Syllabus: First Part Honours (BRABU Muzaffarpur)
Nine questions to be set. Five questions to be answered. Short answer type questions are recommended. There may be several parts in a question and different units may be mixed in questions. While setting questions the entire syllabus may be covered as far as practicable. Two questions to be answered from group 'A' and three questions to be answered from group 'B'
Group: A
Gaseous State:Derivation of Van der Waal equation of state, critical phenomenon, critical constants and their evaluation in terms of van der waals constants. Determination of van der waals constants, law of corresponding states, reduced equation of states, Boyle’s temperature.
Liquid State:
Free volume of a liquid, vapour pressure, Troutons rule, Surface tension, viscosity and their measurements, Molar volume, Parachore, Rheocore and chemical constitution, Kopp’s law, internal pressure, solubility parameters, liquid crystal.
Phase Equilibria:
Phase rule and the definition of terms, involved in it, one component system. Water and sulphur system, two component system Ag-Pb, KI + water, eutectic point, Formation of compounds with congruent melting points, Deliquescence, efflorescence, triple points.
Electrical Transport:
Conductance in electrolytic solutions, Equivalent, specific and Molar conductance, cell constant, effect of dilution on conductance, Ionic mobility, migration of ions, Kohlrausch’s law, transport number and its determination by Hittorff’s methods, Applications of conductance measurements.
Group: B
Periodic Properties:
Atomic and ionic radii, ionization potential, electron affinity and electro negativity, their trends in periodic table and application in explaining and predicting the chemical behaviours.
Chemical Bonding:
Covalent Bond: V.B. theory and its limitations, Directional characteristics of covalent bond, Hybridisation and shape of inorganic molecules and ions VSEPR theory with special reference to bond and electronegativity, M.O. theory. Homonuclear and Heteronuclear diatomic molecules [CO, NO], bond strength, bond energy. Dipole moment: Percentage ionic character in HX, Molecular geometry of polyatomic molecules.
Ionic Solids: Lattice energy, Born Haber cycle, solvation energy solubility of ionic solid, polarizing power and polarizibility of ions, Fajan’s rule. Weak interactions: H-bonding and van der waals forces.
s-Block Elements:
Comparative study, diagonal relationship, hydrides, salvation and complexing tendencies, an introduction to alkyl and aryl organometallics.
p-Block Elements:
comparative study, relationship among metal, nonmetal and metalloids elements of group 13-17, elementary idea of hydrides, oxides and halides. Hydrides of boron. Diborane and Higher boranes. Borazine, boro hydrides, fullerenes.
Nine questions to be set. Five questions to be answered. Short answer type questions are recommended. There may be several parts in a question and different units may be mixed in questions. While setting questions the entire syllabus may be covered as far as practicable. Two questions to be answered from group 'A' and three questions to be answered from group 'B'
Group: A
Atomic Structure:Black body radiation and planck’s quantum theory, Wave-particle, duality for electron and de-Broglie equation, experimental verification of de-Broglie equation by Davission and Germer experiment, de-Broglie wave associated with Bohr orbit in H-atom, Heisenberg uncertainty principle and its importance.
Solid State:
Types of solid, space lattice and unit cell, law of rational indices- Miller and Weiss indices. Interplaner spacing in cubic system, radius ratio and coordination number, packing of particles- Octahedral and tetrahedral voids.
Thermodynamic:
Objectives of thermodynamics, thermodynamic terms, First law and its mathematics formulation, Internal energy, enthalpy, Cp and Cv relation, joule –Thomson effect, Joule Thomson coefficient for ideal an real gases, Inversion temperature, work done in irreversible expansion, Reversible and irreversible adiabatic expansion of an ideal gas.
Thermo-Chemistry:
Exergonic and endergonic compounds, enthalpy of reaction at constant volume and constant pressure, enthalpy of combustion, Bomb calorimeter, enthalpy of neutralization and ionization, Kirchoff’s law, Hess’s law, bond dissociation energy.
Group: B
Estimation of nitrogen and sulphur in an organic compounds. Determination of molecular mass of a caboxylic acid by silver salt method and of an organic base by chloroplatinate salt method.Structure and bonding:
Hybridization and geometry of hydrocarbons bond lengths, bond angles, bond dissociation energy, localized and delocalized chemical bond, Vander Waal interactions and hydrogen bonding resonance,Hyperconjugation, inductive and electromeric effect, their effects on properties of compounds.
Mechanism of Organic Reactions:
Homolysis and heterolysis of covalent bonds. Types of reagents: electrophilic and nucleophilic. Types of organic reactions, energy consideration with reference to activation energy and transition state. Reactive intermediates: carbanions, carbocations and free radicals(generation, structure and stability).
Alcohols:
Classification and nomenclature, Monohydric alcohols: Methods of preparation, physical and properties. Distinction among 1°, 2° and 3° alcohols. Preparation and properties of (i) Ethylene glycol (ii) Glycerol and (iii) Allyl alcohol.
Organometallic Compounds:
Organomagnesium compounds:The Grignard reagent-formation structure and application in organic synthesis. Basic idea about organometallics: Dimethylzinc, Dimethylcadmium, Alkylithium and Lithium dialkylencuprate.
Organosulphur Compounds:
Preliminary idea of organic sulphomides, sulphonics and sulphonic acids, methods of formation and chemical reactions of thiols and thioethers.
Aldehydes and ketones:
Nomenclature, structure of the carbonyl group, general methods of preparation, properties of aldehydes and ketones, An introduction to α-β unsaturated aldehydes and ketones.
Carboxylic acids:
General methods of preparations and properties of monocarboxylic acids and their derivatives such as ester, acid chlorides, amides and anhydrides. Methods of formation and chemical reactions of (i) unsaturated monocarboxylic acids and (ii) dicarboxylic acids.
Organic compounds of nitrogen:
Classification, nomenclature and structure of amines. Preparation and properties of aliphatic amines. Separation and identification of 1° , 2° and 3° amines. Preparation, properties and estimation of urea.
B.Sc. First Part Chemistry MCQs
Physical Chemistry (Paper: 1st & 2nd)
Vander Waal Equation:
Where: P = pressure, R = universal gas constant, T = absolute temperature, a & b = Vander Waal constant, V = molar volume
Unit of Vander Waal Constants:
Unit of 'a': atm lit² mol⁻²Unit of 'b': liter mol⁻¹
Physical Significance of Vander Waal Constants a & b:
Physical Significance of Vander Waal Constants a: Vander Waal constant 'a' represents the magnitude of intermolecular forces of attraction.Physical Significance of Vander Waal Constants b: Vander Waals constant 'b' represents the effective size of the molecules.( or average volume excluded from v by a particle).
