(ii) benzoic acid

(iii) α - hydroxyphenylaceticacid.

(i) Neutralisation of chargers of ion will taken place and hence precipitation will take place (ie) Fe³⁺ and S^2- ion changes are neutralized. No new compounds are formed.
(ii) Fe(OH)₃ is a positive Sol
(iii) As₂S₃ is a negative Sol

(i) Almost all carbohydrates are optically active as they contain one or more chiral carbons.
(ii) The number of optical isomers depends upon the number of chiral carbons (ie) 2ⁿ isomers, where n = total number of chiral carbons.
(iii) Glucose has \(4{ }^{\star} \)C; ∴ It has 2⁴ =16 isomers.

Ea = 200 kJ mol^-1=200 \(\times\) 10³J mol^-1
A = 1.6 \(\times\) 10¹³s^-1; T = 600 K; R = 8.314 JK mol^-1
\(k=A{ e }^{ -\left( \frac { Ea }{ RT } \right) }\)
\(k=1.6\times { 10 }^{ 13 }{ s }^{ -1 }{ e }^{ -\left( \frac { 200\times 10^3}{ 8.314 \times 600 } \right) }\)
\(k=1.6\times { 10 }^{ 13 }{ s }^{ -1 }{ e }^{ -\left( 40.09 \right) }\)
\(k=1.6\times { 10 }^{ 13 }\times 3.8\times { 10 }^{ -18 }{ s }^{ -1 }\)
\(k=6.08\times { 10 }^{ -5 }{ s }^{ -1 }\)

\(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow 12 \mathrm{C}+\mathrm{H}_{2} \mathrm{SO}_{4} \cdot 11 \mathrm{H}_{2} \mathrm{O} \\ \text { Sucrose } \)
\(\underset{\text { Formic acid }}{\mathrm{HCOOH}}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{CO}+\mathrm{H}_{2} \mathrm{SO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\)
\(\underset{\text { Oxalic acid }}{\mathrm{(COOH)}_{2}}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{CO}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{SO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\)

(i) In diborane two BH₂ units are linked by two bridged hydrogens.
(ii) It has eight B-H bonds.
(iii) Diborane has only 12 valance electrons.
(iv) The four terminal B-H- bonds is "2c - 2e" bond (two centre - two electron bond.)
(v) Two three centred B - H - B bonds two electrons each. "(3c - 2e)"
(vi) In diborane, the boron is "sp³" hybridised
(vii) Three of the four "sp³" hydridised orbitals contains single electron and the fourth orbital is empty.

1. This process is called as Hall-Heroult process.
Cathode: In this method, electrolysis is carried out in an iron tank lined with carbon which acts as the cathode.
Anode: The carbon blocks immersed in the electrolyte acts as a anode.
Eletrolyte: A 20% solution of alumina, obtained from the bauxite ore is mixed with molten Cryolite and is taken in the electrolysis chamber.
2. About 10% calcium chloride is also added to the solution.
3. Here Calcium chloride helps to lower the melting point of the mixture.
Temperature: The fused mixture is maintained at a temperature of above 1270 K.
4. The chemical reactions involved in this process as follows
(a) Ionisaiton of alumina: \({ A }l_{ 2 }{ O }_{ 3 }\longrightarrow { 2Al }^{ 3+ }+{ 3O }^{ 2- }\)
(b) Reaction at cathode: \(2{ Al }^{ 3+ }_{(melt)}+{ 6e }^{ - }\longrightarrow { Al }_{ (l) }\)
(c) Reaction at anode: \(6{ O }^{2-}_{(melt)}\longrightarrow { 3O }_{ 2 }+{ 12e }^{ - }\)
5. Since carbon acts as anode the following reaction also takes place
(a) \({ C }_{ (s) }+{ O }^{ 2- }_{(melt)}\longrightarrow CO+{ 2e }^{ - }\)
(b) \({ C }_{ (s) }+{ 2O }^{ 2- }_{(melt)}\longrightarrow { CO }_{ 2 }+{ 4e }^{ - }\)
6. Due to the above two reactions, anodes are slowly consumed during the electrolysis.
7. The pure aluminium is formed at the cathode. The net electrolysis reaction can be written as
\({ 4Al }^{ 3+ }_{(melt)}+{ 6O }^{ 2- }_{(melt)}+{ 3C }_{ (s) }\longrightarrow { 4Al }_{ (l) }+{ 3CO }_{ 2(g) }\)

(i) E is monobasic acid (RCOOH) having molecular weight 60 and it is formed from D on oxidation. So E must be acetic acid and D must be acetaldehyde.
(ii) (B) on oxidation gives CH₃CHO. So (B) must be alcohol (CH₃CH₂OH).
(iii) Acetaldehyde (D) on treating with aqueous alkali (NaOH) gives aldol which on heating gives 2- butenal (E).

(iv) Compound E on catalytic hydrogenation gives butyl alcohol.

(v) Hence compound (A) must be an ester. Ester (A) on reduction with LiAIH₄ yields two alcohols (B) and (C). (A) is ethyl butyrate.

