Chemical Bonding_Final

  1. IIT-JEE Syllabus

Lewis structure; orbital overlap and covalent bond; hybridisation involving s, p & d orbitals only; coordinate covalent bond; ionic bond; orbital energy diagrams for diatomic species; polarity in molecules, dipole moment (qualitative aspects only); VSEPR model and shapes of molecules (linear, angular, planar, pyramidal, tetrahedral, octahedral and square planar.

  1. Introduction

All elements in order to attain the configuration of nearest inert gas combine with each other.

The relationship which exists between two atoms is through a bond. Now to know what a chemical bond is, we can try to explain it with a diagram.

From this diagram it is clear that when the two atoms which are to make a bond are infinite distance apart, there is no interaction between them and the energy of the two bonded atoms are at a level shown by dotted line. Now when they approach each other, distance between them decreases and inter electron – nucleus attraction operates more than inter nuclear and inter electronic repulsion. This net attractive forces reduces the energy of the system. This reduction in potential energy comes to a maximum and potential energy of the system becomes minimum at an equilibrium distance which is called bond length and if the bonding atoms tend to come closer than this equilibrium distance, the inter nuclear repulsion becomes dominant and energy of the system increases and so is its instability. 

Relationship gives shape to a family and so is the bonds in case of molecules. A clear idea about bonding can help to a great extent in predicting the shape of molecules. There are several different theories which explain the electronic structure and shapes of known molecules, and attempt to predict the shapes of molecules whose structures are so far unknown. Each theory has its own virtues and shortcomings. None is rigorous. Theories change in the light of new knowledge. If we knew or could prove what a bond was, we would not need theories, which by definition cannot be proved. The value of a theory lies more in its usefulness than in its truth. Being able to predict the shape of a molecule is important. In many cases all the theories give the correct answers.

  1. The Lewis Theory

The octet rule:  The Lewis theory gave the first explanation of a covalent bond in terms of electrons that was generally accepted. If two electrons are shared between two atoms, this constitutes a bond and binds the atoms together. For many light atoms, a stable arrangement is attained when the atom is surrounded by eight electrons.

This octet can be made up from some electrons which are totally owned and some electrons which are ‘shared’. Thus atoms continue to form bonds until they have made up an octet of electrons. This is called the ‘octet rule’. The octet rule explains the observed valencies in a large number of cases. There are exceptions to the octet rule; for example, hydrogen is stable with only two electrons. Other exceptions are discussed later.Today, the conventional Lewis structure representation of a pair of bonded electrons is by means of a ‘dash’ (-) usually called a ‘bond’. Lone pairs or ‘non-bonded’ electrons are represented by ‘dots’. Some structures are represented below:

Such representations of organic molecules are not usually problematic. However, ‘hit-and-trial’ is generally the method (obviously not very efficient) used by most students in figuring out the structures of inorganic molecules.

Different ways of attaining  the octet give rise to different types of valencies namely 

  1. Electrovalency  or Ionic Valency 
  2. Covalency
  3. Co-ordinate covalency
  1. Electrovalency

This type of valency involves transfer of electrons from one atom to another, whereby each atom may attain octet in their outermost shell. The resulting ions that are formed by gain or  loss of electrons are held together by electrostatic force of  attraction due to opposite nature of their charges. The reaction between potassium and chlorine to form potassium chloride is an example of this type of valency.

Here potassium has one electron excess of it’s octet and chlorine has one deficit of octet. So potassium donates it’s electron to chlorine forming a ionic bond. 

Here the oxygen accepts two electrons from calcium atom. It may be noted that ionic bond is not a true bond as there is no proper overlap of orbitals.

4.1 Criteria for Ionic Bond 

One of the species must have electrons in excess of octet while the other should be deficit of octet. Does this mean that  all substance having surplus electron and species having deficient electron  would form ionic bond? The answer is obviously no. Now you should ask why? The reasoning is that in an ionic bond one of the species is cation and the other is anion. To form a cation from a neutral atom energy must be supplied  to remove the electron and that energy is called ionisation energy. Now it is obvious that lower the ionisation energy of the element the easier it is to remove the electron. To form the anion, an electron adds up to a neutral atom and in this process energy is released. This process is called electron affinity.

So for an ionic bond one of the species must have low ionisation energy and the other should have high electron affinity. Low ionisation energy is mainly exhibited by the
alkali and alkaline earth metals, and high electron affinity by the halogen and chalcogens. Therefore this group of elements are predominant in the field of ionic bonding.

  1. Covalency 

This type of valency involves sharing of electrons between the concerned atoms to attain the octet configuration with the sharing pair being contributed by both species equally. The atoms are then held by this common pair of electrons acting as a bond, known as covalent bond. If two atoms share more than one pair then multiple bonds are formed. Some examples of covalent bonds are 

5.1 Sigma and Pi Bonding 

When two hydrogen atoms form a bond, their atomic orbitals overlap to produce a greater density of electron cloud along the line connecting the two nuclei. In the simplified representations of the formation of H2O and NH3 molecules, the O—H and N—H bonds are also formed in a similar manner, the bonding electron cloud having its maximum density on the lines connecting the two nuclei. Such bonds are called sigma bonds (σ-bond).

A covalent bond established between two atoms having the maximum density of the electron cloud the line connecting the centre of the bonded atoms is called a σ-bond.  A σ-bond is thus said to possess a cylindrical symmetry along the internuclear axis.

Let us now consider the combination of two nitrogen atoms. Of the three singly occupied p-orbitals in each, only one p-orbital from each nitrogen (say, the px may undergo “head –on” overlap to form a σ-bond. The other two p-orbitals on each can no longer enter into a direct overlap. But each p-orbital may undergo lateral overlap with the corresponding p-orbital on the neighbour atom. Thus we have two additional overlaps, one by the two py orbitals, and the other by the two pz orbitals. These overlaps are different from the type of overlap in a σ-bond. For each set of p-orbitals, the overlap results in accummulation of charge cloud on two sides of the internuclear axis. The bonding electron cloud does no more posses an axial symmetry as with the σ-bond; instead, it possess a plane of symmetry. For the overlap of the pz atomic orbital, the xy plane provides this plane of symmetry; for the overlap of the py atomic orbitals, the zx plane serves the purpose. Bonds arising out of such orientation of the bonding electron cloud are designated as π-bonds. The bond formed by lateral overlap of two atomic orbitals having maximum overlapping on both sides of the line connecting the centres of the atoms is called a π-bond. A π-bond possess a plane of symmetry, often referred to as the nodal plane.

  1. General Properties Of Ionic And Covalent Bonds

The different between ionic and covalent binding gives rise to difference in physical and chemical properties. An ionic compound is composed of positve and negative ions –– the ion pairs of opposite charges being held together by electrostatic attraction; the ions arrange themselves in a regular geometrical pattern in the crystal of an ionic compound. X-ray analysis, for example, reveals that the structure of sodium chloride crystal consists of two interlocking face centred cubic lattices, one of sodium ions and the other of chloride ions –– each sodium unit in the crystal is surrounded by six chlorine units, and each chlorine by six sodium units.

An ionic compound does not exist as a definite entity as molecules. There is no evidence, for example, of molecules of sodium chloride as structural units in the lattice of sodium chloride. All truly ionic compounds, e.g.,  common salt,  posses ionic lattice. Owing to the restrictive force in the crystal the ions are not free to move under the influence of an applied electric field. But when an ionic compound is dissolved in water (or a suitable polar solvent which diminishes the attraction between the ions) or melted, the ions acquire mobility and can move in opposite directions in an electric field, i.e., ionic compound is polar and can conduct electricity in the fused state or in solution in a suitable in a suitable solvent.

A purely covalent compound is a non-polar substance and does not conduct an electric current. Anhydrous hydrogen chloride is a non-conductor, but its aqueous solution conducts electricity as a result of chemical reaction which leads to the formation of ions:

HCl + H2O Cl′ + H3O+ (oxonium ion)

The ion-pairs tend to adhere to one another in electrovalent compounds and very powerful forces hold the crystals together. Considerable energy is necessary to break the crystal lattice and separate the ions. The electrovalent compounds are, therefore, solid that are not easily vaporised. They have high melting points and high boiling points. Unlike the electrovalent compounds covalent compounds exist as single molecules which have relatively little attraction to one another –– the only force between them being the weak van der Waals forces; the covalent compounds are therefore either gases, liquids or solids that are easily vaporised. Since the force of attraction between the molecules is not large, covalent compounds melt and boil at comparatively low temperature.

