(a) both assertion and reason are true and reason is the correct explanation of assertion

Isomerism:
Presence of double bond in alkene provides the possibility of both structural and geometrical isomerism.
Structural Isomerism:
The first two member's ethene C₂H₄ and propene C₃H₆ do not have isomers because the carbon atoms in the molecules can be arranged only one distinct way.
However from the third member of alkene family butene C₄H₁₀ structural isomerism exists.
(i) CH₃-CH = CH-CH₃  1-Butene
(ii) CH₂=CH-CH₂-CH₃ 2-Butene
(iii) \({ CH }_{ 2 }=\overset { \underset { | }{ { CH }_{ 3 } } }{ C } -{ CH }_{ 3 }\) 2-Methyl-1-propene
structures (i) & (ii) are position isomers. structures (i) & (iii), (ii) & (iii) are chain isomers. 
Geometrical isomerism:
It is a type of stereoisomerism and it is also called cis-trans isomerism. Such type of isomerism results due to the restricted rotation of doubly bounded carbon atoms.
if the similar groups lie on the same side, then the geometrical isomers are called C is-isomers. When the similar groups lie on the opposite side, it is called a Trans isomer.
for example: the geometrical isomers of 2-Butane is expressed as follow

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.

\(P_A^o\) = 10 torr, P_solution = 9 torr
W_A = 20 g  W_B = 1 g
M_A = 200 g mol^-1  M_B = ?
\({\Delta P\over P_A^o}={W_B\times M_A\over M_B\times W_A}\)
\({10-9\over 10}={1\times 200\over M_B\times 20}\)
\(M_B={200\over 20}\times 10=100\ g\ mol^{-1}\)

The balanced equation for the formation of HCI is,
H₂(g) + CI₂(g) \(\rightarrow\) 2 HCI (g)
As per the stoichiometric equation, under given conditions,
To produce 2 moles of HCI, 1 mole of chlorine gas is required.
To produce 44.8 litres of HCI, 22.4 litres of chlorine gas are required.
\(\therefore\) To produce 11.2 litres of HCI,

= 5.6 litres of chlorine are required.

(i) Classification is a fundamental and essential process in our day-to-day life for the effective utilization of resources, daily events and materials
(ii) In such a way, for the effective utilization of discovered elements becomes fundamentally essential process.
(iii) The periodic classification of the elements is one of the outstanding contributions to the progress of chemistry.

'U' is an extensive property because its magnitude depends on the quantity of material in the system.

a) Hydrogen is the lightest gas in the atmosphere. So it rises up and other gases which are heavier like O₂ & N₂ come down towards the surface of the earth according to Graham's law of diffusions \(\mathrm{r} \alpha \sqrt{\frac{1}{M}}\). Hydrogen diffuses very fast. Its mean speed in greater than the escape velocity from the earth. As a consequence, H₂ would have escaped from the atmosphere long time ago.
b) The acceleration due to gravity 'g' on moon surface is small. The value of escape velocity is also small. The molecules of the atmospheric gases on the moon's surface have thermal velocities greater than the escape velocity.
All the molecules have escaped. So the atmosphere is so thin.

a) Oxidation is performed in acidic medium:
Eg : H₂O₂ Oxidises FeSO₄ to Fe (SO₄)₃
H₂O₂ + 2H⁺+ 2e^- ➝ 2H₂O (E⁰ = +1.77 V)
2FeSO₄ + H₂SO₄ + H₂O₂ ➝ Fe₂(SO₄)₃ + 2H₂O
b) Reduction is performed in basic medium:
Eg : H₂O₂ reduces KMnO₄ to MnO₂
HO₂^- + OH^- ➝ H₂O + 2e^- (E⁰ = +0.08 V)
\(2 \mathrm{KMnO}_{4}+3 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{MnO}_{2}+2 \mathrm{KOH}+2 \mathrm{H}_{2} \mathrm{O}+3 \mathrm{O}_{2}\)

(i) The benzylic free radical is resonance stabilised. It possess more resonance structures.
(ii) Allylic free radical has only two resonating structures, so it has less delocalisation.

