(i) Aromatisation:
Alkanes with six to ten carbon atoms are converted into homologous of benzene at high temperature and in the presence of catalyst. This process is known as aromatization. It occurs by simultaneous cyclisation followed by dehydrogenation of alkanes n-Hexane passed over Cr₂O₃ supported on alumina at 873 K gives benzene.

(ii) Pyrolysis:
Pyrolysis is dened as the thermal decomposition of organic compound into smaller fragments in the absence of air through the application of heat. 'Pyro' means 'fire' and 'lysis' means 'separating'. Pyrolysis of alkanes also named as cracking.
In the absence of air, when alkane vapours are passed through red-hot metal it breaks down into simpler hydrocarbons.
1)                        CH₃-CH₂-CH₃
                                    \(\downarrow 773K\) 
\({ CH }_{ 3 }-CH={ CH }_{ 2 }+{ CH }_{ 2 }={ CH }_{ 2 }+{ H }_{ 2 }+{ CH }_{ 4 }\)
2) 2CH₃-CH₃ \(\overset { 773K }{ \longrightarrow } \) CH₂ = CH₂ + 2CH₄
The products depends upon the nature of alkane, temperature, pressure and presence or absence of catalyst. The ease of cracking in alkanes increases with increase in molecular weight and branching in alkanes. Cracking plays an, important role in petroleum industry.

The chloro fluoro derivatives of methane and ethane are referred by trade name Freons. These Chloro Fluoro Carbon compounds are stable, non-toxic, noncorrosive and noninflammable, easily liquefiable and are used in refrigerators, air- conditioners and ln the production of plastic foams. CFC's are the exhaust of supersonic air craft's and jumbo jets in the upper atmosphere. They slowly pass from troposphere to stratosphere. ,They stay for very longer period of 50 - 100 years. In the presence of uv radiation, CFC's break up into chlorine free radical
CF₂Cl₂ \(\xrightarrow{hv}\) CF₂ CI +Cl.
CFCl₃ \(\xrightarrow{hv}\)CFCI₂+ Cl
Cl^o+O₃ ⟶ C10+O₂
Clo^o+O ⟶ CI+O₂
Chlorine radical is regenerated in the. course of reaction. Due to this continuous attack of Cl⁰ thinning of ozone layer takes place which leads to formation of ozone hole. It is estimated that for every reactive chlorine atom generated in the stratosphere 1,00,000 molecules of ozone are depleted.

Standard characteristics of drinking water.

i) chiral molecule:
The chiral molecule is
ii) S_N2 reaction:
undergoes S_N2 reaction faster as it is a primary alkyl halide.
\( \text { (or) } \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{Cl}\\ 1 - chlorobutane\)

P_solute  = K_H X_{solute in solution}
At pressure 1.5 atm,
p₁ = K_H x_{1       ------}(1)
At pressure 6.0 atm,
p₂ = K_H x₂        -----(2)
Dividing equation (1) by (2)
From equation p₁/p₂ = x₁/x₂
1.5/6.0 = 0.24/x₂
Therefore x₂ = 0.24 x 6.0/1.5 = 0.96 g/L.

\(\Delta n_{g}=\sum n p_{(g)}-\sum n R_{(g)}\)
As \(\Delta n_{p}(g)\) is greater \(\Delta n_{g}=+v e\)
\( \therefore K_{p}=K_{c}(R T)^{+v e} \)
\(\therefore K_{p}>K_{c} \)
So K is smaller than K_p.

The balanced chemical equation is,
\({ CaCO }_{ 3 }\left( s \right) \overset { \triangle }{ \longrightarrow } { CaO }_{ (s) }+{ CO }_{ 2 }(g)\)
As per the stoichiometric equation,
1 mole (100g) CaC0₃ on heating produces 1 mole CO₂

At STP, 1 mole of CO₂ occupies a volume of 22.7 liters
At STP, 50 g of CaCO₃ on heating produces,

= 11.35 liters of CO₂

H₂O \(\rightleftharpoons\) H₂O₍₁₎ \(\triangle_{fus}\) H is 6 kJ mol-1
(a) ΔG=ΔH-TΔS
When the reaction is carried out at 0°K
or ΔS=0
ΔG=ΔH
(b) \(H_2O_{s} \rightleftharpoons H_2O{_{(I)}}\)
\(\Delta_{fus}H= 6\) kJ mol^-1
= 6000 kJ mol^-1
\(\Delta_{fus}H= 6\) kJ mol^-1
= 6000 J mol^-1

