(i) Brominatin of alkene to give bromo alkane.

(iii) In this reaction, benzoyl peroxide acts as a radical initiator. The mechanism involves free radicals.
\({ H }_{ 2 }C=CH+H-Br\overset { \overset { Benzoyl }{ Peroxide } }{ \longrightarrow } C{ H }_{ 3 }-C{ H }_{ 2 }-Br\)

Standard characteristics of drinking water.

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.

Gay Lussac's law:
At constant volume, the pressure of a fixed mass of a gas is directly proportional .to temperature.
\(P\propto T\) (or) \(P\over T\) = Constant
It can be graphically represented as shown here:

Lines in the pressure vs temperature graph are known as isochores ( constant volume) of a gas.

Covalent hydrides are compounds in which hydrogen is attached to another element by sharing of electrons. Covalent hydrides are further divided into three categories, viz., electron precise (CH₄, C₂H₆, SiH₄, GeH₄), electron-deficient (B₂H₆) and electron-rich hydrides (NH₃, H₂O).
Since most of the covalent hydrides consist of discrete, "Small molecules that have relatively weak intermolecular forces, they are generally gases or volatile liquids.

S₂O₃^2-+I²\(\rightarrow\)S₂O₆^2-+I^-

Equalise the increase/decrease in O.N by multiplying the S species by 1 and I species by 3.
S₂O₃^2-+I²\(\rightarrow\)S₂O₆^2-+3I^-
Balance all other atoms except 0 and H
S₂O₃^2-+3I²\(\rightarrow\)S₂O₆^2-+6I^-
Balance 0 atom by adding H₂0 on the side falling short of Oxygen.
S₂O₃^2-+3I²+3H₂O\(\rightarrow\)S₂O₆^2-+6I^-
Balance H atom by adding H+ ion on the side falling short of hydrogen.
S₂O₃^2-+3I²+3H₂O\(\rightarrow\)S₂O₆^2-+6I^- + 6H⁺
Add the equal number of OH- ion on both side since the medium is alkaline
S₂O₃^2-+3I²+3H₂O + 6OH^-\(\rightarrow\)S₂O₆^2-+6I^- + 6H⁺ + 60H^-
S₂O₃^2-+3I²+3H₂O + 6OH^-\(\rightarrow\)S₂O₆^2-+6I^- + 6H₂

Given K_eq= 10
Gas constant R = 8.314 JK^-1 mol^-1
T=300K
The relationship between Free energy change ΔG and equilibrium constant K is ΔG^o=-RTlnK
Since K, T and R are positive values, ΔG^o will be negative.
When ΔG is -ve, the process is spontaneous and feasible

(i) According to aufbau's principle 5th period has nine orbital (one 5s, five 4d and three 6p) to be filled.
(ii) Nine orbitals can accommodate a maximum of 18 electrons. Hence fifth period of the periodic table should has 18 elements from rubidium (2 = 37) to Xenon (Z = 54).

(i) Ethyl methyl ether: Lower halogenated ether reacts with grignard reagent to form higher ether.
\(\underset { Chloro\quad dim\quad ether }{ { CH }_{ 3 }O-{ CH }_{ 2 }Cl } +\underset { Methyl\quad magnesium\\ iodide }{ { CH }_{ 3 }Mgl } \longrightarrow \underset { Ethyl\quad methyl\quad ether }{ { CH }_{ 3 }-O-{ CH }_{ 2 }-{ CH }_{ 3 } } \) 
(ii) Methyl cyanide: Grignard reagent reacts with cyanogen chloride to form alkyl cyanide.

(iii) Methane: Grignard reagent reacts with water to give methane as product.