A covalent bond is rigid and directional (i.e., directed in space), and a covalent compound has its atoms held in definite relative positions so that the different arrangements of the atoms are not easily changed, i.e., a covalent compound may exhibit isomerism. Butane and isobutane for example, are two distinct isomeric compounds. The ionic bond, on the other hand, is non-rigid and non-directional, i.e., the charged particles are free to arrange and pack themselves in chosen ionic lattices.

A polar liquid (with a high dielectric constant) such as water, alcohol and liquid ammonia diminishes the force of attraction between the ions of an ionic compound; it will, therefore, generally be a good solvent for ionic compounds. But a non-polar liquid (with a low dielectric constant), e.g. organic solvents such as hydrocarbons and carbon tetrachlorde, is not good enough to overcome the inter-ionic attraction, and therefore will be a very poor solvent for ionic compounds.

  1. Co-ordinate Covalency 

This is a special type of covalent bond where the sharing pair is contributed by one species only but shared by both. E.g’s are boron trifluoride  / ammonia adduct, sulfur dioxide etc. The boron atom in BCl3 is two short of octet, it receives the lone pair of electrons from the nitrogen in NH3 and  thereby fulfills  the octet.  Let’s take the example of  SO2. It can be represented both as  O←S = O or O = S = O. Now the question that comes to your mind whether both are correct or not? The answer is yes, because a coordinate bond differs from a covalent bond only in the mode of formation. A co-ordinate bond is represented by an arrow pointing from the donor to the acceptor.

Now let us focus our attention to SO2 molecule. Sulfur has six valence electrons which is the same as the valence electrons of oxygen. Now when sulfur forms bonds to oxygen atoms it’s octet is fulfilled along with that of the oxygen. Now if sulfur goes again for a double bond one may seem that sulfur will violate the octet rule. For that reason we can represent the bonding between the sulfur and the other oxygen by a coordinate bond. But we know that sulfur being an element of 3rd period has vacant 3d orbitals therefore it can expand it’s covalency and so SO2 can also be represented by the second structure.

  1. Hybridization

We can explain the formation of four covalent bonds by an atom of carbon by considering promotion of a 2s electron to a 2p orbital. Let us now consider the formation of a molecule of methane, CH4, by such an excited carbon atom. There will be three C—H bonds formed by overlap of the three 2p-orbitals of carbon with the 1s orbitals of three hydrogen atoms. The sp bonds will be mutually perpendicular to one another. The fourth C—H bond would be formed by overlap of the 2s orbital of carbon with an 1s orbital of hydrogen atom. Since the 2s orbital is spherically symmetrical, the direction of the hydrogen atom held by this bond cannot be directly ascertained. At the same time, one should expect this s—s bond to be of lower strength than the other three s—p bonds. But it is contrary to our experience. We know that the four C—H bonds in methane are all alike and they are arranged symmetrically around the central carbon atom directed along the four corners of a tetrahedron. This necessitates a concept of mixing the s-orbital of carbon with its three p-orbitals before overlap. The resulting equivalent orbitals, each having one-fourth s-character and three –fourth p-character, may now undergo overlap with four hydrogen atoms to form four equivalent C—H bonds. We may reasonably extend this concept to interpret the equivalence of the two bonds in BeF2 or the three bonds in BCl3. This procedure of prior mixing of the orbitals has been given rigorous mathematical formulation.

Hybridization is a concept of mixing different atomic orbitals of comparable energy resulting in an equal number of orbitals with mixed character. The resulting hybrid orbitals undergo better overlap and form stronger bonds than the pure orbitals in conformity with the most stable geometry for a molecule. The formation of four equivalent C—H bonds by carbon in forming methane may then be conceived of in terms of the following successive steps:

At first, a 2s electron of the carbon atom gets unpaired and promoted to a 2p orbital. The 2s and the three 2p orbitals are now hybridized to give four equivalent orbitals, each possessing one part s character to three parts p character in their wave function, directed to the corners of a regular tetrahedron. These sp3 hybrid orbitals now form four equivalent
C—H bonds in the methane molecule; the bonds are distributed tetrahedrally around the carbon atom.

Shape Hybridisation

Linear sp

Trigonal planar sp2

Tetrahedral sp3

Trigonal bipyramidal sp3d

Octahedral sp3d2

Pentagonal bipyramidal sp3d3

8.1 VSEPR Theory (Valence Shell Electron Pair Repulsion Theory)

This theory starts from the general principle that valence shell electrons occupy essentially localised orbitals. Mutual interaction among the electrons orient the orbitals in space to an equilibrium position where repulsion becomes minimum. The extent of repulsive interaction then follows the order.

Lone pair – lone pair > lone pair – bond pair> bond pair – bond pair

A lone pair is concentrated around the central atom while a bond pair is pulled out between two bonded atoms. As such repulsion becomes greater when a lone pair is involved. Let’s take an example to illustrate this theory. CH4 contains no lone pairs. The bond pair – bond pair interactions brings about the most stable equilibrium bond angle of 109°28′, the angle predicted from sp3 hybridisation.

Illustration 1: Why the bond angle of H – C – H in methane (CH4) is 109° 28’ while
H – N – H bond angle in NH3 is 107° though both carbon and nitrogen are sp3 hybridized.

Solution: In CH4 there are 4 bond pair of electrons while in NH3 are 3 bond pair of electrons and 1 lone pair of electrons. Since bond pair bond pair repulsion is less than lone pair bond pair repulsion, in NH3 bond angle is reduced from 109°28’ to 107°.

Illustration 2: Why bond angle in NH3 is 107° while in H2O it is 104.5°

Solution: In NH3, central nitrogen atom bears only one lone pair of electrons whereas in H2O central oxygen atom bears two lone pair of electrons.

Since the repulsion between lone pair – lone pair and lone pair – bond pair is more than that between bond pair – bond pair, the repulsion in H2O is much greater than that in NH3 which results in contraction of bond angle from 109°28” to 104.5° in water while in NH3 contraction is less i.e. from 109°28” to 107°.

“If the electronegativity of the peripheral atoms is more, then the bond angle will be less”. For example if we consider NH3 and NF3 F – N – F bond angle will be lower than H – N – H bond angle. This is because in NF3 the bond pair is displaced more towards F and in NH3 it is displaced more towards N. So accordingly the b.p. – b.p. interaction is less in NF3 and more in NH3.

Illustration 3: The bond angle of  H2O is 104°  while that that of F2O is 102°.

Solution: Both H2O and F2O have a lone pair of electrons. But  fluorine being highly electronegative, the bond pair electrons are drawn more towards F in F2O, whereas in H2O it is drawn towards O. So in F2O the bond pairs being displaced away from the central atom has very little tendency to open up the angle. But in H2O this opening up is more as the bond pair electrons are closer to each other. So bond ∠ of F2O is less than H2O.

Exercise 1: Among PCl3 & PF3 which is having greater bond angle and why?

“If the electronegativity of central atom is more then bond angle will be more”. For example in NH3 and PH3, the H – N – H bond angle is more then H – P – H bond angle. The reason for it can be explained in the same way as above.

Illustration 3: Out of H2O and H2S which is having greater bond angle and why?

Solution: H2O is having greater bond angle than H2S since oxygen is more electronegative than S and draws the shared pair of electrons toward itself more than S. Therefore the bond pair interaction is more in case of H2O.

Exercise 2: Out of NH3 and PH3 which is having greater bond angle and why?

Rule for determination  of total  number of hybrid orbitals 
  1. Detect the central atom along with the peripheral atoms. 
  2. Count the valence electron of the central atom and the peripheral atoms. 
  3. Divide the above value  by 8. Then the quotient gives the number of σ bonds and the remainder gives the non-bonded electrons. So number of lone pair
    = .
  4. The number of σ 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 ×7) = 34 

Now  8) 34 (4

∴ Number of hybrid orbitals =  4σ bonds + 1 lone pair 

So 5 hybrid orbitals are necessary and hybridisation mode is sp3d and it is trigonal bipyramidal (TBP).

Both the structures are TBP. But the lone pair is placed in different position. In B it is placed in equatorial position and in A it is in axial.

Now when a lone pair is in equatorial position the repulsion are minimised. So structure (B) is correct.