Boyle's Temperature (TB):
The temperature at which normal gases start to behave like ideal gases(due to the absence of both attractive and repulsive forces at that particular temperature).TB = a/Rb
Derivation of Critical Constants:
We know that-(P + a/V2)(V − b) = RT (for one mole)
or, PV − bP + a/V − ab/V2 = RT
multiplying the above equation by V2/P we get-
V3 − V2b + aV/P −ab/P −RTV2/p = 0
or, V3 − (b + RT/P)V2 + aV/P − ab/P = 0
when, T = Tc and P = Pc then-
V3 − (b + RTc/Pc)V2 + aV/Pc − ab/Pc = 0 --------(equation 1)
at critical points,
V = Vc
or, V − Vc = 0
or, (V − Vc)3 = 0
or, V3 − 3V2Vc + 3Vc2V − Vc3 = 0 --------(equation 2)
on equation the powers of V in equation (1) and equation (2), we get-
3Vc = RTc/Pc + b --------(equation 3)
3Vc2 = a/Pc --------(equation 4)
Vc3 = ab/Pc --------(equation 5)
dividing equation (5) by equation (4) we get-
Vc/3 = b
or, Vc = 3b
putting the value of Vc in equation (4) we get-
27b2 = a/Pc
or, Pc = a/27b2
Now putting the value of Vc and Pc in equation (3) we get-
9b − b = 27b2RTc/a
or, 8a = 27bRTc
or, Tc = 8a/27Rb
Law of Corresponding State or Reduced Equation of State:
According to this law,two different gases behave similarly, if their reduced properties (i.e. P, V and T) are same.
The ratio of P to Pc, V to Vc and T to Tc are called reduced pressure(Pr), reduced volume(Vr) and reduced temperature(Tr) respectively.
so, Pr = P/Pc
or, P = PrPc --------(equation 1)
Vr = V/Vc
or, V = VrVc --------(equation 2)
similarly,
Tr = T/Tc
or, T = TrTc --------(equation 3)
Now replacing the value P, V and T in Vander Waal equation, we get-
(PrPc + a/Vr2Vc2)(VrVc − b) = RTrTc --------(equation 4)
Now putting the values of critical constants in the equation (4) we get-
(Pr a/27b2 + a/Vr29b2)(Vr3b − b) = 8a.RTr/27Rb
or, (Pr a/27b + a/Vr29b)(3Vr − 1) = 8a.Tr/27b
multiply the above equation by 27b/a we get-
(Pr + 3/Vr2) (3Vr − 1) = 8Tr
This equation is completely free from constants such as R, a, b hence this is applicable to all substances n fluid state. This is law of corresponding states.
Trouton's Rule:
The entropy of vaporization of most of the liquids at their boiling points is almost the same and has the value between 85-88 J/K.molQ. Which of the following molecules does not follow Trouton's rule
A. C6H6
B. C6H12
C. CCl4
D. NH3
Q. The liquid that deviates from the Trouton’s rule is
A. Hydrochloric acid
B. Sulphuric acid
C. Phosphoric acid
D. Acetic acid
Molar Volume:
Molar volume of any substance is the volume occupied by one mole of the substance. This is easily determined by dividing the molecular mass by the density of the compound (i.e V = M/D). According to Avogadro's hypothesis, the molar volume of all gases at S.T.P. is 22.4L.It is the volume so expressed in ml. or c.c.. If molecular weight and density of a substance are known, molar volume can easily be calculated.
Kopp's Rule:
Kopp in 1842 state that the molar volume of a liquid at its boiling point is equal to the sum of the atomic volume of its constituent atoms. This is known as Kopp’s rule.According to Kopp's rule it has been found that the molar volumes of two members of a homologous series of organic liquids differ by about 22 mL, for each CH2 group Kopp calculated the volume equivalent of each element by a simple arithmetic means.
Parachor:
Macleod in 1923 gave the following relation between the surface tension (γ) and density (D) for a normal liquid-C = γ1/4/(D −d)
where d is vapour density of the liquid at given temperature and C is constant.
In 1924, Sugden modified the above equation as-
γ1/4/(D −d) = MC = P
where M is the molecular weight of the liquid and P is the parachor.
At ordinary temperature, d is very small in comparision to D then-
M γ1/4/D = P
If γ = 1 at a particular temperature, then-
M/D = P Thus, at a particular temperature, the molar volume of a liquid having surface tension unity is called Parachor.
If two liquids having the same surface tension are taken whose molecular weights are M1 and M2 and their densities are D1 and D2 respectively, then-
M1 γ1/4/D1 = P1
and
M2 γ1/4/D2 = P2
P1/P2 = (M1/D1)/ (M2/D2)
So, the ratio of parachors of two liquids having the same surface tension is equal to the ratio of molar volumes. Parachor is both an additive and constitutive property.
Parachor Value of some Elements and Groups:
| Element | P Value | Group | P Value |
|---|---|---|---|
| C | 8.6 | C=O | 44.4 |
| H | 15.7 | OH | 30.2 |
| N | 12.5 | COOH | 73.7 |
| O | 19.8 | NO2 | 73.8 |
| Cl | 55.2 | Double Bond | 19.9 |
| Br | 68.8 | Triple Bond | 40.6 |
| I | 90.3 | Six Membered Ring | 1.4 |
Application of Parachor:
Parachor data are used to determine the structure of molecules and the nature of bonds.Example:
Two structure proposed for QUINONE-
Structure-1
6C = 6 X 8.6 = 51.6
4H = 4 X 15.7 = 62.8
2O = 2 X 19.8 = 39.6
Four double bonds = 4 X 19.9 = 79.6
one 6 membered ring = 1 X 1.4 = 1.4
So, Calculated Parachor = 235
Structure-2
6C = 6 X 8.6 = 51.6
4H = 4 X 15.7 = 62.8
2O = 2 X 19.8 = 39.6
Three double bonds = 3 X 19.9 = 59.7
Two 6 membered ring = 2 X 1.4 = 2.8
So, Calculated Parachor = 216.5
Since the experimental Parachor is 236.8. Hence structure-1 is correct for QUINONE.
Rheochor:
It was introduced by Newton Friend in 1943. Rheochor is a constant obtained by multiplying molar volume and eighth root of the co-efficient of viscosity. It is denoted by letter Capital 'R'.R = ( M/D ) X η1/8
If η = 1 then-
M/D = R
So, we can say that Rheochor is the molar volume of the liquid at the temperature at which its viscosity is unity. Like Parachor, Rheochor is both additive and constitutive. However it has not proved of much use in solving structural problems.
Black Body Radiation:
When a body is heated, it emits electromagnetic radiation and when temperature is dropped, the energy is absorbed by the body. If a body absorbs all radiations that falls upon it, it is called the Black Body and the radiation emitted by it is called Black Body RadiationNo any object is perfectly black body.
In 1854, Kirchoff proposed the following two laws concerning black body. They are-
1. A black body not only absorbs all radiation falls upon it but also acts as a perfect radiator when heated.
2. The radiation given out by a black body depends upon the temperature of body and is not dependent on the nature of the interior materials.
Plank's Quantum Theory:
This theory explains the spectral distribution of black body radiation. i.e. how the energy distributed among different wavelengths emitted by a black body.Followings are the main points of this theory-
1. Energy emitted or absorbed is not continuous, but is in the form of packets called quanta. Quanta may be taken as behaving like a strem of particles having mass, energy and momentum. The energy of quantum radiation is-
E = hν (where 'h' is Plank's Constant & 'ν' is frequency of radiation)
or, E ∝ ν
2. Each photon carries an energy in discrete level which is directly proportional to the frequency of wavelength. The energy of the nth enenrgy level is given as-
E = nhν
where 'n' is integer having values 0,1,2,3...