Iso electronic species are those which have the same number of electrons.
(i) Ar has 18 electrons. Therefore, the species p^-3, S^-2, Cl^-, K+, Ca⁺² are isoelectronic with Ar.
(ii) F^- has 10 electrons. Therefore the species, N^-3, 0^-2, Ne, Na⁺, Mg⁺² are isoelectronic with F^-
(iii) K⁺ has 18 electrons. Therefore, the species p^-3, S^-2, CI^-, Ar, Ca⁺²are isoelectronic with K⁺
(iv) Sr⁺²has 36 electrons. Therefore Br^-, Kr, Rb⁺ etc are isoelectronic with Sr⁺²

The passenger in a moving bus falls in the forward direction, when brakes are applied suddenly is in accordance with Newton's first law of motion.
When breaks are applied, the automobile comes to rest but the passenger due to inertia of motion tend to continue to move in forward direction hence fall forward.
The movement of helium balloon in the opposite direction is due to the difference in the density of surrounding air. When the car stops suddenly, the air moves forward due to inertia of motion.
As a result the air in the front of the car is more dense that at the rear of the car creating more pressure on the front side of the balloon. This causes the balloon move towards the rear of the car where air is less dense.
The body is at rest but the car is forward motion, due to inertia the passengers are pushed toward the front of the car on forward acceleration.

Elements having same atomic number but different mass numbers are called isotopes. These are variant, of an element which differ in neutron numbers, which contain equal number of protons. They differ in relative atomic mass but not in chemical properties.
The isotopes of hydrogen are:
1. Protium \(\left(\mathrm{H}^{1} \text { or } \mathrm{H}\right)\) [Predominant form]
2. Deuterium (or) Heavy hydrogen \(\left({ }_{1} \mathrm{H}^{2} \text { or }{ }_{1} \mathrm{D}^{2}\right)\)
3. Tritium \(\left({ }_{1} \mathrm{H}^{3} \text { or }{ }_{1} \mathrm{T}^{3}\right)\)

\(\sigma\) - bond is formed by the head on overlap of atomic orbital and \(\pi\) - bond is formed by the side-ways overlap.
The head on overlap of sigma bonds accounts for its higher stability, thereby makes it a stronger bond.

(i) Nucleophilic substitution reaction - Br is substituted by H\(\overset { - }{ s } \)
(ii) Electrophilic addition reaction - HCI is added to the C = C bond.
(iii) β-elimination reaction - Hand Br are eliminated from successive carbon atoms.
(iv) Nucleophilic substitution reaction with rearrangement -\(\overset { - }{ O } \)H us substituted by Br.

\(\underset { (1\times 22.4L) }{ { H }_{ { 2 }_{ (g) } } } +\underset { (1\times 22.4L) }{ Cl_{ { 2 }_{ (g) } } } \rightarrow \underset { (2\times 22.4L) }{ 2HCl_{ _{ (g) } } } \) 
2 x 22.4 L of HCI is produced by 22.4 L of Cl₂.
\(\therefore\) 89.6 L of HCl will be produced by

= 44.8 L of chlorine.

The nuclear charge in Mg⁺² is more than Na+ and therefore electrons are drawn more closely, and hence size decreases

"The total pressure of a mixture of non-reacting gases is the sum of partial pressures of the gases present in the mixture"

(i) ground state :

This type of electronic configuration have ten possible arrangement (i.e. Half filled configuration is More).

A known mass of the organic compound (w) containing phosphorus is heated with fuming HNO₃ in a sealed tube where C is converted into CO₂ and H to H₂0. Phosphorus present in organic compound is oxidized to phosphoric acid which is precipitated, as ammonium phosphomolybdate by heating with cone. HNO₃ and then adding ammonium molybdate.