\({ SO }_{ 4 }^{ 2- }\)contains 6 bond pairs and 10 lone pairs
H₃O⁺ contains 3 bond pairs and 1 lone pair.

Particulate pollutants are small solid particles and liquid droplets suspended in air. Many of particulate pollutants are hazardous.
Examples : dust, pollen, smoke, soot and liquid droplets (aerosols) etc,.
They are blown into the atmosphere by volcanic eruption, blowing of dust, incomplete combustion of fossil fuels induces soot. Combustion of high ash fossil fuels creates fly ash and finishing of metals throws metallic particles into the atmosphere.
The non- viable particulates are small solid particles and liquid droplets suspended in air. They help in the transportation of viable particles. There are four types of non-viable particulates in the atmosphere. They are classified according to their nature and size as follows.
(i) Smoke : Smoke particulate consists of solid particles (or) mixture of solid and liquid particles formed by combustion of organic matter. 
For example, cigarette smoke, oil smoke, smokes from burning of fossil fuel, garbage and dry leaves.
(ii) Dust: Dust composed of fine solid particles produced during crushing and grinding of solid materials.
For example, sand from sand blasting, saw dust from wood works, cement dust from cement factories and fly ash from power generating units.
(iii) Mists : They are formed by particles of spray liquids and condensation of vapours in air.
For example, sulphuric acid mist, herbicides and insecticides sprays can form mists.
(iv) Fumes : Fumes are obtained by condensation of vapours released during sublimation, distillation, boiling and calcination and by several other chemical reactions.
For example : organic solvents, metals and metallic oxides form fume particles.

Functional group :
A functional group is an'atom or a specific combination of bonded atoms that react in a characteristic way, irrespective of the organic molecule in which it is present.
Acetaldehyde - CHO.

\(\underset { 2\ volumes }{ { 2H }_{ { 2 }_{ (g) } } } +\underset { 1\ volumes }{ { O }_{ { 2 }_{ (g) } } } \rightarrow \underset { 2\ volumes }{ { 2H }_{ 2 }{ O }_{ (g) } } \)
Thus 2 volumes of H₂ reacts with 1 volume of O₂ to produce 2 volumes of H₂O (g)
\(\therefore\) 10 volumes of H₂ would react with 5 volumes of O₂ to produce 10 volumes of H₂O(g).
Thus 10 volumes of H₂O will be produced.

(i) For every 10 m of depth, a diver experiences an additional 1 atm of pressure due to the weight of water surrounding him.
(ii) At 20 m, the diver experiences a total pressure on atm. So the most important rule in diving is never hold breath.
(iii) Divers must ascend slowly and breath continuously allowing the regulator to bring the air pressure in their lungs to 1 atm by the time they reach the surface.

The sixth period corresponds to sixth shell. The orbitals present in this shell are 6s, 4f, 5p and 6d. The maximum number of electrons which can be present in these sub-shell is 2 + 14 + 6 + 10 = 32. Since the number of elements in a period corresponds to the number of electrons in the shells, the sixth period should have a maximum of 32 elements.

