CHEMICAL BONDING 

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 

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

(b) Formation of gaseous anions 

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

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

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. 

Born Haber Cycle 

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 ΔHf. 

Example1. Calculate the lattice enthalpy of. Given that 

Enthalpy of formation of = -524 kJ

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

Sublimation energy of Mg = +2187 kJ

Vaporization energy of (I) = +31kJ

Dissociation energy of (g) = +193kJ

Electron gain enthalpy of Br(g) = -331 kJ

Solution:

or

or U = −524 − [2187 − 148 + 31 + 193 + 2 × (−331)]

= =

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.

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

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

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 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 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 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.

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.

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+

Lewis base       Lewis acid  

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

Comparison of ionic, covalent & coordinate compounds

Property	Ionic	Covalent	Coordinate
1. binding force	Between ions strong (coulombic)	Between molecules smaller (Vander Waal’s)	in between
2. mp/bp	High	less than ionic	in between
3. condition	conductor of electricity in fused state & in aqueous solution	bad conductor	Greater than covalent
4. solubility in polar        solvent (H2O)	High	Less	in between
5. Solubility in non polar solvent (ether)	Low	High	in between
6. Physical state	generally solid	liquid & gaseous	solid, liquid gas

 

Example 2. Which of the following statement is/are not true for  σ-bond.

1. It is formed by the overlapping of s − s or s − p orbitals
2. It is weaker than pi bond
3. It is formed when π−bond exists already.
4. It is resulted from partial overlapping of orbitals.

(A) 1, 2, 3, 4 (B) 2, 3 and 4

(C) 2 and 4 (D) 1, 2 and 4

Solution: (B)

 

Example 3. The types of bond present in ZnSO4.7H2O are only 

(A) Electrovalent and covalent

(B) Electrovalent and co-ordinate

(C) Electrovalent, Covalent and co- ordinate

(D) None of these 

Solution: (C)

Example 4. Classify the following bonds as ionic, polar covalent or covalent and give your answer 

(a) Si Si bond in Cl3 SiSiCl3 (b) SiCl bond in Cl3SiSiCl3

(c) CaF bond in CaF2 (d) NH bond in NH3

Solution: (a) Covalent due to identical electronegativity

(b) One electron pair is shared between Si & Cl and thus, covalent bond is expected but electron negativity of Cl is greater than that of Si & some polarity develops giving polar – covalent nature 

(c) Ionic since Ca completes its octet by transfer of two outershell electrons thus, completing their octets

Ca [Ar]4s2, F[He)2s22p5

(d) Polar covalent, explanation as in (b) 

Example 5. Arrange the bonds in order of increasing ionic character in the molecules:

LiF, K2O, N2, SO2 and ClF3

Solution: N2 < ClF3 < SO2 < K2O < LiF

 

Example 6. Arrange the following in order of increasing ionic character:

C−H, F−H, Br−H, Na−I, K−F and Li−Cl

Solution: C−H < Br−H < F−H < Na−I < Li−Cl < K−F

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 π bond as in H2C = CH2 

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

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

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 σ bonds and the   remainder gives the non-bonded electrons. So number of lone pair
  =.

•    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 34/8= 4∴ 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  = 3 σ 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 = 4 σ bonds  Hybridization = sp3 Shape = Tetrahedral
5. SF6 Total valence electrons  = 48 Requirement = 6 σ bonds  hybridization = sp3d2 shape  = octahedral / square  bipyramidal
6. BeF2 Total valence electrons : 16 Requirement : 2 σ bonds  Hybridization : sp Shape : Linear	F – Be – F
7. ClF3 Total valence electrons : 28 Requirement : 3 σ bonds + 2 lone pairs  Hybridization : 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 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 : 5 σ 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

[ 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.

12.     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

14. 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 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= 4 σ bonds  Hybridisation: sp3 Shape: Tetrahedral
17. Total valence electron = 26 Requirement = 3 σ bond + 1 lone pair  Hybridization: sp3 Shape: pyramidal
18. XeO2F2 Total valence electrons : 34 Requirement: 4 σ bonds +1 lone  pairs  Hybridization : sp3d Shape: Distorted TBP (sea-saw geometry)
19. XeO3 Total valence electrons : 26 Requirement: 3 σ bonds + 1 lone  pair Hybridization: sp3 Shape: Pyramidal
20. XeOF4 Total valence electrons : 42 Requirement: 5 σ bonds + 1 lone  pair Hybridization: sp3d2 Shape: square pyramidal.

 

Example 8. Predict the hybridization for the central atom in, ,

Solution: Total No. of V.E. =

So, hybridization =

=

So, hybridization of s =

So, hybridization of I =

 

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) (B)

(C) (D)

 

Solution: In. Xe atom has 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.

(i) SOF2, (ii) SO2, (iii) POCl3, (iv) I3–

 

Exercise 8.