'A' can also be CH₃COOCH₂CH₂CH₂CH₃. This structure will be answering all the above reactions.

\({ CH }_{ 3 }-{ CH }_{ 2 }-OH\overset { { H }_{ 2 }{ SO }_{ 4 } }{ \underset { 443k }{ \longrightarrow } } { CH }_{ 2 }={ CH }_{ 2 }+{ H }_{ 2 }O\)
Mechanism
Primary alcohols undergo dehydration by E₂ mechanism

+ 2H₂O

\(\kappa=\mathrm{C}\left(\frac{l}{\mathrm{~A}}\right)\) (or) k = 1/R.l/A
Where, C - Conductance
\(\frac{l}{A} \Rightarrow\) Cell constant (or)
Sp. Conductance \(\propto\) Cell constant
Sp. Conductance \(\propto\) Concentration
1. Sp. Conductance = concentration x cell constant
i) \(\kappa=0.1 \times 0.5=0.05=5 \times 10^{-2} \mathrm{Scm}^{-1}\)
ii) \(\kappa=0.1 \times 0.25=0.025=2.5 \times 10^{-2} \mathrm{Scm}^{-1}\)
2. So the first case will have greater specific conductance with cell constant 0.05.

(i) Let us consider the reactions between a strong acid, HCI, and a weak base, NH₄OH, to produce a salt, NH₄CI, and water.
HCI_(aq) + NH₄OH_(aq) ⇌ NH₄CI_(aq) + H₂O _(I)
NH₄CI_(aq) ⟶ NH₄⁺ +CI^-_(aq)
(ii) NH₄⁺ is a strong conjugate acid of the weak base NH₄OH and it has a tendency to react with OH^- from water to produce unionised NH₄OH
NH₄⁺ _(aq) + H₂O_(I) ⇌ NH₄OH_(aq) + H⁺_(aq)
(iii) There is no such tendency shown by Cl^- and therefore [H⁺] > [OH^-]; the solution is acidic and the pH is less than 7.
(iv) The K_h and K_b are related by
\(\mathrm{K}_{\mathrm{h}} \cdot \mathrm{K}_{\mathrm{b}}=\mathrm{K}_{\mathrm{w}}\)
(Or)
\(\mathrm{K}_{\mathrm{h}}=\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{b}}}\)
Degree of hydrolysis (h)
\(\underset{(1-h)}{\mathrm{NH}_{4}^{+}(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(j)} \rightleftharpoons \underset{h} {\mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})}} +\mathrm{H}^{+} \underset{h}{(\mathrm{aq})}\\ \)
\(\mathrm{K}_{\mathrm{h}} =\frac{\left[\mathrm{NH}_{4} \mathrm{OH}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{NH}_{4}^{+}\right]} \\ \)
\(=\frac{\mathrm{hc} \times \mathrm{h}}{(1-\mathrm{h}) \mathrm{c}} \)
\(K_{h}=\frac{h^{2} c}{(1-h)} \)
\(\text{If } \mathrm{h}<<1 ; \mathrm{K}_{\mathrm{h}} \simeq \mathrm{h}^{2} \mathrm{c} \)
\(h^{2}=\frac{K_{h}}{c}\)
\(h=\sqrt{\frac{K_{h}}{c}} \ (or) \ h=\sqrt{\frac{K_{w}}{K_{b} \cdot C}} \quad\left(\because K_{h}=\frac{K_{w}}{K_{b}}\right)\)
Also; \(\left[\mathrm{H}^{+}\right]=\sqrt{\mathrm{K}_{\mathrm{h}} \cdot \mathrm{C}} \) (or) \(\left[\mathrm{H}^{+}\right]=\sqrt{\frac{\mathrm{K}_{\mathrm{w}} \cdot \mathrm{C}}{\mathrm{K}_{\mathrm{b}}}}\)
pH = -log [H⁺]

If the monomer is represented by X. Then
\(2 \mathrm{M} \rightarrow(\mathrm{M})_{2}\)
Since the reaction is of second order, the rate of reaction will be given by,
\(\text { Rate }=\mathrm{k}[\mathrm{M}]^{n}  \)
\(7.5 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}=\mathrm{k}\left(0.05 \mathrm{~mol} \mathrm{} \mathrm{~L}^{-1}\right)^{2} \)
\(\mathrm{k} =\frac{Rate}{[M]^{n}}  \)
\(\mathrm{k} =\frac{7.5 \times 10^{-3}}{(0.05)^{2}}  \)
\(=3 \mathrm{~mol}^{-1} \mathrm{~L} \mathrm{~s}^{-1}\)

(i) Fe³⁺ - electronic configuration - [Ar] 3d⁵
(ii) It has exactly half-filled stable electronic configuration.
(iii) Fe²⁺ - electronic configuration -[Ar]3d⁶
(iv) It has only partially filled d-orbitals.
Hence Fe³⁺ is more stable than Fe²⁺.

These two are ionisation isomers. [Co(NH₃)₅Cl]SO₄ gives white precipitate with BaCl₂ solution, but not with AgNO₃ solution. [Co(NH₃)₅SO₄]Cl gives curdy white precipitate with AgNO₃ solution but not with BaCl₂ solution.