(i) Bond formation in Beryllium chloride takes place by sp hybridisation.
(ii) The valence shell of Beryllium has the electronic configuration as follows:

(iii) In BeCl₂ both the Be-Cl bonds are equivalent and it was observed that the molecule is linear. VB theory explains this observed behaviour by sp hybridisation. One of the paired electrons in the 2s orbital gets excited to 2p orbital.
(iv) Now the 2s and 2p orbitals hybridise and produce two equivalent sp hybridised orbitals which have 50% s-character and 50% p-character. These sp hybridised orbitals are oriented in opposite direction.
(v) Each of the sp hybridised orbitals linearly overlap with Pz orbital of the chlorine to form a covalent bond between Be and CI atoms as follow.

w = 0.24g and a = 0.287 g
\( \% \mathrm{Cl} =\frac{35.5}{143.5} \times \frac{\mathrm{a}}{\mathrm{w}} \times 100 \)
\(=\frac{35.5}{143.5} \times \frac{0.287}{0.24} \times 100=.29 .42 \%\)

(i) An alkali metal (A) belongs to period number II and group number I is lithium.
(ii) Lithium reacts with oxygen to form simple oxide lithium oxide (B).
\(4Li + O_2 ⟶ \underset{Lithium\ oxide (B)}{2Li_2O}\)
(iii) Lithium reacts with water to form lithium hydroxide with liberation of hydrogen.
\(2Li + 2H_2O ⟶ \underset{Lithium\ hydroxide (C)}{2LiOH + H_2}\)
(iv) Lithium directly react with carbon to form an ionic compound lithium carbide.
\(2Li + 2C ⟶ \underset{Lithium\ carbide}{Li_2C_2}\)

Step - 1 : To find atoms undergoing change in O.N.
K₂Cr₂O₇ + HI ⟶ KI + Crl₃ + H₂O + I₂
Step - 2 : To find the total increase and decrease in O.N.
K₂Cr₂O7 + CrI₃  (decrease of 3 unit / atom = Total decrease = 6 units / 2 atom)
HI ⟶ I₂ (increase of 1 unit / atom = total increase 1 x 6 = 6)
Step - 3 : To balance the total increase and decrease in O.N, multiply HI by 6.,
K₂Cr₂O₇ + 6 HI ⟶ KI + Crl₃ + H₂O + I₂
Step - 4 : To balance all atoms other than '0' and 'H'
K₂Cr₂O₇ + 6 HI ⟶ 2KI + 2Crl₃ + H₂O + 3I₂
This makes 14 iodine atoms on RHS. (These iodide ions do not undergo any change in O.N). Hence to balance the iodine atoms add 8HI to LHS.
i.e., K₂Cr₂O₇ + 14 HI ⟶ 2KI + 2CrI₃ + H₂O + 3I₂
These, the oxygen atoms are balanced by making 7H₂O as RHS.
K₂Cr₂O₇ + 14 HI ⟶ 2KI + 2CrI₃ + 7H₂O + 3I₂
The hydrogen atoms are balanced by themselves.
Hence the balanced equation is
K₂Cr₂O₇ + 14HI ⟶ 2KI + 2Crl₃ + 7H₂O + 3I₂

Velocity of the electron = 600 ms^-1
Uncertainty in velocity = \(\frac{0.005}{100}\times 600 ms^{-1}\)
= 0.03 ms^-1
= 3 x 10^-2 ms^-1
Now, ( \(\Delta\)x) (m \(\Delta\) v) = \(\frac{h}{4\pi}\)
\(\therefore \Delta x=\frac{h}{4\pi.m.\Delta v}\)
= \(\frac{6.626\times 10^{-34}kgm^2 s^{-1}}{4\times 3.14\times 9.1\times10^{-31}kg\times 3\times 10^{-2}ms^{-1}}\)
= 1.93 x 10^-3 m.

The required equation is
C₆H₁₂O_6(s)+6O_2(g) ➝ 6CO_2(g)+6H₂O_(l)
(i)  C_(graphite)+O_2(g) ➝ CO_2(g); ΔH= -395.0 KJ
(ii) H_2(g)+\(\frac{1}{2}\)O_2(g) ➝ H₂O_(l);  ΔH= -269.4 KJ
(iii) 6C_(graphite)+6H_2(g)+3O_2(g) ➝ C₆H₁₂O_6(s);  ΔH= -1169.8 KJ
Multiply equation (i) and (ii) by 6 and add them up
(iv) 6C_(graphite)+6H_2(g)+9O_2(g) ➝ 6CO_2(g)+6H₂O_(l) ; ΔH= -3984.6 KJ
Subtracting equation (iii) from (iv)
C₆H₁₂O_6(s)+6O_2(g) ➝ 6CO_2(g)+6H₂O_(l) ΔH= -2816.6 KJ
∴ Enthalpy of combustion of glucose = -2816.6 KJ.

