1styear CHEMISTRY Notes Chapter-6

Chapter-6
CHEMICAL EQUILIBRIUM

Chemical Equilibrium
Reversible Reactions
Those chemical reactions which take place in both the directions and never proceed to completion are called Reversible reaction.
For these type of reaction both the forward and reverse reaction occur at the same time so these reaction are generally represented as
Reactant ≅Product
The double arrow ≅indicates that the reaction is reversible and that both the forward and reverse reaction can occur simultaneously.
Some examples of reversible reactions are given below
1. 2Hl ≅H2 + l2
2. N2 + 2 H2 ≅2 NH3

Irreversible Reactions
Those reactions in which reactants are completely converted into product are called Irreversible reaction.
These reaction proceed only in one direction. Examples of such type of reaction are given below
1. NaCl + AgNO3 ----> AgCl + NaNO3
2. Cu + H2SO4 ----> CuSO4 + H2

Equilibrium State
The state at which the rate of forward reaction becomes equal to the rate of reverse reaction is called Equilibrium state.

Explanation
Consider the following reaction
A + B ≅C + D
It is a reversible reaction. In this reaction both the changes (i.e. forward & backward) occur simultaneously. At initial stage reactant A & B are separated from each other therefore the concentration of C and D is zero.
When the reaction is started and the molecules of A and B react with each other the concentration of reactant is decreased while the concentration of product is increased. With the formation of product, the rate of forward reaction decreased with time but the rate of reverse reaction is increased with the formation of product C & D.
Ultimately a stage reaches when the number of reacting molecules in the forward reaction equalizes the number of reacting molecules in the reverse direction, so this state at which the rate of forward reaction becomes equal to the rate of reverse reaction is called equilibrium state.

Law of Mass Action


Statement
The rate at which a substance reacts is proportional to its active mass and the rate of a chemical reaction is proportional to the product of the active masses of the reactant.
The term "active mass" means the concentration in terms of moles/dm3

.Derivation of Equilibrium Constant Expression
Consider in a reversible reaction "m" mole of A and "n" moles of B reacts to give "x" moles of C and "y" moles of D as shown in equation.
mA + nB ≅xC + yD
In this process
The rate of forward reaction 8 [A]m [B]n
Or
The rate of forward reactin = Kf [A]m [B]n
&
The rate of reverse reaction 8 [C]x [D]y
Or
The rate of reverse reaction = Kf [C]x [D]y
But at equilibrium state
Rate of forward reaction = Rate of reverse reaction
Therefore,
Kf [A]m [B]n = Kf [C]x [D]y
Or
Kf / Kr = [C]x [D]y / [A]m [B]n
Or
Ke = [C]x [D]y / [A]m n
This is the expression for equilibrium constant which is denoted by Ke and defined as
The ratio of multiplication of active masses of the products to the product of active masses of reactant is called equilibrium constant.

Equilibrium Constant for a Gaseous System
Consider in a reversible process, the reactants and product are gases as shown
A(g) + B(g) ? C(g) + D(g)
When the reactants and products are in gaseous state, their partial pressures are used instead of their concentration, so according to law of mass action.

Determination of Equilibrium Constant
The value of equilibrium constant K(C) does not depend upon the initial concentration of reactants. In order to find out the value of K(C) we have to find out the equilibrium concentration of reactant and product.

1. Ethyl Acetate Equilibrium
Acetic acid reacts with ethyl alcohol to form ethyl acetate and water as shown
CH3COOH + C2H5OH ≅CH3COOC2H5 + H2O
Suppose 'a' moles of acetic acid and 'b' moles of alcohol are mixed in this reaction. After some time when the state of equilibrium is established suppose 'x' moles of H2O and 'x' moles of ethyl acetate are formed while the number of moles of acetic acid and alcohol are a-x and b-x respectively at equilibrium.

According to law of mass action
K(C) = [CH3COOC2H5] [H2O] / [CH3COOH] [C2H5OH]
K(C) = [x/V] [x/V] / [a-x/V] [b-x/V]
K(C) = (x) (x) / (a-x) (b-x)
K(C) = x2 / (a-x) (b-x)

2. Hydrogen Iodide Equilibrium
For the reaction between hydrogen and iodine suppose a mole of hydrogen and 'b' moles of iodine are mixed in a scaled bulb at 444ºC in the boiling sulphur for some time. The equilibrium mixture is then cooled and the bulbs are opened in the solution of NaOH. Let the amount of hydrogen consumed at equilibrium be 'x' moles which means that the amount of hydrogen left at equilibrium is a-x moles. Since 1 mole of hydrogen reacts with 1 mole of iodine 'o' form two moles of hydrogen iodide hence the amount of iodine used is also x moles so its moles at equilibrium are b-x and the moles of hydrogen iodide at equilibrium are 2x.

According to law of mass action
K(C) = [Hl]2 / [H2] [l2]
K(C) = [2x/V]2 / [a-x/V] [b-x/V]
K(C) = 4x2 / (a-x) (b-x)

Applications of Law of Mass Action
There are two important applications of equilibrium constant.
1. It is used to predict the direction of reaction.
2. K(C) is also used to predict the extent of reaction.