Compton Effect:
When x-rays fall on a crystal, they are scatterd. Compton in 1923, observed that the wave length of the scattered radiation(λ') is always greater that the wavelength of the incident radiation(λ)i.e. λ' > λ
The change in wavelength is independent of the wavelength of the incident radiation as well as that of scatterer. The change in wavelength of the scattered radiation is called Compton effect.
de-Broglie Wave Equation:
We know that Einstein's equation-E = mc2 ---------(eq.1)
where, E = energy, m = mass the particles and c is the velocity of light
According to Plank's rdiation theory-
E = hν = h.c/λ ---------(eq.2)
where, h = Plank's constant and λ = wave length of radiation
From equation 1 and 2 we have-
mc2 = h.c/λ
or, mc = h/λ
or, p = h/λ [as mass(m) X velocity(c) = momentum(p)]
or, λ = h/p ---------(eq.3)
Equation '3' is called de-Broglie Wave Equation.
Heisenberg Uncertainty Principle:
It is not possible to determine precisely and simultaneously the momentum and position of small moving particles.If position of te particle is known then momentum is unknown and vice-versa.
Δx. Δp ≥ ℏ/2 = h/4π
where, Δx = uncertain position, Δp = uncertain momentum, ℏ = h/2π & h = Plank's constant
Δx. mΔv ≥ ℏ/2
Δx. Δv ≥ ℏ/2m
Δx. Δv ≥ h/4πm
Question: A particle is moving with constant momentum. The uncertainty in the momentum of the particle is 3.3 x 10-2 kg ms-1. Calculate the uncertainty position.
[Hints: Δx = ℏ/2.Δp
Δx = 5.27 ☓ 10-35/3.3 ☓ 10-2
1.59 ☓ 10-33m]
[Hints: Δx = ℏ/2.Δp
Δx = 5.27 ☓ 10-35/3.3 ☓ 10-2
1.59 ☓ 10-33m]
Crystal Lattice and Unit Cell:
Crystal Lattice: Crystal Lattice is a three-dimensional representation of constituent particles (atoms, molecules or ions) arranged in a specific order. Or the geometric arrangement of constituent particles of crystalline solids as point in space is called crystal lattice. There are total 14 possible three-dimensional lattices. Crystal lattices are also known by Bravais Lattices.Characteristics of crystal lattice:
☛ Each constituent particle is represented by one point in a crystal lattice.
☛ These points are known as lattice point or lattice site.
☛ Lattice points in a crystal lattice are joined together by straight lines.
☛ By joining the lattice points with straight lines the geometry of the crystal lattice is formed.
Unit Cell: The smallest portion of a crystal lattice is called Unit Cell. By repeating in different directions unit cell generates the entire lattice.
Characteristics of Unit Cell:
☛ A unit cell has three edges a, b and c and three angles α, β and γ between the respective edges.
☛ The a, b and c may or may not be mutually perpendicular.
☛ The angle between edge b and c is α, a and c is β and that of between a and b is γ.
Number of Particles Per Unit Cell (Z):
☛ For Simple Cubic Unit Cell:In Simple cubic unit cell, particles are present at corners only. In a crystal lattice every corner is shared by eight adjacent unit cells. Therefore, only 1/8 of constituent particles, belong to a particular unit cell. Therefore, Z = 8 X 1/8 = 1
So, the number of particles per unit cell for simple cubic cell is 1.
☛ For Body Centered Cubic (BCC) Unit Cell:
In Body Centered Cubic unit cell, particles are present at corners as well as at the center of the body. Therefore,
Z = (8 X 1/8) + 1 = 2
So, the number of particles per unit cell for BCC Unit cell is 2.
☛ For Face Centered Cubic (FCC) Unit Cell:
In Face Centered Cubic unit cell, particles are present at corners as well as at the center of the face which is shared between adjacent two particles. Therefore,
Z = (8 X 1/8) + (6 X 1/2) = 4
So, the number of particles per unit cell for FCC Unit cell is 4.
☛ For End Centered Cubic (ECC) Unit Cell:
In Eace Centered Cubic unit cell, particles are present at corners as well as at the center of two face which is opposite to each other. Therefore,
Z = (8 X 1/8) + (2 X 1/2) = 2
So, the number of particles per unit cell for ECC Unit cell is 2.
Limiting Radius Ratio:
The limiting radius ratio is the minimum allowable value for the ratio of cationic radii to anionic radii (ρ=r+/r-) for the structure to be stable. Here, r+ is the radius of the cation and r- is the radius of the surrounding anions.Bragg's Equation:
Whem a beam of light falls on a crystal plane composed of regularly arranged particles, the X- rays are diffrected. If the waves are in phase after reflection,the difference in distance travelled by the two rays must be eqaul to an integral number of wavelength, nλ for constructive interference.Thus, Path difference = WY + YZ
= XY sinθ + XY sinθ
= 2XY sinθ
= 2d sinθ
So, nλ = 2d sinθ
This equation is Known as Bragg's equation.
where, n = 1,2,3...(diffrection order)
λ = wavelength of X-rays
d = distance between planes
θ = angle at which interference occurs.
Law of rational indices:
The law of rational indices was deduced by Hauy.The law of rational indices states that the intercepts of any face of a crystal along the crystallographic axes are either equal to the unit intercepts or some simple whole number multiples of them.
The faces of crystals and also planes within crystals can be characterized by means of a suitable set of coordinates.
Let consider the three axes OX, OY, OZ which are cut by a crystal face ABC at distances OA, OB, and OC from the origin. Let OX, OY, and OZ represent the three crystallographic axes and let ABC be a unit plane. The unit intercepts will then be a, b and c.
According to the above law, the intercepts of any face such as KLM, on the same three Axes will be simple whole number multiples of a,b and c respectively. These distances at called intercepts. It is found that if the axes are suitably chosen, the intercepts of different faces upon them bear a simple ratio to each other or a given face may cut an axis at infinity. This is called the law of rational intercepts or indices.
System and Surrounding:
The part of universe which is under study is called system and the rest part of the universe is called surrounding. That means the universe is the combination of system and surrounding.Types of System:
Open System:The system which can exchange both heat and matter with the surrounding is called open system. Hot water in a beaker is an example of this system.
Closed System:
The system which can exchange only heat but not matter with the surrounding is called closed system. Hot water in a sealed tube is an example of this system.
Isolated System:
The system which can exchange neither energy nor matter with the surrounding is called Isolated system. Hot water in a thermos flask is an example of this system.
Thermodynamic Process:
When a system changes itself from one to another state, the operation is called procss.Isothermal Process:
The process which takes place at constant temperature is called isothermal process.
i.e. ΔT=O
Adiabatic Process:
The process in which no heat change occurs is called adiabatic process.
i.e. ΔQ=O
Isochoric Process:
The process which takes place at constant volume is called isochoric process.
i.e. ΔV=O
Isobaric Process:
The process which takes place at constant pressure is called isobaric process.
i.e. ΔP=O
Reversible Process:
The process which takes place infinitesimally slowly and whose direction at any point can be reversed by applying an infinitesimal change in the state of the system is called reversible Process.