The type of hybridisation of orbitals employed in the formation of SF6 molecule is ………….. 

Exercise 9.

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

 

Exercise 10.

The bond angle in 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

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
3	θ = 120°	θ = 120°
4
5
6

Irregular Geometry

Molecule Type	No. of Bonding pairs	No. of lone pair	Arrangement of electrons pairs	Shape (Geometry)	Examples
	2	1		Bent
	3	1		Trigonal pyramidal
	2	2		Bent
	4	1		See saw
	3	2		T – shaped
	5	1		Square pyramidal
	4	2		Square planar

Example 13. 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 14. 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 15. 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 ∠ of F2O is less than H2O.

 

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 in which the true structure cannot be written at all.

Some other examples

(i) CO32– ion

(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.

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 17. 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).

 

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++ 2N3–) 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 18. 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  19. 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 20. 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

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. 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, 

=

As q is in the order of esu and d is in the order of cm, is the order of. Dipole moment is measured in ‘Debye’ unit (D)

 

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, 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, etc. 

Calculation of Resultant Bond Moments

Let AB and AC are two polar bonds inclined at an angle their dipole moments are and. Resultant dipole moment may be calculated using vectorial method.

when θ = 0 the resultant is maximum when,  = 180°, the resultant is minimum

 

Example 21. The compound which has zero dipole moment is

(A) CH2Cl2 (B) NF3

(C) PCl3F2 (D) ClO2

Solution: (C)

 

Example 22. Sketch the bond moments and resultant dipole moment in 

(i) (ii) and                 (iii)

Solution:

Resultant μ = 0

Example 23. 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 24. 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 25. 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 26. The molecule having largest dipole moment among the following is

(A) CH4 (B) CHCl3

(C) CCl4 (D) CHI3

Solution: (B)

Example 27. 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. 

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 =

Example 28. 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

μKCl = 3.336 × 10–29 Cm

∴ % ionic character of KCl = = 80.09%

Example 29. Calculate the % of ionic character of a bond having length = 0.83 Å and 1.82 D as it’s observed dipole moment.

Solution: To calculate 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 =

 

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.

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 units or picometer (pm). 

Bond length in ionic compound =

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) =

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. 

                         

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

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

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

Important features

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

(decreasing bond length)

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

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

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

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

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

(viii) Bond energy increase in the following order: 

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

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. 

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,

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.

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

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

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

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.
•    Induced dipole interaction: These generate from the polarisation of a neutral molecule by a charge or ion.

•    Instantaneous dipole-induced dipole interaction: In non polar molecules dipoles may generate due to temporary fluctuations in electron density. These transient dipole can now induce dipole in neighbouring molecules producing a weak temporary interaction.

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 

=

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 e–s so diamagnetic

B2 molecule : Total no. of electrons = 10

Arrangement :

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 33. Compare the bond energies of O2, &

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

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 >

Example  34. 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  = σ1s2

Bond order = 1

H2– = σ1s2σ*1s1

Bond order = 0.5

He2 = σ1s2σ1s*2

Bond order = 0

He2– = σ1s2σ*1s2σ2s1

Bond order = = 0.5

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

 

Example 35. Which of the following species have the bond order same as?

(A) (B)

(C) (D)

 

Solution. In no. of bonding electrons and anti−bonding electrons are 10 and 4 respectively. Therefore the bond order is = 3. Out of those given only is isoelectronic with. Therefore has the same bond order as

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 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

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 36. 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)

 

MISCELLANEOUS EXERCISES 

 

Exercise 1: Compare the bond energies of N2, &

 

Exercise 2: Though the electronegativities of nitrogen and chlorine are same, NH3 exists as liquid whereas HCl as gas. Why?

 

Exercise 3: Explain giving reason:

is linear, but the ion is bent

 

Exercise 4: Explain with reason:

Two different bond lengths are observed in PF5 but only one bond length observed in SF6.

 

Exercise 5: (a) Amongst BBr3 and BF3 which is a stronger acid and why?

 

Exercise 6: Explain why the measured resultant dipole moment for FNO is 1.81 D, is so much higher than the value for nitryl fluoride FNO2 (0.47 D).

 

Exercise7: Why

is liquid at room temperature while

is a high melting solid?

 

Exercise 8: (i) Among the compounds CH3COOH, NH3, HF and CH4 in which maximum hydrogen bonding is present?

(ii) Which one of the following has strongest bond?

HF, HCl, HBr, HI

 

Exercise 9: Why BeF2 and BF3 are stable though Be and B have less than 8 electrons? Which one is more stable?

 

Exercise10: Ether

and water

have same hybridisation at oxygen. What angle would you expect for them?

