(i) (A) on oxidation with air and V₂O₅ gives (B).
\(\underset{(A)}{{ C }_{ 6 }{ H }_{ 5 }{ CH }_{ 3 }}\overset { \left( O \right) }{ \underset { { V }_{ 2 }{ O }_{ 5 } }{ \longrightarrow } } \underset{(B)}{{ C }_{ 6 }{ H }_{ 5 }CHO}\) 
(ii) (B) on reduction with LiAIH₄ gives C.
\({ C }_{ 6 }{ H }_{ 5 }CH\overset { \left[ H \right] }{ \underset { LiA/{ H }_{ 4 } }{ \longrightarrow } } \underset{(C)}{{ C }_{ 6 }{ H }_{ 5 }CHO}\)

(i) Hormones are classified according to the distance over which they act as, endocrine, paracrine and autocrine hormones.
(ii) Endocrine hormones act on cells distant from the site of their release. Example: insulin and epinephrine are synthesized and released in the bloodstream by specialized ductless endocrine glands.
(iii) Paracrine hormones (alternatively, local mediators) act only on cells close to the cell that released them. For example, interleukin -1 (IL-1) Autocrine hormones act on the same cell that released them. For example, protein growth factor interleukin-2 (IL-2).

(i) Structure:
(a) Linear polymers (long continuous chain) E.g. HDPE, PVC
(b) Branched polymers (one main chain with small chains as branches) E.g. polypropylene, LDPE.
(c) Cross linked polymers (linking of chain polymers) E.g. bakelite, melamine, formaldehyde
(ii) Mode of synthesis:
(a) Addition polymers. Formed by polymerisation of monomers without the elimination of byproduct. E.g. polyethylene, PVC, teflon.
(b) Condensation Polymer formed by the condensation of two or more monomers with the elimination of simple molecules like H₂O, NH₃, etc., E.g. Nylon:6-6, polyester.

(i) An organic compound (A) C₆H₆O gives violet colour with neutral FeCl₃ solution.
(ii) With NH₃ in the presence of anhydrous ZnCI₂,(A) gives (B) (C₆H₇N).

(i) Primary amine react with nitrous acid to form alcohols and nitrogen gas
\(\underset { primary \ amine }{ { CH }_{ 3 }NH_{ 2 } } \rightarrow \underset { unstable }{ { [{ CH } }_{ 3 }-N=N-OH] } \rightarrow { CH }_{ 3 }OH+{ N }_{ 2 }\)
Aliphatic diazonium compound is unstable because of absence of resonance stabilisation.
(ii) Secondary amines react with nitrous acid to form N-nitroso amines which are water insoluble yellow oils.
\(\underset { Secondary \ amine }{ { { (CH }_{ 3 } })_{ 2 }NH } +HO-N=O\rightarrow \underset { N-nitroso \ dimethy \ amine-yellow \ oil\\ (insolube \ in \ water) }{ { (CH }_{ 3 })_{ 2 }N-N=O } \)
(iii) Tertiary amine react with nitrous acid to form trialkyl ammonium nitrite salts which are soluble in water
\(\underset { Tertiary \ amine }{ { (CH }_{ 3 })_{ 2 }N } +HONO\rightarrow \underset { trimethyl \ ammonium \ nitrite\\ (salt \ soluble \ in \ water) }{ { { (CH }_{ 3 }) }_{ 3 }{ NH }^{ + }{ NO }_{ 2 }^{ - } } \)

Condensation Methods:
Various chemical methods for the formation of colloidal particles.
(i) Oxidation:
Sols of some non-metals are prepared by this method. (a) When hydroiodic acid is treated with iodic acid, 12 sol is obtained.
\({ HIO }_{ 3 }+5HI\longrightarrow { 3H }_{ 2 }O+{ I }_{ 2 }\) (Sol)
(ii) Reduction:
Many organic reagents like phenyl hydrazine, formaldehyde, etc are used for the formation of sols. For example: Gold sol is prepared by reduction of auric chloride using formaldehyde.
\(2{ AuCl }_{ 3 }+3HCHO+{ 3H }_{ 2 }O\longrightarrow 2Au\left( sol \right) +6HCl+3HCOOH\) 
(iii) Hydrolysis: 
Sols of hydroxides of metals like chromium and aluminium can be produced by this method.
For example,
\({ FeCl }_{ 3 }+3H_{ 2 }O\longrightarrow Fe(OH)_{ 3 }+3HCl\) 
(iv) Double decomposition: 
For the preparation of water insoluble sols this method can be used. When hydrogen sulphide gas is passed through a solution of arsenic oxide, a yellow coloured arsenic sulphide is obtained as a colloidal solution.
\({ As }_{ 2 }{ O }_{ 3 }+3{ H }_{ 2 }S\longrightarrow { AS }_{ 2 }{ S }_{ 3 }+{ 3H }_{ 2 }O\) 
(v) Decomposition:
When few drops of an acid is added to a dilute solution of sodium thio sulphate, the insoluble free sulphur produced by decomposition of sodium thiosulphate accumulates into small, clusters which impart various colours blue, yellow and even red to the system. depending on their growth within the size of colloidal dimensions.
\({ S }_{ 2 }{ O }_{ 3 }^{ 2- }+{ 2H }^{ + }\longrightarrow \underset{sol}S+{ H }_{ 2 }O+{ SO }_{ 2 }\) 
By exchange of solvent:
(i) Colloidal solution of few substances like phosphorous or sulphur is obtained by preparing the solutions in alcohol and pouring them into water.
(ii) As they are insoluble in water, they form colloidal solution.
P in alcohol + water ⟶ Psol.

