\(\text { (ii) } \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}+\mathrm{KOH}(a q) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\mathrm{KCl}\)
        Ethyl chloride                               Ethyl alcohol       

(i) Clemmensen reduction It involves reduction of carbonyl group of an aldehyde or ketone to methylene group to form a hydrocarbon. Clemmensen reduction is carried out in presence of Zn amalgam and conc. HCl.

It is widely used for the reduction of aldehydes and ketones which are sensitive to alkalies.
(ii) Cannizzaro reaction Aldehydes which do not have ∝-H atoms undergo Cannizzaro reaction on treatment with cone. alkali. In this reaction, one molecule of aldehyde is reduced to alcohol while another molecule is oxidised to salt of carboxylic acid.

i) Iodoform test: Ethanal gives this test due to the presence of CH₃CO :
CH₃HO + 4NaOH + 3I₂ \(\longrightarrow \) CHI₃ + HCOONa + 3Nal + 3H₂0
Ethanal                                  (yellow ppt)
                                               loadform
CH₃CH₂CHO + 4NaOH + 3I₂ \(\longrightarrow \) No reaction
(ii) Sodium bicarbonate test: Benzoic acid gives brisk effervescence of CO₂ when react with NaHCO₃ while phenol does not give it, as it is a weak acid.

(i) pentaamminechloridocobalt (III) chloride
(ii) chloridobis (ethane-1, 2-diamine) nitritocobalt (III) ion.

Here,  \({ K }_{ H }=4.27\times { 10 }^{ 5 }mm,\ p=760 \ mm\)
Applying Henry's law, \(p={ K }_{ H }x, we \ have \ x=\frac { p }{ { K }_{ H } } =\frac { 760 \ mm }{ 4.27 \times{ 10 }^{ 5 }mm } =1.78 \times{ 10 }^{ -3 }\)
i.e., mole fraction of methane in benzene = 1.78 x 10^-3

(i) It is due to -1 and -R effect of -NO₂ group and +I and +R effect of CH₃ group, p-nitrophenoxide ion is more stable than \(\rho\)-methyl phenoxide ion.
(ii) It is due to double bond character in phenol due to resonance but not in CH₃ OH.
(iii) 3° halides undergo \(\beta\)-elimination with strong nucleophile (CH₃O^-).

Positive deviation from Raoult's law occurs when the total vapour pressure of the solution is more than corresponding vapour pressure in case of ideal solution. 
\(P={P}_{{A}}+{P}_{{B}}>{P}_{{A}}^{\circ} {X}_{{A}}+{P}_{{B}}^{\circ} {X}_{{B}}\)
Negative deviation from Raoult's law occurs when the total vapour pressure of the solution is less than corresponding vapour pressure in case of the ideal solution.
\({P}={P}_{{A}}+{P}_{{B}}<{P}_{{A}}^{\circ} {X}_{{A}}+{P}_{{B}}^{\circ} {X}_{{B}}\)
For positive deviation from Raoult's law, Δ _{mix ,}H has a positive sign.
For negative deviation from Raoult's law, Δ _{mix .}H has a negative sign.

(i) Linkage isomerism: The isomerism in which a ligand can- form linkage with metal through different atoms, e.g. nitro group (-NO₂) can link to metal either through nitrogen atom or through oxygen atom, e.g. [Co(NH₃)₅ONO]CI₂ (Pentaamminenitrito-O-cobalt(III) chloride) and [Co(NH₃)₅NO₂]CI₂  (Pentaamminenitrito-N cobalt(llI) chloride) are linkage isomers.
(ii) When 4s, 4p and 4d orbitals take part in hybridization, it is called outer orbital complex.                                                           or
An element which involves ns, np and nd orbitals for bonding is called outer orbital complex.
 e.g. [CoF₆ ]^3- 
Co(27) : [Ar] 4S²3d⁷
C0³⁺ : [Ar] 4s⁰3d6
 
It has sp³d²  hybridisation, therefore, it is an outer orbital complex.
(iii) A ligand which can form two c bonds with central metal ion is called bidentate ligand. It has two donor atoms. e.g:-
 COO^-
 I
 COO^-

(i) ΔT_f = 273 K - 272.4 K = 0.6 K.
Also, ΔT_f = K_f x m
⇒ 0.6 K = 1.86 x m
⇒ m = 0.322 mol/kg
(ii) ∆T_b = m x Kb = 0.322 x 0.512
= 0.165 K
T_b = 373 K + 0.165 K = 373.165 K
(iii)\({p_A^0-p_A\over p_A^0}=X_B={m\over m+{1000\over 18}}={0.322\over 0.322+55.55}\)
\(⇒ 1-{p_A\over p_A^0}={0.322\over 55.87}\)
\(⇒ {p_A\over 23.757}=1-{0.322\over 55.87}={55.585\over 55.87}\)
\(⇒ p_A={23.757\times55.55\over 55.87}\)
= 23.62 mm of Hg

Higher the solubility, lower is K_H. As CO₂ has maximum solubility, its K_H is lowest.

(b) (i) Add CHCl₃ and KOH. Aniline will form offensive smelling compound where as N-methyl aniline will not react.
(ii) Add HNO₂ (CH₃)₂NH will form yellow oily compound where as (CH₃)₃N will form salt soluble in water.

