(i) From the molecular formula, (A) is identified as benzaldehyde and it is called as oil of bitter almonds.
(ii) Benzaldehyde is oxidised to benzoic acid by alkaline permanganate which gives brisk effervescence with NaHCO₃.
 \(\underset { (A) }{ { C }_{ 6 }{ H }_{ 5 }CHO } \overset { (O) }{ \longrightarrow } \underset { (B) }{ { C }_{ 6 }{ H }_{ 5 } } -COOH\)
(iii) When benzaldehyde is refluxed with aqueous alcoholic KCN, benzoin (C) is formed.
\({ C }_{ 6 }{ H }_{ 5 }CH=O+H-\overset { \underset { || }{ O } }{ C } -{ C }_{ 6 }{ H }_{ 5 }\overset { alc }{\underset{KCN} \longrightarrow } { C }_{ 6 }{ H }_{ 5 }-\underset { \overset { | }{ OH\\ (C) } }{ CH } -\overset { \underset { || }{ O } }{ C } -{ C }_{ 6 }{ H }_{ 5 }\)

(i) An aromatic compound (A) possess the smell of bitter almonds is benzaldehyde. From the molecular formula (A) is identified as C₆H₅CHO benzaldehyde.
(ii) Benzaldehyde on treatment with Cl₂ in the absence of catalyst gives benzoyl chloride C₆H_sCOCl and it is (B).

(iii) Benzaldehyde on treatment with C in the presence of catalyst gives m-cblorobenzaldehyde and it is (C).

Compound	Compound Name	Formula
A	Benzaldehyde
B	Benzoyl chloride
C	m-Chloro benzaldehyde

Cannizaro reaction involves three steps.
Step 1:  Attack of OH- on the carbonyl carbons

Step 2: Hydride ion transfer

Step 3: Acid - base reaction
\({ C }_{ 6 }{ H }_{ 5 }-\overset { \underset { || }{ O } }{ C } -OH+{ C }_{ 6 }{ H }_{ 5 }H_{ 2 }{ O }^{ - }+{ Na }^{ + }\overset { Proton }{ \underset { exchange }{ \longrightarrow } } \underset { Sodium\quad benzoate }{ { C }_{ 6 }{ H }_{ 5 }-\overset { \underset { || }{ O } }{ C } -ONa } +\underset { Benzyl \ alcohol }{ { C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }OH } \)
Cannizaro reaction is a characteristic of aldehyde having no a - hydrogen

(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).

Ethers are a class of organic compound in which an oxygen atom is connected to two alkyl/aryl groups Ethers can be considered as the derivatives of hydrocarbon in which one hydrogen atom is replaced by an alkoxy (-OR) or an aryloxy (-OAr ) group. The general formula of aliphatic ether is C_nH_2n+2O.
Classification:

(i) When ethylamines is treated with chloroform and an alkali, carbylamine reaction takes place and a foul smelling substance called carbylamlne (or) alkyl isocyanide is formed.
C₂H₅NH₂ + CHCl₃ + 3NaOH ➝ \(\underset { Ethyl \ isocyanide }{ { C }_{ 2 }{ H }_{ 5 }NC+3NaCl+3{ H }_{ 2 }O } \)
(ii) When ethylamine is warmed with CS₂ and mercuric chloride, alkyl isothiocyanate having a pungent mustard like odour is obtained. This reaction is called mustard oil reaction.
C₂H₂NH₂ + S = C-S \(\underrightarrow { Hg{ Cl }_{ 2 } } \) C₂H₅ - N = C = S + H₂S
(iii) When ethylamine condense with aromatic aldehyde (benzaldehyde) Schiff's base is formed.
C₆H₅CHO + H₂NC₂H₅ ➝ \(\underset { Schiffi's \ base \ Benzal-ethylamine }{ { C }_{ 6 }{ H }_{ 5 }CH=N{ C }_{ 2 }{ H }_{ 5 }+{ H }_{ 2 }O } \)

An organic compound A (C₂H₆O₂) liberates hydrogen with metallic sodium.

