(i) Compound (A) with molecular formula C₂H₄O reduces Tollen'sreagent.
(ii) (A) on treatment with HCN gives compound (B).
\(\underset { (A) }{ { CH }_{ 3 }-CHO } +HCN\longrightarrow { CH }_{ 3 }-\underset { \overset { | }{ OH\\ (B) } }{ CH } -CN\) 
(iii) Compound (B) on hydrolysis with an acid gives compound (C) with molecular formula C₃H₆O₃ which is an optically active compound.
\({ CH }_{ 3 }-\underset { \overset { | }{ OH\\ (B) } }{ CH } -CN\overset { { H }^{ + }{ H }_{ 2 }O }{ \underset { \Delta }{ \longrightarrow } } CH_{ 3 }-\underset { \overset { | }{ OH\\ (C) } }{ CH } -COOH\) 
(iv) Compound (A) on reduction with N₂H/C₂H₅ONa gives a hydrocarbon (D) of molecular formula C₂H_6.
\(\underset { (A) }{ { CH }_{ 3 }CHO } \overset { [H] }{ \underset { { N }_{ 2 }{ H }_{ 4 }/{ C }_{ 2 }{ H }_{ 5 }ONa }{ \longrightarrow } } \underset { (D) }{ CH_{ 3 }{ CH }_{ 3 } } \)

(i) An organic compound (A) with formula C₇H₆O form bisulphite means it must be benzaldehyde C₆H₅CHO.
(ii) Benzaldehyde on treatment with KCN form benzoin.
\(\underset { (A) }{ { C }_{ 6 }{ H }_{ 5 }CHO+{ C }_{ 6 }{ H }_{ 5 }CHO } \overset { KCN }{ \longrightarrow } \underset { (B) }{ { C }_{ 6 }{ H }_{ 5 }-\underset { \overset { | }{ HO } }{ CH } -\underset { \overset { || }{ O } }{ C } -{ C }_{ 6 }{ H }_{ 5 } } \) 
(iii) Benzaldehyde on treatment with sodium acetate and acetic anhydride form an acid, cinnamic acid it is (C).
\(\underset { (A) }{ { C }_{ 6 }{ H }_{ 5 }CHO } +\left( { CH }_{ 3 }{ CO } \right) O\overset { { CH }_{ 3 }COONa }{ \longrightarrow } \underset { (C) }{ { C }_{ 6 }{ H }_{ 5 }CH } =CH-COOH+CH_{ 3 }COOH\)

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

Structure of glucose: Glucose is an aldohexose. It is optically active with four asymmetric carbons. Its solution is dextrorotatory and hence it is also called as dextrose. The proposed structure of glucose is shown in the figure which was derived based on the following evidences

(i) Elemental analysis and molecular weight determination show that the molecular formula of glucose is C6H120 6'
(ii) On reduction with concentrated HI and red phosphorus at 373K, glucose gives a mixture of n hexane and 2, iodohexane indicating that the six carbon atoms are bonded linearly.

(iii) Glucose reacts with hydroxylamine to form oxime and with HCN to form: cyanohydrin. These reactions indicate the I presence of carbonyl group in glucose

(v) Glucose is oxidised to gluconic acid with ammonical silver nitrate (Tollen's reagent) and alkaline copper sulphate (Fehling's solution). Tollens reagent is I reduced to metallic silver and Fehlings I solution to cuprous oxide which appears as red precipitate. These reactions further: confirm the presence of an aldehyde group


(vi) Glucose forms penta acetate with acetic anhydride suggesting the presence of five alcohol groups.

(vii) Glucoseis a stable compound and does not undergo dehydration easily. It indicates that not more than orie hydroxyl group is bonded to a single carbon atom. Thus the five hydroxyl groups are attached to five different carbon atoms and the sixth carbon is an aldehyde group.

(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) The organic compound (A) which is sodium salt of phenolic acid is sodium salicylate.
(ii) (A) on heating with soda lime gives Compound (B) phenol. Phenol gives violet colouration with neutral perchloride.

(iii) Phenol on treatment with C₆H₅COCl in the presence of NaOH gives on ester 'C':
C₆H₅OH + C₆H₅COCl \(\overset { NaOH }{ \longrightarrow } \underset { (C) }{ { C }_{ 6 }{ H }_{ 5 }OCO{ C }_{ 6 }{ H }_{ 5 }+HCl } \)

Compound	Compound Name	Formula
A	Sodium benzoate'
B	Phenol	C6H5OH
C	Phenyl benzoate	C6H5OCO C6H5

Inter molecular dehydration of alcohol: We have already learnt that when ethanol is treated with con.HSO₂ 4 at 443K, elimination takes place to form ethene. If the same reaction is carried out at 413K, substitution competes over elimination to form ethers.