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

Illustration 4:

1. NCl3  Total valence electrons = 26 

Requirement  = 3 σ bonds + 1 lone pair 

Hybridsation = sp3

Shape  = pyramidal

2. BBr3 Total valence electron  = 24

Requirement  = 3σ bonds 

Hybridisation = sp2

Shape  = planar trigonal 

3.  SiCl4 Total valence electrons = 32

Requirement  = 4σ bonds 

Hybridisation =  sp3

Shape  = Tetrahedral 

4. CI4 Total valence electron = 32

Requirements = 4 σ bonds 

Hybridisation = sp3

Shape = Tetrahedral 

5. SF6 Total valence electrons  = 48

Requirement = 6 σ bonds 

hybridisation = sp3d2

shape  = octahedral / square  bipyramidal

6. BeF2 Total valence electrons : 16

Requirement : 2 σ bonds 

Hybridisation : sp

Shape : Linear

F – Be – F
7. ClF3 Total valence electrons : 28Requirement : 3 σ bonds + 2 lone pairs 

Hybridisation : sp3d

Shape : T – shaped 

We have already discussed that whenever there are lone pairs they should be placed in equatorial positions. Now a question that may come to your mind that though the hybridisation is sp3d, so the shape should be T.B.P. But when all the bonds are present the actual shape is TBP. But when instead of bond there are lone pairs in TBP the actual geometry is determined by the bonds not by the lone pairs. Here in ClF3 the bond present (2 in axial and 1 in equatorial) gives the impression of T shape.

8. PF5  Total valence electrons : 40

Requirement : 5 σ bonds

Hybridisation : sp3d

Shape :  Trigonal bipyramidal (TBP)

9. XeF4 Total valence electrons : 36

Requirement : 4 σ 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 inimize 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 inimize repulsion. So both structure (A) and (B) are correct.

10. XeF2 Total valence electrons : 22 

Requirements : 2σ bonds + 3 lone pairs 

Hybridisation: sp3d

Shape : Linear 

[ l.p. are present in equatorial position and ultimate shape  is due to the bonds that are formed]

11. PF2Br3 Total valence electrons : 40

Requirements : 5 σ bonds 

Hybridisation: sp3d

Shape : trigonal bipyramidal

Here we see that fluorine is placed in axial position whereas bromine is placed in equatorial position. It is the more electronegative element that is placed in axial position and less electronegative element is placed in equatorial position. Fluorine, being more electronegative pulls away bonded electron towards itself more than that is done by bromine atom which results in decrease in bp – bp repulsion and hence it is placed in axial position.

In this context it can also be noted that in T.B.P. shape the bond lengths are not same. The equatorial bonds are smaller than axial bonds. But in square bipyramidal shape, all bond lengths are same.


Total valence electrons : 32 

Requirement : 4 σ bonds 

Hybridisation: sp3

Shape: tetrahedral

Here all the structures drawn are resonating structures with O resonating with double bonded oxygen.

13. NO2 Total valence electron: 18 

Requirement : 2σ bonds + 1 lone pair

Hybridisation: sp2

Shape: angular 

  1. CO32– Total valence electrons : 24

Requirement = 3 σ bonds 

Hybrdisation = sp2

Shape: planar trigonal 

But C has 4 valence electrons of these 3 form σ bonds the rest will form a π bond.

In the structure one bond is a double bond and the other 2 are single. The position of the double bonds keep changing in the figure. Since peripheral atoms are isovalent, so contribution of the resonant structures are equal. Thus it is seen that none of the bonds are actually single or double. The actual state is 

15. CO2 Total valence electrons : 16

Requirement: 2σ bonds 

Hybridisation: sp

Shape: linear

O = C = O
16. Total  valence electrons = 32

Requirement= 4 σ bonds 

Hybridisation: sp3

Shape: Tetrahedral

17. Total valence electron = 26

Requirement = 3 σ bond + 1 lone pair 

Hybridisation: sp3

Shape: pyramidal

18. XeO2F2 Total valence electrons : 34

Requirement: 4 σ bonds +1 lone  pairs 

Hybridisation : sp3d

Shape: Distorted TBP (sea-saw geometry)

19. XeO3 Total valence electrons : 26

Requirement: 3 σ bonds + 1 lone  pair

Hybridisation: sp3

Shape: Pyramidal

20. XeOF4 Total valence electrons : 42

Requirement: 5 σ bonds + 1 lone  pair

Hybridisation: sp3d2

Shape: square pyramidal.

Exercise 3 Draw the structure the following indicating the hybridisation of the central atom.

  1. i) SOF2, ii) SO2,  (iii) POCl3,  (iv) I3

8.2 Maximum Covalency 

Elements which have vacant d-orbital can expand their octet by transferring electrons, which arise after unpairing, to these vacant d-orbital e.g. in sulphur.

In excited state sulphur has six unpaired electrons and shows a valency of six e.g. in SF6. Thus an element can show a maximum covalency equal to is group number e.g. chlorine shows maximum covalency of seven.

  1. Deviation from Ideal Behaviour 

Still now we have discussed all about those compounds which are either cent percent ionic or covalent from the Lewis theory i.e. the formation of the compound from the respective elements takes place either by mutual sharing or by complete transfer of electrons resulting either covalent species or ionic species.

Now we know that NaBr is an ionic crystal. Similarly AlBr3 is also an ionic crystal as its formation takes place by complete transfer of electrons. So as both are ionic compounds and as it is a well known fact that ionic compounds have high melting points, so we should expect a high melting point for both. But contrary to our expectations the melting point of NaBr is 775°C and that of AlBr3 is 97.5°C. Similarly CaF2 has got a melting point of 1400°C while that of CaI2 is 575°C. This fact indicates that all those compounds which are ionic from the concept of transfer of electrons are not 100% ionic but have some percent of covalent character induced in it. So it is better to say a compound to be more of ionic rather than totally ionic.

9.1 Transition from Ionic to Covalent Character

In an ideal ionic lattice, the ions are supposed to be hard inelastic spheres just touching each other at equilibrium. But properties of most ionic solids deviate from ideal ionic character because lattice may undergo some distortion. Cations are usually smaller than anions and have higher effective nuclear charge than the latter. The outer electron cloud in a cation is more firmly held than in an anion. Electron cloud in an anion is rather loosely held and may be attracted by the neighbouring cation in the lattice. Hence the anion may not retain its spherical symmetry and may be distorted. Such a phenomenon is known as polarisability of the anion and the capacity of the cation to distort the electron cloud of anion is known as polarization power of the cation. As the anion shift’s slightly its electrons towards cation, the position charge over cation decreases and negative charge over anion also decreases imparting partial ionic character to the bond. In other words a completely ionic bond becomes partial ionic bond and shifting of electrons towards cation imparts covalent nature to the bond i.e. bond starts behaving like polar covalent bond.

The polarising power of a cation is expressed by an index, φ defined as

where φ is called ionic potential.

Again the extent to which an anion is polarised is given by its polarisability. This is governed by the charge and size of the anion.

9.2 Factors Governing Polarization and Polarisability  (Fajan’s Rule)

  1. Cation Size: Smaller is the cation more is the value of φ and hence more its polarising power. As a result more covalent character will develop. Let us take the
    example of the chlorides of the alkaline earth metals. As we go down from Be to Ba the cation size increases and the value of φ decreases which indicates that BaCl2 is less covalent i.e. more ionic. This is well reflected in their melting points. Melting points of  BeCl2 = 405°C and  BaCl2 = 960°C.
  2. Cationic Charge: More is the charge on the cation, the higher is the value of φ and higher is the polarising power. This can be well illustrated by the example already given, NaBr and AlBr3. Here the charge on Na is +1 while that of Al in +3, hence polarising power of Al is higher which in turn means a higher degree of covalency resulting in a lowering of melting point  of  AlBr3 as compared to NaBr.
  1. Noble Gas vs Pseudo Noble Gas Cation: A Pseudo noble gas cation consists of a noble gas core surrounded by electron cloud due to filled d-subshell. Since
    d-electrons provide inadequate shielding from the nuclei charge due to relatively less penetration of orbitals into the inner electron core, the effective nuclear charge (ENC) is relatively larger than that of a noble gas cation of the same period. NaCl has got a melting point of 800°C while CuCl has got melting point of 425°C. The configuration of Cu+ = [Ar] 3d10 while that Na+ = [Ne]. Due to presence of d electrons ENC is more and therefore Cl is more polarised in CuCl leading to a higher degree of covalency and lower melting point. 
  1. Anion Size: Larger is the anion, more is the polarisability and hence more covalent character is expected. An e.g. of this is CaF2 and CaI2, the former has melting
    point of 1400°C and latter has 575°C. The larger size of I ion compared to F causes more polarization of the molecule leading to a lowering of covalency and increasing in melting point. 
  2. Anionic Charge: Larger is the anionic charge, the more is the polarisability. A well illustrated example is the much higher degree of covalency in magnesium nitride (3Mg++ N3–) compared to magnesium fluoride (Mg++ 2F). This is due to higher charge of nitride compare to fluoride. 

These five factors are  collectively known as Fajan’s Rule.