To Predict the Direction of Reaction
The value of equilibrium constant K(C) is used to predict the direction of reaction. For a reversible process.
Reactant ≅Product
With respect to the ratio of initial concentration of the reagent.
There are three possibilities for the value of K
1. It is greater than K(C)
2. It is less than K(C)
3. It is equal to K(C)

Case I
If [Reactant]initial / [Product]initial > K(C) the reaction will shift towards the reverse direction.

Case II
If [Reactant]initial / [Product]initial > K(C) the reaction will shift towards the forward direction.

Case III
If [Reactant]initial / [Product]initial > K(C) this is equilibrium state for the reaction.

To Predict the Extent of Reaction
From the value of K(C) we can predict the extent of the reaction.
If the value of K(C) is very large e.g.
For 2 O3 ≅3 O2 ........... K(C) = 10(55)

From this large value of K(C) it is predicted that the forward reaction is almost complete.
When the value of K(C) is very low e.g.,
2 HF ≅H2 + F2 ........... K(C) = 10(-13)

From this value it is predicted that the forward reaction proceeds with negligible speed.
But if the value of K(C) is moderate, the reaction occurs in both the direction and equilibrium will be attained after certain period of time e.g., K(C) for
N2 + 3 H2 ≅2 NH3 ............. is 10
So the reaction occurs in both the direction.

Le Chatelier's Principle
Statement
When a stress is applied to a system at equilibrium the equilibrium position changes so as to minimize the effect of applied stress.
The equilibrium state of a chemical reaction is altered by changing concentration pressure or temperature. The effect of these changes is explained by Le Chatelier.

Effect of Concentration
By changing the concentration of any substance present in the equilibrium mixture, the balance of chemical equilibrium is disturbed. For the reaction,
A + B ? C + D
K(C) = [C][D] / [A][ B ]
If the concentration of a reactant A or B is increased the equilibrium state shifts tc right and yield of products increases.
But if the concentration of C or D is increased then the reaction proceed in the backward direction with a greater rate and more A & B are formed.

Effect of Temperature
The effect of temperature is different for different type of reaction.
For an exothermic reaction the value of K(C) decreased with the increase of temperature so the concentration of products decreases.
For a endothermic reaction heat is absorbed for the conversion of reactant into product so if temperature during the reaction is increased then the reaction will proceed with a greater rate in forward direction.

ENDOTHERMIC REACTION
Temperature increase ----> More products are formed
Temperature decrease ----> More reactants are formed

EXOTHERMIC REACTION
Temperature increase ----> More reactants are formed
Temperature decrease ----> More products are formed

Effect of Pressure
The state of equilibrium of gaseous reaction is distributed by the change of pressure. There are three types of reactions which show the effect of pressure change.

1. When the Number of Moles of Product are Greater
In a reaction such as
PCl5 <----> PCl3 + Cl2
The increase of pressure shifts the equilibrium towards reactant side.

2. When the Number of Moles of Reactant are Greater
In a reaction such as
N2 + 3H2 <----> 2NH3
The increase of pressure shifts the equilibrium towards product side because the no. of moles of product are less than the no. of moles of reactant.

3. When Number of Moles of Reactants and Products are Equal
In these reactions where the number of moles of reactant are equal to the number of moles of product the change of pressure does not change the equilibrium state e.g.,
H2 + l2 ≅2 Hl
Since the number of moles of reactants and products are equal in this reaction so the increase of pressure does not affect the yield of Hl.

Important Industrial Application of Le Chatelier's Principle
Haber's Process
This process is used for the production of NH3 by the reaction of nitrogen and hydrogen. In this process 1 volume of nitrogen is mixed with three volumes of hydrogen at 500ºC and 200 to 1000 atm pressure in presence of a catalyst
N2 + 3 H2 ≅2 NH3 ............... ΔH = -46.2 kJ/mole

1. Effect of Concentration
The value of K(C) for this reaction is
K(C) = [NH3]2 / [N2] [H2]3
Increase in concentration of reactants which are nitrogen and hydrogen the equilibrium of the process shifts towards the right so as to keep the value of K(C) constant. Hence the formation of NH3 increases with the increase of the concentration of N2 or hydrogen.

2. Effect of Temperature
It is an exothermic process, so heat is liberated with the formation of product. Therefore, according to Le Chatelier's principle at low temperature the equilibrium shifts towards right to balance the equilibrium state so low temperature favours the formation of NH3

3. Effect of Pressure
The formation of NH3 proceeds with the decrease in volume, therefore, the reaction is carried out under high pressure or in other words high pressure is favourable for the production of NH3.

Contact Process
The process is used to manufacture H2SO4 on large scale. In this process the most important step is the oxidation of SO2 to SO3 in presence of a catalyst vanadium pentoxide.
2 SO2 + O2 ≅2 SO3 ................... ΔH = - 395 kJ/mole

1. Effect of Concentration
The value of K(C) for this reaction is
K(C) = [SO3]2 / [SO2]2 [O2]
Increase in concentration of SO2 or O2 shifts the equilibrium towards the right and more SO3 is formed.