ANSWER TO MISCELLANEOUS EXERCISES 

 

Exercise 1: The configuration of N2 is

Now means removal of an electron from a bonding 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 electronic repulsion and thereby shortens the bond length. So the bond energy becomes more than compared to

∴N2 > >

 

Exercise 2: The size of nitrogen is less than the size of chlorine. Therefore, electron density in nitrogen is more than that of chlorine. So, nitrogen forms hydrogen bonding leading to association of molecules. Hence, NH3 is liquid. Hydrogen bonding is not possible with chlorine.

Exercise 3: In central chorine atom involves sp3d hybridisation, to have minimum electronic repulsion three lone pairs should be in equitorial position as follows; giving linear shape to the ion. Whereas, in case of ion central atom Cl involves sp3 hybridisation having two lone pairs, resulting in bent shape for the ion, (bond angle less than 109°28′ due to repulsion of bond pair by lone pair) 

Exercise 4: PCl5 has trigonal pyramidal structure (sp3d hybridisation of central atom) in which bond angles are 90° and 120° respectively and there are two types of bond axial and equatorial. In case of SF6 the structure is octahedral (sp3d2 hybridization of the central atom – S) resulting only one type of bond, bond angle (90°) and one type of bond length.

Exercise 5: (a) BBr3 since back bonding is present in BF3.

 

Exercise 6: Molecular symmetry (in terms of bond angles) leads to lesser dipole moment.

 

Exercise 7: (i) shows intramolecular hydrogen bonding while 

(ii) shows intermolecular hydrogen bonding.

 

Exercise 8: (i) HF due to maximum electronegativity of fluorine.

(ii) HF

 

Exercise 9: The stability is explained by symmetrical linear structure for BeF2 and triangular planar structure for BF3. BeF2 is more stable because its greater bond angle (180o).

 

Exercise 10: In H2O bond angle is less than 109o.28’ due to lone pair and bond pair repulsion. But in ether, due to strong mutual repulsion between two alkyl groups bond angle becomes greater than 109o.28’.

 

SOLVED PROBLEMS

 

Subjective: 

 

Board Type Questions

 

Prob 2. Arrange the bonds in order of increasing ionic character in the molecules:

LiF, K2O, N2, SO2 and ClF3

 

Sol. N2 < ClF3 < SO2 < K2O < LiF

 

Prob 3. Arrange the following in order of increasing ionic character:

C−H, F−H, Br−H, Na−I, K−F and Li−Cl

 

Sol. C−H < Br−H < F−H < Na−I < Li−Cl < K−F

 

IIT Level Questions 

 

Prob 6. Predict the shapes of the following ions

(a) BeF3–                  (b) BF4–

(c) IF4–                     (d) IBr2–

 

Sol. (a) Triangular

(b) Tetrahedral 

(c) Square planar 

(d) Linear

Prob 8. Arrange the following in increasing order of stability O2, O2+, O2–, O22- 

 

Sol. O22- < O2– < O2 < O2+

Calculate first the bond order which is as follows 

O2 → 2, O2+ → 2.5, O2– → 1.5, O22- →1 & then arrange according to increasing bond order.

 

Prob 11. Arrange the following:

(i) N2, O2, F2, Cl2 in increasing order of bond dissociation energy.

(ii) Increasing strength of hydrogen bonding (X – H – X): 

O, S, F, Cl, N

Sol. (i) F2 < Cl2 < O2 < N2  (ii) Cl < S < N < O < F

Prob 12. Explain the following

o – hydroxy benzaldehyde is liquid at room temperature while p – hydroxy benzaldehyde is high melting solid.

Sol. There is intramolecular H bonding in o – hydroxy benzaldehyde while intermolecular hydrogen bonding in p-hydroxy benzaldehyde

 

Prob 14. Explain why ClF2– is linear but ClF2+ is a bent molecular ion?

 

Sol. Chlorine atom lies in sp3d hybrid state. Three lone pairs are oriented along the corners of triangular plane 

Chlorine atom lies in sp3 hybrid state. Two lone pairs are oriented along two corners of tetrahedral  

 

Prob 16. AlF3 is ionic while AlCl3 is covalent.

 

Sol. 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.

 

Prob 18. Write down the resonance structure of nitrous oxide

 

Sol.

 

Prob 20. Explain why BeH2 molecule has zero dipole moment although the Be−H bonds are polar.

Sol. BeH2 is a linear molecule (H−Be−H) with bond angle equal to 180o. Although the Be−H bonds are polar on account of electronegativity difference between Be and H atoms, the bond polarities cancel with each other. The molecule has resultant dipole moment of zero.

 

1. What are the factors influencing ionic bond formation?
2. Out of MgO and NaCl, whech has higher lattice energy and why?
3. BaSO4 being an electrovalent compound and still it does not pass into solution state in water.
4. Why an ionic bond is formed between two elements having large difference in their electronegativity?