(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 ➝ \(\underset { Schiffi's \ base \ Benzal-ethylamine }{ { C }_{ 6 }{ H }_{ 5 }CH=N{ C }_{ 2 }{ H }_{ 5 }+{ H }_{ 2 }O } \)

(i) Compound A is phenol. (C₆H₅OH)
Phenols on reaction with metallic Na compound (B) ie., Sodium Phenoxide
\(\underset { (A) }{ { 3C }_{ 6 }{ H }_{ 5 }OH+2Na } \longrightarrow \underset { (B) }{ { 2C }_{ 6 }{ H }_{ 5 }ONa+{ H }_{ 2 } } \) 
(ii) (B) is heated with CO₂ at 400K under pressure, to give (C). This is decomposed by dil.HCl, and (D) is formed

Compound	Compound Name	Formula
A	Phenol	C6H5OH
B	Sodium Phenoxide	C6H5ONa
C	Sodium Salicylate
D	Salicylicacid

(i) For a chemical reaction, catalyst is needed in very small quantity.
(ii) There may be some physical changes, but the catalyst remains unchanged in mass and chemical composition in a chemical reaction.
(iii) A catalyst itself cannot initiate a reaction. 
(iv) A solid catalyst will be more effective if it is taken in a finely divided form.
(v) A catalyst are specific in nature.
(vi) In an equilibrium reaction, presence of catalyst reduces the time for attainment of equilibrium and hence it does not affect the position of equilibrium and the value of equilibrium constant.
(vii) A catalyst is highly effective at a particular temperature called as optimum temperature.
(viii) Presence of a catalyst generally does not change the nature of products

Substances like AgCl, PbSO₄ etc., are sparingly soluble in water. The solubility product of such substances can be determined using conductivity measurements.
Let us consider AgCl as an example
AgCl(s) ⇌ Ag⁺ + Cl^-
K_sp = [Ag⁺][Cl^-]
Let the concentration of [Ag⁺] be 'C' mol L^-1.
As per the stoichiometry, if [Ag⁺] = C, then [Cl^-] also equal to 'C' mol L^-1.
Ksp = C.C
⇒ K =C²
We know that the concentration (in mol dm^-3) is related to the molar and sepcific conductance by the following expressions
\({ \Lambda }_{ 0 }=\frac { k\times { 10 }^{ -3 } }{ C(in \ mol \ { L }^{ -1 }) } \)
(or)
\(C=\frac { k\times { 10 }^{ -3 } }{ \Lambda } \)
Substitute the concentration value in the relation K_sp =C²
\({ K }_{ sp }={ \left( \frac { k\times { 10 }^{ -3 } }{ \Lambda } \right) }^{ 2 }\)

Given: E = 0.301 V; [Cu²⁺] = 0.01M
Formula:
\({ E }_{ { Cu }^{ 2+ }/Cu }^{ o }={ E }_{ { Cu }^{ 2+ }/Cu }^{ }+\frac { 2.303Rt }{ nF } \log\frac { [{ Cu }^{ 2+ }] }{ [Cu] } \)
Solution:
\(=+0.301+\frac { 0.0591 }{ 2 } \log\frac { 0.01 }{ 1 } \)
\({ E }^{ o }=0.301+\frac { 0.059 }{ 2 } \times 2=0.3591V\)
E^o = 0.36 V

Given: Electrochemical
equivalent = Z = 2.35 g amp^-1 sec^-1
Time = t = 10 sec.
Current strength = 1 = 5 ampere
Formula: m = Zlt
Solution: 2.35 x 10 x 5 = 117.5 gm
The amount of the substance deposited
= 117.5 g

Let us explain the buffer action in a solution containing CH₃COOH and CH₃COONa.
The dissociation of the buffer components occurs as below.
\( \mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{sq})} \rightleftharpoons \mathrm{CH}_{3}-\mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{H}_{3} \mathrm{O}_{(\mathrm{aq})}^{+} \)
\(\mathrm{CH}_{3} \mathrm{COONa}_{(\mathrm{s})} \stackrel{\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}}{\longrightarrow} \mathrm{CH}_{3}-\mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{Na}_{(\mathrm{aq})}^{+}\)
If an acid is added to this mixture, it will be consumed by the conjugate baseCH₃COO^- to form the undissociated weak acid i.e, the increase in the concentration of H⁺ does not reduce the pH significantly.
\({ { CH }_{ 3 }COO }_{ (aq) }^{ - }+{ H }_{ (aq) }^{ + }\rightarrow { CH }_{ 3 }{ COOH }_{ (aq) }\)
If a base is added, it will be neutralized by H₃O⁺, and the acetic acid is dissociated to maintain the equilibrium. Hence the pH is not significantly altered.

