CHEMICAL BONDING AND MOLECULAR STRUCTURE

INTRODUCTION

A molecule is formed if it is more stable and has lower energy than the individual atoms. Normally only electrons in the outermost shell of an atom are involved in bond formation and in this process each atom attains a stable electronic configuration of inert gas. Atoms may attain stable electronic configuration in three different ways by loosing or gaining electrons by sharing electrons. The attractive forces which hold various constituents (atoms, ions etc) together in different chemical species are called chemical bonds. 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.

Three different types of bond may be formed depending on electropositive or electronegative character of atoms involved.

Electropositive element + Electronegative element = Ionic bond (electrovalent bond)

Electronegative element + Electronegative element = Covalent bond

or less electro positive + Electronegative element = Covalent bond

Electropositive + Electropositive element = Metallic bond.

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 an 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.

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 ionization energy. Now it is obvious that lower the ionization 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 ionization energy and the other should have high electron affinity. Low ionization 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.

Energy Change During the Formation of Ionic Bond

The formation of ionic bond can be consider to proceed in three steps 

(a)    Formation of gaseous cations
        A(g) + I.E. → A⁺ (g) + e^– 

        The energy required for this step is called ionization energy (I.E)

  (b)    Formation of gaseous anions

        X(g) + e^– → X^–(g) + E.A  

        The energy released from this step is called electron affinity (E.A.)

  (c)    Packing of ions of opposite charges to form ionic solid

         A+(g) + X–(g) → AX(s) + energy

 The energy released in this step is called lattice energy.

 Now for stable ionic bonding the total energy released should be more than the energy required.

 From the above discussion we can develop the factors which favour formation of ionic bond and also determine its strength. These factors have been discussed below:

 (a) Ionization energy: In the formation of ionic bond a metal atom loses electron to form cation. This process required energy equal to the ionization energy. Lesser the value of ionization energy, greater is the tendency of the atom to form cation. For example, alkali metals form cations quite easily because of the low values of ionization energies.  (b) Electron affinity: Electron affinity is the energy released when gaseous atom accepts electron to form a negative ion. Thus, the value of electron affinity gives the tendency of an atom to form anion. Now greater the value of electron affinity more is the tendency of an atom to form anion. For example, halogens having highest electron affinities within their respective periods to form ionic compounds with metals very easily.

(c) Lattice energy: Once the gaseous ions are formed, the ions of opposite charges come close together and pack up three dimensionally in a definite geometric pattern to form ionic crystal.

  Since the packing of ions of opposite charges takes place as a result of attractive force between them, the process is accompanied with the release of energy referred to as lattice energy. Lattice energy may be defined as the amount of energy released when one mole of ionic solid is formed by the close packing of gaseous ion. 

  In short, the conditions for the stable ionic bonding are:

(a)    I.E. of cation forming atom should be low:

(b)    E.A. of anion – forming atom should be high;

(c)    Lattice energy should be high.

Determination of Lattice Energy

The direct calculation of lattice enthalpy is quite difficult because the required data is often not available. Therefore lattice enthalpy is determined indirectly by the use of the Born – Haber cycle. The cycle uses ionization enthalpies, electron gain enthalpies and other data for the calculation of lattice enthalpies. The procedure is based on the Hess’s law, which states that the enthalpy of a reaction is the same, whether it takes place in a single step or in more than one step. In order to understand it let us consider the energy changes during the formation of sodium chloride from metallic sodium and chlorine gas. The net energy change during the process is represented by ΔH_f. 

Example 1: 

Calculate the lattice enthalpy of MgBr₂. Given that

Enthalpy of formation of MgBr₂ = -524 kJ mol^–1

Some of first & second ionization enthalpy (IE₁ + IE₂ ) = 148 kJ mol^-1

Sublimation energy of Mg = +2187 kJ mol^–1

Vaporization energy of Br₂(I) = +31kJ mol^–1

Dissociation energy of Br₂(g) = +193kJ mol^–1

Electron gain enthalpy of Br(g) = -331 kJ mol^–1

Solution:
 

          ΔH^o_f = S + I.E. + ΔH_vap + D + 2 × E.A. + U

Or     U = ΔH^o_f – (S + I.E. + ΔH_vap + D + 2 × E.A.)