Role of φ (Ionic Potential)

  1. Prediction of degree of covalency in an ionic compound.
  2. Tendency of a cation to form complexes can be estimated.
  3. Tendency of cations towards solvation.

E.g.: [Li(H2O)6]+, [Na(H2O)4]+, [Cs(H2O)]+

  1. Nature of oxides: Emperically for an oxide of the type
  2. a) <2 basic
  3. b) lying between 2.2 – 3.2 amphoteric
  4. c) >2 acidic

i.e. higher is the value of φ greater is acidic nature of oxide.

Illustration 5: The melting point of KCl is higher than that of AgCl though the crystal radii of Ag+ and K+ ions are almost same.

Solution: Now  whenever any comparison is asked about the melting point of the compounds which are fully ionic from the electron transfer concept it means that the compound having lower melting point has got lesser amount of ionic character than the other one. To analyse such a question first find out the difference between the 2 given compounds. Here in both the compounds the anion is the same. So the deciding factor would be the cation. Now if the cation is different, then the answer should be from the variation of the cation. Now in the above example, the difference of the cation is their electronic configuration. K+ = [Ar]; Ag+ = [Kr] 4d10. This is now a comparison between a noble gas core and pseudo noble gas core, the analysis of which we have already done.  So try to finish off this answer.

Exercise 4: Na2CO3 does not decompose on heating whereas CaCO3 decomposes, why?

Illustration 6: The solubility of the hydroxides of the alkaline earth metals increases i.e. Ba(OH)2 has got a higher solubility in water compared to Mg(OH)2

Solution: Here both the cations Ba2+ and Mg2+ have the same charge, but as the radius of Ba2+ is more therefore φ of Ba2+ is less which implies that Ba(OH)2 having higher degree of ionic character is more soluble in polar solvents like water.

But now if I ask to predict the solubility of MgSO4 & BaSO4.  The answer seems to be quite similar to the earlier one and BaSO4 turns out to be the one having higher solubility. But contrary to our expectation the trend is reversed here. BaSO4 is sparingly soluble in water. The question comes why? In case of hydroxide it is something. In case of sulfate it’s the other way around. Is there any way by which we can a predict the solubility trend? The answer is yes. 

When a lattice is dissolved in water, the ions are solvated and the solvated ions are more stable than a free ion and due to this stability energy is released. This energy released is called solvation energy and if this overcomes the lattice energy then it is soluble. The lattice energy of NaCl  is 778 kJ mol–1 and the heats of hydration of Na+ and Cl is –787 kJ mol–1.  As it is more than the lattice energy of NaCl therefore it is soluble.

Now we should focus our attention to the solubility trend in a given series. For a comparison of the solubility both the lattice energy and hydration energy factors have to be taken into account. If in a series the decrease of lattice energy is more compared to the decrease in hydration energy then the substance becomes soluble.

Now the hydration enthalpy of a salt is given by

where k1 and  k2 = constant

& lattice energy k3 = constant

Case (i): When r+ << r the contribution of the anion to the hydration enthalpy is small so the total  ΔHhydration would be dominated by the cation alone. In a series of salts of a large anion, the hydrational enthalpy will decrease in magnitude with increasing cation size. Now how does the lattice energy respond to this changing cation radius? The lattice energy is inversely proportional to (r+ + r). Since r >> r+, the sum will not change significantly as r+ increases. Consequently the lattice energy will not decrease as fast as the hydration energy with increasing cationic size. The more quickly diminishing hydration energy results in a decrease in solubility. 

E.g. Solubility of LiI > NaI > KI…

MgSO4 > CaSO4 > SrSo4 > BaSO4

Case (ii): r+ r

Here the lattice energy decreases with increasing cationic size more rapidly than the hydration energy which therefore results in an enhanced solubility in a series.

E.g. Solubility of LiF < NaF < KF

Mg(OH)2 < Ca(OH)2 < Ba(OH)2

9.3 Deviation  from Covalent to Ionic Character

In the previous section we discussed about those compounds which deviate from fully ionic to some degree of covalency. A similar trend can also be observed with pure covalent molecules which can change to a partially ionic bond.

This happens when the electronegativities of the two atoms which form the covalent bond are not the same. The atom having higher electronegativity will draw the bonded electron pair more towards itself resulting in a partial charge separation. The distribution of the electron cloud in the bond does not remain uniform and shifts towards the more eletronegative one. Such bonds are called polar covalent bonds. For example the bond formed between hydrogen and chlorine or between hydrogen and oxygen in water is of this type. 

Molecules like HCl, H2O, NH3 i.e. molecules of the type H – X having two polar ends (positive and negative) are known as polar molecules. The extent of polar character or the degree of polarity in a compound is given by it’s dipole moment which is defined as the product of the net positive or negative charge and the distance of separation of the charges i.e. the bond length. The symbol of dipole moment is μ

μ = electronic charge (e) × distance

The unit of dipole moment is Debye (D)

1D = 3.33 × 10–30 Cm = 10–18 esucm

Dipole moment is indicated by an arrow having a symbol () pointing towards the negative end. Dipole moment has both magnitude and direction and therefore it is a vector quantity.

To calculate the dipole moment of a molecule we should calculate the net dipole due to all bonds and for lone pair if any. Diatomic molecules like HCl, HF have the dipole moment of the bond (called bond dipole) equal to the molecular dipole as the structure has one bond only. But for poly atomic molecules the net dipole is the resultant of the individual bond dipoles. A compound having a zero dipole moment indicates that the compound is a symmetrical one.

Illustration 7: CO2 has got dipole moment of zero why?

Solutions: The structure of CO2 is. This is a highly symmetrical structure with a plane of symmetry passing through the carbon. The bond dipole of C–O is directed towards oxygen as it is the negative end. Here two equal dipoles acting in opposite direction cancel each other and therefore the dipole moment is zero.

Illustration 8: Dipole moment of CCl4 is zero while that of CHCl3 is non zero.

Solution: Both CCl4 & CHCl3 have tetrahedral structure but CCl4 is symmetrical while CHCl3 is non-symmetrical. 

Due to the symmetrical structure of CCl4 the resultant of bond dipoles comes out to be zero. But in case of CHCl3 it is not possible as the presence of hydrogen introduces some dissymmetry.

Illustration 9: Compare the dipole moment of H2O  and F2O.

Solutions: Let’s draw the structure of both two compounds and then analyse it.

In both H2O and F2O the structure is quite the same. In H2O as O is more electronegative than hydrogen so the resultant bond dipole is towards O, which means both the lone pair and bond pair dipole are acting in the same direction and  dipole moment of H2O is high. In case of F2O the bond dipole is acting towards fluorine, so in F2O the lone pair and bond pair dipole are acting in opposition resulting in a low dipole.

Exercise 5: Compare the dipole moment of NH3 and NF3

In C-H, carbon being more electronegative the dipole is projected towards C. Now the question comes whether hybridization has anything to do with the dipole moment. The answer is obviously yes. If yes, why? Depending on the hybridization state the electronegativity of carbon changes and therefore the dipole moment of C-H bond will change. As  the s character in the hybridized state increases, the electronegativity of C increases due to which C attracts the electron pair of C-H bond more towards itself resulting in a high bond dipoles.

Now as we have said about carbon hydrogen bonds, the question that is coming to your mind is whether we would be dealing with organic compounds or not. Yes we would be dealing with the organic compounds.

For instance but -2- ene. It exists in two forms Cis and Trans.

The trans isomer is symmetrical with the 2 methyl groups in anti position. So the bond dipoles the two Me– C bonds acting in opposition cancel each other result in a zero dipole. Whereas in cis isomer the dipoles do not cancel each other resulting in a net dipole.

Illustration 10: Compare the dipole moment of Cis 1,2 dichloroethylene and
trans 1,2 dichloroethylene.


In the trans compound the C-Cl bond dipoles are equal and at the same time acting in opposition cancel each other while Cis compound the dipoles do not cancel each other resulting in a higher value.

Generally all Trans compounds have a lower dipole moment corresponding to Cis isomer, when both the substituents attached to carbon atom are either electron releasing or electron withdrawing. 

9.4 Dipole Moment in Aromatic Ring System

The dipole moments of the aromatic compounds present a very good illustration of dipole moment. We all know when a substituted benzene is treated with a reagent different products namely ortho, meta and para products are formed. The dipole moments of these products are different since the orientation of the groups are different at ortho, meta and para position. Let us take an example which will make it easily digestive for you. Suppose we have three isomers of o-nitrophenol, m-nitrophenol and p-nitrophenol. We have also the e.g. o-aminophenol, m-aminophenol and p-aminophenol.