The degree of ionisation and the dissociation constant of the weak acid are related by the equation.
\({ K }_{ a }=\frac { { \alpha }^{ 2 }{ C } }{ 1-\alpha } \cong { \alpha }^{ 2 }C\)
α = 5% (or) 0.05
C = 0.02M
K_a = (0.05)2 x 0.02 = 0.00005
K_a = 5 x 10^-5

Formic acid is a weak acid. Let it be represented as HA.
\(\underset { C(1-\alpha ) }{ HA } \rightleftharpoons \underset { C\alpha }{ { H }^{ + }+ } \underset { C\alpha }{ { A }^{ - } } \)
According to Ostwald's dilution law;
\({ K }_{ a }={ C\alpha }^{ 2 }\)
\(\therefore \alpha =\sqrt { \frac { { K }_{ a } }{ C } } \)
K_a = 1.8 x 10^-4; C = 0.2 M = 2 x 10^-1 M
\(\therefore \alpha =\sqrt { \frac { 1.8\times { 10 }^{ -4 } }{ 2\times { 10 }^{ -1 } } } =\sqrt { 9\times { 10 }^{ -4 } } =3\times { 10 }^{ -2 }\)
Percentage of ionisation = 100 α
= 10² x 3 x 10^-2 = 3

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%

Rate = k [x]² = ka²
[x] = a mol^-1
If the concentration of x is in cross three time, then
(x) = 3a mol L^-1
Rate = R(3a)² = 9 ka²
Hence, the rate of formation will increase by 9 times.

(i) The first transition metal Scandium exhibits only +3 oxidation state, but all other transition elements exhibit variable oxidation states by loosing electrons from (n-1)d orbital and ns orbital as the energy difference between them is very small.
(ii) At the beginning of the series, +3 oxidation state is stable but towards the end +2 oxidation state becomes stable.
(iii) The number of oxidation states increases with the number of electrons available, and it decreases as the number of paired electrons increases.
(iv) Hence, the first and last elements show less number of oxidation states and the middle elements with more number of oxidation states.
(v) For example, the first element Sc has only one oxidation state +3; the middle element Mn has six different oxidation states from +2 to +7. The last element Cu shows +1 and +2 oxidation states only.
(vi) The relative stability of different oxidation - states of 3d metals is correlated with the extra stability of half filled and fully filled electronic configurations. Example: Mn²⁺(3d⁵) is more stable than Mn⁴⁺(3d³).

(i) The element which occupies group number 15 and period number 3 is phosphorus. Therefore A is phosphorus. Phosphorus reacts with chlorine to give PCI₃There fore compound B is phosphorus trichloride and it has a pyramidal shape.
P₄ + 6Cl₂ \(\longrightarrow \) 4PCl₃
PCl₃ further reacts with Cl₂ to give PCl₅
Therefore, the compound C is phosphorus pentachloride and it has a trigonal bipyramidal shape.
PCI + CI\(\longrightarrow \) PCI
Thus, A= Phosphorus
B = Phosphorus trichloride (PCI₃)
C = Phosphorous pentachloride (PCI₅)

Oxidising property : 
Sulphur dioxide, oxidises hydrogen sulphide to sulphur and magnesium to magnesium oxide.
\({ 2H }_{ 2 }S+{ SO }_{ 2 }\longrightarrow 3S+{ 2H }_{ 2 }O\)
\(2Mg+{ SO }_{ 2 }\longrightarrow 2MgO+S\)
Reducing property : 
As it can readily be oxidised, it acts as a reducing agent. It reduces chlorine into hydrochloric acid.
\({ SO }_{ 2 }+2{ H }_{ 2 }O+{ { Cl }_{ 2 }\longrightarrow { H }_{ 2 }{ SO }_{ 4 }+2HCl }\)
It also reduces potassium permanganate and dichromate to Mn²⁺ and Cr³⁺ respectively.
\({ 2KMnO }_{ 4 }+5{ SO }_{ 2 }+2{ H }_{ 2 }O\longrightarrow { K }_{ 2 }{ SO }_{ 4 }+{ 2MnSO }_{ 4 }+2{ H }_{ 2 }{ SO }_{ 4 }\)
\({ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }+{ 3SO }_{ 2 }+{ H }_{ 2 }{ SO }_{ 4 }\longrightarrow { K }_{ 2 }{ SO }_{ 4 }+{ Cr }_{ 2 }\left( SO_{ 4 } \right) _{ 3 }+{ H }_{ 2 }O\)

(i) O₂ molecules are held together by weak Vander Waal's force because of small size and high electronegativity of oxygen.
(ii) In contrast, sulphur shows catenation and forms stronger S-S bonds.
(iii) Due to catenation, sulphur forms octa-atomic S₈ molecules having eight membered puckered ring structure.
(iv) Because of its bigger size the force of attraction holding S₈ molecules are much stronger.
(v) Hence sulphur is a solid at room temperature or in other words, that is why there is a large difference between the boiling point (also melting points) of the two elements.