This method is based upon the fact that nitrogenous compound when heated with cupric oxide in an atmosphere of CO₂ yields free nitrogen. Thus
\({ C }_{ x }{ H }_{ y }{ N }_{ z }+\left( 2x+\frac { y }{ 2 } \right) CuO\longrightarrow xCO_{ 2 }+\frac { y }{ 2 } H_{ 2 }O+\frac { Z }{ 2 } { N }_{ 2 }+\left( 2x+\frac { y }{ 2 } \right) Cu\)
Traces of oxide of nitrogen, which may be formed in some cases, are reduced to elemental nitrogen by passing over heated copper spiral. The apparatus used in Dumas method consists of CO₂ generator, combustion tube, Schiffs nitrometer.
CO₂ generator:
CO₂ needed in this process is prepared by heating magnetite or sodium bicarbonate contained in a hard glass tube or by the action of dil. HCI on marble in a Kipps apparatus. The gas is passed through the combustion tube after being dried-by bubbling through cone. H₂SO₄,
Combustion Tube:
The combustion tube is heated in a furnace is charged with a) a roll of oxidized copper gauze to prevent the back diffusion of the products of combustion and to heat the organic substance mixed with CuO by radiation b) a weighed amount of the organic substance mixed with excess of CuO, c) a layer of course CuO packed in about 2/3 of the entire length of the tube and kept in position by loose asbestos plug on either side; this oxidizes the organic vapors passing through it, and d) a reduced copper spiral which reduces any oxides of nitrogen formed during combustion to nitrogen.
Schiff's nitro meter: 
The nitrogen gas obtained by the decomposition of the substance in the combustion tube is mixed with considerable excess of CO₂ It is estimated by passing nitrometer when CO₂ is absorbed by KOH and the nitrogen gets collected in the upper part of graduated tube.
Procedure:
To start with the tap of nitrometer is left open CO₂ is passed through the combustion tube to expel the air in it. When the gas bubbles risin through, the potash solution fails to reach the top of it and is completely absorbed it shows that only CO₂ is coming and that all air has been expelled from the combustion tube. The nitrometer is then and the tap is closed. The combustion tube is now heated in the furnace and the temperature rises gradually. The nitrogen set free from the compound collects in the nitrometer. When the combustion is complete a strong current of CO₂ is sent through, the apparatus in order to sweep the last trace of nitrogen from it. The volume of the gas gets collected is noted after adjusting the reservoir so that the solution in it and the graduated tube is the same. The atmospheric pressure and the temperature are also recorded.

Calulations: 
Weight of the substance taken = wg
Volume of nitrogen = V₁L
Room temperature =T₁K
Atmospheric pressure = P mm of Hg 
Agueen tension at
room temperature = p¹ mm ofHg
Pressure of dry nitrogen = (P - p¹) = PI mm of Hg.
Let P_o V_o and T_o be the pressure, Volume and temperature respectively of dry nitrogen at STP,
Then, \(\frac { { P }_{ 0 }{ V }_{ 0 } }{ { T }_{ 0 } } =\frac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } \)
\(\therefore { V }_{ 0 }=\frac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } \times \frac { { T }_{ 0 } }{ { P }_{ 0 } } \)
\({ V }_{ 0 }=\left( \frac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } \times \frac { 273K }{ 760 } \right) \)
Calculation of percentage of nitrogen. 22.4 L of N₂ at STP weigh 28g of N₂
\(\therefore \) V0 L of N2 at S.T.P weigh \(\frac { 28 }{ 22.4 } \times { V }_{ 0 }\)
wg of organic compound contain \(\left( \frac { 28 }{ 22.4 } \times \frac { { V }_{ 0 } }{ W } \right) \)
\(\therefore \) Percentage of nitrogen= \(\left( \frac { 28 }{ 22.4 } \times \frac { { V }_{ 0 } }{ W } \right) \times 100\)

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.

K_p1 = 8.19 x 10² T₁ = 298K
K_p2 = 4.6 x 10^-1 T₂ = 498K
\(\log({K_{p_2}\over K_{P_1}})={\Delta H^o\over 2.303\times 8.314}({T_2-T_1\over T_1T_2})\)
\(\log({4.6\times 10^{-1}\over 8.19\times 10^2})={\Delta H^o\over 2.303\times 8.314}({498-298\over 498\times 298})\)
\({-3.2505\times 2.303\times 8.314\times 498\times 298\over 200}=\Delta H^o\)
\(\Delta H^o=\) -46181 J mol^-1
\(\Delta H^o=\) -46.18 J mol^-1.