In the case  X = Y, the para  isomer becomes symmetrical  and  have zero dipole.

Now the obvious question that is peeping through your mind is that which isomer in which case has got higher dipole moment. The answer lies in the nature of the groups linked to the benzene ring. In nitrophenol groups one group is electron pushing and the other is electron withdrawing while in the second case both the groups attached are electron pushing. So depending on the nature of the groups attached one of the isomer, o, m or p has the largest dipole moment.

Case (i): Now when X & Y are both electron pushing or electron withdrawing.

Suppose, bond dipole of C – X = μ1

And that of C – Y = μ2

Here we have assumed  a sign of + when groups are electron pushing and–when groups are electron withdrawing. The net dipole is the resultant of two bond dipoles at different orientations.

When both X & Y are electron pushing or electron withdrawing.

μortho =


μ0 =

μmeta =

μm =

μpara =


μp = μ1μ2

From the above expression of μ0, μm & μp it is clear that when both  X & Y are of the same nature i.e. both are electron withdrawing or both are electron pushing the para product has the least dipole moment and ortho product has the highest. Now when  X = Y, μp = μ1μ2 = μ1μ1 = 0

Which we have already discussed.

Case (ii) When X is electron pushing and Y is electron withdrawing or vice versa.

Let C – X dipole = μ1

& C – Y dipole = μ2

∴μ0 =   =


μmeta = =


μpara =  


= μ1 + μ2

Now if you see the expressions, it is very clear that the para isomer has the highest dipole moment and ortho is the least.

So to calculate the dipole moments of disubstituted benzene one should  consider about the nature of the groups linked and then only one can predict the dipole moment of the molecule.

Illustration 11:

Solution: Para isomer is having highest dipole moment since two groups attached to benzene ring have dipole moment directed in the same direction thereby they reinforce one another in this case.

Exercise 6:

10. Percentage of Ionic Character

Every ionic compound having some percentage of covalent character according to Fajan’s rule. The percentage of ionic character in a compound having some covalent character can be calculated by the following equation.

The percent ionic character =

Illustration 12: Calculate the % of ionic character of a bond having length = 0.92  Å and 1.91 D as it’s observed dipole moment.

Solution: Calculated μ considering 100% ionic bond

= 4.8×10–10× 0.92 ×10–8esu cm

= 4.8 × 0.92  × 10–18 esu cm  =    4.416 D

∴ % ionic character = = 43.25

The example given above is of a very familiar compound called HF. The % ionic character is nearly 43.25%, so the % covalent character is (100 – 43.25) = 56.75%. But from the octet rule HF should have been a purely covalent compound but actually it has some amount of ionic character in it which is due to the electronegativity difference of H and F. Similarly knowing the bond length and observed dipole moment of HCl, the % ionic character can be known. It was found that HCl has 17% ionic character. Thus it can be clearly seen that although we call HCl and HF as covalent compounds but it has got appreciable  amount of ionic character. So from now onwards we should call a compound having more of ionic  less of covalent and vice versa rather than fully ionic  or covalent.

Exercise 7: Calculate the percentage of ionic character of HI having bond length  = 1.62Å and observed dipole moment = 0.39 D.

11. Hydrogen Bonding


11.1 Introduction

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 electronagative 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

11.2 Conditions for Hydrogen Bonding

  1. Hydrogen should be linked to a highly electronegative element.
  2. 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.

11.3 Types of Hydrogen Bonding

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

Inter molecular 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 H-bond.

  1. 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

  1. a) State of the substance
  2. b) Solubility of the substance
  3. c) Boiling point
  4. d) Acidity of different isomers

These can be evident from the following examples.

Illustration -13: 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 that of an isolated H2S molecule and therefore it is a gas.

Illustration-14: Ethyl alcohol (C2H5OH) has got a higher boiling point than dimethyl either (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.

C2H5 C2H5


—–O ⎯ H ———-O ⎯ H ——

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

Exercise 9: O-nitrophenol is more volatile than p-nitrophenol.

Exercise 10: Why is salicylic acid more acidic than p-hydroxy benzoic acid?

11.4 Importance of Hydrogen Bonding in Biological Systems

Hydrogen bonding plays a vital role in physiological systems. Proteins contain chains of amino acids. The amino acid units are arranged in a spiral form somewhat like a stretched coil spring (forming a helix). The N-H group of each amino acid unit and the fourth C=O group following it along the chain, establishes the N–H—O hydrogen bonds. These  bonds are partly responsible for the stability of the spiral structure. Double helix structure of DNA also consists of two strands forming a double helix and are joined to each other through hydrogen bond.

12. Intermolecular Forces


12.1 Introduction

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 liquefaction 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 (1813) 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.

12.2 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.

12.3 Origin of Intermolecular Forces 

Intermolecular forces may have a wide variety of origin.

  1. a) Dipole-dipole interaction: This forces would exist in any molecule having a permanent dipole e.g. HF, HCl, H2O etc.
  2. b) Ion-dipole interaction: These interaction are operative in solvation and dissolution of ionic compounds in polar solvents.
  3. c) Induced dipole interaction: These generate from the polarisation of a neutral molecule by a charge or ion.
  1. 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.

13. Molecular Orbital Theory

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 denotion of MO’s, σ indicates head on overlap and  π 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 =

Application of MOT to homonuclear diatomic molecules.

H2 molecule : Total no. of electrons = 2

Arrangement :

Bond order : ½ (2 – 0) = 1

molecule : Total no. of electrons = 1

Arrangement :

Bond order : ½ (1 – 0) = 1/2

He2 molecule : Total no. of electrons = 4

Arrangement :

Bond order : ½ (2 – 2) = 0

∴He2 molecule does not exist.

molecule : Total no. of electrons = 3

Arrangement :

Bond order : ½ (2 – 1) = 1/2

So exists and has been detected in discharge tubes.

Li2 molecule : Total no. of electrons = 6

Arrangement :

Bond order : ½ (4 – 2) = 1

No unpaired e’s so diamagnetic

Be2 molecule : Total no. of electrons = 8

Arrangement :

Bond order : ½ (4 – 4) = 0

No unpaired es so diamagnetic

B2 molecule : Total no. of electrons = 10

Arrangement :

Bond order : ½ (6 – 4) = 1


But observed Boron is paramagnetic

C2 molecule : Total no. of electrons = 12

Arrangement :

Bond order : ½ (4 – 0) = 2

It is paramagnetic

But observed C2 is diamagnetic 

N2 molecule : Total no. of electrons = 14

Arrangement :

Bond order : ½ (6 – 0) = 3

∴ It is diamagnetic 

O2 molecule : Total no. of electrons = 16

Arrangement :

Bond order : ½ (6 – 2) = 2

It is paramagnetic

F2 molecule : Total no. of electrons = 18

Arrangement :

Bond order : ½ (6 – 4) = 1

It has been seen that in case of B2,C2 & N2 the order of filling the e’s is different from the normal sequence.

B2 :

It is paramagnetic

C2 :

It is diamagnetic

N2 :

It is diamagnetic

Illustration 15: Compare the bond energies of O2, &

Solutions : Higher the bond order greater will be the bond energies.

Now configuration of O2  =

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 >

Exercise 11: Compare the bond energies of N2, &

13.1 M.O. of Some Diatomic Heteronuclei Molecules

The molecular orbitals of hetronuclei 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 is electronegativity and so will their corresponding atomic orbital. 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 is A.O. and the chlorine 1s, 2s, 2p & 3s orbitals can be ruled out because their energies are too low. The combination of H 1s1 and 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 heteronucleai 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

Inert Pair Effect: 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 0.5, +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 0.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)

Exercise 12: Explain why CCl2 is unstable while PbCl2 is stable?

Back bonding: Sometime surrounding atom in order to make up electron deficiency in the central atom forms a π-bond which is one sided i.e. shared electrons are contributed by surrounding atom only which is having lone pair of electrons. Overlap involves vacant orbital on central atom and filled orbital on surrounding atom.

Exercise 13: Trisilyl amine (SiH3)3N is weaker base than (CH3)3N, why?


  1. Solution to Exercises

Exercise 1: PCl3 will be having greater bond angle since ‘F’ is more electronegative and drags the shared paired towards itself thereby reducing the b.p. – b.p. repulsion in PF3.

Exercise 2: NH3 will have greater bond angle since ‘N’ is more electronegtaive and drags the shared pair towards itself and thereby increases the b.p. – b.p. repulsion.