Or     U = -524 - [2187 - 148 + 31 + 193 + 2 × (-331)]

            = – 524 – 1897 = –2421 kJ mol^–1

Characteristics of ionic compounds :  

The following are some of the general properties shown by these compounds

(i) Crystalline nature :  These compounds are usually crystalline in nature with constituent units as ions. Force of attraction between the ions is non-directional and extends in all directions. Each ion is surrounded by a number of oppositely charged ions and this number is called co-ordination number. Hence they form three dimensional solid aggregates. Since electrostatic forces of attraction act in all directions, therefore, the ionic compounds do not posses directional characteristic and hence do not show stereoisomerism.

(ii) Due to strong electrostatic attraction between these ions, the ionic compounds have high melting and boiling points.

(iii) In solid state the ions are strongly attracted and hence are not free to move. Therefore, in solid state, ionic compounds do not conduct electricity. However, in fused state or in aqueous solution, the ions are free to move and hence conduct electricity.

(iv) Solubility : Ionic compounds are fairly soluble in polar solvents and insoluble in non-polar solvents. This is because the polar solvents have high values of dielectric constant which defined as the capacity of the solvent to weaken the force of attraction between the electrical charges immersed in that solvent. This is why water, having high value of dielectric constant, is one of the best solvents.

The solubility in polar solvents like water can also be explained by the dipole nature of water where the oxygen of water is the negative and hydrogen being positive, water molecules pull the ions of the ionic compound from the crystal lattice. These ions are then surrounded by water dipoles with the oppositely charged ends directed towards them. These solvated ions lead an independent existence and are thus dissolved in water. The electrovalent compound dissolves in the solvent if the value of the salvation energy is higher than the   lattice energy of that compounds.

        AB + Lattice energy → A⁺ + B^–

These ions are surrounded by solvent molecules. This process is exothermic and is called salvation.

     A⁺ + x(solv.) → [A(solv.)_x]⁺ + energy

     B^– + y(solv.) → [B(solv.)_y]^– + energy

The value of solvation energy depend on the relative size of the ions. Smaller the ions more is the solvation. The non-polar solvents do not solvate ions and thus do not release energy due to which they do not dissolve ionic compounds.

 (v) Ionic reactions: Ionic compound furnish ions in solutions. Chemical reactions are due to the presence of these ions. For example 

           Na₂SO₄ → 2Na⁺ + SO₄^2–

           BaCl₂ → Ba²⁺ + 2Cl^–

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

                    

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 along 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 p_x may undergo “head –on” overlap to form a s-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 p_y orbitals, and the other by the two p_z orbitals. These overlaps are different from the type of overlap in as-bond. For each set of p-orbitals, the overlap results in accumulation of charge cloud on two sides of the internuclear axis. The bonding electron cloud does no more posses an axial symmetry as with the s-bond; instead, it possess a plane of symmetry. For the overlap of the p_z atomic orbital, the xy plane provides this plane of symmetry; for the overlap of the p_yatomic 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.

σ-Bond

(a)    s-s         overlapping
(b)    s-p         overlapping
(c)    p-p         overlapping

π - Bond:

This type of bond is formed by the sidewise or lateral overlapping of two half filled atomic orbitals.

                

HYBRIDIZATION 
 

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 p₂C = CH₂ bond as in H

(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

Example 7:

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.

CO-ORDINATE COVALENCY

A covalent bond results from the sharing of pair of electrons between two atoms where each atom contributes one electron to the bond. It is also possible to have an electron pair bond where both electrons originate from one atom and none from the other. Such bonds are called coordinate bond or dative bonds. Since in coordinate bonds two electrons are shared by two atoms, they differ from normal covalent-bond only in the way they are formed and once formed they are identical to normal covalent –bond.

It is represented as [——→]

Atom/ion/molecule donating electron pair is called Donor or Lewis base. Atom / ion / molecule accepting electron pair is called Acceptor or Lewis acid [——→] points donor to acceptor

NH₄⁺, NH₃ has three (N – H) bond & one lone pair on N – atom. In NH₄⁺ formation this lone pair is donated to H⁺ (having no electron)

NH₃ + H⁺ → NH₄⁺

                    

Properties of the coordinate compounds are intermediates of ionic and covalent compounds.