(ii) Compound A when heated with anhydrous zinc chloride ultimately gives B (C₂H₄O)

(iii) Compound A when heated with conc. phosphoric acid gives C (C₄H₁₀O₃).

(iv) Compound A on oxidation with acidified K₂Cr₂O₇ gives compound D (CH₂O₂).

Compound	Compound Name	Formula
A	Ethylene glycol	\(\overset { { CH }_{ 2 }OH }{ \underset { { CH }_{ 2 }OH }{ | } } \)
B	Acetaldehyde	CH3-CHO
C	Diethylene glycol	\(\overset { HCOOH }{ \underset { HCOOH }{ + } } \)
D	Formic acid

(i) Compound (A) is carbonyl compound, it is acetone
(ii) (A) on reduction with LiAlH₄ gives (B) it gives blue colour in Victor Meyer'stest.
\({ CH }_{ 3 }-\underset { \overset { || }{ \underset { (A) }{ O } } }{ C } -{ CH }_{ 3 }\overset { { LiAIH }_{ 4 } }{ \underset { \left[ H \right] }{ \longrightarrow } } { CH }_{ 3 }-{ CH }_{ 3 }-\underset { \overset { | }{ \underset { (B) }{ OH } } }{ CH } -{ CH }_{ 3 }\) 
(iii) (B) reacts with SOCl₂to give (C).

(iv) (C) on treatment with alcoholic KOH, forms (D) by elimination reaction.
\({ CH }_{ 3 }-\underset { \overset { | }{ \underset { (C) }{ Cl } } }{ C } H-{ CH }_{ 3 }\overset { alc.KOH }{ \longrightarrow } \underset { (D) }{ { CH }_{ 3 }CH={ CH }_{ 2 }+HCl } \) 

Freundlich adsorption isotherm:
According to Freundlinch
\(\frac { x }{ m } =kp^{ \frac { 1 }{ n } }\) 
where x is the amount of adsorbate or adsorbed on 'm' gm of adsorbent at a pressure of p. K and n are constants Value is always less than unity.
This equation is applicable for adsorption of gases on solid surfaces. The same equation becomes  \(\frac { x }{ m } =Kc^{ \frac { 1 }{ n } }\) when used for adsorption in solutions with c as concentration.
These equation quantitively predict the effect of pressure(or concentration) on the adsorption of gases(or adsorbates) at constant temperature.
Taking log on both sides of equation \(\frac { x }{ m } ={ kp }^{ \frac { 1 }{ n } }\) 
\(log\frac { x }{ m } =logK+\frac { 1 }{ n } logP\) 

Hence the intercept represents the value of log k and the slope \(\frac { b }{ q } \) gives \(\frac { 1 }{ n } \)
This equation explains the increase of \(\frac { x }{ m } \) with increase in pressure. But experimental values show the deviation at low pressure.

Standard Hydrogen Electrode (SHE) is used as the reference electrode. It has been assigned an arbitrarily emf of zero volt. It consists of a platinum electrode in contact with 1M HCI solution and 1atm hydrogen gas. The hydrogen gas is bubbled through the solution at 25°C. SHE can act as a cathode as well as an anode. The Half cell reactions are given below.
If SHE is used as a cathode, the reduction reactions is
2H⁺ (aq,1M) + 2e^- ⟶ H₂ (g, 1 atm)  E^o= 0 volt
If SHE is used as an anode, the oxidation reaction is
H² (g.1 atm) ⟶ 2H⁺ (aq, 1M) + 2e^- E^o= volt
Illustration: Let us calculate the reduction potential of zinc electrode dipped in zinc sulphate solution using SHE.
Step 1: The following galvanic cell is constructed using SHE
Zn_(s) | Zn²⁺ (aq,1M) || H⁺ (aq, 1M) | H₂ (g, 1 atm)|pt_(s)
Step 2: The emf of the above galvanic cell is measured using a volt meter. In this case, the measured emf of the above galvanic cell is 0.76V.
Calculation
We know that,
E^o_cell = (E^o_ox)_Zn|Zn²⁺ + (E^o_red)_SHE
E^o_cell = 0.79 and (E^o_red)_SHE = 0V.
Substitute these values in the above equation
⇒ 0.76V = (E^o_ox) _Zn|Zn2+ + 0V
⇒ (E^o_ox)_Zn|Zn²⁺ = 0.76V
This oxidation potential corresponds to the below mentioned half cell reaction which takes place at the cathode.
Zn ⇾ Zn²⁺ + 2e^- (Oxidation)
The emf for the reverse reaction will give the reduction potential
Zn²⁺ + 2e^- ⇾ Zn; E^o= - 0.76V
∴ (E^o_ox)_Zn²⁺|Zn = - 0.76V