 

COVALENT BOND :

6. Which compound has 3 carbon atoms, 4 double bonds and the total no. of atoms are five?
7. Which compound has 3 carbon atoms. 4 σ and 4π bonds but not 4 double bonds and the total no. of atoms are 5. Writes its IUPAC 
8. Element A has 3 electrons in the valency shell and its principal quantum no. for the last electron is 3 and element B has 4 electrons in the valency shell and its principal quantum to. For the last electrons is 2. Identify the compound and it’s nature of bonding.
9. The compound X has 8 atoms, 4 σ bonds, no π bonds, no. ionic bonds, no coordinate bonds, no H-bond. Explain its structure.
10. The compound X has 8 atoms, 0 σ bonds, no ionic or π bonds. Explain its structure.
11. SnCl4 has melting point –15ºC where as SnCl2 has melting point 535ºC. Why?
12. Inorganic benzene is more reactive than organic benzene. Why?

 

FAJAN’S RULE :

13. SnCl2 is white but SnI2 is red. Why?
14. Explain the least melting point and highest solubility in H2O.

(i) LiCl NaCl KCl

(ii) NaCl NaBr Nal

(iii) LiCl BeCl2 BCl3

(iv) beSO4 MgSO4 CaSO4 SrSO4 BaSO4

(v) CaF2 CaCl2 CaBr2

15. Which one has highest and lowest melting point and why?

NaCl KCl RbCl CsCl

16. LiOH and carbonates decomposes on heating in I-group. Other hydroxides and carbonates of this group will not. Why?

 

LEWIS STRUCTURE AND FORMAL CHARGE :

17. Draw the Lewis structures of the following molecules and ions.

PH3, H2S, BeF2, SiCl4, N2O4, H2SO4, O22, IO65-.

 

V.B.T. & HYBRIDISATION :

20. Explain hydrbidisation in

(1) XeF2 (2) XeF6 (3) PCl3 (4) IF3

(5) IF5 (6) IF7 (7) CCl4 (8) SiCl4

(9) SlH4 (10) H2O

22. PH5 is not possible but PCI5 is possible. Why?

 

VSEPR THEORY :

24. Which one has highest and least bond angle in the following –

(1) NH3 PH3 AsH3 SbH3

(2) CH4 PH3 AsH3 SbH3

(3) H2O H2S H2Se H2Te

(4) CH4 PH3 AsH3 H2Te

(5) CH4 SiH4 CCl4 SiCl4

(6) NCl3 PCl3 NBr3 PBr3

(7) PF3 PH3

(8) As3F3 AsH3

 

MOT :

 

29. Super oxide are coloured and paramagnetic why?
30. Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
31. Exalain why NO+ is more stable towards dissociation into its atoms than NO, where as CO+ is less stable than CO.

 

BACK BONDING AND HYDROLYSIS :

40. Arrange the following boron trihalides in the increasing order of their ease of hydrolysis. Also give the reason for the same.

BF3, BCl3, BBr3

 

DIPOLE MOMENT :

53. Why NH3 is having more dipole moment than NF3.
54. Why  is having more dipole moment than /
55. Why p-dichloro benzene in having zero dipole moment while hydroquinone is having some dipole moment?
56. Why CH3 Cl is having high dipole moment than CH3F?
57. Why ortho fluoro Phenol have greater dipole moment than ortho chloro phenol?
58. Why trans–1, 2 sichloro ethene in having zero dipole moment than cis form?
59. Write the order of dipole moments of 1, 2–1.3– and 1, 4–dichlorobenzene.
60. Arrange in increasing order of dipole moment ; H2O, H2S, BF3.
61. BcF2 has zero dipole moment whereas H2O has a dipole moment?
62. CCl4 having zero dipole momnet but CHCl3 having some dipole moment?
63. While down the resonance structure (S) for :

(i) N3– (ii) O3 (iii) CO2

(iv) N2O4 (v) SCO22–

 

1. Arrange the following in increasing order of property given-

(i) O, F, S, Cl, N strength of H-bonding (X-H_X)