Oxidation number method:
This method is based on the fact that
Number of electrons lost by atoms = Number of electrons gained by atoms
Steps to be followed while balancing Redox reactions by Oxidation Number method:
1. Write skeleton equation representing redox reaction
2. Write the oxidation number of atoms undergoing oxidation and reduction.
3. Calculate the increase or decrease in oxidation numbers per atom.
4, Make increase in oxidation number equal to decrease in oxidation number by multiplying the formula of oxidant and reductant by suitable numbers.
5. Balance the equation atomically on both sides except O and H atoms.
6. Balance oxygen atoms by adding required number of water molecules to the side deficient in oxygen atoms.
7. Add required number of H⁺ ions to the side deficient in hydrogen atom if the reaction is in acidic medium.
8. For reactions in basic medium, add H₂O molecules to the side deficient in hydrogen atoms and simultaneously add equal number of OH^- ions on the other side of the equation.
9. Finally, balance the equation by cancelling common species present on both sides of the equation.

Initial temperature T₁ of the gas = 0°C = 0 + 273
= 273°C
Let the Initial volume of the gas V₁ = x ml
The final volume V₂ of the gas = 1.25 \(\times\) xml
According to Charles' law
\(\frac { { V }_{ 1 } }{ { T }_{ 1 } } =\frac { { V }_{ 2 } }{ { T }_{ 2 } } \)
\(\frac { x }{ 273 } =\frac { 1.25\times x }{ { T }_{ 2 } } \)

= 341.25 K
The final temperature = 341.25 K 

Characteristics of internal energy (U) :

• The internal energy of a system is an extensive property.
• The internal energy of a system is a state function.
• The change in internal energy of a system is expressed as ΔU= U_f - U_i
• In a cyclic process,' there is no internal energy change. ΔU_(cyclic)= 0
• If the internal energy of the system in the final state (U_f) is less than the internal energy of the system in its initial state (U_i), then ΔU would be negative.
• ΔU = U_f- Ui = -ve (U_f < U_i)
• If the internal energy of the system in the final state (U_f) is greater than the internal energy of the system in its initial state (U_i), then ΔU would be positive.
• ΔU = U_f - U_i = +ve (U_f > U_i)

Hydrogen atom has only one electron in its valence shell. So, to achieve the stable noble gas configuration of helium, it requires only one electron. For this it shares its one electron with another hydrogen atom to form a stable diatomic molecule.

(i) Alkaline earth metal (A) belonging to 3rd period is magnesium.
(ii) So A is Magnesium. Magnesium reacts with oxygen and nitrogen as follows.
\(2Mg+O_2⟶\underset{(B)}{2MgO}\)
\(3Mg+N_2⟶\underset{(C)}{Mg_3N_2}\)
So B is Magnesium oxide and C is magnesium nitride.
(iii) Magnesium undergoes metal displacement reaction with AgNO₃ as follow to give D as follows :
\(Mg+2AgNO_3⟶\underset{D}{Mg(NO_3)_2}+2Ag\)
So D is Magnesium nitrate.
Result  :

r(K⁺)+r(Cl^-) = d(K^+-Cl-) = 3.14 Å.
The effective nuclear charge for K+ and CI- can be calculated as follows.
K+ = (1s²)          (2s²2p⁶)        (3s²3p⁶)
        inner shell    (n-1)th shell   n^th shell
Z*(K-) = Z-S
= 19 - [(0.35 x 7) + (0.85 x 8) + (1 x 2)]
= 19 - 11.25 = 7.75
Z*(Cl-) = 17- [(0.35 x 7) + (0.85 x 8) + (1 x 2)]
= 17-11.25 = 5.75
∴ \(\frac { r({ K }^{ + }) }{ r(Cl^{ - }) } =\frac { Z*(Cl^{ - }) }{ Z*(K^{ + }) } =\frac { 5.75 }{ 7.75 } \)=0.74
∴ r(K⁺) = 0.74 r(Cl^-)
Substitute (2) in (1)
0.74 r(Cl^-) + r(Cl^-) = 3.14 Å.
1.74 r(Cl^-) = 3.14 Å
r(Cl^-) = \(\frac { 3.14\overset { 0 }{ A } }{ 1.74 } \)=1.81.Å.