2. Effect of Temperature
Since the process is exothermic, so low temperature will favour the formation of SO3. The optimum temperature for this reaction is 400 to 450ºC.

3. Effect of Pressure
In this reaction decrease in volume takes place so high pressure is favourable for the formation of SO3.

Common Ion Effect
Statement
The process in which precipitation of an electrolyte is caused by lowering the degree of ionization of a weak electrolyte when a common ion is added is known as common ion effect.

Explanation
In the solution of an electrolyte in water, there exist an equilibrium between the ions and the undissociated molecules to which the law of mass action can be applied.
Considering the dissociation of an electrolyte AB we have
AB ℘A+ + B-
And
[A+][B-] / [AB] = K (dissociation constant)
If now another electrolyte yielding A+ or B- ions be added to the above solution, it will result in the increase of concentration of the ions A+ or B- and in order that K may remain the same, the concentration AB must evidently increase. In other words the degree of dissociation of an electrolyte is suppressed by the addition of another electrolyte containing a common ion. This phenomenon is known as common ion effect.

Application of Common Ion Effect in Salt Analysis
An electrolyte is precipitated from its solution only when the concentration of its ions exceed from the solubility product. The precipitates are obtained when the concentration of any one ion is increased. Thus by adding the common ion, the solubility product can be exceeded.
In this solution Ou(OH)2 is a weak base while H2SO3 is a strong acid so the pH of the solution is changed towards acidic medium.
When Na2CO3 is dissolved in water, it reacts with water such as
Na2CO3 + 2 H2O ℘2 NaOH + H2CO3
In this solution H2CO3 which is weak acid an NaOH which is a strong base are formed. Due to presence of strong base the medium is changed towards basic nature.

Solubility Product
When a slightly soluble ionic solid such as silver chloride is dissolved in water, it decompose into its ions
AgCl ℘Ag+ + Cl-
These Ag+ and Cl- ions from solid phase pass into solution till the solution becomes saturated. Now there exists an equilibrium between the ions present in the saturated solution and the ions present in the solid phase, thus
AgCl ℘Ag+ + Cl-
Applying the law of mass action
K(C) = [Ag+][Cl-] / [AgCl]
Since the concentration of solid AgCl in the solid phase is fixed, no matter how much solid is present in contact with solution, so we can write.
K(C) = [Ag+][Cl-] / K
Or
K(C) x K = [Ag+][Cl-]
Or
K(S.P) = [Ag+][Cl-]
Where K(S.P) is known as solubility product and defined as
The product of the concentration of ions in the saturated solution of a sparingly soluble salt is called solubility product.
the value of solubility product is constant for a given temperature.

Calculation of Solubility Product From Solubility
The mass of a solute present in a saturated solution with a fixed volume of solvent is called solubility, which is generally represented in the unit of gm/dm3. With the help of solubility we can calculate the solubility product of a substance e.g., the solubility of Mg(OH)2 at 25ºC is 0.00764 gm/dm3. To calculate the K(S.P) of Mg(OH)2, first of all we will calculate the concentration of Mg(OH)2 present in the solution.
Mass of Mg(OH)2 = 0.00764 gm/dm3
Moles of Mg(OH)2 = 0.00764 / 58 moles / dm3
= 1.31 x 10(-4) moles/dm3
The ionization of Mg(OH)2 in the solution is as follows.
Mg(OH)2 ℘Mg(+2) + 2 OH-
And the solubility product for Mg(OH)2 may be written as,
K(S.P) = [Mg(+2)] [OH-]2
Since in one mole of Mg(OH2) solution one mole of Mg++ ions are present while two moles of OH- ions are present, therefore in 1.31 x 10(-4) mole/dm3 solution of Mg(OH)2, the concentration of Mg(+2) is 1.31 x 10(-4) moles/dm3 while the concentration of OH- is 2. 62 x 10(-8) moles/dm3. By substituting these values
K(S.P) = [Mg(+2)][OH-]2
= [1.31 x 10(-4)] [2.62 x 10(-4)]2
= 9.0 x 10(-12) mole3 / dm9
So in this way the solubility product of a substance may be calculated with the help of solubility.

Calculation of Solubility from Solubility Product
If we know the value of solubility product, we can calculate the solubility of the salt.
For example, the solubility of PbCrO4at 25ºC is 2.8 x 10(-13) moles/dm3.
m = n2 / w1 in kg
m = (w2 / m2) / (w1 / 1000)
m = w2 / m2) x (1000 / w1)