Irreversible Process:
The process which takes place in one step and can not be reversed. This is a fast process.
Extensive and Intensive Properties:
Properties which depend upon the amount of the substance present in the system are called extensive properties. Mas, Volume, Number of moles, Enthalpy, Entropy, Free energy etc are the example of extensive properties. These properties are additive. If mass of the gas is changed, the volume is also changed and so is the number of moles and their internal energy of the system.Properties which don't depend upon the amount of the substance present in the system are called intensive properties. Temperature, Pressure, Boiling Point, Melting Point etc are the example of intensive properties. If temperature of a glass of water is 25oC, then each and every drop of water in this glass has the temperatue of 25oC. These properties are non additive.
Internal Energy:
All forms of energy associated with a system is called internal energy or simply energy of the system (E).This is expressed in Joule. This arises due to movement of molecules, arrangement of atoms in molecules, number of arrangement of electrons in atoms etc.
It is neither possible nor necessary to calculate the absolute value of internal energy of a system. It is a state function so depend only on the initial and final state of the system.
Heat Capacity:
The amount of heat required to change its temperature by one degree of a substance.Q = CΔT
or, C = Q/ΔT
Molar Heat Capacity:
The amount of heat required to change its temperature by one degree of one mole of a substance.Heat Capacity at Constant Pressure:
The amount of heat required to change its temperature by one degree of a substance at constant pressure.Heat Capacity at Constant Volume:
The amount of heat required to change its temperature by one degree of a substance at constant volume.Relation between CP and CV:
We know thatH = E + PV
or, H = E + RT (as PV = RT for one mole)
differentiating the above equation w.r.t T, we get-
dH/dT = dE/dT + R(dT/dT)
or, CP = CV + R
or, CP − CV = R
First law of Thermodynamics:
It is also called energy conservation principle. According to this principle, energy can neither be created nor be destroyed, it can only be transfer or change fron one form to another form.In other way-
Heat absorbed (Q) by the system is equal to change in internal energy (ΔE) plus work done by the system (−W).
Q = ΔE + (−W)
or, ΔE = Q + W
Limitations:
First law does not indicate whether heat can flow from a cold body to a hot body or not.
First law does not specify that process is feasible or not.
Practically it is not possible to convert the heat energy into an equivalent amount of work.
Work done in Isothermal and Reversible Expansion:
Let us consider 'n' moles of an ideal gas enclosed in a cylinder fitted with a frictionless, weightless and movable piston. Let P be the pressure of the gas and P-dP be the external pressure under which volume of the gas increased by dV, then work done in this expansion isdw = −(P − dP)dV = − PdV (as dP.dV is very small).
For a infinite volume change from V1 to V2, the total work done during expansion-
where P1 and P2 are the initial and final pressure respectively.
Question: 3 moles mole of and ideal gas are expanded isothermally and reversibly from volume of 10 m3 to the volume 20 m3 at 300 K. Claculate the work done by the system. (Answer: 5.178 KJ)
Joule-Thomson Effect:
When a gas is allowed to expand from high to low pressure through a porous plug under adiabatic conditions, the gas gets cooled. The drop in temperature (dT) produced by fall in pressure (dP) under adiabatic condition is called J-T Effect. The fall in temperature is due to decrease in kinetic energy of the gas molecules. Since a portion of it is used up in overcoming the vander waal force of attraction existing among them during expansion. Since ideal gas has no such forces, therefore, there is no expenditure of energy in overcoming these forces during expansion.Inversion Temperature:
The temperature at which Joule - Thomson Coefficient becomes zero is called Inversion Temperature(Ti).Ti = 2a/Rb
Joule Thomson Coefficient for Ideal and Real gas:
Calorimeter:
Heat involved in a chemical reaction is measured in a suitable apparatus is called calorimeter.Bomb Calorimeter:
This apparatus was devised by Berthelot in 1881 to measure the heat of combustion of organic compounds. A modified form of the apparatus is shown below is consists of a sealed combustion chamber, called a bomb, containing a weighed quantity of the substance in a dish along with oxygen under about 20 atm. Pressure.The bomb is lowered in water contained in an insulated copper vessel. This vessel is provided with a stirrer and a thermometer reading up to 1/100th of a degree. It is also surrounded by an outer jacket to ensure complete insulation from the atmosphere. The temperature of water is noted before the substance is ignited by an electric current. After combustion, the rise in temperature of the system is noted on the thermometer and heat of combustion can be calculated from the heat gained by water and the calorimeter.
Bond Energy and Bond Dissociation Energy:
Whenever a bond is formed energy is evolved and bond is broken, energy is absorbed. Bond energy may be defined as, The amount of energy released when 1 mole of bonds are formed from isolated gaseous atoms. Whereas Bond Dissociation energy may be defined as, The amount of energy required to break one mole bond present between the atoms of a gaseous molecule.Hess' Law of Constant Heat Summation:
Hess Law of Constant Heat Summation is also known as Second Law of Thermochemistry. The law states, “That the total heat change accompanying a chemical reaction is the same whether the reaction takes place in one step or multiple steps.”Example: Carbon dioxide can be formed directly from carbon and also from carbon via carbon monoxide. The heat change involved in both the process are found to be same. Formation of CO2 directly:
C (s) + O2 (g) → CO2 (g) ΔH = – 94 K Cal.
Formation of CO2 Via Carbon Monoxide:
C (s) + 1/2 O2 (g) → CO (g) ΔH1 = – 26.4 K Cal.
CO (g) + 1/2 O2 (g) → CO2 (g) ΔH2 = – 67.6 K Cal.
ΔH = ΔH1 + ΔH2
ΔH = –26.4 + (– 67.6) K Cal.
ΔH = – 26.4 – 67.6 K Cal.
ΔH = – 94 K Cal.
Application of Hess Law
Hess' law finds its application in determining the heat of changes for reactions for which experimental determination is not possible. Some important applications are given below:1. Calculation of Enthalpy of Formation
2. Calculation of Enthalpy of Allotropic Transformation
3. Calculation of Calorific Value
Enthalpy of Combustion:
The enthalpy of combustion of a substance is defined as the heat exchange when 1 mole of substance is completely burnt or oxidized in oxygen.Standard enthalpy of combustion is the enthalpy change when 1 mole of a substance burns under standard state conditions.
For example, the enthalpy of combustion of ethanol, −1366.8 kJ/mol, is the amount of heat produced when one mole of ethanol undergoes complete combustion at 25 °C and 1 atmosphere pressure.
C2H5OH(l) + 3O2(g) ⟶ 2CO2 + 3H2O(l) ΔH∘298 = −1366.8kJ.
Enthalpy of Neutralization:
Neutralisation is the reaction between an acid and a base to form a salt and water. During neutralisation reaction, hydrogen ions from acid react with hydroxide ions from alkali to form water.H+(aq) + OH–(aq) → H2O(aq)
The heat of neutralisation is the heat produced when one mole of water is formed from the reaction between an acid and an alkali.
Enthalpy of neutralization is always constant for a strong acid and a strong base: this is because all strong acids and strong bases are completely ionized in dilute solution.