Examples of Hybridization

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 hybridization 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 : 5σ bonds Hybridization : sp3d Shape : Trigonal bipyramidal (TBP)
9. XeF4 Total valence electrons : 36 Requirement:4σ 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.

10. XeF2   Total valence electrons : 22 Requirements : 2σ bonds + 3 lone pairs Hybridisation: sp3d Shape : Linear

 [Q 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 : 5σ 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.

12.  Total valence electrons : 32 Requirement : 4σbonds 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

 14. CO₃^2– Total valence electrons: 24

 Requirement = 3σ bonds

 Hybrdisation = sp²

  Shape: planar trigonal

  But C has 4 valence electrons of these 3 form s bonds \ the rest will form a p bond.

                  

 In the structure one bond is a double bond and the other 2 are single. The position of the double bonds keeps changing in the figure. Since peripheral atoms are isovalent, so contribution of the resonanting 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= 4σ bonds Hybridisation: sp3 Shape: Tetrahedral
17. Total valence electron = 26 Requirement = 3σ bond + 1 lone pair Hybridization: sp3 Shape: pyramidal
18. XeO2F2 Total valence electrons : 34 Requirement: 4σ bonds +1 lone pairs Hybridization : sp3d Shape: Distorted TBP (sea-saw geometry)
19. XeO3 Total valence electrons : 26 Requirement: 3σ bonds + 1 lone pair Hybridization: sp3 Shape: Pyramidal
20. XeOF4 Total valence electrons : 42 Requirement: 5σ bonds + 1 lone pair Hybridization: sp3d2 Shape: square pyramidal.

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 × 7) = 34

Now 34/8 = 4 2/8 

∴ Number of hybrid orbitals = 4σ 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 = 3σ bonds + 1 lone pair Hybridization = sp3 Shape = pyramidal
2. BBr3 Total valence electron = 24 Requirement = 3σ bonds Hybridization = sp2 Shape = planar trigonal
3. SiCl4 Total valence electrons = 32 Requirement = 4σ bonds Hybridization = sp3 Shape = Tetrahedral
4. CI4 Total valence electron = 32 Requirements = 4σ bonds Hybridization = sp3 Shape = Tetrahedral
5. SF6 Total valence electrons = 48 Requirement = 6σ bonds Hybridization = sp3d2 Shape = octahedral/square bipyramidal
6. BeF2 Total valence electrons : 16 Requirement : 2σ bonds Hybridization : sp Shape : Linear	F – Be – F
7. ClF3 Total valence electrons : 28 Requirement: 3σ bonds + 2 lone pairs Hybridization : sp3d Shape : T – shaped

Solved Examples Based on Hybridization

 Example 8:

Predict the hybridization for the central atom in POCl₃, OSF₄, OIF₅.

 Solution:

 Total No. of V.E. = 5+6+21/8 = 32/8 = 4

 So, hybridization = sp³

 OSF₄ = 6+6+28/8 = 40/8 = 5

 So, hybridization of s = dsp³

 OIF₅ = 6+7+35/8 = 48/8 = 6

So, hybridization of I = d²sp³

 Example 9:

 Out of the three molecules XeF₄, SF₄ and SiF₄ one which has tetrahedral structures is

 (A) All of three                  (B) Only SiF₄

 (C) Both SF₄ and XeF₄        (D) Only SF₄ and XeF₄

 Solution:

 Hybridization of XeF₄ = sp³d², SF₄ = sp³d, SiF₄ = sp³

 Hence (B) is correct.

 Example 10:

 Among the following compounds in which case central element uses d-orbital to make bonds with attached atom

 (A) BeF₂                             (B) XeF₂ 
 (C) SiF₄                              (D) BF₃

 Solution:

 In XeF₂. Xe atom has sp³d hybridisation. Hence (B) is correct.

 Example 11:

 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.

                  

 Hence (B) is the correct answer.

 Exercise 4: Among the following which has bond angle very near to 90°

 (A) NH₃                   (B) XeF₄

 (C) BF₃                   (D) H₂O

Exercise 5: Homolytic fission of C – C bond in hexafluoroethane gives an intermediate in which hybridization state of carbon is

(A) sp²                    (B) sp³

(C) sp                     (D) cannot be determined 

 Exercise 6: A molecule XY₂ contains two σ, two π bonds and one lone pair of electrons in valence shell of X. The arrangement of lone pair as well as bond pairs is

 (A) Square pyramidal                (B) Linear
 (C) Trigonal planar                   (D) Unpredictable

 Exercise 7: Draw the structure the following indicating the hybridisation of the central atom.