Hydration
Addition of water or association of water molecules with a substance without dissociation is called Hydration.
Water is a good solvent and its polar nature plays very important part in dissolving substances. It dissolves ionic compounds readily.
When an ionic compound is dissolved in water, the partial negatively charged oxygen of water molecule is attracted towards the cation ion similarly the partial positively charged hydrogen of water molecule is attracted towards the anions so hydrated ions are formed.
In solution, the number of water molecules which surround the ions is indefinite, but when an aqueous solution of a salt is evaporated the salt crystallizes with a definite number of water molecules which is called as water of crystallization E.g., when CuSO4 recrystallized from its solution the crystallized salt has the composition CuSO4. 5H2O. Similarly when magnesium chloride is recrystallized from the solution, it has the composition MgCl2.6H2O. This composition indicates that each magnesium ion in the crystal is surrounded by six molecules. This type of salts is called hydrated salts.
It is observed experimentally that the oxygen atom of water molecule is attached with the cation of salt through co-ordinate covalent bond so it is more better to write the molecular formulas of the hydrated salts as given below.
[Cu(H2O)5]SO4 ................. [Mg(H2O)6]Cl2
It is also observed that these compound exist with a definite geometrical structure e.g., the structure of [Mg(H2O)6]Cl2 is octahedral and [Cu(H2O)4]+2 is a square planar.

Factors for Hydration
The ability of hydration of an ion depend upon its charge density.
For example the charge density of Na+ is greater than K+ because of its smaller size, so the ability of hydration for Na+ is greater than K+ ion. Similarly small positive ions with multiple charges such as Cu(+2), Al(+3), Cr(+3) posses great attraction for water molecules.

Hydrolysis
Addition of water with a substance with dissociation into ions is called Hydrolysis.
OR
The reaction of cation or anion with water so as to change its pH is known as Hydrolysis.
Theoritically it is expected that the solution of salts like CuSO4 or Na2CO3 are neutral because these solutions contain neither H+ ion nor OH-, but it is experimentally observed that the solution of CuSO4 is acidic while the solution of Na2CO3 is basic. This acidic or basic nature of solution indicate but H+ ions or OH- ions are present in their solutions which can be produced only by the dissociation of water molecules.

Theory of Ionization
1n 1880, a Swedish chemist Svante August Arrhenius put forward a theory known as theory of ionization, in order to account for the conductivity of electrolytes, electrolysis and certain properties of electrolytic solutions. According to this theory.
1. Acids, Bases and Salts when dissolved in water yield two kinds of ions, one carry positive charge and the other carry negative charge. The positively charged ions are called cations which are derived from metals or it may be H+ ion but the negatively charged ions which are known as anions are derived from non-metals
NaCl ----> Na+ + Cl-
H2SO4 ----> 2 H+ + SO4(-2)
KOH ----> K+ + OH-

2. Ions in the solution also recombine with each other to form neutral molecules and this process continues till an equilibrium state between an ionized and unionized solid is attained.

1styear CHEMISTRY Notes Chapter-5

Chapter-5
ENERGETICS OF CHEMICAL REACTION

Energetics Of Chemical Reaction
Thermodynamics
Definition
It is branch of chemistry which deals with the heat energy change during a chemical reaction.

Types of Thermochemical Reactions
Thermo-chemical reactions are of two types.
1. Exothermic Reactions
2. Endothermic Reactions

1. Exothermic Reaction
A chemical reaction in which heat energy is evolved with the formation of product is known as Exothermic Reaction.
An exothermic process is generally represented as
Reactants ----> Products + Heat

2. Endothermic Reaction
A chemical reaction in which heat energy is absorbed during the formation of product is known as endothermic reaction.
Endothermic reaction is generally represented as
Reactants + Heat ----> Products



Thermodynamic Terms
1. System
Any real or imaginary portion of the universe which is under consideration is called system.

2. Surroundings
All the remaining portion of the universe which is present around a system is called surroundings.

3. State
The state of a system is described by the properties such as temperature, pressure and volume when a system undergoes a change of state, it means that the final description of the system is different from the initial description of temperature, pressure or volume.

Properties of System
The properties of a system may be divided into two main types.

1. Intensive Properties
Those properties which are independent of the quantity of matter are called intensive properties.
e.g. melting point, boiling point, density, viscosity, surface, tension, refractive index etc.

2. Extensive Properties
Those properties which depends upon the quantity of matter are called extensive properties.
e.g. mass, volume, enthalpy, entropy etc.

First Law of Thermodynamics
This law was given by Helmheltz in 1847. According to this law
Energy can neither be created nor destroyed but it can be changed from one form to another.
In other words the total energy of a system and surroundings must remain constant.

Mathematical Derivation of First Law of Thermodynamics
Consider a gas is present in a cylinder which contain a frictionless piston as shown.

Diagram Coming Soon

Let a quantity of heat q is provided to the system from the surrounding. Suppose the internal energy of the system is E1 and after absorption of q amount of heat it changes to E2. Due to the increase of this internal energy the collisions offered by the molecules also increases or in other words the internal pressure of the system is increased after the addition of q amount of heat. With the increase of internal pressure the piston of the cylinder moves in the upward direction to maintain the pressure constant so a work is also done by the system.
Therefore if we apply first law of thermodynamics on this system we can write
q = E2 - E1 + W
OR
q = ΔE + W
OR
ΔE = q - W
This is the mathematical representation of first law of thermodynamics.