(ii) N2, O2, F2, Cl2 bond dissociation energy

(iii) MCl, MCl2, MCl3 ionic nature

(iv) HI, HBr, HCl, HF dipole moment

(v) AsH3, PH3, NH3 bond angle

5. Explain the structure hybridisation and oxidation state of S i sulphuric acid, marshall’s acid, caro’s acid, oleum.
6. Boric acid is monobasic acid. Why?
7. Boron has exceptionally high melting point. Why?
8. BCl3 is more acidic than BF3. Why?
9. CCl4 is not dissolved in H2O but SiCl4 dissolves. Why?
10. Trimethylamine (CH3)3 N, is pyramidal but trisylyamine (SiH3)3 N is planer. Why?
11. SnCl4 has melting point –15ºC Where as SnCl2 has melting point 535ºC.Why?
12. PbCl4 is possible but PbBr4 and Pbl4 are not. Why?
13. Pb+4, Bi+5 and Tl+3 act as oxidising agent. Why?
14. NCl3 & PCl3 on hydrolysis will give different products. Why?
15. ClO2 does not forms dimer but NO2 forms. Why?
16. How many σ and π bonds are presents in hexacyanoethane and tetra cyanoethylene?
17. Explain the structure of ClF3 on the basis of bent rule.
18. All bonds length of PCl5 are not equal but PF5 has same bond lengths. Why?
19. The experimentally determined N – F bond length in NF3 is greater than sum of single bond covalent radii of N and F.
20. The dipole moment of HBr is 7.95 debye and the intermolecular separation is 1.94 × 10-10 m Find the % ionic character in HBr molecule.
21. HBr has dipole moment 2.6 × 10-30 cm. If the ioninc character of the bond is 11.5%, calculate the interatomic spacing.
22. Dipole moment of LiF was experimentally determined and was found to be 6.32 D. Calculate percentage ionic character in LiF molecule Li – F bond length is 0.156 pm.
23. A diatomic molecule has a dipole moment of 1.2 D. If bond length is 1.0 A, what percentage of an electronic charge exists on each atom.

 

SECTION (B) FILL IN THE BLANKS :

1. Two atoms of similar electronegativity are expected to form __________ compound.
2. When two atoms approach each other, potential energy ______ and a ___________ is formed between them.
3. Conversion of a neutral atom into a cation is _________ process.
4. The strongest hydrogen bond is formed between _________ and hydrogen.
5. Low ionization potential of electropositive element and high electron affinity of electronegative element favours the formation of __________ bond.
6. NaCl is soluble in water due to its low ___________ energy.
7. The ________ value of lattice energy of a crystal favours the formation of an ionic compound.
8. Solid NaCl ___________ conductor of electricity.
9. For dissolution of an ionic solid, ________ energy should be low and hydration energy should be _________________.
10. ___________ cation and ___________ anion favour covalency.
11. Anhydrous AlCl3 is a _____________ compound while hydrated AlCl3 is ___________.
12. Covalent compounds are generally _________ conductors of electricity.
13. There are ___________ π bonds in a nitroge molecule.
14. A double bond is shorter than ___________ bond.
15. Axial overlapping of half-filled atomic orbitals results in __________ bond.
16. ____________ and ___________ bonds are present in N2O5 molecule.
17. The angle between two covalent bonds is maximum in _________ (CH4, H2O, CO2).
18. ___________ hybrid orbital of nitroger atom are involved in the formation of ammonium ion.
19. The hybridization state of oxygen in water molecule is ___________.
20. In the ion [Cu (H2O)4]++, copper is in dsp2 state of hybridization. The shape of the ion is ________.

 

TRUE OR FALSE :

5. A diatomic molecule has a dipole moment of 1.2 D. If bond length is 1.0 a, what percentage of an electronic charge exists on each atom.
6. The size of negative ion decreases with increasing magnitude of negative charge.
7. Ca2+ is smaller in size than K+ because the effective nuclear charge is greater.
8. The higher the latteice energy of an ionic solid, the greater will be its stability.
9. Molten sodium chloride conducts electricity due to the presence of free ions.
10. Linear overlap of two atomic p-rbitals leads to a σ-bond.
11. The H-N bond angle in NH3 is greater than the H-AS-H bond angle in AsH3.
12. sp2 hybrid orbitals have equal s and p-character.
13. The tetrahedral geometry in SiF4 is due to sp3 hybridization of Si atom.
14. SnCl2 is a non-linear molecule.
15. There are seven electron bond pairs in lF7 molecules.
16. Dipole moment of CHF3 is greater than CHCl3.
17. Dipole moment of NF3 is lesser than NH3.
18. Among HF, HCl, HBr and HI, HF has highest dipole moment.
19. All molecules with polar bonds have dipole moment.
20. The presence of polar bonds in a polyatomic molecule suggests that the molecule has non zero dipole moment.
21. AgCl is more covalent than NaC.

 

REASONING AND ASSERTION :

Direction: These quaestions consist of two statements each printed as Assertion and Reason.While answering these questions you are required to choose any one of the following four responses to encircle (A, B, C, D) as per instructions given below:

(A) If both Assertion and Reason are true and Reason is a correct explanation of Assertion.

(B) If both Assertion and Reason are true and Reason is not a correct explanation of Assertion.

(C) If Assertion is true but Reason is false.

(D) If Assertion is false but Reason is true.

3. Assertion : Na2SO4 is soluble in water while BaSO4 is insoluble.

Reason :  Lattice energy of BaSO4 exceeds its hydration energy.

(a) A (b) b (c) C (d) D

4. Assertion : py and pz cannot combine to give molecular orbital.

Reason :  Both py and pz are dumb-bell shaped.