When an acid and alkali react, heat is given out. So, Neutralisation is an exothermic reaction and always produces heat. Therefore, enthalpy of neutralisation, ΔH is always negative.
Heat of Reaction:
Heat of reaction may be defined as the difference in enthalpy or heat content, ∆H, between the products and the reactants.If heat is evolved the reaction is said to be exothermic and when heat is absorbed the reaction is an endothermic.
Heat of Reaction at Constant Pressure:
The difference between the sum of enthalpies of products and the sum of enthalpies of reactants at a given temperature and constant pressure is called the heat of reaction at the constant pressure at a given temperature. It is denoted by ΔH.ΔH = ∑ΔHProducts – ∑ΔHReactants
ΔH is negative for exothermic reaction and positive for an endothermic reaction.
Heat of Reaction at Constant Volume:
The difference between the sum of internal energies of products and the sum of internal energies of reactants at a given temperature and constant volume is called the heat of reaction at the constant volume at a given temperature. It is denoted by ΔE.ΔE = ΔE = ∑ΔEProducts – ∑ΔEReactants
Factors Affecting Enthalpy of Reaction:
The various factors on which enthalpy of reaction depend are given below:Physical State of Reactants and Products:
The enthalpy of reaction changes with change in physical state because as the physical state changes, extent of heat is evolved.Quantity of Reactants:
The change in enthalpy of reaction depends upon the quantity of reactants used. When the number of moles of reactants are doubled, the enthalpy change also becomes double.Allotropic Modification:
For elements existing in different allotropic modifications, the heat of reaction is different if different allotropic form is involved in reaction.Temperature and Pressure:
The enthalpy of reaction depends upon the temperature and pressure of reaction. Therefore, the values are generally expressed under standard conditions of temperature (298K) and pressure(1 atm.)Inorganic Chemistry (First Paper)
Molecular Orbital Theory:
This theory was put froth by F. Hund and R.S. Mulliken in 1930 and later on developed by I.F. Lennard Jones and Charles Coulson.The basic posulates of this theory is as follows:
1. When nuclei of two atoms come close to each other, their atomic orbitals interact leading to the formation of molecular orbitals. After the formation of molecular orbitals, atomic orbitals lose their identity.
2. Each molecular orbital is described by a wave function (Ψ) called molecular orbital wave function.
3. The molecular orbital wave function (Ψ) is such that Ψ2 represents the probability density or electron charge density.
4. Each molecular orbital wave function is associated with a set of four quantum numbers which determine the energy and the shape of the molecular orbital.
5. Each wave function asssociated with a definite energy value and the total energy of the moleculeis the sum of the energies of the occupied molecular orbitals.
6. Electron fill the molecular orbitals in the same way as they fill the atomic orbitals.
7. Each electron in a molecular orbital belongs to all the nuclei present in the molecules.
8. Each electron moving in a molecular orbital has a spin of +1/2 and -1/2.
What are Alkaline Earth Metals ? Why are they so called ?
Elements of group 2 of the periodic table are called alkaline earth metals. They are Be, Mg, Ca, Sr, Ba and Ra. Their valence shell configuration is ns2.They are earth like minerals and their aqueous solutions are alkaline in nature. Hence they are called alkaline earth metals.
Thermal stability of alkaline earth metals carbonates (MCO3):
Thermal stability of alkaline earth metals carbonates (MCO3) are increases down the group. As we go down the group, more heat energy is required to thermally decompose the carbonate. So, down the group the carbonates become more thermally stable. This is explained by the polarizing ability of group IIA metal ions. Smaller the cation, greater the ability to polarize the anion. Hence the polarising ability of the Group IIA ions decreases down the group as the size of cation increases down the group. Group IIA carbonates thermally decompose they produce the metal oxide and carbon dioxide.MCO3 -----> MO + CO2
So the correct order of stability of carbonates of Group IIA is
BaCO3 > SrCO3 > CaCO 3 > MgCO3 > BeCO3
How does Be reacts with NaOH and H2SO4 ?
Strong NaOH solutions attack slowly whereas fused NaOH attack it readily forming beryllates and liberates hydrogen gas.Be + 2NaOH -----> Na2BeO2 + H2
Be displaces hydrogen gas from dilute sulphuric acid but with hot concentrated sulphuric acid, it liberates SO2 gas
Be + H2SO4 -----> BeSO4 + H2
Be + 2H2SO4 -----> BeSO4 + SO2 + 2H2O
Alkaline Earth Metals are not obtained by the electrolysis of aqueous solutions. Explain:
The aqueous solutions of alkaline earth metals upon electrolysis give metal hydroxide instead of pure metal. That's why alkaline earth metals are not obtained by electrolysis of their aqueous solutions.MX2 -----> M+2 + 2X-
2X- -----> X2 + 2e
2H2O + 2e -----> 2OH- + H2
-------------------------------------------------
MX2 + 2H2O -----> M(OH)2 + X2 + H2
CaO is basic oxide but ZnO is amphoteric oxide. Why?
CaO is soluble in acids only but ZnO is soluble in acids as well as alkalis. That's why CaO is basic oxide but ZnO is amphoteric oxide.CaO + 2HCl -----> CaCl2 + H2O
ZnO + 2HCl -----> ZnCl2 + H2O
ZnO + 2NaOH -----> Na2ZnO2 + H2O
Be atom has no unpaired electrons, yet it forms BeF2 Why?
4Be: 1s22s2In ground state, Be does not have any unpaired elecron but in excited state it has two unpaired electrons. One electron from 2s2 goes into the 2p orbital. These two orbitals (i.e. 2s and 2p)get hybridized and forms two sp hybrid orbitals and arranged linearly in 3D space. Hence, with two F atoms it forms BeF2 molecule.
Solubility of Alkaline Earth Metal (IIA) Hydroxides in Water:
The solubility of alkaline earth metal(IIA) hydroxides in water increases down the group as the lattice energy decreases down the group due to increase in size of the alkaline earth metals cation whereas the hydration energy of the cation remains almost unchanged. The resultant of two effects i.e.ΔHSolution = ΔHLattice - ΔHHydration
Becomes more negative as we move from Be(OH)2 to Ba(OH)2 which accounts for increase in solubility.
Solubility of Alkaline Earth Metal (IIA) Sulphates in Water:
The solubility of alkaline earth metal(IIA) sulphates decreases down the groups i.e. BeSO4 > MgSO4 > CaSO4 > SrSO4 > BaSO4. Thus BeSO4 and MgSO4 are highly soluble, CaSO4 is sparingly soluble but SrSO4 > BaSO4 are virtually insoluble.The magnitude of the lattice energy remains almost constant because the sulphate is so big. However the hydration energy decreases from Be+2 to Ba+2 appreciably as the size of the cation increase down the group. Hence, the solubilities of sulphates of alkaline earth metals decrease down the group mainly due to the decreasing hydration energies from Be+2 to Ba+2. The high solubility of BeSO4 and MgSO4 is due to high hydration energies due to smaller size of Be+2 and Mg+2 ions.