Given:
E₁ = 0.5 V
Cr³⁺+ 3e^- ➝ Cr_(s).......(1)
E₂ = 0.41 V
Cr³⁺+ e^- ➝ Cr²⁺  .....(2)
The required reaction is,
Cr²⁺+ 2 e^- ➝ Cr_(s)
Then,
Formula:
\({ E }_{ 3 }=\frac { 3{ E }_{ 1 }+{ E }_{ 2 } }{ 2 } \)
Solution:
= \(\frac { 3(0.5)+(0.41) }{ 2 } =\frac { 1.5+0.41 }{ 2 } \)
= 0.955 V
E₃ = 0.955 V

Initial concentration of
AgNO₃ = 1M= IN
Quantity of current 0.75 F
Formula:
1Faraday = 1equivalent mass
Solution:
For IF current In AgNO₃ will be liberated.
For 0.75 F current 0.75 N AgNO₃ will be liberated
The concentration of AgNO₃ remaining
= 1.0 - 0.75 = 0.25 N
ஃ The concentration of AgNO₃ remaining

HCI ⟶ H⁺ + Cl^-. HCI is a strong acid.
[H⁺] from HCI is very much greater than [H⁺] from water which is 1 x 10^-7 M.
∴ [H⁺] = [HCI] = 0.001 M
∴ pH = -log (0.001) = 3.0
∴ That is acidic solution.

Given: Half life period (t_1/2) = 10 mins
Formula: \(k=\frac { 2.303 }{ t } \log { \frac { a }{ a-x } } \)
Solution:
\(k=\frac { 0.693 }{ { t }_{ \frac { 1 }{ 2 } } } =\frac { 0.693 }{ 10 } \)
= 0.693 min^-1 = 6.93 x 10^-2 min^-1
Time taken = 1 hour = 60 minutes
a = 100%
x = ?
\(k=\frac { 2.303 }{ t } log\frac { a }{ a-x } \)
\(=\frac { 2.303 }{ 60 } \) [log 100 - log(a - x)]
log 100 - log(a-x) = \(\frac { 6.93\times { 10 }^{ -2 }\times 60 }{ 2.303 } \)
log 100 - log(a-x) = 1.8060
log (a - x) = log 100 - 1.8060
= 2 - 1.8060
log (a - x) = 0.1940
a - x = Antilog of 0.1949
a - x = 1.563%

Average rate = \(-\frac { \triangle \left( R \right) }{ \triangle t } =-\frac { { \left[ R \right] }_{ 2 }-{ \left[ R \right] }_{ 1 } }{ { t }_{ 2 }-{ t }_{ 1 } } \)
\(=-\frac { 0.02M-0.03M }{ 25min } =\frac { -0.01M }{ 25min } \)
= 4 x 10^-4 M min^-1 and 
= \(-\frac { -0.01m }{ 25\times 60 } \) = 6.66 x 10^-6 Ms^-1