The precipitate of ammonium phosphomolybdate thus formed is filtered washed, dried and weighed. In an alternative method, the phosphoric acid is precipitated as magnesium-ammonium phosphate by adding magnesia mixture (a mixture containing MgCI₂, NH₄CI and ammonia) This ppt is washed, dried and ignited to get magnesium pyrophosphate which is washed, dried a weighed. The following are
the reaction that takes place. By knowing the mass of the organic compound and the mass of ammonium phosphomolybdate or magnesium pyrophosphate formed, the percentage of P is calculated.
Mass of organic compound is wg
Weight of ammonium phosphomolybdate = xg
Weight of magnesium pyrophosphate = yg
Mole mass of (NH₄)₃PO₄.l2Mo0₃ is 877g
[3 x(14 + 4) + 31 +4(16)] + 12 (96+3x16)
Molar mass of Mg₃P₂O₇ is 222 g
(2x24) + (31x2) + (7x16)
1877g of (NH₄)₃PO₄.12Mo O₃contains 31g of P 
xg of(NH₄)₃PO₄.12 MoO₃ in w g of organic
compound contains \(\left( \frac { 32 }{ 1877 } \times \frac { x }{ w } \right) \)of phosphorous
Percentage of phosphorous = \(\left( \frac { 32 }{ 1877 } \times \frac { x }{ w } \times 100 \right) \)
(or)
227 of Mg₂P₂O₇ contains 62 g of P
Y g of of Mg₂P₂O₇ in w g of organic compound contains \(\left( \frac { 62 }{ 222 } \times \frac { y }{ w } \right) \)of P.
Percentage Phosphorous= \(\left( \frac { 62 }{ 222 } \times \frac { y }{ w } \times 100 \right) \)

Reaction between iso-butylene and HCI are as follows:

(ii) When CH₃- Br is treated with KCN, methyl cyanide is formed.
\(\underset { Methyl\\ Bromide }{ { CH }_{ 3 } } -Br+\underset { Potcyanide }{ KCN } \longrightarrow \underset { Methyl\\ Cyanide }{ { CH }_{ 3 } } -CN+\underset { Pot\\ Bromide }{ KBr } \)
(iii)  is a chiral molecule, as four atoms attached to *C - atom are different
(iv) 
This order is due to steric reasons, the order of reactivity in S_N2 follows the order: 1°>2°> s 3°.

This type of isomers have same molecular formula. but differ in their bonding sequence. Structural or constitutional isomerism is further classfied into following types.
(a) Chain or nuclear or skeletal isomerism: These isomers differ in the way in which the carbon atoms are bonded to each other in a carbon chain or in other words isomers have similar molecular formula but differ in the nature of the carbon skeleton (ie. Straight or branched)
\({ CH }_{ 3 }-{ CH }_{ 2 }-{ CH }_{ 2 }-CH_{ 2 }-{ CH }_{ 3 }\)
n-Pentane
\(\underset { \underset { CH_{ 3 } }{ | } }{ { CH }_{ 3 }-{ CH }-{ CH }_{ 2 }-{ CH }_{ 3 } } \)
Isopentane 2-methyl butane
(b) Position isomerism:
If different compounds belonging to same homologous series with the same molecular formula and carbon skeleton, but differ in the position of substituent or functional group or an unsaturated linkage are said to exhibit position isomerism.
CH₃-CH₂-CH₂-CH = CH₂
Pent-1ene and
CH₃-CH₂-CH = CH-CH₃
pent-2-ene
(c) Functional isomerism: Different compounds having same molecular formula but different functional groups are said to exhibit functional isomerism.
CH₃-CH₂.CHO
Propanal
(aldehyde group)
\(\overset { \overset { O }{ || } }{ { CH }_{ 3 }-{ CH }-{ CH }_{ 2 }-{ CH }_{ 3 } } \)
Propanone
(Keto group)
(a) Metamerism:This type of isomerism is a special kind of structural isomerism arises due to different alkyl groups attached to . the either side ofthe same functional group and having same molecular formula.
CH₃ - O - C₃H₇
Methyl propyl ether
1-methoxypropane
C₂H₅-OC₂H₅
diethyl ether
ethoxyethane
(b) Tautomerism : It is a special type of functional isomerism in which a single compound exists in two readily inter convertible structures that differ markedly in the relative position of atleast one atomic nucleus, generally hydrogen. The two different structures are known as tautomers. There are slveral types of tautomerism and the two important types are dyad and triad systems.