Crystal defects are classified as follows
(i) Point defects
(ii) Line defects
(iii) Interstitial defects
(iv) Volume defects
Point defects are further classified as follows

(i) It is given that the atoms of A are present at the corners of the cube.
(ii) Therefore, number of atoms of Q in one unit cell = 8 x \(\frac{1}{8}\) = 1
(iii) It is also given that the atoms of P are present at the body - centre.
(iv) Therefore, number of atoms of P in one unit cell.
(v) This means that the ratio of the number of P atoms to the number of Q atoms, P: Q = 1:1.
(vi) Hence, the formula of the compound is PQ.
(vii) The coordination number of both P and Q is 8.

(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

Valance bond theory helps us to visualize the bonding in complexes. However, it has limitations as mentioned above. Hence Crystal Field Theory to explain some of the properties, like colour, magnetic behavior, etc., This theory I was originally used to explain the nature of bonding in ionic crystals. Later on, it is used to explain the properties of transition metals and their complexes. The salient features of this theory are as follows.
(i) Crystal Field Theory (CFT) assumes that the bond between the ligand and the central metal atom is purely ionic. i.e. the bond is formed due to the electrostatic attraction between the electron rich ligand and the electron deficient metal.
(ii) In the coordination compounds, the central metal atom/ion and the ligands are considered as point charges (in case of I charged metal ions or ligands) or electric dipoles (in case of neutral metal atoms or ligands).
(iii) According to crystal field theory, the complex formation is considered as the following series of hypothetical steps.
Step 1: In an isolated gaseous state, all the five d orbitals of the central metal ion are degenerate. Initially, the ligands form a spherical field of negative charge around the metal. In this filed, the energies of all the five d orbitals will increase due to the repulsion between the electrons of the metal and the ligand.
Step 2: The ligands are approaching the metal atom in actual bond directions. To illustrate this let us consider an octahedral field, in which the I central metal ion is located at the origin and the six ligands are coming from the +x, -x, +y, -y, +z  and -z directions as shown below.

As shown in the figure, the orbitals lying along the axes dx²-y² and dz² orbitals will experience strong repulsion and raise in energy to a greater extent than the orbitals with lobes directed between the axes (d_xy, d_yz, and d_zx). Thus the degenerate d orbitals now split into two sets and the process is called crystal field splitting.
Step 3: Up to this point the complex formation would not be favored. However, when the ligands approach further, there will be an attraction between the negatively charged electron and the positively charged metal ion, that results in a net decrease in energy. This decrease in energy is the driving force for the complex formation.
Crystal field splitting in octahedral complexes: During crystal field splitting in octahedral field, in order to maintain the average energy of the orbitals (barycentre) constant, the energy of the orbitals d_x2-_y2 and d _z2 (represented as e_g orbitals) will increase by 3/5 \({ \triangle }_{ o }\) while that of the other three orbitals d_{xy '} d_yz and d_zx (represented as t_2g orbitals) decrease by 2/5  \({ \triangle }_{ o }\) , Here, \({ \triangle }_{ o }\)   represents the crystal field splitting energy in the octahedral field.

The postulates of valence bond theory
(i) The central metal atom/ion makes available a number of vacant orbitals equal to its coordination number.
(ii) These vacant orbitals form covalent bonds with the ligand orbitals.
(iii) A covalent bond is formed by the overlap of a vacant metal orbital and filled ligand orbitals. This complete overlap leads to the formation of a metal ligand, σ (sigma) bond.
(iv) A strong covalent bond is formed only when the orbitals overlap to the maximum extent.
(v) This maximum overlapping is possible only when the metal vacant orbitals undergo a process called 'hybridisation'.
(vi) A hybridised orbital has a better directional characteristic than an unhybridized one. The following table gives the coordination number, orbital hybridisation, and geometry of the complexes.

Magnetic moment 
The paramagnetic moment is given by the following spin-only formula.
\({ \mu }_{ s }=\sqrt { n(n+2) } \) BM
BM = Bohr magneton
\({ \mu }_{ s }\) = spin -only magnetic moment
n = number of unpaired electrons.