(i) It oxidises ferrous salts to ferric salts.
 Cr₂O₇^2- + 6Fe²⁺+ 14H⁺ ⟶ 2Cr³⁺ + 6Fe³⁺ + 7H₂O
(ii) It oxidises iodide ions to iodine
Cr₂O₇^2- + 6I^-+ 14H⁺ ⟶ 2Cr³⁺ + 3I₂ + 7H₂O
(iii) It oxidises sulphide ion to sulphur
Cr₂O₇^2- + 3S^2-+ 14H⁺ ⟶ 2Cr³⁺ + 3S + 7H₂O
(iv) It oxidises sulphur dioxide to sulphate ion
Cr₂O₇^2- + 3SO₂+ 2H⁺ ⟶ 2Cr³⁺ + 3SO₄^2- + H₂O
(v) It oxidises stannous salts to stannic salt
 Cr₂O₇^2- + 3Sn²⁺ + 14H⁺ ⟶ 2Cr³⁺ + 3Sn⁴⁺ + 7H₂O
(vi) It oxidises alcohols to acids
2K₂Cr₂O₇ + 8H₂SO₄ + 3CH₃CH₂OH ⟶ 2K₂SO₄ +2Cr₂(SO₄)₃+ 3CH₃COOH + 11H₂O
(vii) It oxidises oxalic acid to CO2
2MnO₄^- + 5(COO)^2- + 6H⁺ ⟶ 2Mn²⁺+ 10CO₂ + 8H₂O
(viii) It oxidises iodide ions to iodine
2MnO4^- + 10I^- + 16H⁺ ⟶ 2Mn²⁺+ 5I₂+ 8H₂O
(ix) It oxidises sulphide ion to sulphur
2MnO₄^- + 5S^2-+ 16H⁺ ⟶ 2Mn²⁺+ 5S + 8H₂O
(x) It oxidises nitrites to nitrates
 2MnO₄^- + 5NO^2- + 6H⁺ ⟶ 2Mn²⁺+ 5NO₃^- + 3H₂O
(xi) It oxidises alcohols to aldehydes.
 2KMnO₄ + 3H₂SO₄ + 5CH₃CH₂OH ⟶ 2K₂SO₄ + 2MnSO₄ + 5CH₃CHO + 8H₂O
(xii) It oxidises sulphite to sulphate
2MnO₄^- + 5SO₃^2- + 6H⁺ ⟶ 2Mn²⁺ + 5SO₄^2- + 3H₂O

(i) In this process a mixture of air and hydrochloric acid is passed up a chamber containing a number of shelves, pumice stones soaked in cuprous chloride are placed.
(ii) Hot gases at about 723 K are passed through a jacket that surrounds the chamber.
\(4 \mathrm{HCl}+\mathrm{O}_{2} \frac{400^{\circ} \mathrm{C}}{\mathrm{Cu}_{2} \mathrm{Cl}_{2}} 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{Cl}_{2} \uparrow\)
The chlorine obtained by this method is dilute and is employed for the manufacture of bleaching powder. The catalysed reaction is,
\(2 \mathrm{Cu}_{2} \mathrm{Cl}_{2}+\mathrm{O}_{2} \rightarrow 2 \mathrm{Cu}_{2} \mathrm{OCl}_{2}\\ \quad \quad \quad \quad \quad \quad \quad \text { cuprous oxychloride } \)
\(\mathrm{Cu}_{2} \mathrm{OCl}_{2}+2 \mathrm{HCl} \rightarrow 2 \mathrm{CuCl}_{2}+\mathrm{H}_{2} \mathrm{O}\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad\text { Cupric chloride } \)
\(2 \mathrm{CuCl}_{2} \rightarrow \mathrm{Cu}_{2} \mathrm{Cl}_{2}+\mathrm{Cl}_{2}\\ \quad \quad \quad \quad \text { cuprous chloride }\)

(a) Rhombic Sulphur (α - Sulphur):
(a) It is yellow in colour.
(b) Its melting point is 385.8K and specific gravity is 2.06
(c) It is stable form of sulphur at room temperature.
(d) It is formed on evaporating the solution of sulphur in CS₂.
(e) It in insoluble in water, readily soluble in CS₂ and dissolves to some extent in benzene, alcohol and ether.
(b) Monoclinic sulphur \(\left( \beta -sulphur \right) \):
(a) Its melting point is 393K and specific gravity is 1.98
(b) It is prepared by melting rhombic sulphur in a dish and cooling, till crust is formed. Two holes are made in crust and remaining liquid is powered out. On removing crust, colourless needle - shaped crystals of β - sulphur is formed.
(c) Monoclinic sulphur is stable above 369K and below 369K α - sulphur is stable.
(d) At 369K both forms are stable and this temperature is called transition temperature.
(e) Both rhombic and monoclinic sulphur have S₈ molecules, these are packed to give different crystal structure S₈ form is puckered and has crown shape.
Several other modifications containing 6-20 sulphur atoms per ring are synthesised 

(f) In Cyclo-S₆ the ng adopts chair form.