(c) Ring chain isomerism: In this type of isomerism, compounds having same molecular formula but differ in terms of bonding of carbon atom to form open chain and cyclic structures for eg:

Given \(m_{N_2}\) = 28 g \(m_{H_2}\) = 6g
V = 1 L
\((n_{N_2})_{initial}={28\over 28}=1\ mol\)
\((n_{H_2})_{initial}={6\over 2}=3\ mol\)
N₂(g) + 3H₂(g) ⇌ 2 NH₃(g)

\([NH_3]=({17\over 17})=1\ mol=1\ mol\)
Weight of N₂ = (no. of moles of N₂) × molar mass of N₂
= 0.5 x 28 = 14 g
Weight of H₂ = (no. of moles of H₂) × molar mass of H₂
= 1.5 x 2 = 3 g

\(\therefore\) The empirical formula = CH₂Cl

(a) 5.076 atm
(b) 3.495 atm

(i) Water molecule undergoes Sp³ hybridization.
(ii) The O - atom in H₂O molecule contains two bonding pairs and two lone pairs of electrons.
(iii) The overall arrangement for four electron pairs is tetrahedral but the Ip-Ip repulsions being greater than Ip - bp rep in H₂O.
(iv) The HOH angle is reduced to 104.5° than 109°28'.
(v) The molecule has a bent shape.

Born - Haber cycle for the formation of CaCl₂
Born - Haber cycle is used to calculate the lattice enthalpy of CaCl₂ 

Δ^oH₁ - Enthalpy change for the sublimation of Ca_(s) to Ca_(g)
Δ^oH₂ - Enthalpy change for dissociation of Cl_2(g) to 2Cl_(g)
Δ^oH₃ - Ionisation energy for Ca_(g) to Ca²⁺_(g)
Δ^oH₄ - Electron affinity for the conversion of 2Cl_(g) to 2Cl^-_(g)
Δ^oH₅ - Lattice enthalpy for the formation of solid CaCl₂·
Δ^oH_f =Δ^oH₁+Δ^oH₂+Δ^oH₃+Δ^oH₄+Δ^oH₅

(i) Electron configuration of nitrogen
(Z = 7) 1s² 2s² 2p³.
Electron configuration of oxygen
(2= 8) 1s² 2s² 2p⁴.
Nitrogen has a half filled electronic configuration which is much more stable than an incomplete p-orbital of oxygen which would need to give up one of it's electrons to attain the stability of nitrogen. Hence nitrogen would require more ionization energy to remove an electron from it's outer shell than oxygen.
(ii) Electron configuration of carbon
(Z = 6) 1s²2s²2p².
Electron configuration of Boron
(Z = 5) Is²2s²2p¹
The size of a carbon atom is smaller than boron So the valence electron of carbon has greater nuclear charge than that of boron. Hence the first I.E of carbon is greater than that of boron. However, the second ionization enthalpy of boron is higher than that of carbon. This is because after losing electron, Boron has a fully filled orbital (2s²) than carbon (2p¹). Fully filled orbitals have more stability than partially filled orbitals so greater amount of energy will be needed to remove an electron from boron. So in this case, the second I.E of boron is higher than that of carbon.
(iii) The electron affinities of Be, Mg and noble gases are almost zero because both Be (Z = 4; 1s²2s²) and Mg (Z = 12; Is²2s²2p⁶3s²) are having s orbital fully filled in their valence shell. Fully filled orbitals are most stable due to symmetry. Therefore, these elements would be having least tendency to accept electron. Hence, Be and Mg would be having zero electron affinity. Whereas N (Z = 7; 1s²2s²2px¹2py¹2pz¹ and P (Z = 15) Is² 2s² 2p⁶ 3s² 3p³ is having half filled 2p-subshell. Half filled sub shells are most stable due to symmetry (Hund's rule). Thus, nitrogen and phosphorous are having least tendency to accept electron. Hence, have low electron affinity.
(iv) Fluorine is highly electro negative in nature therefore as it gains the electron its octet become stable and releases the energy so exothermic. while in oxygen the addition of first electron is exothermic in nature but addition of second electron experiences high repulsive force. So needs extra external energy to enter outer shell, hence endothermic in nature.

i) Both (A) and R are correct and (R) is the correct explanation of (A)