Pressure - Volume Work
Consider a cylinder of a gas which contain a frictionless and weightless piston, as shown above. Let the area of cross-section of the piston = a
Pressure on the piston = P
The initial volume of the gases = V1
And the final volume of the gases = V2
The distance through which piston moves = 1
So the change in volume = ΔV = V2 - V1
OR ΔV = a x 1
The word done by the system W = force x distance
W = Pressure x area x distance
W = P x a x 1
W = P Δ V
By substituting the value of work the first law of thermodynamics may be written as
q = ΔE + P Δ V
The absorption or evolution of heat during chemical reaction may take place in two ways.

1. Process at Constant Volume
Let qv be the amount of heat absorbed at constant volume.
According to first law qv = ΔE + P ΔV
But for constant volume ΔV = O
Therefore,
P ΔV = P x O = O
So,
qv = ΔE + 0
Or
qv = ΔE
Thus in the process carried at constant volume the heat absorbed or evolved is equal to the energy ?E.

2. Process at Constant Pressure
Let qp is the amount of heat energy provided to a system at constant pressure. Due to this addition of heat the internal energy of the gas is increased from E1 to E2 and volume is changed from V1 to V2, so according to first law.
qp = E2 - E1 + P(V2 - V1)
Or
qp = E2 - E1 + PV2 - PV1
Or
qp = E2 + PV2- E1 - PV1
Or
qp = (E2 + PV2) - (E1 - PV1)
But we known that
H = E + PV
So
E1 + PV1 = H1
And
E2 + PV2 = H2
Therefore the above equation may be written as
qp = H2 - H1
Or
qp = ΔH
This relation indicates that the amount of heat absorbed at constant pressure is used in the enthalpy change.

Sign of ΔH
ΔH represent the change of enthalpy. It is a characteristic property of a system which depends upon the initial and final state of the system.
For all exothermic processes ?H is negative and for all endothermic reactions ?H is positive.

Thermochemistry
It is a branch of chemistry which deals with the measurement of heat evolved or absorbed during a chemical reaction.
The unit of heat energy which are generally used are Calorie and kilo Calorie or Joules and kilo Joules.
1 Cal = 4.184 J
OR
1 Joule = 0.239 Cal

Hess's Law of Constant Heat Summation
Statement
If a chemical reaction is completed in a single step or in several steps the total enthalpy change for the reaction is always constant.
OR
The amount of heat absorbed or evolved during a chemical reaction must be independent of the particular manner in which the reaction takes place.

Explanation
Suppose in a chemical reactant A changes to the product D in a single step with the enthalpy change ΔH
Diagram Coming Soon
This reaction may proceed through different intermediate stages i.e., A first changes to B with enthalpy change ΔH1 then B changes to C with enthalpy change ΔH2 and finally C changes to D with enthalpy ΔH3.
According to Hess's law
ΔH = ΔH1 + ΔH2 + ΔH3

Verification of Hess's Law
When CO2 reacts with excess of NaOH sodium carbonate is formed with the enthalpy change of 90 kJ/mole. This reaction may take place in two steps via sodium bicarbonate.
In the first step for the formation of NaHCO3 the enthalpy change is -49 kJ/mole and in the second step the enthalpy change is -41 kJ/mole.

According to Hess's Law
ΔH = ΔH1 + ΔH2
ΔH = -41 -49 = -90 kJ/mole
The total enthalpy change when the reaction is completed in a single step is -90 kJ/mole which is equal to the enthalpy change when the reaction is completed into two steps. Thus the Hess's law is verified from this example.

1styear CHEMISTRY Notes Chapter-4

Chapter-4
CHEMICAL BOND

Chemical Bond
Introduction
Atoms of all the elements except noble gases have incomplete outermost orbits and tends to complete them by chemical combination with the other atoms.
In 1916, W Kossel described the ionic bond which is formed by the transfer of electron from one atom to another and also in 1916 G.N Lewis described about the formation of covalent bond which is formed by the mutual sharing of electrons between two atoms.
Both these scientists based their ideas on the fact that atoms greatest stability when they acquire an inert gas electronic configuration.

Definition
When two or more than two atoms are combined with each other in order to complete their octet a link between them is produced which is known as chemical bond.
OR
The force of attraction which holds atoms together in the molecule of a compound is called chemical bond.

Types of Chemical Bond
There are three main types of chemical bond.
1. Ionic bond or electrovalent bond
2. Covalent bond
3. Co-ordinate covalent bond or Dative covalent bond

Ionic Bond OR Electrovalent Bond
Definition
A chemical bond which is formed by the complete shifting of electron between two atoms is called ionic bond or electrovalent bond.
OR
The electrostatic attraction between positive and negative ions is called ionic bond.

Conditions for the Ionic Bond Formation
1. Electronegativity
Ionic bond is formed between the element having a difference of electronegativity more than 1.7 or equal to 1.7 eV.
Therefore ionic bond is generally formed between metals (low electronegative) and non-metal (high electronegative) elements.