Fullerenes:
This was discoved in 1985. The real name of Fullerene is buckminster fullerene. Fullerene is third allotrope form of carbon material after graphite and diamond whose molecule consists of 20 hexagonal and 12 pentagonal rings as the basis of an icosohedral symmetry closed cage structure. Each carbon atom is bonded to three others and is sp2 hybridised.The most stable and common of fullerenes is the C60 molecule. C60 molecule has two bond lengths - the 6:6 ring bonds can be considered "double bonds" and are shorter than the 6:5 bonds. C60 is not "superaromatic" as it tends to avoid double bonds in the pentagonal rings, resulting in poor electron delocalisation. As a result, C60 behaves like an electron deficient alkene, and reacts readily with electron rich species. Fullerene is harder than diamond. Also, the ductility is 100 times stronger than steel. Fullerene has amazing electrical conductivity and has more stronger conductivity than copper. Fullere's weighs only 1/6th of copper material. The geodesic and electronic bonding factors in the structure account for the stability of the molecule. Theoratically, an infinite number of fullerenes can exist, their structure based on pentagonal and hexagonal rings, constructed according to rules for making icosahedra.
USES:
One of the most important uses of Fullerene is medicine.
Fullerenes are active molecules.
Fullerene molecule can be used as an antioxidant because it can easily react with radicals due to the high affinity of the electron.
It is used as an anti-aging and anti-damage agent in the cosmetic sector.
Fullerenes are used as antiviral agents. This use is provided by its unique molecular structure, antioxidant effect and biological compatibility.
May be used for osteoporosis treatment because of its preferential localization.
Inert Pair Effect:
The reluctance of two ns-electrons to take part in bond formation due to the high energy needed for unpairing them is called inert pair effect. This was proposed by Sidgwick.Some p-block elements, such as thullium(Tl), polonium(Po), tin( Sn ), lead(Pb), bismuth( Bi ) etc exhibit Inert pair effect. The '5s' electron of tin and '6s' electrons of lead and bismuth have a tendency to remain inert due to inert pair effect.
From their electronic arrangement, we see that these elements contain inner 'd' or 'f' electrons and due to poor shielding effect of 'd' or 'f' orbitals, the effective nuclear charge on outer most 's'-electrons act strongly. So, the nucleus pull the 's'-electrons strongly. Consequently, these two 's'-electrons show a less tendency to participate in chemical reaction and these two 's'-electrons remain inert.
Consequences:
# Due to inert pair effect, many physical and chemical properties of the relating elements become changes.
SnCl2 is more stable than SnCl4 because, in the formation of SnCl2, 'Sn' metal utilize two '5p'-electrons only. While, in the formation of SnCl4, 'Sn' metal used both '5s' and '5p' electrons. But, '5s' electron of 'Sn' is not available for formation of molecule due to inert pair effect. Hence, SnCl4 becomes unstable.
# Inert pair effect cause the variable valency of elements.
Sn & Pb shows +2 & +4 oxidation states while Tl shows +3 & +1 oxidation state due to inert pair effect.
# Inert pair effect, affect melting and boiling point of elements. For example, the melting and boiling point of 'Po' is less than that of 'Te'. Generally, the melting and boiling point of elements increase down the group. Now, 'Po'-element contain maximum number of 'd' and 'f' electrons. So, the inert pair effect becomes more effective for 'Po' than 'Te'. As a result, the '6s'-electron of 'Po' is less available than '5s'-electron of 'Te'. Hence, in case of Po, the inter-atomic Vander waal's force of attraction is less than Te. Consequently, the melting and boiling point of Po is less than Te.
# Inert pair effect, influence the oxidizing and reducing properties of compounds. For example, +5 oxidation state of Bi is less stable than +3 oxidation state. Due to this, BiF5 act as a strong oxidizing agent. It oxidized the other reducing substance readily, and changes it's oxidation state from +5 to more stable +3.
Structure of Peroxydisulphuric Acid:
Peroxydisulfuric acid is the inorganic compound with the chemical formula H2S2O8 also called Marshall's acid after its inventor Professor Hugh Marshall. Oxidation state of Sulfur is +6. It is a powerful oxidizing agents.Which is more acidic H2SO4 or H2SO3 ?
The oxidation state of S is +6 in H2SO4 and +4 in H2SO3. Greater is the oxidation number of central atom, more it attracts the electrons towards itself and also, more it stabilises the conjugate base. So, H2SO4 is stronger acid than H2SO3Borazine or Borazole (Inorganic Benzene):
Borazine, also known as borazole, is a polar inorganic compound with the chemical formula B3N3H6. The compound is isoelectronic and isostructural with benzene. For this reason borazine is sometimes referred to as “inorganic benzene.Preparation:
When B2H6 and NH3 in the ratio of 1:3 are heated borazole is obtained.
B2H6 + 6NH3 -----> 2B3N3H6 + 12H2
Properties:
1. It is colorless liquid
2. It shows unsaturated reaction.
B3N3H6 + 3HCl -----> B3N3H9Cl3
B3N3H6 + 3Br2 -----> B3N3H6Br6
B3N3H6 + 3H2O -----> B3N3H12O3
3. It forms two series of methyl derivatives-
a. N- methyl derivatives
b. B- methyl derivatives
4. It is isoelectronic with benzene.
5. When borazine is pyrolysed above 340°C, B3N6H10 and B5N5H8 are produced. These products are boron- nitrogen analogues of diphenyl and naphthalene respectively.
6. Borazine gets slowly hydrolysed by water to produce boric acid, ammonia and Hydrogen.Hydrolysis is favoured by the increase in temperature.
B3N3H6 + H2O -----> H3BO3 + NH3 + H2
Struture:
In Borazine both Boron and Nitrogen are sp2 hybridised. Each N-atom has one lone pair of electrons, while each B-atom has an empty p-orbital. (B-N) bond in borazine is a Dative bond, which arises from the sidewise overlapping between the filled p-orbitals of N-atom and empty p-orbitals of B-atom. In this cyclic compound, the three BH units and three NH units are alternate to each-other.
Preparation and Preperties of Diborane:
Preparation:1. It is prepared by the action of H3PO4 on Mg-boride-
H3PO4 + Mg3B2 = B2H6 + Mg3(PO4)2
2. When BF3 is reduced by sodium borohydride or LiH, diborane is formed.
4BF3 + 3NaBH4 = 2B2H6 + 3NaBF4
Properties:
Physical Propeeties:
* It is air sensitive, volatile and reactive gas with repelling smell.
* It is fairly stable in absence of moisture and grease.
Chemical Propeeties:
1. Reaction with dry HCl: it react with dry HCl in presence of Al2X6, forms B2H5Cl
B2H6 + HCl = B2H5Cl + H2
2. Reaction with concentrated KOH solution:
With concentrated KOH solution, it forms potassium hypoborite.
B2H6 + 2KOH = K2(BH2OH)2 + H2
3. Hydrolysis: On hydrolysis, diborane forms boric acid.
B2H6 + 6H2O = 2H3BO3 + 6H2
4. Burn in Air: It burns in air release large amount of heat.
B2H6 + 3O2 = B2O3 + 3 H2O
5. Reaction with chlorine: it violently react with chlorine.
B2H6 + 6Cl2 = 2BCl3 + 6HCl
6. Reaction with Ammonia:
Diborane react with ammonia but the product is depend upon the experimental condition.