(a) A (b) b (c) C (d) D

7. Assertion : Although PF5, PCl5 and PBr5 are known, the pentahalides of nitrogen have not been observed.

Reason :  Phosphorus has lower electronegativity than nitrogen.

(a) A (b) b (c) C (d) D

9. Assertion : NO3– is planar while NH3 is pyramidal.

Reason :  N in NO3– is sp2 and in NH3 it is sp3 hybridized.

(a) A (b) b (c) C (d) D

11. Assertion : s-orbital cannot accommodate more than two electrons.

Reason :  s-orbitals are extremely poor shielders.

(a) A (b) b (c) C (d) D

 

MATCH THE FOLLOWING :

1. Column – I Column – II

(A) Size of secondary layer of hydrated ions (P)Maximum in solid and minimum in gaseous state.

(B) Magnitude of hydrogen bonding (Q) Strength of ion – dipole attraction

(C) Mobility of ions in water (R) Inversely proportional to the size of metal ion

(D) Degree of polarity of a bond (S) Dipole moment

(T) Directly proprotional to the size of metal ion

2. Column – I Column – II

(A) SO3 (gas) (P)Polar with pπ – dπ bonds and identical S – O bond, lengths.

(B) OSF4 (Q) One lone pair and pπ – dπ bond.

(C) SO3F– (R) non – polar with pπ –pπ and pπ – dπ bonds. Identical S–O bod lengths.

(D) ClOF3 (S) Polar with pπ – dπ bond.

(T) Hybridisation of central atom in ClO2F3.

3. Column – I Column – II

Molecule/ion Hybridisation of central atom

(A) IO2F2– (P) sp3d

(B) F2SeO (Q)  sp3

(C) ClOF3 (R)  sp2

(D) XeF5+ (S) sp2

 

Only one Answer Correct:

1. In which molecule (s) is/are the vander waals force likely to be most important in determining m.p. and b.p.

(a) ICI (b) Br2 (c) H2S (d) CO

2. Which is the most ionic

(a) LiF (b) Li2O (c) Li3N (d) All same

3. The correct order of the increasing ionic character is-

(a) BeCl2 < MgCl2 < CaCl2 < BaCl2 (b) BeCl2 < MgCl2 < BaCl2 < CaCl2

  (c) BeCl2 < BaCL2 < MgCl2 < CaCl2 (d) BaCl2 < MgCl2 < CaCl2 < BeCl2

4. An ionic bond A+ B– is most likely to be formed when :

(a) the ionization energy of A is high and the electron affinity of B is low

(b) the ionization energy of A is high and the electron affinity of B is low

  (c) the ionization energy of A and the electron affinity of B is high

(d) the ionization energy of A and the electron affinity of B is low

5. Which of the following compounds of elements in group IV is expected to the most ionic?

(a) PbCl2 (b) PbCl4 (c) CCl4 (d) SiCl4

6. The molecule BF3 borth are covalent compounds. But BF, is non,polar and NF3 is polar. The reason is that-

(a) boron is a metal and nitrogen is a gas in uncombined state

(b) B – F bond have no dipole moment whereas N – F bond have dipole moment

  (c) atomic size of boron is smaller than that of nitrogen

(d) BF3 is symmetrical molecule where as NF3 is unsymmetrical

7. Least melting point is shown by the compound-

(a) PbCl2 (b) SnCl4 (c) NaCl (d) AlCl3

8. Which of the following is in rder of increasing covalent character?

(a) CCl4 < BeCl2 < BCl3 < LiCl (b) LiCl < CCl4 < BeCl2 < BCl3

  (c) LiCl < BeCl2 < BCl3 < CCl4 (d) LiCl < BeCl2 < CCl4 < BCl3

9. Which of the following compounds contain/s both ionic and covalent bonds?

(a) NH4Cl (b) KCN (c) CuSO4.5H2O (d) NaOH

10. The tyes of bonds present in CuSO4.5H2O are

(a) electrovalent and covalent (b) electrovalent and coordinate

  (c) covalent and coordinate (d) electrovalent, covalent and coordinate

11. Three centre-two electron bonds exist in :

(a) B2H6 (b) Al2(CH3)6 (c) BeH2(s) (d) BeCl2(s)

 

Octet Rule :

12. Example of super octet molecule is :

(a) SF6 (b) PCl5 (c) IF7 (d) All of these

13. To which of the following species octet rule is not applicable :

(a) BrF5 (b) SF6 (c) IF7 (d) CO

14. NH3 and BF3 combine readily because of the formation of :

(a) a covalent bond (b) hydrogen bond (c) acoordinate bond (d) an ionic bond

15. Which of the following species cotain covalent and ccordincate bond.

(a) AlCl3 (b) CO (c) [Fe(CN)6]4– (d) N–3

 