2. Ionization Potential
We know that ionic bond is formed by the transference of electron from one atom to another, so in the formation of ionic bond an element is required which can lose its electrons from the outer most shell. It is possible to remove electron from the outermost shell of metals because of their low ionization potential values.

3. Electron Affinity
In the formation of ionic bond an element is also required which can gain an element is also required which can gain electron, since non-metals can attract electrons with a greater force due to high electronegativity. So a non-metal is also involved in the formation of ionic bond due to high electron affinity.

Example of Ionic Bond
In order to understand ionic bond consider the example of NaCl. During the formation of Ionic bond between Na and Cl2, Sodium loses one electron to form Na+ ion while chlorine atom gains this electron to form Cl- ion. When Na+ ion and Cl- ion attract to each other NaCl is formed. The stability of NaCl is due to the decrease in the energy. These energy change which are involved in the formation of ionic bond between Na and Cl are as follows.
i. Sodium has one valence electron. In order to complete its octet Na loses its valence electron. The loss of the valence electron required 495 kJ/mole.
Na ----> Na+ + e- ....................... ?H = 495 kJ/mole

ii. Chlorine atom has seven electrons in its valence shell. It require only one electron to complete its octet, so chlorine gains this electron of sodium and release 348 kJ/mole energy.
Cl + e- ----> Cl- ...................... ?H = -348 kJ/mole
Here the energy difference is 147 kJ/mole (495 - 348 = 147). This loss of energy is balanced when oppositely charged ions are associated to form a crystal lattice.

iii. In third step, positively charged Na+ ion and negatively charged Cl- ion attract to each other and a crystal lattice is formed with a definite pattern.
Na+(g) + Cl-(g) ----> Na+Cl- ........... ?H = - 788 kJ/mole
This energy which is released when one mole of gaseous ions arrange themselves in definite pattern to form lattice is called lattice energy.
From this example, we can conclude that it is essential for the formation of ionic bond that the sum of energies released in the second and third steps must be greater than the energy required for the first step.

Characteristics of Ionic Compounds
1. An ionic compounds, the oppositely charged ions are tightly packed with each other, so these compounds exist in solid state.
2. Due to strong attractive forces between ions a larger amount of energy is required to melt or to boil the compound and hence the melting and boiling point of the ionic compound are generally high.
3. Ionic compounds are soluble in water but insoluble in organic solvents like benzene, CCl4. etc.
4. In the aqueous solution, the ionic compounds are good electrolytes, because in water the interionic forces are so weakened that the ions are separated and free to move under the influence of electric current. Due to this free movement of ions, the ionic compounds conduct electricity in their solutions.

Covalent Bond
Definition
A link which is formed by the mutual sharing of electrons between two atoms is called covalent bond.

Explanation
In the formation of covalent bond, mutual sharing of electron takes place. This mutual sharing is possible in non-metals, therefore covalent bond is generally formed between the atoms of non-metals. For example
In Cl2 molecule, two atoms of chlorine are combined with each other to form Cl2 molecule. Each atom of chlorine having seven electrons in its valencies shell. These atoms are united with each other by sharing one of its valence electron as shown.
Cl Cl: ----> :Cl :Cl OR Cl - Cl
In this molecule, one shared pair of electrons forms a single covalent bond between two chlorine the atoms. With the formation of a covalent bond the energy of the system is also decreased.
Cl + Cl ----> Cl - Cl .............. ?H = - 242 kJ / mole
This released energy lowered the energy of the molecule and the stability of the compound is also increased.

Types of Covalent Bond
There are three main types of covalent bond.

1. Single Covalent Bond
When a covalent bond is formed by sharing of one electron from each atom, that it is called single covalent bond and denoted by (-) single line between the two bonded atoms e.g.
Cl - Cl, H - H, H - Br etc.

2. Double Covalent Bond
In a covalent bond, if two electrons are shared from each of the bonded atom then this covalent bond is called double covalent bond and denoted by (=) two lines e.g.
O = O, O : : O

3. Triple Covalent Bond
When a covalent bond is formed by sharing of three electrons from each atom then this type of covalent bond is called triple covalent bond, and denoted by (=) three lines between the two bonded atoms e.g.
N : : N :, N = N
The bond distance of multiple bonds are shorter and the bond energies are higher.

Characteristics of Covalent Compounds
The main characteristics properties of covalent compounds are as follows
1. The covalent compounds exist as separate covalent molecules, because the particles are electrically neutral so they passes solid, liquid or gaseous state. This intermolecular force of attraction among the molecules.
2. Since the covalent compound exist in all the three states of matter so their melting points and boiling point may be high or low.
3. Covalent compounds are non-electrolytes so they do not conduct electricity from their aqueous solution.
4. Covalent compounds are generally insoluble in water and similar polar solvent but soluble in the organic solvents.

Co-Ordinate OR Dative Covalent Bond
Definition
It is a type of covalent bond in which both the shared electrons are donated only be one atom, this type is called co-ordinate covalent bond.
The 8 ordinate covalent bond between two atoms is denoted by an arrow (?). The atom which donates an electron pair is called as a donor of electron and the other atom involved in this bond is called acceptor. E.g.
A + B ----> A : B OR A ? B

Dipole Moment
Definition
The product of the charge and the distance present in a polar molecules is called dipole moment and represented by µ.
OR
The extent of tendency of a molecule to be oriented under the influence of an electric field is called dipole moment.