Exercise 3 i) ii)
iii) iv)

Exercise 4: The solution lies in the thermal stability of Na2CO3 compared to CaCO3. The decomposition of a carbonate means formation of an oxide along with CO2

CaCO3 CaO + CO2 (g)

Now more the carbonate ion is polarised by the cation more is the chance of formation of CO2 and therefore higher is the probability of decomposition. Now let’s see the case of CaCO3. The polarising power of Ca2+ is more than Na+ due to higher cationic charge of calcium ion. Therefore it polarises the carbonate ion more compared to sodium whose φ is insufficient to polarise and so CaCO3 decomposes but not Na2CO3.

⎯⎯→ No polarization

Exercise 5: Let’s draw the structure of both two compounds and then analyse it.

In both NH3 and NF3 the structure is quite the same. In NH3 as N is more electronegative than hydrogen so the resultant bond dipole is towards N, which means both the lone pair and bond pair dipole are acting in the same direction and  dipole moment of NH3 is high. In case of NF3 the bond dipole is acting towards fluorine, so in NF3 the lone pair and bond pair dipole are acting in opposition resulting in a low dipole.

Exercise 6: p-isomer will be having highest dipole moment since dipole moment of – CF3 group is reinforced to maximum extent in this case by – CH3 group. (One is electron withdrawing and other is electron releasing).

Exercise 7: Calculate μ considering 100% ionic bond

= 4.8 × 10–14 esu × 1.62 × 10–8 cm = 7.776 D

% ionic character = = 5

% covalency = (100 – 5) = 95%

Exercise 9: More volatility means it has got lower boiling point. Now p-nitrophenol remains associated through intermolecular hydrogen bonding. But in o-nitrophenol only intramolecular H-bonding  formation takes places, as a result of which there is no association. So p-nitrophenol which remains as an associated species has got higher boiling point and so less volatile.

Exercise 10: Salicylate ion forms a 6-membered chelate ring through hydrogen bond formation which results in stabilization of ion which is not possible in p-hydroxy benzloic acid.


Exercise 11: The configuration of N2 is

Now means removal of an electron from a body M.O. This will decrease the B.O.

∴B.O. of = ½ (5 – 0) = 2.5

Now again for bond order is ½ (6–1) = 2.5

So from the bond order it may seem that both & may have the same bond energy. But removal of an electron  from a diatomic species tend to decrease the inter electron repulsion and thereby shorters the bond length. So the bond energy becomes more than compared to

∴N2 > >

Exercise 12: In PbCl2 + II oxidation is more stable than +IV state due to inert pair effect i.e. ns2 electrons do not participate in bonding.

Exercise 13: Due to pπ – dπ bonding that exists between fully filled p-orbital of nitrogen and vacant d-orbital of silicon, availability of lone pair of electrons on nitrogen decreases.


  1. Solved problems


15.1 Subjective

Problem 1: In (CH3)3N nitrogen is sp3 hybridized whereas in (SiH3)3N it is sp2 hybridised why?

Solution: In (SiH3)3N pπ-dπ bonding exists between fully filled p-orbital of nitrogen and vacant d-orbital of silicon which requires planarity in molecule that is possible only when nitrogen is sp2 hybridised.

Problem 2: Why does Al2(CO3)3 does not exist?

Solution: Al+++ due to high positive charge density (small size and greater charge) polarizes CO32– ion in such a way that results in formation of CO2 and oxide of aluminium.

Problem 3: OsO4 exists but OsF8 does not exist. Why?

Solution: In both cases O.S. of Os in +8 but in OsO4, Os is bonded to oxygen by 4 double bonds whereas in OsF8, Os will have to form 8 single bonds which require much more space around central atom.

Problem 4: 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)

Problem 5: Why can HgCl2 and SnCl2 not exist together in an aqueous solution?

Solution: SnCl2 is a reducing agent which first reduces HgCl2 to Hg2Cl2 (white) and then to Hg (black)

SnCl2  + 2HgCl2 ⎯→ SnCl4 + Hg2Cl2

SnCl2 + Hg2Cl2 ⎯→ 2Hg + SnCl4

Problem 6: AlCl3 is covalent but AlCl3⋅6H2O is ionic, Why?

Solution: Hydration energy helps in formation of Al3+ ion which otherwise is difficult to form. Polar water molecules also stabilize Al3+.

Problem 7: Why Sn+2 is a reducing agent but Pb2+ is not.

Solution: In Pb inert pair effect is more pronounced than Sn and therefore stability of +II O.S. is higher than that in Sn.

Problem 8: White phosphorus is more reactive than red phosphorus.

Solution: White phosphorus has following strained structure where each P atom is bonded to three other phosphorus atoms by sigma bonds.

In red phosphorus the molecule is somewhat relieved from this strained structure by forming a chain structure (polymer like) where one intra molecular p-p bond is replaced by two intermolecular p-p bonds

Problem 9: Though Cs is most electropositive element in periodic table Li has highest oxidation potential why?

Solution: A metal ionizes in following way in gaseous state

M ⎯→ M+(g) + 1e ΔH = I.E. …(1)

But in water cation undergoes hydration

M+(g) + nH2O ⎯→ [M(H2O)n]+ ΔH = hydration energy …(2)

I.E. has positive value but hydration energy has –ve value. For Cs I.E. is less than Li but for Li hydration energy is more than Cs as Li+ has higher charge density. The resultant of these two values is more –ve for Li rather than Cs. Therefore in aqueous solution Li ionizes more than Cs.

Problem 10: Though Li+ ion is smaller than Cs+ ion its mobility in aqueous solution is more than Cs+ ion.

Solution: Due to higher +ve charge density on Li+(smaller size) hydration of Li+ takes place to a greater extent than Cs+ ion. As Li+ gets attached to greater number of water molecules than Cs+ its mobility decreases.

Problem 11: BaCO3 has got a higher decomposition temperature than MgCO3.

Solution: BaCO3 has got a higher decomposition temperature the MgCO3 which illustrates its high thermal stability. The thermal stability depends on the polarising power of the cation which in turn is inversely proportional to the radius Ba2+ having a larger radius than Mg2+ has got less polarising power and hence thermaly more stable.

Problem 12: Dipole moment of BF3 is zero.

Solution: BF3 is a planar molecule with the bond angles being 120°.

In each B-F bond the bond dipole is projected towards Fluorine. Now the direction of the resultant bond dipoles of these two bonds is shown in the diagram as R. If R is equal to the dipole of the 3rd B-F bond then R and B-F dipole will neutralize each other as equal dipoles acting in opposite direction cancel each other. Now let’s see the magnitude of R. As dipole moment is a vector quantity, so the resultant of two dipoles can be obtained from the law of vector addition. Suppose the
bond dipole is μ1.

R = a & b are the individual vectors

= θ = angle between vectors

= cos120° =

= = μ1 = μB–F

R is equal in magnitude to B-F bond dipole. Similar is the case with the other two resultants. So the net dipole of BF3 is zero.

Problem 13: Compare the dipole moments of CH3Cl, CH2Cl2, CHCl3, CCl4

Solution: All the molecules have tetrahedral geometry with CCl4 having symmetrical structure resulting in a zero dipole.

The net dipole of the molecules is the resultant of the bond dipoles. 

In CHCl3 due to the interaction of the non bonded electrons of chlorine the bond angle increases whereas in CH3Cl this interaction in less and so the bond angle is less compared to CHCl3. Now the resultant dipole is dependent on cosθ. Now higher the value of θ less is cosθ and therefore less is the dipole moment. On that basis CH3Cl has highest dipole and CHCl3 has lowest with CCl4 having zero dipole.

Problem 14: SO2 is angular and CO2 is linear why?

Solution: This is because, by the method discussed above, S in SO2 having sp2 hybridisation with one lone pair of S occupying the sp2 hybridised orbital gives angular shape whereas C in CO2 is sp hybridised and is thus linear.

Problem1 5: AlF3 is ionic while AlCl3 is covalent.

Solution: Since F is smaller in size, its polarisability is less and therefore it is having more ionic character. Whereas Cl being large in size is having more polarisability and hence more covalent character.

Problem 16: Ice is lighter than water.

Solution: Ice is lighter then water because ice has an open cage like structure in which lesser molecules are packed per cc. Whereas in liquid water they are packed closely together. Consequently the density of water is more than ice. 

Problem 17: PH5 is not possible whereas PF5 is possible.

Solution: Since fluorine is more electronegative it contracts the size of d-orbitals of P and reduces its energy so that it can undergo hybridisation with s and p-orbitals expanding its covalency.

Problem 8: In water the H – O – H bond angle is 105°. Determine the magnitude of the charge on the oxygen atom in the water molecule.