Fajans Rule

16. Which of the following combination of ion will have highest polarisation

(a) Fe2+, Br– (b) Ni4+, Br– (c) Ni2+, Br– (d) Fe, Br–

17. Which of the following cannot be explained on the basis of Fajan’s Rules.

(a) Ag2S is much less soluble than Ag2O

(b) Fe(OH)3 is much less soluble than Fe(OH)2

  (c) BaCO3 is much less soluble than MgCO3

(d) Melting point of AlCl3 is much less than that of NaCl

18. The correct order of decreasing polarizability of ion is :

(a) Cr–, Br–, I–, F– (b) F–, I–, Br–, Cl– (c) I–, Br–, Cl–, F– (d) F–, Cl–, Br–, I–

19. Which ion has a higher polarising power

(a) Mg2+ (b) Al3+ (c) Ca2+ (d) Na+

20. Which combination will show maxinum polarising power & maxinum polarisability

(a) Mn2+.F– (b) Mn7+, I– (c) Mn2+, I– (d) Mn7+, F–

 

Lewis Structure :

21. The possible structure (s) of monothiocarbonate ion is :

 

(a) (b)  (c) (d) 

 

22. Pick out among the following species isoelectronic with CO2.

(a) N–3 (b) (CNO)– (c) (NCN)2– (d) NO–2

 

V.B.T. & Hybridisation :

23. The strength of bonds by s – s, p – p, s – p overlap is in the order :

(a) s – s < s – o < p – p (b) s – s < p – p < s – p

  (c) s – p < s – s < p – p (d) p – p < s – s < s – p.

24. Number and ype of bonds between two carbon stoms in CaC2 are :

(a) one sigm3 (σ) and one pi (π) bond (b) one σ and two π bonds

  (c) one σ and one and a half π bond (d) one σ bond

25. In the context of carbon, which of the follwoing is arranged in the correct order of electronegativity:

(a) sp > sp2 > sp3 (b) sp3 > sp2 > sp (c) sp2 > sp > sp3 (d) sp3 > sp > sp2

26. Which of the following oxyacids of sulphur contain ‘S – S bonds?

(a) H2S2O8 (b) H2S2O6 (c) H2S2O4 (d) H2S2O5

27. Which of the following represent the given mode of hybridisation sp2 – sp2 – sp – sp from left to right

(a) H2C = C = c = CH2 (b) HC ≡ C – C ≡ CH 

(c) H2C = CH – C ≡ N (d) H2C = CH – C ≡ CH

28. For BF3 molecule which of the following is true.

(a) B-atom is sp2 hybridised

(b) There is a Pπ –Pπ back bonding in this molecule

  (c) Observed B–F bond length is found to be less than the expected bond length.

(d) All of these

29. Which of the following is isoelectronic as well as isostructura’ with N2O

(a) N3H (b) H2O (c) NO2 (d) CO2

30. The hybridization state of B in B2H6 is –

(a) sp (b) sp2 (c) sp3 (d) sp3d

31. What is not true for SiH4 molecule –

(a) Tetrahedral hybridisation (b) 109º angle

  (c) 4σ bond (d) 4-lone pair of electrons

32. Which of the following has a geometry different from the other three species (having the same geometry)?

(a) (b)  (c) XeF4 (d) 

33. In C – C bond C2H6 undergoes heterolytic fission, the hybridisation of two resulting carbon atoms is/are

(a) sp2 both (b) sp3 both (c) sp2, sp3 (d) sp, sp2

34. Which of the following statements are not correct?

(a) Hybridization is the mixing of atomic orbitals of large energy difference.

(b) sp2 – hybrid orbitals are formed form two p – atomic orbitals and one s-atomic orbitals

  (c) dsp2 – hybrid orbitals are all at 90º to one another

(d) d2sp3 – hybrid orbitals are directed towards the corners of a regular octahedron

35. pπ – dπ back bonding occurs between oxygen and

(a) phosphorus in P4O10 (b) chlorine in HClO4 

(c) nitrogen in N2O5 (d) carbon in CO2

36. In which of the following groups all the the members have linear shape

(a) NO2+, N3–, H – C – H (b) N3–, I3–, H – C – H

  (c) XeF2, C2H2, SO2 (d) CO2, BeCl2, SnCl2

37. Consider the following molecules :

H2O        H2S         H2Se          H2Te

  I            II            III            IV

Arrange these molecules in increasing order of bond angles.