B2H6 + 2NH3 ---L.T.----> B2H6.2NH3
B2H6 + 2NH3 ---H.T.---->2BN + 6H2
3B2H6 + 6NH3 ---H.T.---->2B3N3H6+ 12H2
Organic Chemistry (Second Paper)
Determination of Molecular Mass of Acids by Silver Salt Method:
This method is based on the fact that they form insoluble silver salts, which upon heating decompose to leave a residue of metallic silver.Procedure: The unknown acid is dissolved in water and treated with a slight excess of ammonium hydroxide. The excess of ammonia is boiled off. To this sufficient quantity of silver nitrate is added when a white precipitate of silver salt is obtained. The precipitate is separated by filtration, washed successively with water, alcohol and ether and dried in the steam oven. About 0.2 g of the dry silver salt is weighed into a crucible and ignited until the decomposition is complete. Ignition is repeated until the crucible with the residue of silver has attained constant weight. The molecular mass of the acid is then calculated from the mass of the silver salt taken and the mass of the residue of metallic silver obtained from it.
Calculations:
Weight of unknown carboxylic acid = W g
Weight of silver salt of that acid = X1 g
Weight of metallic silver = X2 g
i.e. X2 g of silver are obtained from X1 g of silver salt
Then 108 g of silver are obtained from = (X1/X2) X 108 g of silver salt
Thus, molar mass of silver salt of carboxylic acid = (X1/X2) X 108 g
For monocarboxylic acid, molar mass = (molar mass of Ag-salt) - (atomic mass of Ag) + (Atomic mass of H)
= (108 X1/X2) − 108 + 1
= (108 X1/X2) − 107
For polybasic carboxylic acid of basicity (n), molar mass =n(Molar mass of Ag-salt) - n(atomic mass of Ag) + n(atomic mass of H)
= (n108 X1/X2) − 108n + 1n
= (n108 X1/X2) − 107n
Molar mass of Acid = [wt. of Ag salt/wt. of Ag) X 108 − 107] Basicity of Acid
Q. About 0.759 g of silver salt of a dibasic acid was ignited when a residue of 0.463 g of metallic silver was left. Calculate the molecular mass of the acid.
Solution: 0.463 g of metallic silver is left by silver salt = 0.759 g216 g of metallic silver is left by silver salt =( 0.759/0.463) X 216 = 354 g
so, molecular mass of dibasic acid = (mol. wt of Ag salt − 2) X 108 + 2 = 140 g
Detection of Nitrogen:
Sodalime Test:
The given substance is mixed with double the amount of sodalime and heated in a test tube. The vapour of ammonia evolved show the presence of nitrogen.Sodium Test (Lassaigne's Test):
This is a golden test for the detection of nitrogen in all classes of nitrogenous compounds. It involves the following steps:The substance is heated strongly with Na metal
Na + C + N → NaCN
The water extract of the fused mass is boiled with ferrous sulphate solution.
FeSO4 + 2NaOH → Fe(OH)2 + Na2SO4
6NaCN + Fe(OH)2 → Na4[Fe(CN)6] + 2NaOH
To the cooled solution is then added a little ferric chloride solution and excess of concentrated HCl.
Na4[Fe(CN)6] + 4FeCl3 → Fe4[Fe(CN)6]3 + 12NaCl
prussian blue
The formation of prussian blue or green coloration confirms the presence of nitrogen in given organic compound.
Detection of Halogens:
Sodium Test:
Upon fussion with sodium, the halogens in the organic compound are converted to the corresponding sodium halides. Thus,Cl + Na → NaCl
Br + Na → NaBr
I + Na → NaI
Acidify a portion of sodium extract with dilute HNO3 and add to it AgNO3 solution.
White ppt. soluble in ammonia indicates Chlorine
Yellowish ppt. sparingly soluble in ammonia indicates Bromine
Yellow ppt. insoluble in ammonia indicates Iodine.
Detection of Sulphur
Sodalime Test:
If sulphur is present in the given organic compound, upon fussion with Na reacts to form sodium sulphide.2Na + S → Na2S
Thus, the sodium extract obtained from the fused mass may be tested as:
A. To a portion, add freshly prepared sodium nitropruside solution. A deep violet coloration indicates the presence of sulphur.
B. Acidify a second portion of the extract with acetic acid and then add lead acetate solution. A black ppt. of lead sulphide confirm the presence of sulphur.
Pb(CH3COO)2 + N2S → PbS + 2CH3COONa
Lead Acetate Black ppt.
Detection of Phosphorus
The solid substance is heated with an oxidizing agent such as concentrated nitric acid or a mixture of sodium carbonate and potassium nitrate. The phosphorus present in the substance is thus oxidized to phosphate. the residue is extracted with water, boiled with some nitric acid and then a hot solution of ammonium molybdate is added to it in excess. A yellow coloration or ppt. indiccates the presence of phosphurs.Estimation of Nitrogen
Kjeldahl Method
This method is based on the fact that when an organic compound containing nitrogen is heated with concentrated sulphuric acid, the nitrogen in it is quantitatively converted into ammonium sulphate. The resultant liquid is then treated with excess of alkali and the liberated ammonia gas absorbed in excess of standard acid. The amount of ammonia is determined by finding the ammount of acid neutralized by back titration with some statndard alkali.Calculation:
Let the weight of organic substance be x gm.
V volume of N HCl is required for complete neutralization of NH3 evolved.
V ml. N HCl ≡ V ml. N NH3
1000ml N NH3 contains 17 gm of NH3 or 14 gm N
Ammount of N present in V mol of N NH3 = (14/1000). V . N gm
% of nitrogen = wt. of nitrogen (100/wt. of substance)
= (14/1000).V.N.(100/x) = 1.4VN/x
where: N = Normality and V = Volume of the acid used.
Estimation of Sulphur
Carius Method
Known weight of organic compound is heated with fuming nitric acid in carius tube, sulphur present in organic compound oxidized to sulphuric acid. The whole solution is then transfered in a beaker and BaCl2 solution in excess is added to it for complete ppt. The ppt. of BaSO4 is then washed, dried and ignited to obtain the constant wt. of BaSO4. Now by knowing the weight of organic compound and BaSO4, the percentage of sulphur in the organic compound is easily calculated.Organic Compound + HNO3 → CO2 + H2 + H2SO4
H2SO4 + BaCl2 → BaSO4 + 2HCl
Let wt. of an organic compound = a gm
wt. of BaSO4 = b gm
Calculation:
molecular weight of BaSO4 = 233
∵ 233 gm BaSO4 contains 32 gm S.
∴ b gm BaSO4 contains (32 X b)/233 gm S.
Again,
∵ a gm of organic compound contains (32 X b)/233 gm S.
∴ 100 gm of organic compound contains (32 X b X 100)/233 X a gm S.
∴ % of Sulphur in organic compound = (32 X b X 100)/233 X a
Grignard Reagent/Alkyl magnesium halide (R-Mg-X)
Method of Preparation:Grignard Reagent may be prepared by the reaction of alkyl halide with Mg metal in presennce of dry ether.