 (A) SOF₂                                  (B) SO₂

 (C) POCl₃                            (D) I₃^–

 Exercise 8: The type of hybridisation of orbitals employed in the formation of SF₆molecule is …………..

 Exercise 9:

 The angle between two covalent bonds is maximum for……………(CH₄, H₂O, CO₂)

  Exercise 10:The bond angle in SO^2–₄ ion is ………………..

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 its group number e.g. chlorine shows maximum covalency of seven.

VALENCE SHELL ELECTRON PAIR REPULSION (VSEPR) THEORY (SHAPES AND GEOMETRY OF MOLECULES)

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

RESONANCE

There may be many molecules and ions for which it is not possible to draw a single Lewis structure. For example we can write two electronic structures of O₃.

 In (A) the oxygen - oxygen bond on the left is a double bond and the oxygen-oxygen bond on the right is a single bond. In B the situation is just opposite. Experiment shows however, that the two bonds are identical. Therefore neither structure A nor B can be correct.

One of the bonding pairs in ozone is spread over the region of all the three atom rather than associated with particular oxygen-oxygen bond. This delocalised bonding is a type of bonding in which bonding pair of electrons is spread over a number of atoms rather than localised between two.

 Structures (A) and (B) are called resonating or canonical structures and C is the resonance hybrid. This phenomenon is called resonance, a situation in which more than one plausible structure can be written for a species and i Chemical bonding is the major part of chemistry which is an interaction between two or more atoms that holds them together by reducing the potential energy of their electrons. In other words Bonds are the like chemists "glue" - which hold atoms together in molecules or ions. Valence electrons are the outer shell electrons of an atom which take part in chemical bonding.

Atoms gain or lose electrons to attain a more stable noble gas - like electron configuration (octet rule). There are two ways in which atoms can share electrons to satisfy the octet rule:

Ionic Bonding - occurs when two or more ions combine to form an electrically-neutral compound

The positive cation "loses" an electron (or 2 or 3)
The negative anion "gains" the electron (or 2 or 3)
 The anion steals the electrons from the cation. 

 Covalent Bonding - occurs when two or more atoms combine to form an electrically-neutral compound

The electrons are shared between the two atoms.

Both atoms don't have charge in the beginning and the compound remains with zero charge.

The chemical activity of an atom is determined by the number of electrons in its valence shell. With the help of concept of chemical bonding one can define the structure of a compound and is used in many industries for manufacturing products.

n which the true structure cannot be written at all.

Some other examples

(i) CO₃^2– ion

        

      Example
 (ii) Carbon-oxygen bond lengths in carboxylate ion are equal due to resonance.

  

 (iii) Benzene

         

  (iv) Vinyl Chloride

          

Difference in the energies of the canonical forms and resonance hybrid is called resonance stabilization energy and provides stability to species.

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

Regular Geometry

Number of electron pairs	Arrangement of electrons	Molecular geometry	Examples
2		B – A – B Linear	BeCl2, HgCl2
3	θ = 120°	θ = 120°	BF3, AlCl3
4			CH4, NH4+, SiF4
5			PCl5, PF5
6			SF6

 Irregular Geometry

Molecule Type	No. of Bonding pairs	No. of lone pair	Arrangement of electrons pairs	Shape (Geometry)	Examples
AB2E	2	1		Bent	SO2, O3
AB3E	3	1		Trigonal pyramidal	NH3
AB2E2	2	2		Bent	H2O
AB4E	4	1		See saw	SF4
AB3E2	3	2		T – shaped	CIF3
AB5E	5	1		Square pyramidal	BrF5
AB4E2	4	2		Square planar	XeF4

DIPOLE MOMENT

Difference in polarities of bonds is expressed on a numerical scale. The polarity of a molecule is indicated in terms of dipole moment (μ). To measure dipole moment, a sample of the substance is placed between two electrically charged plates. Polar molecules orient themselves in the electric field causing the measured voltage between the plates to change.