Mathematical Representation of Dipole Moment
Suppose the charge present on a polar molecule is denoted by e and the separation between the two oppositely charged poles of the molecules is d, then the product of these two may be written as
e x d = µ
Where µ is dipole moment.

Dipole Moment in Diatomic Molecules
The diatomic molecules which are made up of similar atoms will be non-polar and their dipole moment is zero but the diatomic molecules made up of two different atoms e.g. HCl or Hl are polar and have some dipole moment. The value of the dipole moment depends upon the difference of electronegativities of the two bonded atom. If the difference of electronegativity between the atoms is greater, the polarity and also the dipole moment of the molecule is greater e.g.
The dipole moment of HCl = 1.03 debye
Whereas dipole moment of HF = 1.90 debye

Dipole Moment of Poly Atomic Molecules
In poly atomic molecules, the dipole moment of molecules depends upon the polarity of the bond as well as the geometry of the molecule.

Ionic Character of Covalent Bond
In homonuclear diatomic molecules like Cl2, O2, l2, H2 both the atoms are identical so the shared electrons are equally attracted due to identical electronegativities and hence the molecules are non-polar.
When two dissimilar atoms are linked by a covalent bond the shared electrons are not attracted equally by the two bonded atoms. Due to unsymmetrical distribution of electrons one end of the molecules acquire partial positive charge and the other end acquire a partial negative charge. This character of a covalent bond is called Ionic character of a covalent bond.
The ionic character of a covalent bond depends upon the difference of electronegativity of the two dissimilar atoms joined with each other in a covalent bond. E.g., the H-F bond is 43% ionic whereas the H-Cl bond is 17% ionic. The ionic character greatly affects the properties of a molecules e.g., melting point, boiling point of polar molecules are high and they are soluble in polar solvent like H2O. Similarly the presence of partial polar character shortens the covalent bond and increases the bond energies.

Bond Energy
Definition
The amount of energy required to break a bond between two atoms in a diatomic molecule is known as Bond Energy.
OR
The energy released in forming a bond from the free atoms is also known as Bond Energy.
It is expressed in kilo Joules per mole or kCal/mole.
Examples
i. The bond energy for hydrogen molecule is
H - H(g) ----> 2 H(g) .......................... ?H = 435 kJ/mole
OR
H(g) + H(g) ----> H - H ....................... ?H = 435 kJ/mole
It can be observed from this example that the breaking of bond is endothermic whereas the formation of the bond is exothermic.

ii. The bond energy for oxygen molecule is
O = O(g) ----> 2 O(g) ........................ ?H = 498 kJ/mole
OR
O(g) + O(g) ----> O = O .................... ?H = -498 kJ/mole
Bond energy of a molecule also measure the strength of the bond. Generally bond energies of polar bond are greater than pure covalent bond.
E.g.
Cl - Cl ----> 2 Cl ........................ ?H = 244 kJ/mole
H - Cl ----> H+ + Cl- ................... ?H = 431 kJ/mole
The value of bond energy e.g., triple bonds are usually shorter than the double bond therefore the bond energy for triple bond is greater than double bond.

Sigma & PI Bond
Sigma Bond Definition
When the two orbitals which are involved in a covalent bond are symmetric about an axis, then the bond formed between these orbitals is called Sigma Bond.
OR
A bond which is formed by head to head overlap of atomic orbitals is called Sigma Bond.

Explanation
In the formation of a sigma bond the atomic orbital lies on the same axis and the overlapping of these orbital is maximum therefore, all such bonds, in which regions of highest density around the bond axis are termed as sigma bond.

Types of Overlapping in Sigma Bond
There are three types of overlapping in the formation of sigma bond.
1. s-s orbitals overlapping
2. s-p orbitals overlapping
3. p-p orbitals overlapping
In all the three types, when the two atomic orbitals are overlapped with each other two molecular orbitals are formed. In these two molecular orbitals the energy of one orbital is greater than the the atomic orbitals which is known as sigma antibonding orbital while the energy of the other orbital is less than the atomic orbital this orbital of lower energy is called sigma bonding orbital and the shared electron are always present in the sigma bonding orbitals.