(a) I < II < III < IV (b) IV < III < II < I (c) I < II < IV < III (d) II < IV < III < I

38. Which has the smallest bond angle (X – O – X) in the given molecules?

(a) OSF2 (b) OSCl2 (c) OSBr2 (d) OSI2

39. Consider the following iodides :

  PI3              AsI3            SbI3

102º            100.2º           99º

The bond angle is maximum in Pl3, which is

(a) due to small size of phosphorus (b) due to more bp–bp repulsion in PI3 

(c) due to less electronegativity of P (d) none of these

 

Bonds angles & Bond Length :

40. The correct order of increasing X – O – X bond angle is (X = H, F or Cl):

(a) H2O > Cl2 > F2O (b) Cl2O > H2O > F2O

  (c) F2O > Cl2O > H2O (d) F2O > H2O > Cl2O

41. Which of the following compounds have bond angle as nearly 90º?

(a) NH3 (b) H2S (c) H2O (d) SF6

42. Number of non bonding electrons in N2 is :

(a) 4 (b) 10 (c) 12 (d) 14

43. Which of the following species is paramagnetic ?

(a) NO– (b)  (c) CN– (d) CO

44. Tho bond order depends on the number of electrons in the bonding and non bonding orbitals. Which of the following statements is/are correct about bond order?

(a) Bond order cannot have a negative value

(b) It always has an integral value

  (c) It is a nonzero quantity

(d) It can assume any value-positive or negative, integral or fractional, including zero

45. In the formation of  from N2, the electron is removed from :

(a) σ orbital (b) π orbital (c) σ* orbital (d) π* orbital

46. Which of the following has fractional bond order :

(a) (b)  (c) (d) 

47. How many unpaired electrons are present in  :

(a) 1 (b) 2 (c) 3 (d) 4

48. Which of the following have identical bond order?

(a) (b)  (c) NO (d) 

49. Which of the following have identical bond order?

(a) (b) NO+ (c) CN– (d) CN+

50. Given the species : N2, CO, CN– and NO+, Which of the following statements are true for these

(a) All species are paramagnetic (b) The species are isoelectronic

  (c) All the species have dipole moment (d) All the species are linear

51. Which of the following are paramangetic?

(a) B2 (b) O2 (c) N2 (d) He2

52. Which of the following species have a bond order of 3?

(a) CO (b) CN– (c) NO+ (d) 

 

53. Which of the following pairs have identical values of bond order?

(a)  and (b) F2 and Ne2 (c) O2 and B2 (d) C2 and N2

54. Find out the bond order of :

(a) H2 (b)  (c) He2 (d) Li2               (f) B2

55. The correct order of boiling point is :

(a) H2O < H2S < H2Se < H2Te (b) H2O > H2Se > H2Te > H2S

  (c) H2O > H2S > H2Se > H2Te (d) H2O > H2Te > H2Se > H2S

56. Which of the following models best describes the bonding within a layer of the graphite structure?

(a) metallic bonding (b) ionic bonding 

(c) non-metallic covalent bonding (d) van der Waals forces

57. The critical temperature of water is higher than that of O2 because the H2O molecule has :

(a) fewer electrons than O2 (b) two covalet bonds

  (c) V – shape (d) dipole moment

58. Arrange the following in order of decreasing boiling point :

(I) n-Butane (II) n-Butanol (III) n-Butyl chloride (IV) Isobutane

(a) IV > III > II > I (b) IV > II > III > I (c) I > II > III > IV (d) II > III > I > IV

59. For H2O2, H2S, H2O and HF, the correct order of increasing extent of hydrogen bonding is :

(a) H2O > HF > H2O2 > H2S (b) H2O > HF > H2S > H2O2

  (c) HF > H2O > H2O2 > H2S (d) H2O2 > H2O > HF > H2S

60. Which one of the following does not have intermolecular H-bonding?

(a) H2O (b) o-nitro phenol (c) HF (d) CH3COOH

61.

(a) has intermolecular H – bonding (b) has intramolecular H-bonding

  (c) has low boilng point (d) is steam-volatile

62. Which of the following is/are observed in metallic bonds?

(a) Mobile valence electrons (b) Overlapping valence orbitals

  (c) Highly directed bond (d) Delocalized electrons

63. Intermolecular hydrogen bonding increases the enthalpy of vapourization of a liquid due to the 

(a) decrease in the attraction betwwen molecules

(b) increase in the attraction between molecules

  (c) decrease in the molar mass of unassociated liquid molecules

(d) increase in the effective molar mass of hydrogen – bonded molecules

64. Of the following molecules, the one, which has permanent diple moment, is-

(a) SiF4 (b) BF3 (c) PF3 (d) PF5

65. The dipole moments of the given molecules are such that-

(a) BF3 > NF3 > NH3 (b) NF3 > BF3 > NH3 (c) NH3 > NF3 > BF3 (d) NH3 > BF3 > NF3

66. Which of the following has the least dipole moment

(a) NF3 (b) CO2 (c) SO2 (d) NH3

67. Which of the following compounds possesses zero dipole moment?

(a) Water (b) Benzene 

(c) Carbon tetrachloride (d) Boron trifluoride