The dipole moment is defined as the product of the distance separating charges of equal magnitude and opposite sign, with the magnitude of the charge. The distance between the positive and negative centres called the bond length.

Thus, = μ = electric charge × bond length = q × d

As q is in the order of 10^–10 esu and d is in the order of 10^–8 cm, μ is the order of10^–18 esu cm. Dipole moment is measured in ‘Debye’ unit (D)

1D = 10^–18 es cm = 3.33 × 10^–30 coulomb metre

Note:

(i) Generally as electronegativity difference increase in diatomic molecules, polarity of bond between the atoms increases therefore value of dipole moment increases.  

(ii) Dipole moment is a vector quantity  

(iii) A symmetrical molecule is non-polar even though it contains polar bonds. For example, CO₂, BF₃, CCl₄ etc. because summation of all bond moments present in the molecules cancel each other.

(iv) Unsymmetrical non-linear polyatomic molecules have net value of dipole moment. For example, H₂O, CH₃OH, NH₃ etc.

Solved Examples

Example: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°.

Example: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.

Example: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 of 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 Ð of F₂O is less than H₂O.

 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 = Observed dipole moment/Calculated dipole moment assuming 100% ionic bond × 100

Example: 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^–10 m. 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: Dipole moment μ = e × d coulomb metre

              For KCl d = 2.6 × 10^–10 m

              For complete separation of unit charge

              e = 1.602 × 10^–19 C

              Hence μ = 1.602 × 10^–19 × 2.6 × 10^–10 = 4.1652 × 10^–29 Cm

              μ_KCl = 3.336 × 10^–29 Cm

              ∴ % ionic character of KCl = 3.336×10^–29/4.165×10^–29  = 80.09%

Example. Calculate the % of ionic character of a bond having length = 0.83 Å and 1.82 D as it’s observed dipole moment.

Solution: Tocalculate μ considering 100% ionic bond

                 = 4.8 × 10^–10 × 0.83 × 10^–8esu cm

              = 4.8 × 0.83 × 10^–18 esu cm = 3.984 D

              ∴ % ionic character = 1.82/3.984 × 100 = 45.68

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.

Rules for Writing Resonating Structures

Only electrons (not atoms) may be shifted and they may be shifted only to adjacent atoms or bond positions.
The number of unpaired electrons should be same in all the canonical form.

The positive charge should reside as far as possible on less electronegative atom and negative charge on more electronegative atom.

Like charge should not reside on adjacent atom

The larger the number of the resonating structures greater the stability of species. Greater number of covalency add to the stability of the molecule.

Example:
Out of the following resonating structures for CO₂ molecule, which are important for describing the bonding in the molecule and why?

         

 Solution:

Out of the structures listed above, the structure (III) is wrong since the number of electron pairs on oxygen atoms are not permissible. Similarly, the structures (II) has very little contribution towards the hybrid because one of the oxygen atoms (electronegative) is show to have positive charge. Carbon dioxide is best represented by structures (I) and (IV).

BOND CHARACTERISTICS

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).

[1Å = 10^–8 cm; 1 pm = 10^–12 m]

 Bond length in ionic compound = r_c⁺ + r_a^–

 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) = r_A + r_B

 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.

             sp³C – H = 1.112Å:                    sp²C – H = 1.103Å;                   spC – H = 1.08Å;                                                 

 (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 mol^–1 (200-400 kJ mol^–1). 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.

     C ≡ 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 thefollowing order:

       s < p < sp < sp² < sp³

           C – C          >         N – N         >          O – O

      (No lone pair)        (One lone pair)         (Two lone pair)

3. 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      109^o28'      120^o        180^o

        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.

FACTORS GOVERNING POLARIZATION AND POLARISABILITY (FAJAN’S RULE)

Cation Size: Smaller is the cation more is the value of charge density (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.

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 on 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.  

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 of 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.

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.
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⁺⁺ 2N^3–) 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.
Example:
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.

Example: 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 larger in size is having more polarisability and hence more covalent character.

Example:

Which compound from each of the following pairs is more covalent and why? 

(a) CuO or CuS                 (b) AgCl or AgI

(c) PbCl₂ or PbCl₄              (d) BeCl₂ or MgCl₂
Solution:
(a) CuS                           (b) AgI

(c) PbCl₄                         (d) BeCl­₂

HYDROGEN BONDING

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 – H^…Y 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,

     H            H            H

     |            |            |

     O — H — O — H — O — H —

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 
H-bond.

