(i) The principle behind chromatography is selective distribution of the mixture of organic substances between two phases - a stationary phase and a moving phase. The stationary phase can be a solid or liquid, while the moving phase is a liquid or a gas. When the stationary phase is a solid, the moving phase is a liquid or a gas. If the stationary phase is solid, the basis is adsorption, and when it is a liquid, the basis is partition. So the chromatography is dened as a technique for the separation of a mixture brought about by differential movement of the individual compound through porous medium under the influence of moving solvent.
(ii) Thin layer chromatography : This method is an another type of adsorption chromatography with this method it is possible to separate even minute quantities of mixtures. A sheet of a glass is coated with a thin layer of adsorbent
(cellulose, silica gel or alumina). This sheet of glass' is called chromoplate or thin layer chromatography plate. After drying the plate, a drop of the mixture is placed just above one edge and the plate is then placed in a closed jar
containing eluent (solvent). The eluent is drawn "up the adsorbent layer by capillary action. The components of the mixture move up along with the eluent to different distances depending upon their degree of adsorption of each component of the mixture. It is expressed in terms of its retention factor (ie) R_f value 
\({ R }_{ f }=\frac { Distance\ moved\ by\ the\ substance\ from\ base\ line(x) }{ Distance\ moved\ by\ the\ solvent\ from\ base\ line(y) } \)
The spots of colored compounds are visible on TLC plate due to their original color. The colorless compounds are viewed under uv light or in another method using iodine crystals or by using appropriate reagent.

Elimination reactions may proceed through two different mechanisms namely E₁ and E₂

(i) The rate of E₂ reaction depends on the concentration of alkyl halide and base Rate = k [alkyl halide] [base]
(ii) It is therefore, a second order reaction. Generally primary alkyl halide undergoes this reaction in the presence of alcoholic KOH. It is a one step process in which the abstraction of the proton from the . p carbon and expulsion of halide from the a carbon occur simultaneously. The mechanism is shown below.


(iii) Generally, tertiary alkyl halide which undergoes elimination reaction by this mechanism in the presence of alcoholic KOH. It follows first order kinetics. Let us. consider the following elimination reaction.
Step - 1: Heterolytic fission to yield a carbocation

Step - 2 Elimination of a proton from the \(\beta\)- carbon to produce an alkene.

Partition Chromatography:
Paper chromatography (PC) is an example of partition chromatography. The same procedure is followed as in thin layer chromatography except that a strip of paper acts as an adsorbent is method involves continues differential portioning of  components of a mixture between stationary and mobile phase. In paper chromatography, a special quality paper known as chromatography paper is used. This paper act as a stationary phase.
A strip of chromatographic paper spotted at the base with the solution of the mixture is suspended in a suitable solvent which act as the mobile phase solvent rises up and flows over the spot. The paper selectively retains different components according to their different partition in the two phases where a chromatogram is developed. The spots of the separated colored compounds are visible at different heights from the position of initial spots on the chromatogram. The spots of the separated colorless compounds may be observed either under ultraviolent light or by the use of an appropriate spray reagent.

\(\Delta T_f=0.093^oC=0.093K\)
m = ?
K_f = 1.86K Kg mol^-1
\(\Delta T_f=K_f.m\)
\(\therefore m={\Delta T_f\over K_f}\)
\(={0.093K\over 1.86\ K\ Kg\ mol^{-1}}\)
= 0.05 mol Kg^-1
= 0.05 m.

Given data:
[A] = [B] = [C]= [D] =1 M
K_c = 100
[D]_eq = ?
Solution:
Let x be the no moles of reactants reacted

\(K_c={[C][D]\over [A][B]}\)
\(100={(1+x)(1+x)\over (1-x)(1-x)}\)
\(\sqrt{100}=\sqrt{{(1+x)(1+x)\over (1-x)(1-x)}}\)
\(10={1+x\over 1-x}\)
10(1 - x) = 1 + x
10 - 10x - 1 - x = 0
9 - 11x = 0
11x = 9
\(x={9\over 11}=0.818\)
[D]eq = 1+x = 1 + 0.818 = 1.818M.

(a) Hydrogen peroxide can be made by adding a metal peroxide to dilute acid.
BaO_2(s) + H₂SO_4(aq) ⟶ BaSO_4(S) + H₂O_2(aq)
(b) Structure of H₂O₂

(i) H₂O₂ has a non-polar structure. The molecular dimensions in the gas phase and solid phase differ as shown in the figure.
(ii) Both in gas phase and solid phase, the H₂O₂ molecule adopt a skew configuration due to repulsive interaction of the -OH bonds with lone pairs of electrons on each oxygen atom.
(iii) Indeed, it is the smallest molecule known to show hindrance rotation about a single bond. In solid phase, the dihedral angle is sensitive and hydrogen bonding decreasing from 111.5° in the gas phase to 90.2°, in the solid phase.
(iv) Structurally, H₂O₂ is represented by the dihydroxyl formula in which the two O-H groups do not lie in the same plane. In the solid phase of molecule, the dihedral angle reduces to 90.2° due to hydrogen bonding and the O-O-H angle expands from 94.8° to 101.9°.
(v) One way of explaining the shape of hydrogen peroxide is that the hydrogen atoms would lie on the pages of a partly opened book, and the oxygen atoms along the spin.