R-X + Mg → R-Mg-X
Some Important Reactions:
Important Questions For Examination-2020
1. Explain the followings:
a. Kohlrausch’s Law
b. Cell Constant
c. Boyle’s Temperature
d. Vapour Pressure
e. Ionic mobility
2. a. What is Azeotropic mixture ?
b. Why vapour pressure of pure solvent is more than that of solution ?
3. State Vander Waal equation and Explain the significant of Vander Waal constants a and b.
4. Write short notes on the followings:
a. Trouton’s Rule
b. Viscosity
c. Specific conductance
5. What is conductivity? Explain the effect of dilution on molar, equivalent and specific conductivity.
6. What is critical constants ? Write its value.
7. What is transport Number ? why transport number of Li+
ion is less than that of Na+ ?
8. Explain the deviation of real gas from ideal gas behaviour.
9. Derive the Vander Waal equation or real gas equation.
1. Explain the followings:
a. Limiting Radius Ratio
b. De Broglie Hypothesis
c. Heisenberg Uncertainty Principle
d. Internal Energy and Enthalpy
e. Exothermic and Endothermic Reaction
2.a. Find out an expression for the work done in an isothermal reversible expansion
b. Prove that CP – CV = R
3. Explain the first law of thermodynamics and its limitations.
4. Write short notes on the followings:
a. Black Body Radiation
b. Adiabatic Process
c. Entropy
5. Calculate the number of atoms per unit cell in
a. Simple cubic
b. Face centred cubic
c. Body centred
6. What is Bragg’s Equation.
7. Explain the Law of rotational Indices.
8. Explain the Bond dissociation Enthalpy.
Please Wait...
Please Wait...
Previous Year Questions
Year-2018
1. a. Describe the deviation of real gas from ideal behavior.
b. State vander waal equation and explain the significance of vander waal constant.
c. What is Boyle’s temperature.
2. a. What do you mean by vapour pressure of a liquid. Give a method for its determination.
b. Explain the term parachore. Discuss its utility in the elucidation of structure of molecules.
3. a. Define
i. Number of phases
ii. Degree of freedom
iii.Number of component with suitable example
b. State and explain phase rule. Give example.
4. a. Explain any two of the following: i. Kohlrausch’s law
ii. Cell constant
iii. Specific constant
b. The molar conductivities of NaCl, HCl and sodium acetate are 126.4, 425.9 and 91.0 scm2mol-1 respectively at infinite dilution. Calculate the molar conductance of CH3COOH at infinite dilution.
Year-2018
1. a. Discuss
i. de broglie hypothesis
ii. Heigenberg uncertainty principle
b. Calculate the wave length associated with an electron (m = 9.11 x 10-28g) travelling with a velocity of 50% of the velocity of light.
2. a. Explain
i. Exothermic and Endothermic reaction
ii. Internal energy and Enthalpy
b. Prove that Cp − Cv = R for an ideal gas
3. a. What do you mean by
i. Bravias Lattices
ii. Millar Indices
b. Discuss the crystal structure of CsCl
4. a. Find out an expression for the work done in isothermal reversible expansion.
b. What do you mean by limiting radius ratio. Calculate it for coordination number six.
Year-2018
1. a. Name all the alkaline earth metals and write down their atomic number and electronic configuration.
b. How will you account for any three of the following:
i. KOH is stronger base than Ba(OH)2
ii. Lattice energy varies as LiF > NaF > KF > RbF > CsF
iii. The conductivity of Li+ in aqueous solution is less than that of Cs+
iv. Na gives golden yellow color in flame
2. a. Draw molecular orbital energy level diagram for oxygen molecule.
b. Arrange the following in order of increasing bond length-
O2+, O2 and O2-
Identify the species which are paramagnetic among the above.
c. What is difference between σ and π bond.
3. a. Identify the group and period of the following elements in the periodic table-
Mg(12), Sn(50), Ag(79) and Ce(58)
To which block of the periodic table they belong.
b. Which among Ti(22), V(23), Cr(24) and Mn(25) has the maximum second ionization energy.
c. Define electronegativity.
4. a. Give the structure difference of diamond and graphite.
b. Describe the structure of fullerenes.
c. Why CO2 is gas but SiO2 is solid ?
5. a. Explain any two of the following-
i. Oxyacids of phosphorus
ii. Inert pair effect
iii. Inter halogen compounds
b. Give the hybridization, structure and shape of PCl3 and PCl5
Year-2019
1. a. Define ionization potential and explain why ?
i. Ionization potential of Li is greater than Na
ii. Ionization potential of Na is less than Mg
iii. First ionization potential of K is less than Ca but second ionization potential is high.
2. a. Name alkali metals and give their electronic configuration.
b. Discuss the variation of thermal stability of BeCO3, MgCO3, CaCO3 and SrCO3
c. The solubilities of sulphates of alkaline earth metals decreases on moving down the group. Justify.
3. a. What do you mean by dipole moment.
b. Which among the following pairs has high dipole moment
i. CO2 and NO2
ii. HF and HCl
iii. o-dichlorobenzene and p-dichlorobenzene
c. What is unit of dipole moment
4. a. Sulphur forms SF4 and SF6 but oxygen does not form OF4 and OF6, why ?
b. Discuss the variation in b.p. of H2O, H2S, H2Se and H2Te
c. Which among H2SO3 and H2SO4 are more acidic ?
d. Write down the structure of peroxodisulphuric acid
5. a. Give the shape, structure and hybridization of the following-
XeF2, CO3-2, BrF3, SF4 and AF5
b. Write the resonating structure of NO3-
Year-2019
Discuss the different types of hybridization used in organic molecules with reference to bonding in C2H2 and C6H6.
Why acetylene is more acidic than ethylene ?
How will you determine molecular mass of a carboxylic acid by silver salt method.
How will you synthesis glycerol from propene ?
How glycerol reacts with the followings:
a. a mixture of acetic acid and acitic anhydride
b. Hydrogen Iodide
c. PCl3
d. P2O5
Write down all possible isomeres of CH3CH2CH2CH2NH2.
How a Primary, Secondary and Tertiary animes reacts with Nitrous acid
Give formula of Lithium dialkyl cuprate. What is its use.
Arrange the following
a. In increasig order of stability: CH3CH2CH2+, CH3CH+CH3, (CH3)3C+
b. In increasing order of acidic strength:
HCOOH, CH3COOH, CH3CH2COOH
c. In increasing order of basic strength:
CH3C(NH)NH2, CH3CH2NH2, (CH3)2NH, CH3CONH2
d. In increasing order of stability 1,3-pentadiene, 1,4-pentadiene,Benzene
Explain Nucleophoile, Electrophile and Free Radical
What do you mean by SN1 and SN2 reaction
Give an example of electrophylic substitution reaction
Join Maxbrain Chemistry for Study.
Classroom Course: XI-XII(CBSE & ICSE) │ B.Sc. All Parts (Hon's & Sub.) │ M.Sc. │ IIT-JEE │ NEET │ IIT-JAM │ GATE │ PAT │ SET │ CSIR │ Biochemistry | Microbiology