Given dipole moment of water = 1.85 D

Covalent radii = Hydrogen = 0.28Å

Oxygen = 0.66Å


cos 52.5 =

δ = 0.572Å

δ =  = 3.2 × 10–10 esu

or 0.67 times the electronic charge.

Problem1 9: What order of C – H bond length do you expect in C2H6, C2H4 and C2H2
and why?

Solution: The order of increasing C – H bond length is

C2H2 < C2H4 < C2H6

This is because the percentage of s-character is maximum in C2H2 (50%) and least in C2H6 (25%). As the % of s-character increases the bond length decreases as the extent of overlap increases.

Problem 20: Why Li2CO3 decomposes on heating but not other alkali metal carbonates?

Solution: Since Li is very small in size, it has high polarising power which polarises CO32– in CO2 and O2– and also due to large size difference of Li+ and CO32–, the lattice is very much unstable and decomposes on heating.

15.2 Objective

Problem 1: I3 (tri-iodide ion) has

(A) Linear shape (B) Angular shape

(C) Trigonal planar shape (D) T-shape

Solution: Central I atom in I3 is sp3d hybridised. Lone pairs occupy equatorial position

∴ (A)

Problem 2: S – O bond length in SOCl2

(A) Equal C – O bond length in COCl2

(B) Greater than C – O bond length in COCl2

(C) Less than C – O bond length in COCl2

(D) None of the above

Solution: It is less than C – O bond length due to pπ – dπ bonding

∴ (C)

Problem 4: When NH3 is treated with HCl, state of hybridisation on central nitrogen

(A) Changes from sp3 to sp2 (B) Remains unchanged

(C) Changes from sp3 to sp3d (D) Changes from sp3 to sp

Solution: On NH4+ state of hybridisation on central nitrogen atom is sp3 as in NH3.

∴ (B)

Problem 4: Lanthanoid ion which is most likely to be reduced by Cr(+II) is

(A) Sm (B) Yu

(C) Yb (D) All of the above

Solution: All of them on being reduced will get either stable half filled orbitals or stable completely filled orbitals.

∴ (D)

Problem 5: The correct bond order of S – O bond in decreasing order is

(A) OSF2 > OSCl2 > OSBr2 (B) OSCl2 > OSBr2 > OSF2

(C) OSF2 > OSBr2 > OSCl2 (D) OSBr2 > OSCl2 > OSF2

Solution: As electronegativity of halogen attached with sulphur increases, sulphur become more electron deficient and hence its tendency to get electrons from oxygen through pπ – dπ bonding also increases i.e. extent of pπ – dπ bonding increases and hence bond order also increases.


Problem  6: The species which does not show paramagnetism is

(A) O2 (B) O2+

(C) O22– (D) H2+

Solution: O22– has no unpaired electron

∴ (C)

Problem  7: The shape of a molecule which has three bond pairs and one lone pairs is

(A)  Octahedral (B) Triangular planar

(C) Pyramidal (D) Tetrahedral

Solution: i.e. NH3 has three bond pair and one lone pair.

∴ (C)

Problem 8: Which molecule is T-shaped.

(A) BeF2 (B) BCl3

(C) NH3 (D) ClF3


Solution: ClF3 has sp3d hybridisation with two lone pairs of electron on central

∴ (D)

Problem 9: The pair having similar geometry is

(A) BF3, NH3 (B) BF3, AlF3

(C) BeF2, H2O (D) BCl3, PCl3

Solution: both have sp2 hybridisation of central atom.

∴ (B)

Problem 10: How many unpaired electrons are present in N2+ 

(A) 1 (B) 2

(C) 3 (D) 4

Solution: In N2 all electrons are paired. Thus N2+ has one unpaired electron.

∴ (A)

Problem 11: PCl5 exist, but NCl5 does not exist because

(A) Nitrogen has no vacant 2-d orbital

(B) NCl5 is unstable

(C) N-atom is much smaller than P

(D) Nitrogen is highly inert

Solution: The excitation of 2s electron is not possible

∴ (A)


Problem 12: The following compounds have been arranged in order of their increasing thermal stabilities identify the correct order

K2CO3(I) MgCO3(II)


(A) I < II < III < IV (B) IV < II < III < I

(C) IV < II < I < III (D) II < IV < III < I


Solution: The stability of carbonates increases with increasing electropositive character of metal.

∴ (B)

Problem 13: Among the following compounds the one that is polar and has the central atoms with sp2 hybridisation is

(A) H2CO3 (B) SiF4

(C) BF3 (D) HClO2

Solution: Carbon in H2CO3 has sp2 hybridisation and also polar. BF3 has sp2 hybridisation but non-polar. SiF4 has sp3 hybridisation. HClO2 has sp3 hybridisation.

∴ (A)

Problem 14: In the formation of N2+ from N2 the electron is removed from

(A) A σ orbital (B) A π-orbital

(C) A σ* M orbital (D) A π-M orbital

Solution: See the electronic configuration of N2

∴ (A)

Problem 15: CO2 has the same geometry as

(A) HgCl2 (B) NO2

(C) C2H2 (D) Both (A) and (C)

Solution: both HgCl2 and C2H2 are linear like CO2 because of sp hybridisation.


Problem  16: Among the following species identify the isostructural pairs

NF3, NO3, BF3 H3O+, HN3

(A) [NF3, NO3 and [BF3H3O+] (B) [NF3, HN3] and [NO3, BF3]

(C) [NF3, H3O+] and [NO3, BF3] (D) [NF3, H3O+] and [HN3, BF3]

Solution: NF3 and H3O+ have sp3 hybridisation; NO3 and BF3 have sp2 hybridisation

∴ (C)

Problem 17: Among KO2, AlO2, BaO2 and NO2+ one unpaired electron is present in

(A) O2+ and BaO2 (B) KO2 and AlO2

(C) KO2 only (D) BaO2 only

Solution: KO2 has K+O2 structure having on unpaired electron

∴ (C)

Problem 18: Which ion has the higher polarising power

(A) Mg2+ (B) Al3+

(C) Ca2+ (D) Na+

Solution: small cation having higher positive charge has more polarising power

∴ (B)

Problem  19 : The correct order of decreasing polarisability of ion is

(A) Cl > Br > I > F (B) F > I > Br > Cl

(C) I > Br > Cl > F (D) F > Cl > Br > I

Solution: Larger the ion , more is its polarisability.


Problem 20: Amongst LiCl, RbCl, BeCl2 and MgCl2 the compounds with greatest and least ionic character respectively are

(A) LiCl and RbCl (B) RbCl and BeCl2

(C) RbCl and MgCl2 (D) MgCl2 and BeCl2

Solution: See Fajan’s rule

∴ (A)


  1. Assignments  (Subjective Problems)



  1. Why BF3 is weaker Lewis acid than BCl3?
  2. Why [CuCl4]2– exists but [CuI4]2– does not exist?
  3. SF6 is known but SCl6 is not known. Why?
  4. Li3N exists but Na3N does not exist. Why?
  5. [Al(OH)6]3– exists but [B(OH)6]3– does not exist. Why?
  6. CCl4 is has zero dipole moment but CH2Cl2 has an appreciable dipole moment.
  7. CaCO3 dissolves in HCl but not in water.
  8. HF from stronger H-bonds than H2O still ΔHvap of HF is lower than water.
  9. Valency of Fluorine is generally one whereas chlorine shows valencies of one, three, five and seven.
  10. Draw the structure of the following


  1. ii) XeF6

iii) XeO2F2

  1. O-nitrophenol is less soluble in water than p-nitrophenol.
  2. NF3 is pyramidal but BF3 is planar.
  3. Dipole moment of KCl is 3.336 × 10–29 coulomb metre which indicates that it is highly polar molecule. The interatomic distance between K+ and Cl is 2.6 × 10–10m. 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.
  4. PbCl4 exists but PbI4 is not possible.
  5. PCl5 exists in the solid state in the form of [PCl4]+ [PCl6] yet it is a non conductor of electricity.