Molar mass of O₂, = 32 g mol^-1
∴ 8 g of O₂=\(\frac{8}{32}\)mol = 0.25 mol
molar mass of H₂ = 2 g mol^-1
∴ 4 g of H₂=\(\frac{4}{42}\)mol = 2 mol
Total number of mol (n) = 0.25 + 2 = 2.25
Volume (V) = 1 dm³; Temperature (T) = 27° C = 300K
R = 0.083 bar dm³K^-1mol^-1
PV = nRT (or) \(P=\frac{nRT}{V}\)
(or)    \(P=\frac{(2.25 mol)(0.083 bar dm^{3}K^{-1}mol^{-1})(300 K)}{1 dm^{3}}\)
= 56.025 bar

(i) Free energy is defined as G = H - TS. 'G' is a state function.
(ii) G- Extensive property; ΔG - intensive property. When mass remains constant between initial and final states of system.
(iii) 'G' has a single value for the thermodynamic state of the system.
(iv) G and ΔG values correspond to the system only.

(v) Gibbs free energy and the net work done by the system:
For any system at constant pressure and temperature
ΔG = ΔH - TΔS .....(1)
We know that,
ΔH = ΔU + PΔV
ΔG =ΔU + PΔV-TΔS
from first law of thermodynamics
ΔU = q +w
from second law of thermodynamics
Δ S=\(\frac{q}{T}\) Δ G=q+w+PΔ V-T\((\frac{q}{T}) \)
Δ G = w+PΔV
-ΔG = -w - PΔV ......(2)
But -PΔV represents the work done due to expansion against a constant external pressure.

Displacement reaction: Redox reactions in which an ion (or an atom) in a compound is replaced by an ion (or atom) of another element are called displacement reactions. They are further classified into
(i) metal displacement reactions
(ii) non-metal displacement reactions.

(i) Metal displacement reactions :
Place a zinc metal strip in an aqueous copper sulphate solution taken in a beaker. Observe the solution, the intensity of the blue colour of the solution slowly reduced and finally disappeared.
The zinc metal strip became coated with brownish metallic copper. This is due to the following metal displacement reaction.

(ii) Non-metal displacement:

Variation along a period: Ionisation energy usually increases along a period. This is due to increase of nuclear charge and decrease in size as we move from left to right in a period.
Periodic variation in group: Ionisation energy decreases down a group. As we move down a group, the valence electron occupies new shells, the distance between the nucleus and the valence electron increases. So, the nuclear forces of attraction on valence electron decreases and hence ionisation energy also decreases down a group.

Among anthracene and cyclopentadiene - anthracene is aromatic in nature.

Anthracene		Cyclopentadiene
1.	Planar structure	Planar structure
2.	Contains 14 delocalised \(\pi\) electrons (4n + 2 = 14\(\pi\)e-)	Contains 4 non-delocalised \(\pi\) electrons
3.	Hence aromatic	Hence non-aromatic

This method is to purify liquids from non-volatile impurities, and used for separating the constituents of a liquid mixture which differ in their boiling points.
There are various methods of distillation depending upon the difference in the boiling points of the constituents. The methods are
(i) simple distillation
(ii) fractional distillation and
(iii) steam distillation.
The process of distillation involves the impure liquid when boiled gives out vapour and the vapour so formed is collected and condensed to give back the pure liquid in the receiver. is method is called simple distillation. Liquids with large difference in boiling point (about 40K) and do not decompose under ordinary pressure can be purified by simply distillation Eg. The mixture of C₆H₅NO₂ (b.p 484K) & C₆H₆(354K) and mixture of diethyl ether (b.p 308K) and ethyl alcohol (b.p 351K)

(i) (a) Carbon Monoxide:
Carbon monoxide is a poisonous gas produced as a result of incomplete combustion of coal are firewood. It is released into the air mainly by automobile exhaust. It binds with haemoglobin and form carboxy haemoglobin which impairs normal oxygen transport by blood and hence the oxygen carrying capacity of blood is reduced. This oxygen deficiency results in headache, dizziness, tension, Loss of consciousness, blurring of eye sight and cardiac arrest.
(b) Carbon dioxide:
Carbon dioxide is released into the atmosphere mainly by the process of respiration, burning of fossil fuels, forest fire, decomposition of limestone in cement industry etc.Green plants can convert CO₂ gas in the atmosphere into carbohydrate and oxygen through a process called photosynthesis. The increased CO₂ level in the atmosphere is responsible for global warming. It causes headache and nausea.
ii) a) Carbon Monoxide :
It binds with haemoglobin and form carbory haemoglobin which impairs normal oxygen transport by blood and hence the oxygen carrying capacity of blood is reduced. This oxygen deficiency results in headache, dizziness, tension, Loss of consciousness, blurring of eye sight and cardiac arrest.
b) Carbon dioxide :
It causes headache and nausea.

