and the 2 electrons occupy the bonding M.O. leaving the anti-bonding MO empty.

NO Molecule: The M.O. of NO is also quite complicated due to energy difference of the atomic orbitals of N and O.

As the M.O.’s of the heteronuclei species are quite complicated, so we should concentrate in knowing the bond order and the magnetic behaviour.

Molecules/Ions Total No. of electrons

Magnetic behaviour

CO 14 Diamagnetic
NO 15 Paramagnetic
NO+ 14 Diamagnetic
NO 16 Diamagnetic
CN 13 Paramagnetic
CN 14 Diamagnetic


Heavier p-block and d-block elements show two oxidation states. One is equal to group number and second is group number minus two. For example Pb(5s25p2) shows two OS, +II and +IV. Here +II is more stable than +IV which arises after loss of all four valence electrons.

Reason given for more stability of +II O.S. that 5s2 electrons are reluctant to participate in chemical bonding because bond energy released after the bond formation is less than that required to unpair these electrons (lead forms a weak covalent bond because of greater bond length).

Question.         Why does PbI4 not exist?

Solution:         Pb(+IV) is less table than Pb(+II) due to inert pair effect and therefore Pb(+IV) is reduced to Pb(+II) by I which changes to I2(I is a good reducing agent)

  Characteristics of covalent compounds are:

(1) Covalent compounds have low melting points and boiling points. As such they may be gases, liquids or low melting solids.

(2) They are insoluble in water but soluble in organic (non-polar) solvents

(3) They do not conduct electricity in the solution or the molten state.

(4) Since covalent bonds are rigid and possess directional characteristics, therefore, they show stereoisomerism.

(5) Their reactions are slow and molecular in nature and never proceed to completion.

Co-ordinate or Dative Bond

This type of bond formation occurs by one sided sharing of electrons, i.e., one atom donates a pair of electrons while the other simply shares it so as to complete its octet.

The atom that donates a pair of electrons is called the donor while the other which accepts these electrons is called the acceptor. The coordinate bond is usually represented by an arrow pointing from the donor towards the accept

          For example,       

In terms of VB theory, a coordinate bond is formed by overlap of a fully filled orbital containing a lone pair of electrons with an empty orbital of another atom.

Examples of molecules in which all the three types of bonds, i.e., ionic, covalent and coordinate bonds are present :

\(CuS{{O}_{4}}.\,\,\,5{{H}_{2}}O,\,\,N{{H}_{4}}Cl,\,\,{{K}_{4}}\left[ {Fe{{{\left( {CN} \right)}}_{6}}} \right],\,\,\,\left[ {Cu{{{\left( {N{{H}_{3}}} \right)}}_{4}}} \right]S{{O}_{4}}\) etc.

Besides these bonds, \(CuS{{O}_{4}}.\,\,5{{H}_{2}}O\) also contains a H-bond.

Characteristics of coordinate compounds are :

Since coordinate compounds are in fact covalent compound, therefore, their properties are almost similar to those of covalent compounds. For example,

(1) Like covalent compound, they are insoluble in H2O but are soluble in organic solvents.

(2) They usually do not conduct electricity.

(3) Their melting and boiling points are higher than those of covalent compounds but lower than those of ionic compounds.

(4) Like covalent bonds, coordinate bonds are directional and hence these compounds also exhibit stereoisomerism.

Question.       The \(\Delta H{{{}^\circ }_{f}}CaB{{r}_{{2(s)}}}=-675\,kJ\,mo{{l}^{{-1}}}\) The first and second ionization energies of Ca are 590 and 1145 kJ mol–1. The enthalpy of formation of Ca(g) is 178 kJ mol–1. The bond enthalpy of Br2 is 193 kJ mol–1 and the enthalpy of vaporization of Br2 is 31 kJ mol–1. The electron affinity of Br(g) is 325 kJ mol–1. Calculate the lattice energy of CaBr2(s).              

Solution :       The given data are as follows:

  1. \(C{{a}_{{(s)}}}+B{{r}_{2}}(l)\to CaB{{r}_{{2(a)}}};\Delta {{H}_{f}}=-675\,kJ\,mo{{l}^{{-1}}}\)
  2. \(C{{a}_{{(s)}}}+B{{r}_{2}}(l)\to CaB{{r}_{{2(a)}}};\Delta {{H}_{f}}=-675\,kJ\,mo{{l}^{{-1}}}\)
  3. \(C{{a}_{{(g)}}}\to C{{a}^{+}}_{{(g)}};\Delta H{}^\circ {{I}_{1}}=590\,kJ\,mo{{l}^{{-1}}}\)
  4. \(\displaystyle C{{a}^{+}}_{{(s)}}\to C{{a}^{{2+}}}_{{(g)}};\Delta H{}^\circ \,{{I}_{2}}=1145\,kJ\,mo{{l}^{{-1}}}\)
  5. \(B{{r}_{2}}_{{(1)}}\to B{{r}_{2}}_{{(g)}};\Delta H{{{}^\circ }_{{vap}}}=31\,kJ\,mo{{l}^{{-1}}}\)
  6. \(2B{{r}_{{(g)}}}+2{{e}^{-}}\to 2B{{r}^{-}}_{{(g)}};\,\Delta H{}^\circ =2\times 325=-\,650kJ\)
  7. \(B{{r}_{{2\left( g \right)}}}\to 2B{{r}_{{\left( s \right)}}};\,\Delta {{H}^{o}}93\,kJ\,mo{{l}^{{-1}}}\)
  8. \(C{{a}^{{2+}}}_{{(g)}}\to 2B{{r}^{-}}_{{(g)}}\to CaB{{r}_{{2(s)}}}\Delta H{}^\circ =?\)

On adding all the above thermochemical equations, we get

\(\Delta H{{{}^\circ }_{f}}=\Delta H{{{}^\circ }_{{(sub)}}}+{{I}_{1}}+{{I}_{2}}+\Delta H{{{}^\circ }_{{vap}}}+\Delta H{}^\circ Dissociation+(E.A)+(U)\)

On substituting the given values:

\(-675=(+178)+(590)+(1145)+31+193+(-2\times 325)+U\)

On substituting the given values:

= 178 + 590 + 1145 + 31 + 193 – 650 + (U) (lattice energy

= +1487 + U

or U = – 675 – 1487 = –2162 kJ mol–1

Thus the lattice energy of CaBr2(s) is –2162 kJ mol–1.