  1. NCl5 and BiCl5 do not exist whereas PCl5 exists. Why?
  2. Though electronegativity difference is maximum between hydrogen and fluorine yet HI is stronger acid than HF. Why?
  3. Why HF (hydrofluoric acid) is used for etching glass?
  4. F2 bond strength is lower than Cl2 bond strength.
  5. POCl3 molecule has the shape of an irregular tetrahedran with P atom located centrally. The Cl – P – Cl angle is found to be 103.5° give a qualitative explanation for the deviation of this structure from a regular tetrahedran.
  6. Li+ ion has got the smallest radius among the alkali metals, yet its mobility is less than other alkali metal ions in aqueous solution.
  7. Bond angle of F2O is 102° while that of Cl2O is 110°.
  8. Explain why the solubility in water of halides of Aluminium follows the order

AlF3 > AlCl3 > AlBr3 > AlI3

  1. SnF4 boils at 705°C while SnCl4 boils at 114°.
  2. Which bond is stronger F – F or Cl – Cl and why?
  3. Formula of nitrogen is N2 while that of phosphorus is P4.
  4. What is the Lewis acidity order of BF3, BCl3, BBr3 and BI3 and why?
  5. Why Al(OH)3 is amphoteric in nature?
  6. In ICl5, the four chlorine atoms on square planar base and the central atom I does not remain in same plane. Justify the statement.
  7. o-hydroxy benzaldehyde is more soluble in water than p-hydroxy benzaldehyde.



  1. In (CH3)3N nitrogen is sp3 hybridized whereas in (SiH3)3N it is sp2 hybridised why?
  2. Why H3SiNCO is linear in vapour phase (excluding hydrogen atoms) but H3CNCO is not linear?
  3. Gold which is insoluble in all acids be is soluble in aqua-regia. Why?
  4. Why SnI4 is orange in colour?
  5. Explain why C – O bond length in CO is 1.128Å whereas in metal carbonyls C – O bond length is 1.15Å?
  6. Out of FeCl3 and FeCl2 which have got greater melting point and why?
  7. Explain why dipole movement of p-dimethoxy benzene is non-zero?
  8. Why is ortho-hydroxy benzoic acid more acidic than para-hydroxy benzoic acid?
  9. CO2 is a gas while SiO2 in a solid.
  10. The geometry of trimethyl amine and trisilyl amine is not the same.
  11. Li + O2 ⎯→ Li2O

2Na + O2 ⎯→  Na2O

Why the number of oxygen atoms combining with each metal increases.

  1. AgI is least soluble of the various silver halides in water. 
  2. The product of hydrolysis of NCl3 and PCl3 are different. Why?
  3. SF6 is possible but SH6 is not possible. Why?
  4. KHF2 is possible but not KHCl2. Why?


  1. Assignments  (Objective Problems)


  1. Which of the following compounds is (are) non – polar 

(A) HCl (B) CH2Cl2

(C) CHCl3 (D) CCl4

  1. The types of bonds present in CuSO4.5H2O are only 

(A) Electrovalent and covalent

(B) Electrovalent and co-ordinate covalent

(C) Covalent and co-ordinate covalent 

(D) Electrovalent 

  1. Donor properties of hydrides of VA group in decreasing order is

(A) NH3 > PH3 > AsH3 > SbH3 > BiH3 (B) BiH3 > SbH3 > AsH3 > PH3 > NH3

(C) SbH3 > BiH3 > AsH3 > PH3 > NH3 (D) PH3 > NH3 > AsH3 > SbH3 > BiH3

  1. In which of the following central atom does not use sp3 hybrid orbitals in its bonding?

(A) (B)

(C) (D) NF3

  1. XeF2 involves hybridisation 

(A) sp3 (B) sp2

(C) sp3d2 (D) None of these 

  1. The type of hybrid orbitals used by the chlorine atom in is 

(A) sp3 (B) sp3d

(C) sp (D) None of these 

  1. Bond angle is minimum for 

(A) H2O (B) H2S

(C) H2Se (D) H2Te

  1. For which of the following hybridisation the bond angle is maximum 

(A) sp2 (B) sp

(C) sp3 (D) dsp2

  1. The species having one unpaired electron is 

(A) NO (B) CO

(C) CN (D) O2

  1. Compound involving sp3d hybridisation is 

(A) BF3 (B) PF5

(C) SF6 (D) IF7

  1. Which of the following does not have dipole moment?

(A) ClO2 (B) CO2

(C) NO2 (D) SO2

  1. The molecule that will have dipole moment

(A) 2,2-Dimethyl propane (B) trans -2- butane

(C) cis-3-Hexene (B) 2,2,3,3-Tetramethyl butane 

  1. Which one has minimum dipole moment 

(A) Butene -1 (B) Cis-2-Butene 

(C) trans-2-Butene (D) 2-Methylpropene

  1. XeF4 has shape of 

(A) Trigonal bipyramidal (B) square planar 

(C) Tetrahedral (D) none of these 

  1. Which of the following has minimum bond length 

(A) O2 (B)

(C) (D)

  1. Which of the following is diamagnetic 

(A) O2 (B)

(C) (D)

  1. Which of the following species is paramagnetic 

(A) (B) CN

(C) CO (D) NO+

  1. Dipole moment is shown by 

(A) 1,4-Dichlorobenzene (B) Cis-1,2,Dichloroethene 

(C) trans -1,2-Dichloroethene (D) 1,4-dimethyl benzene

  1. Which of the following has highest bond order 

(A) O2 (B)

(C) (D)

  1. Which of the following has lowest bond order?

(A) O2 (B)

(C) (D)



  1. The ion which is isoelectronic with CO is 

(A) CN (B)

(C) (D)

  1. Which of the following structures are isoelectronc 

(I) (II) H3O+ (III) NH3 (IV)

(A) I and II (B) III and IV

(C) I and III (D) II,III and IV

  1. CO2 is isostructural with 

(A) NO2+ (B) SnCl2

(C) C2H4 (D) NO2

  1. Which of the following has tetrahedral disposition?

(A) SO2 (B) SO2

(C) (D)

  1. Which is incorrect statement?

(A) CO2 is a monomer while SiO2 is a three dimensional giant molecule.

(B) Graphite is anisotropic with respect to electricity

(C) (CH3)3N has C – N – C bond angle of 107.9° whereas in (SiH3)3N has Si – N – Si a bond angle of 120°.

(D) Bond lengths of four apical P – O bonds are greater than calculated.

  1. Which of the following cannot exist on the basis of MO theory 

(A) (B)

(C) He2 (D) O2

  1. Which of the following has a bond order of 2.5?

(A) CO (B) NO

(C) He2+ (D)

  1. Which of the following have identical bond order 

(I) CN     (II)   (III) NO+        (IV) CN+

(A) I & II (B) I & III

(C) II & III (D) III & IV

  1. Which of the following species is paramagnetic 

(A) N2 (B) O22–

(C) O2 (D) F2

  1. Which has got highest bond angle among NO2, NO2+ and NO2

(A) NO2 (B) NO2+

(C) NO2 (D) All are equal

  1. Highest covalent character is found  in 

(A) CaF2 (B) CaCl2

(C) CaBr2 (D) CaI2

  1. (CH3)3SiOH is stronger acid than (CH3)3COH because

(A) Conjugate base anion in (CH3)3SiO is stabilized by Opπ – Sidπ bond formation

(B) Bulkiness of (CH3)3SiO anion inhibits its stabilization

(C) Electronegativity of Si is greater than C

(D) All of the above

  1. Two elements X and Y have 2, and 7 electrons respectively in their valence shell. The expected compound formed by combination of X and Y 

(A) XY2 (B) X5Y2

(C) X2Y5 (D) XY5

  1. When NH3 is treated with HCl H – N – H bond angle

(A) Increases (B) Decreases

(C) Remains same (D) Depends upon temperature

  1. o-Nitrophenol is more volatile than p-nitro phenol because of 

(A) Resonance

(B) Presence of intermolecular hydrogen bonding in the o-isomer 

(C) Absence of intermolecular hydrogen bonding in the o-isomer

(D) none of these 

  1. The correct bond order among halogens is

(A) F2 > Cl2 > Br2 > I2 (B) Cl2 > F2 > Br2 > I2

(C) Br2 > Cl2 > F2 > I2 (D) I2 > Cl2 > Br2 > F2

  1. Among the following compounds the one that is non-polar and has central atom sp2 hybridised is 

(A) H2CO3 (B) SiF4

(C) BF3 (D) HClO2

  1. Among the following compounds which has the largest dipole moment 

(A) CH3OH (B) CH4

(C) CF4 (D) CH3F

  1. Resonance is not shown by

(A) C6H6 (B) NO2

(C) (D) SiO2

  1. Of the following species, the one having planar structure is 

(A) (B)

(C) XeF4 (D) CCl4


  1. Answers to  Objective Assignments



  1. D 2. A
  2. A 4. A
  3. B 6. A
  4. D 8. B
  5. A 10. B
  6. B 12. C
  7. C 14. B
  8. B 16. D
  9. A 18. B
  10. B 20. D



  1. A 2. D
  2. A 4. C
  3. D 6. C
  4. B 8. B
  5. C 10. B
  6. D 12. A
  7. A 14. A
  8. C 16. B
  9. C 18. D
  10. D 20. C