(b) Add I₂ and NaOH. Acetophenone will give yellow ppt of iodoform whereas benzophenone will not react.
(c) It is because carbanion formed by losing H⁺ from \(\alpha\)-carbon is stabilised by resonance.
​​​​​​

(a) (i) It is because ethanal is more polar than acetone due to only one methyl group, whereas in acetone, there are two methyl groups which are electron releasing and reduce positive charge on carbonyl carbon.
(ii) It is because it does not have alpha (a) hydrogen.
(iii) It is due to greater intermolecular H-bonding in carboxylic acids than alcohols and they exist as dimers.
(b) (i) Add I₂ and NaOH. Acetophenone will give yellow ppt. of iodoform, whereas Benzophenone will not react.
(ii) Add I₂ and NaOH. Ethanal will give yellow ppt. of iodoform, whereas benzaldehyde will not react.
Alternathe Method: Add Fehling's solution A and B to each of these and heat. Ethanal will give brick red ppt., whereas benzaldehyde will not react.

(a) (i) It is because \(\mathrm{CH}_{3} \mathrm{O}^{\Theta} \) is strong nucleophile, 3° halide undergo elimination more readily than nucleophilic substitution reaction.
(ii) It is due to double bond character in phenol due to resonance but not in ethanol.
(b) (i) Add neutral FeCI₃ . Ethanol does not react. Phenol gives violet colour.
(ii) Add l₂ and NaOH. Methanol will not react. Propan-2-ol will give yellow ppt. of Iodoform.
(c) (i) 3-Phenoxy heptane.
(ii) 3,3-Dimethyl pentan-2-ol.

(i) (a) Metal carbonyls: Metal carbonyls are the compounds in which carbon monoxide (CO) acts as the ligand. The metal-carbon bond in metal carbonyls possesses both \(\sigma\) and \(\pi\) characters, e.g. N(CO)₄[nickeltetracarbonyl]. Its IUPAC name is tetra carbonyl nickel(0).
(b) Chelate: An inorganic metal complex in which there is a close ring of atoms caused by attachment of a ligand to a metal atom at two points. An example is the complex ion formed between ethylene diamine and cupric ion,[Cu(NH₂CH₂CH₂NH₂)₂]²⁺.

(ii) (a) Add BaCl₂ solution. [Co(NH₃)]₅Cl]SO₄ will give white ppt. of BaSO₄.  
 \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{SO}_{4}+\mathrm{BaCl}_{2}(a q) \)\( \longrightarrow\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}+\mathrm{BaSO}_{4} \downarrow \)
                                                                    \(\text { (White ppt.) }\)
Add AgNO₃(aq) solution [Co(NH₃)₅SO₄] Cl will give white ppt. of AgCl.
\( {\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{Cl}+\mathrm{AgNO}_{3}(a q)} \) \(\longrightarrow \mathrm{AgCl}+\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{NO}_{3} \)
                                                                                          \(\text { (White ppt) }\)
(c) Fe³⁺ (4s⁰3d⁵) In strong field \( \Delta_{0}>P t_{2 g}^{5} e g^{0} \)
In weak field \(\Delta_{0}

(i) It shows positive deviation.
It is due to weaker interaction between acetone and ethanol than ethanol-ethanol interactions.
(ii) Given : W_B  = 10g, W_s = 100 g, W_A = 90 g, M_B = 180 g.mol and d = 1.2 g/mL
\(M=\frac { Wt%\times density\times 10 }{ Mol.wt } \)
\(M=\frac { 10\times 1.2\times 10 }{ 180 } \)
= 0.66 M or 0.66 mol/L
\(m=\frac { { W }_{ B }\times 1000 }{ { M }_{ B }\times { W }_{ A }(in\quad g) } \)
\(m=\frac { 10\times 1000 }{ 180\times 90 } \)
= 0.61 m or 0.61 mol/kg
(or any other suitable method)

(b) If both assertion and reason are true but the reason is not the correct explanation of the assertion.

(b): The tendency to show H-bonding decreases with increasing hydrophobic character of carbon chain.

(b): The complex ion[Al(NH₃)₆]³⁺ undergoes the change into new complex ion [Al(H₂O)₆]³⁺ in aqueous medium due to higher heat of hydration of aluminium ion on account of its small size.
\(\left[\mathrm{Al}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}+6 \mathrm{H}_{2} \mathrm{O} \rightarrow\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}+6 \mathrm{NH}_{3}\)

(a) When the divers come towards the surface, pressure gradually decreases. This releases the dissolved gases and leads to the formation of bubbles of nitrogen in the blood. Hence, both (A) and (R) are correct and (R) is the correct explanation of (A).

(a) (CH₃)₃N does not react with Hinsberg reagent.
(b) 

(c) 
 cetyltrimethyl ammonium bromide.
(d) 

(e) Add NaNO₂ and cone, HCI. Cool it to 0 - 5°C. Then add alkaline solution of phenol. Aniline will give orange, azo dye, whereas Benzyl amine does not.

(i) (d): Oxidation state of Mn in [Mn₂(CO)₁₀] is zero.
(ii) (d): In \(\left[\mathrm{V}(\mathrm{CO})_{6}\right]^{-}\), the anionic carbonyl complex can delocalise more electron density to antibonding  \(\pi\)-orbital (d\(\pi\)-p\(\pi\) back bonding) of CO and thus lowers the bond order.
(iii) (c): K[CO(CO)₄]
+ 1 + (x) + 4(0) = 0 or x = -1
(iv) (d): Mn₂(CO)₁₀ is made up of two square pyramidal Mn(CO)₅ units joined by Mn - Mn bond.