1. s-s Orbitals Overlapping
In order to explain s-s overlapping consider the example of H2 molecule. In this molecule is orbital of one hydrogen overlaps with is orbital of other hydrogen to form sigma bonding orbitals. Due to this bonding a single covalent bond is formed between the two hydrogen atoms.
Diagram Coming Soon

2. s-p Orbitals Overlapping
This type of overlapping takes place in H-Cl molecule. 1s orbital of hydrogen overlaps with 1p orbital of chlorine to form a single covalent bond. In this overlapping two molecular orbitals are formed, one of the lower energy while the other orbital is of higher energy. The shapes of these orbitals are as follows.
Diagram Coming Soon

3. p-p Orbitals Overlapping
This type of overlapping takes place in fluorine molecule. In this mole 1p orbital of a fluorine atom is overlapped with 1p orbital of the other fluorine atom. The molecular orbitals formed in this overlapping are given in figure
Diagram Coming Soon

PI Bond
When the two atomic orbital involved in a covalent bond are parallel to each other then the bond formed between them is called pi bond.
In this overlapping, two molecular orbitals are also formed. The lower energy molecular orbitals is called p bonding orbital while the higher energy molecular orbital is called p antibonding orbital. The shape of these molecular orbitals are as follows.
Diagram Coming Soon

Hybridization
Definition
The process in which atomic orbitals of different energy and shape are mixed together to form new set of equivalent orbitals of the same energy and same shape.
There are many different types of orbital hybridization but we will discuss here only three main types.

1. sp3 Hybridization
The mixing of one s and three p orbitals to form four equivalent sp3 hybrid orbitals is called sp3 hybridization. These sp3 orbitals are directed from the center of a regular tetrahedron to its four corners. The angles between tetrahedrally arranged orbitals are 109.5º.
It has two partially filled 2p orbitals which indicate that it is divalent, but carbon behaves as tetravalent in most of its compounds. It is only possible if one electron from 2s orbital is promoted to an empty 2pz orbital to get four equivalent sp3 hybridized orbitals.
Diagram Coming Soon
The four sp3 hybrid orbitals of the carbon atom overlap with 1s orbitals of four hydrogen atoms to form a methane CH4 molecule.
The methane molecule contains four sigma bonds and each H-C-H bond angle is 109.5º.

2. sp2 Hybridization
The mixing of one s and two p orbitals to form three orbitals of equal energy is called sp2 or 3sp2 hybridization. Each sp2 orbital consists of s and p in the ratio of 1:2. These three orbitals are co-planar and at 120º angle as shown
Diagram Coming Soon
A typical example of this type of hybridization is of ethane molecule. In ethylene, two sp2 hybrid orbitals of each carbon atom share and overlap with 1s orbitals of two hydrogen atoms to form two s bonds. While the remaining sp2 orbital on each carbon atom overlaps to form a s bond. The remaining two unhybridized p orbitals (one of each) are parallel and perpendicular to the axis joining the two carbon nuclei. These generates a parallel overlap and results in the formation of 2 p orbitals. Thus a molecule of ethylene contain five s bonds and one p bond.
Diagram Coming Soon

3. sp Hybridization
When one s and one p orbitals combine to give two hybrid orbitals the process is called sp hybridization. The sp hybrid orbitals has two lobes, one with greater extension in shape than the other and the lobes are at an angle of 180º from each other. It means that the axis of the two orbitals form a single straight line as shown.
Now consider the formation of acetylene molecule HC = CH. The two C-H s bonds are formed due to sp-s overlap and a triple bond between two carbon atoms consist of a s bond and two p bond. The sigma bond is due to sp-sp overlap whereas p bonds are formed as a result of parallel overlap between the unhybridized four 2p orbitals of the two carbon.
Diagram Coming Soon

Valence Shell Electron Pair Repulsion Theory
The covalent bonds are directed in space to give definite shapes to the molecules. The electrons pairs forming the bonds are distributed in space around the central atom along definite directions. The shared electron pairs as well as the lone pair of electrons are responsible for the shape of molecules.
Sidwick and Powell in 1940 pointed out that the shapes of the molecules could be explained on the basis of electron pairs present in the outermost shell of the central atom. Pairs of electrons around the central atom are arranged in space in such a way so that the distances between them are maximum and coulombie repulsion of electronic cloud are minimized.
The known geometries of many molecules based upon measurement of bond angles shows that lone pairs of electrons occupy more space than bonding pairs. The repulsion between electronic pairs in valence shell, decreases in the following order.
Lone Pair - Lone Pair > Lone Pair - Bond Pair > Bond Pair - Bond Pair
When we apply this theory we can see the variation of angle in the molecular structures.
Consider the molecular structures of NH3, OH & H2O.
Diagram Coming Soon
Variation from ideal bond angles are caused by multiple covalent bonds and lone electron pairs both of which require more space than single covalent bonds and therefore cause compression of surrounding bond angles.
Thus the number of pairs of electrons in the valency shell determine the overall molecular shape.

Structure of BeCl2
The two bond pairs of electrons in BeCl2 arrange themselves as far apart as possible in order to minimize the repulsion between them.

Structure of BF3 OR BCl3
In this molecule three bond pair are present around boron to arrange themselves as far apart as possible a trigonal structure is formed.

Hydrogen Bond
When hydrogen is bonded with a highly electronegative element such as nitrogen oxygen, fluorine, the molecule will be polarized and a dipole is produced. The slightly positive hydrogen atom is attracted by the slightly negatively charged electronegative atom. An electrostatic attraction between the neighbouring molecules is set up when the positive pole of one molecule attracts the negative pole of the neighbouring molecule. This type of attractive force which involves hydrogen is known as hydrogen bonding.