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.

Calculation of Resultant Bond Moments

Let AB and AC are two polar bonds inclined at an angle  θ their dipole moments are μ1 and μ2. Resultant dipole moment may be calculated using vectorial method. μ = √μ12 + μ22 + 2μ1μ2 cos θ
when θ = 0 the resultant is maximum μR = μ1 + μ2 when, θ = 180°, theresultant is minimum μR = μ1 ∼ μ2

Example. The compound which has zero dipole moment is

               (A) CH₂Cl₂        (B) NF₃          (C) PCl₃F₂        (D) ClO₂

Solution: (C)

Example. Sketch the bond moments and resultant dipole moment in

                (i) SO₂        (ii) Cis–C₂H₂Cl₂ and         (iii) trans-C₂H₂Cl₂

Solution:

              Resultant m = 0

Example. 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.

Example. 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.

Example: Compare the dipole moment of H₂O and F₂O.

Solution: Let’s draw the structure of both the 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.

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 results into a zero dipole. Whereas in cis isomer the dipoles do not cancel each other resulting in a net dipole.

Example: The molecule having largest dipole moment among the following is

               (A) CH₄               (B) CHCl₃

               (C) CCl₄              (D) CHI₃

Solution: (B)

Example. Compare the dipole moment of cis 1, 2 dichloroethylene and trans 1, 2 dichloroethylene.

Solution: 

In the trans compound the C-Cl bond dipoles are equal and at the same time acting in opposition cancel each other while in 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.

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

These can be evident from the following examples.

Example. 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.

Example.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.               

— H — O — H — O — H — O —

          |            |           |

        C₂H₅       C₂H₅      C₂H₅

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

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.

INTERMOLECULAR FORCES

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.

• : 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.

METALLIC BONDING

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.    

Example 32: 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)

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 denotation 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 = 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               :        σ_1s²

           Bond order                  :       ½ (2 – 0) = 1

           molecule                     :       Total no. of electrons = 1    

           Arrangement               :        σ_1s¹

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

He₂      molecule                     :       Total no. of electrons = 4    

           Arrangement               :       σ_1s²σ_1s^*2

           Bond order                  :       ½ (2 – 2) = 0

                                                   ∴ He₂ molecule does not exist.

           molecule                    :       Total no. of electrons = 3    

           Arrangement               :       σ_1s²σ_1s^*1

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

                                                   So He₂⁺  exists and has been detected in discharge tubes.

Li₂        molecule                    :       Total no. of electrons = 6    

           Arrangement               :       σ_1s²σ_1s^*2σ_2s²

           Bond order                  :       ½ (4 – 2) = 1

                                                   No unpaired e’s so diamagnetic

Be₂        molecule                    :       Total no. of electrons = 8    

           Arrangement               :       σ_1s²σ_1s^*2σ_1s²σ_1s^*2

           Bond order                  :       ½ (4 – 4) = 0

                                                   No unpaired e^–s so diamagnetic

B₂        molecule                    :       Total no. of electrons = 10   

           Arrangement               :       σ_1s²σ_1s^*2σ_2s²σ_2s^*2σ_2px²

           Bond order                  :       ½ (6 – 4) = 1 ∴diamagnetic

                                                   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

Example. 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  = σ_1s²                             

                 Bond order = 1

              H₂^– = σ_1s²σ_*1s¹             

              Bond order = 0.5

              He₂ = σ_1s²σ_1s^*2                   

                 Bond order = 0

              He₂^– = σ_1s²σ^*_1s²σ_2s¹       

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

              H₂^– and He₂^– are expected to have same stabilities.

Example: Which of the following species have the bond order same as ?

              (A) CN^–                                (B) OH^–

              (C) NO^–                                (D) CO⁺

Solution: In  no. of bonding electrons and anti-bonding electrons are 10 and 4 respectively. Therefore the bond order is 10–4/2  = 3. Out of those given only CN^– is isoelectronic with N₂. Therefore CN^– has the same bond order as N₂.

Hence (A) is correct

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 3p¹_x 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

               Alt txt : heteronuclei species

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 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).

Example: 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)