(i) Schottky defect arises due to the missing of equal number of cations and anions from the crystal lattice. This effect does not change the stoichiometry of the crystal.
(ii) Ionic solids in which the cation and anion are of almost of similar size show schottky defect.
Example: NaCl.
(iii) Presence of large number of schottky defects in a crystal, lowers its density.
(iv) Presence of Schottky defect in the crystal provides a simple way by which atoms or ions can move within the crystal lattice.

Aniline does not undergo Friedel - Craft's reaction:
Aniline does not undergo Friedel - Craft's reaction (alkylation and acetylation). Aniline is basic in nature and it donates its lone pair of electrons to the lewis acid AlCl₃ to form an adduct which inhibits further electrophilic substitution reaction.
Diazonium salts of aromatic amines are more stable than those of aliphatic amines:
This is due to resonance
Resonance Structure:
The stability of arene diazonium salt is due to the dispersal of the positive charge over the benzene ring.

pK_b of aniline is more than that of methylamine:
pK_b - methylamine -3.35
pK_b - aniline -9.376
In aniline the lone pair of electrons on N - atom is delocalized over the benzene ring. So, the electron density on the N - atom decreases. In methylamine + 1 effect to CH₃ group increases the electron density on the nitrogen atom Hence aniline is a weaker base than methylamine. Due to this, the pK_b value for aniline is more than that of methylamine.
(iv) Gabriel phthalimide synthesis is preferred for synthesising primary amines:
In this method alkyl halides react with pottassium phthalimide to give pure primary amine by nucleophilic substitution. In contrast, Aniline (Aromatic primary amine) can not be prepared by this method because Aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide. Therefore, this method used for the Aliphatic primary. amines only. Aryl halides do not undergo S_N2 mechanism with the ion formed by the phthalimide.
(v) Ethylamine is soluble in water whereas aniline is not:
(a) Ethylamine is soluble in water, as it can form intermolecular H - bonds with water molecules. In aqueous solution, the substituted ammonium cation get stabilized not only by electron releasing (+I) effect of the alkyl group but also by solvation with water molecules. The greater the size of the ion, the lower will be the solvation.
(b) Amiline doesn't form H - bond with water to a very large extent due to the presence of a large hydrophobic -C₆H₅ group.
(vi) Amines are more basic than amides:
This is because, in amides, the carbonyl group is highly electro negative It has a greater power to attract the electrons towards it. It makes the lone pair of electrons on amide nitrogen (-CONH₂) less available to accept a proton.
(vii) Although amino group is o - and p - directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m - nitro aniline:
In strong acid medium, aniline is protonated to form anilinium ion which is m - directing and hence m - nitro aniline is formed.
 

(i) To understand how a soap works as a cleansing agent, let us consider sodium palmitate an example of a soap. The cleansing action of soap is directly related to the structure of carboxylate ions (palmitate ion) present in soap. The structure of palmitate exhibit dual polarity. The hydrocarbon portion is non polar and the carboxyl portion is polar.

(ii) The nonpolar portion is hydrophobic while the polar end is hydrophilic. The hydrophobic hydro carbon portion is soluble in oils and greases, but not in water. The hydrophilic carboxylate group is soluble in water.
(iii) The dirt in the cloth is due to the presence of dust particles intact or grease which stick. When the soap is added to an oily or greasy part of the cloth, the hydrocarbon part of the soap dissolve in the grease, leaving the negatively charged carboxylate end exposed on the grease surface.
(iv) At the same time the negatively charged carboxylate groups are strongly attracted by water, thus leading to the formation of small droplets called micelles and grease is floated away from the solid object. When the water is rinsed away, the grease goes with it. As a result, the cloth gets free from dirt and the droplets are washed away with water. The micelles do not combine into large drops because their surfaces are all negatively charged and repel each other. The cleansing ability of a soap depends upon its tendency to act as a emulsifying agent between water and water insoluble greases.

The cleansing action of detergents are similar to cleansing action of soap. Eg: the structure of a cationic detergents is:

(i) The dissociation of a weak acid (CH₃COOH) is suppressed in the presence of a salt (CH₃COONa) containing an ion common to the weak electrolyte. It is called the common ion effect.
(ii) Consider the dissociation of a weak acid, acetic acid whose ionisation is incomplete.
CH₃COOH_(aq) ⇌ H⁺_(aq) + CH₃COO^-_(aq)
(iii) If the salt sodium acetate with common ion CH₃COO^- is added to the above equilibrium, it dissociates completely increasing.
CH₃COONa_(aq)⟶Na⁺_(aq) + CH₃COO^-_(aq)
(iv) Hence, the overall concentration of CH₃COO^- is increased, and the acid dissociation equilibrium is disturbed.
(v) So, in order to maintain the equilibrium, the excess CH₃COO^- ions combines with H⁺ ions to produce much more unionized CH₃COOH i.e, the equilibrium will shift towards the left. In other words, the dissociation of CH₃COOH is suppressed.