(g) At elevated temperatures (～1000K), S₂ is dominant species and is, paramagnetic like O₂

(i) By knowing the density of an unknown metal and the dimension of its unit cell, the atomic mass of the metal can be: determined.
(ii) Let 'a' be the edge length of a unit cell of a crystal, 'd' be the density of the metal, 'm' be the atomic mass of the metal and 'z' be the number of atoms in the unit cell.
(iii) Now,
Density of the unit cell
\(=\frac{Mass\ of\ the\ unit\ cell}{Volume\ of\ the\ unit\ cell}\)
\(d=\frac{Z\times m}{a^3}\) ...(1)
[Since, mass of the unit cell = Number of atoms in the unit cell x Atomic mass]
[Volume of the unit cell = (edge length of the cubic unit cell)³]
(iv) From equation (1), We have
\(m=\frac{d\times a^3}{Z}\) ....(2)
(v) Now, 
Mass of the metal (M) \(=\frac{Atomic\ mass(M)}{Avogadro's\ number(N_A)}\)
M=\(\frac{d\times a^3 \times N_A}{Z}\)
(vi) From equation (3), we can determine the atomic mass of the unknown metal.

(i) Carbon monoxide can be prepared by the reaction of carbon with limited amount of oxygen.
2C + O₂ ⟶ 2CO
(ii) (a) On industrial scale carbon monoxide is produced by the reaction of carbon with air.
(b) The carbon monoxide formed will contain nitrogen gas also and the mixture of nitrogen and carbon monoxide is called producer gas.
(c) \(2C+{ O }_{ 2 }/{ N }_{ 2 }(air)\longrightarrow \underset { Producers \ Gas }{ 2CO } +{ N }_{ 2 }\)
(d) The producer gas is then passed through a solution of copper(I) chloride under pressure which results in the formation of CuCI(CO).2H₂O.
(e) At reduced pressures this solution releases the pure carbon monoxide.
(iii) Pure carbon monoxide is prepared by warming methanoic acid with concentrated sulphuric acid which acts as a dehydrating agent.
HCOOH + H₂SO₄ ⟶ CO + H₂O + H₂SO₄

(i) The impure nickel is heated in a stream of carbon monoxide at around 350 K.
(ii) The nickel reacts with the CO to form a highly volatile nickel tetracarbonyl.
(iii) The solid impurities are left behind 
\({ Ni }_{ (s) }+4{ CO }_{ (g) }\longrightarrow { Ni(CO) }_{ 4(g) }\)
(iv) On heating the nickel tetracarbonyl around 460 K, the complex decomposes to give pure metal.
\({ Ni(CO) }_{ 4(g) }\longrightarrow { Ni }_{ (s) }+{ 4CO }_{ (g) }\)

The structures of the binuclear metal carbonyls involve either metal-metal bonds or bridging CO groups, or both. The carbonyl ligands that are attached to only one metal atom are referred to as terminal carbonyl groups, whereas those attached to two metal atoms simultaneously are called bridging carbonyls. Depending upon the structures, metal carbonyls are classified as follows.
Non-bridged metal carbonyls:
These metal carbonyls do not contain any bridging carbonyl ligands. They may be of two types.
(i) Non- bridged metal carbonyls which contain only terminal carbonyls. Examples: [Ni (CO)₄], [Fe (CO)₅] and [Cr (CO)₆]
(ii) Non- bridged metal carbonyls which contain terminal carbonyls as well as Metal- Metal bonds. For examples, The structure of Mn₂(CO)₁₀ actually involve only a metal-metal bond, so the formula is more correctly represented as (CO)₅Mn-Mn(CO)₅ 
Other examples of this type are, Tc₂(CO) 10, and Re₂(CO)₁₀.