According to Ostwald dilution Law,
\({ K }_{ a }=\frac { { \alpha }^{ 2 }C }{ (1-\alpha ) } \)  ...(1)
Substitute a value in the above expression (1)
\({ K }_{ a }=\frac { { \Lambda }_{ m }^{ 2 }C }{ { \Lambda }_{ m }^{ 2 }\left( 1-\frac { { \Lambda }_{ m } }{ { \Lambda }_{ m }^{ o } } \right) } \)
\({ K }_{ a }=\frac { { \Lambda }_{ m }^{ 2 }C }{ { \Lambda }_{ m }^{ 2 }\frac { { \Lambda }_{ m }^{ o }-{ \Lambda }_{ m } }{ { \Lambda }_{ m }^{ o } } } \)
⇒ \({ K }_{ a }=\frac { { \Lambda }_{ m }^{ 2 }C }{ { \Lambda }_{ m }^{ 2 }{ (\Lambda }_{ m }^{ o }-{ \Lambda }_{ m }) } \)

(i) The concentration of hydronium ion in an acidic buffer solution depends on the ratio of the concentration of the weak acid to the concentration of its conjugate base present in the solution i.e.,
\(\left[ { H }_{ 3 }{ O }^{ + } \right] ={ K }_{ a }\frac { [{ acid] }_{ aq } }{ [{ base] }_{ aq } } \)
(ii) The weak acid is dissociated only to a small extent. Moreover, due to common ion effect, the dissociation is further suppressed and hence the equilibrium concentration of the acid is nearly equal to the initial concentration of the unionised acid. Similarly, the concentration of the conjugate base is nearly equal to the initial concentration of the added salt.
\(\left[ { H }_{ 3 }{ O }^{ + } \right] ={ K }_{ a }\frac { [{ acid] } }{ [{ salt] } } \)
(iii) Here [acid] and [salt] represent the initial concentration of the acid and salt, respectively used to prepare the buffer solution
Taking logarithm on both sides of the equation
\(\log\left[ { H }_{ 3 }{ O }^{ + } \right] ={ \log K }_{ a }+\log\frac { [{ acid] } }{ [{ salt] } } \)
reverse the sign on both sides
- \(\log\left[ { H }_{ 3 }{ O }^{ + } \right] ={- \log K }_{ a }-\log\frac { [{ acid] } }{ [{ salt] } } \)
We know that
pH = -log [H₃O⁺] and pK_a = -log K_a
\(\Rightarrow pH={ pK }_{ a }-\log\frac { [{ acid] } }{ [{ salt] } } \)
\(​​\Rightarrow pH={ pK }_{ a }+\log\frac { [{ acid] } }{ [{ salt] } } \)
Similarly for a basic buffer,
pOH = \({ pK }_{ a }+\log\frac { [{ acid] } }{ [{ salt] } } \)

Given data: Time taken for 25% completion of the reaction = 100 mins
a = 100; x = 25 and a - x = 75; t = 100min
Formula: \(k=\frac { 2.303 }{ t } \log { \frac { a }{ a-x } } ;{ t }_{ \frac { 1 }{ 2 } }=\frac { 0.693 }{ k } \)
Solution:
\(k=\frac { 2.303 }{ 100 } \log\frac { 100 }{ 100-25 } \)
\(=\frac { 2.303 }{ 100 } \log\frac { 100 }{ 75 } =\frac { 2.303 }{ 100 } \log\frac { 4 }{ 3 } \)
\(=\frac { 2.303 }{ 100 } \times 0.1249\)
Rate constant(k) = 2.8773 x 10^-3 min^-1
\({ t }_{ \frac { 1 }{ 2 } }=\frac { 0.693 }{ k } =\frac { 0.693 }{ 2.8773\times { 10 }^{ -3 }{ ,min }^{ -1 } } \)
\(=\frac { 693 }{ 2.8773 } =240.85\)
Half-life period (t_1/2) = 240.85 minutes

Density of the element, d = 2.7 x 10³ kg m^-3
Molar mass, M = 2.7 x 10^-2 kg mol^-1
Edge length, a = 405 pm
= 405 x 10^-12 m
= 4.05 x 10^-10 m
Avogadro's number, N_A= 6.022 x 10²³ mol^-1
\(\therefore d=\frac { Z\times M }{ { a }^{ 3 }\times { N }_{ A } } \)
\(\Rightarrow Z=\frac { d\times { a }^{ 3 }{ N }_{ A } }{ M } \)
\(=\frac { 2.7\times { 10 }^{ 3 }kg\quad { m }^{ -3 }{ (4.05\times { 10 }^{ -10 }m) }^{ 3 }\times 6.022\times { 10 }^{ 23 }{ mol }^{ -1 } }{ 2.7\times { 10 }^{ -2 }kg\quad { mol }^{ -1 } } \)
= 4.004 = 4.
This implies that four atoms of the element are present per unit cell. Hence the unit cell is face centred cubic.