The electronic configuration of K atom is
K₁₉ = (1s²)(2s²p⁶)(3s²3p⁶)4s¹
Effective nuclear charge (Z*) = Z - S
Z* = 19 - [(0.85 \(\times\) No. of electrons in (n -1)t^h shell) + (1.00 total number of electrons in the inner shells)]
= 19-[0.85 \(\times\) (8) + (1.00 \(\times\) 10)]
Z* =  2.20

The element which occupies group number (16) and period number (2) is oxygen. (B) is D₂O which is used as a moderator in nuclear reactions.
So (A) must be deuterium, which is an isotope of hydrogen
    \(2\underset { (A) }{ { D }_{ 2 } } +{ O }_{ 2 }\rightarrow 2\underset { (B) }{ { D }_{ 2 }O } \)
So (B) is D₂O
(A) adds to (C) as follows :
\(3 \mathrm{D}_{2}+\mathrm{C}_{3} \mathrm{H}_{6} \rightarrow \mathrm{CH}_{3}-\mathrm{CH}-\mathrm{CH}_{2}\)
So (D) is 1,2 - dideutero propane. 

Statement : "No two electrons in an atom can have the same set of values of all four quantum numbers"
Explanation : It means that, each electron must have unique values for the four quantum numbers (n, l, m and s).
For the lone electron present in hydrogen atom, the four quantum numbers are: n = 1; l = 0; m = 0 and s = +1/2. For the two electrons present in helium, one electron has the quantum numbers same as the electron of hydrogen atom, n = 1.
l = 0, m = 0 and s = +1/2. For other electron, the fourth quantum number is different i.e., n = 1, l = 0, m = 0 and s = -1/2.
As we know that the spin quantum number can have only two values +1/2 and - 1/2, only two electrons can be accommodated in a given orbital in accordance with pauli exclusion principle.

Halo arenes do not undergo nucleophilic substitution reaction readily is due to C-X bond in aryl halide is short and strong and also the aromatic ring isa centre of high electron density.
The halogen of haloarenes can be substituted by OH^-, NH₂^-, or CN^- with appropriate nucleophilic reagents at high temperature and pressure.

The equilibrium constant is a constant and it is for equilibrium condition. But 'Q', the reaction quotient is not a constant as it is for non - equilibrium condition. 'Q' is the ratio of the product of active masses of a reaction products raised to the respective stoichiometric coefficients in the balanced chemical equation to that of the reactants, under non - equilibrium conditions.
i) If Q = K_c; it is equilibrium
ii) If Q > K_c.; the reaction will proceed in reverse direction
iii) If Q < K_c ; the reaction will proceed in forward direction

In the reaction 2NaHCO₃(s) \(\rightarrow\) Na₂CO₃(s) +CO₂(g) + H₂O(g) the number of gaseous components increase and hence entropy of the system will increase.

Molar Mass of NaOH = 23 + 16 + 1 = 40
Acidity = 1
Equivalent Mass = \(\frac { Molar\ mass }{ acidity } =\frac { 40 }{ 1 } =40geq^{ -1 }\)

The pressure decreases with the increase in altitude because there are fewer molecules per unit volume of air. Above 9200 m (30000 ft.) the pressure is so low that one could pass out for lack of oxygen. For this reason most airplanes cabins are artificially pressurized.

The solubility product of NaCl is lower than that of NaOH. The more soluble a substance is, the higher the K_sp value it has In aqueous solution NaOH gives OH^- ions. It can be solvated by establishing H-bonds with water molecules. So it is more water soluble.

Efflorescence is the spontaneous loss of water by a hydrated salt, which occurs when the aqueous vapor pressure of the hydrate is greater than the partial pressure of the water vapour in the air. This is the property of salts
Ex: Glauber's salt- Na₂SO₄·10H₂O
Epsom salt - MgSO₄·7H₂O

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.

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