a) The growth of algae in extreme abundance covers the water surface and reduces the oxygen concentration in water. Thus, bloom-infested water inhibits the growth of other living organisms in the water body. This process in which the nutrient rich water bodies support a dense plant population, kills animal life by depriving it of orygen and results in loss of biodiversity is known as eutrophication.
b) Chemicals from industries
c) Toxic pesticides
d) Detergents and oil floats
e) Acids from mine drianage and salts from various sources.

CH_4(g) + 2O₂ \(\rightarrow\) CO₂ + 2H₂O
16g                          (2x18)g
As per stoichiometric equation,
16 g of methane produces 36 g of H₂O
\(\therefore\) 32 g of methane will produce = \(\frac { 36 }{ 16 } \times 32=72\) g of water.

The elements beyond fermium are called as transfermium elements.

The larger the size of the molecule, the greater will be Van der Waals' attraction. Therefore greater the deviation from ideal behaviour. So bromine will deviate more from ideal behaviour because it has bigger atoms.

(i) 2KMnO₄ + 3H₂O₂ ➝ 2MnO₂ + 2KOH + 2H₂O + 3O₂ (Redox reaction)
(ii) CrCl₃ + 6H₂O ➝ [Cr Cl₂ (H₂O)CI. 2H₂O [Hydration reaction]
(iii) CaO + H₂O ➝ Ca(OH)₂ (Hydration reaction)

\(C^+_2:(\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2 (\pi_2p_x)^2=(\pi_2p_y)^1\)
Bond order = \({1\over2}(8-4)=2\)
\(C^+_2:(\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2 (\pi_2p_x)^2=(\pi_2p_y)^1\)
Bond order = \({1\over2}(7-4)=1.5\)
\(\therefore\)The Bond order decreases by 1.5 during the process.

During the addition reaction the hydridisation of the substrate changes (from sp² ⟶ sp³ in the addition reaction of alkenes or sp ⟶ sp² in the addition reaction of alkynes) as only one bond breaks and two new bonds are formed.

The rate of the following SN₁ reaction depends upon the concentration of alkyl halide (RX) and is independent of the concentration of the nucleophile (OH-).
Hence Rate of the reaction = k [alkyl halide]
SN₁ reaction mechanism by taking a reaction between tertiary butyl bromide with aqueous KOH.
\({ CH }_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ \underset { \overset { | }{ { CH }_{ 3 } } }{ C } } -Br\overset { { OH }^{ - }(aq) }{ \underset { -Br }{ \longrightarrow } } { CH }_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ \underset { \overset { | }{ { CH }_{ 3 } } }{ C } } -OH\)
Tert-Butyl bromide                           Tert-Butyl alcohol
This reaction takes place in two steps as shown below
Step-1 Formation of carbocation :
The polar C - Br bond breaks forming a carbocation and bromide ion. is step is slow and hence it is the rate determining step

The carbocation has 2 equivalent lobes of the vacant 2p orbital, so it can react equally rapidly from either face
Ste - 2
The nucleophile immediately reacts with the carbocation. This step is fast and hence does not affect the rate of the reactions.

As shown above, the nucleophilic reagent OH^- can attack carbocation from both the sides.

The long form of periodic table need clarification about the following:
(i) Position of hydrogen is not defined till now.
(ii) Lanthanides and actinides still find place in the bottom of the table

rate of diffusion \(\propto \frac{1}{\sqrt{M}}\)

Energy electrons symmetry
This is due to the symmetrical distribution and exchange energy of given d- electrons. Symmetry leads to stability.
Exchange energy:
If two or more electrons with the same spin are present in degenerate orbitals, there is a possibility for exchanging their positions. During exchange process, the energy is released and the released energy is called exchange energy. If more number of exchanges are possible, more exchange energy in released. More number of exchanges are possible only in case of half filled and fully filled configurations.
For example, in chromium the electronic configuration is [Ar]3d⁵ 4s¹. The 3d orbital is half filled and there are ten possible exchanges as shown in figure. On the other hand only six exchanges are possible for [Ar]3d⁴ 4s² configuration. Hence, exchange energy for the half filled configuration is more. This increases the stability of half filled 3d orbitals.

The exchange energy is the basis for Hund's rule, which allows maximum multiplicity, that is electron pairing is possible only when all the degenerate orbitals contain one electron each.