Dichloromethane has highest dipole moment among CH₂​Cl₂​, CHCl₃​ and CCl₄​. The decreasing order of dipole moments is CH₂​Cl₂​>CHCl₃​>CCl₄​. These molecules have tetrahedral geometry due to sp3 hybridization of carbon atom.In CCl4​, the individual C−Cl bond dipoles cancel each other which results in zero dipole moment. 
Hence, CCl₄​ is non polar.

From Arrhenius equation, we obtain
\(\log \frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303 R}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right)\) 
Also, k₁ = 4.5 × 10³ s⁻¹
T₁ = 273 + 10 = 283 K
k₂ = 1.5 × 10⁴ s⁻¹
_Ea = 60 kJ mol⁻¹ = 6.0 × 10⁴ J mol⁻¹
Then,
\(\log \frac{1.5 \times 10^{4}}{4.5 \times 10^{3}}=\frac{6.0 \times 10^{4} \mathrm{Jmol}^{-1}}{2.303 \times 8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}}\left(\frac{T_{2}-283}{283 T_{2}}\right)\) 
\(\Rightarrow 0.5229=3133.627\left(\frac{T_{2}-283}{283 T_{2}}\right)\) 
\(\Rightarrow 0.0427 T_{2}=T_{2}-283\) 
 \(\Rightarrow 0.9528 T_{2}=283\) 
\(\Rightarrow T_{2}=297.019 K \text { (approximately) }\) 
= 297K
= 24º C
Hence, k would be 1.5 × 10⁴ s⁻¹ at 24°C.

A nucleoside contains only two basic components of nucleic acids namely a pentose sugar and a nitrogenous base.
A nucleotide contains all the three basic components of nucleic acids namely a phosphoric acid group, a pentose sugar and a nitrogenous base.

(i) No (ii) Yes (iii)Yes

The solubility of gas is related to the mole fraction in aqueous solution. The mole fraction of the gas in the solution is calculated by applying Henry’s law. Thus:
\(x(\text { Nitrogen })=\frac{p \text { (nitrogen) }}{K_{\mathrm{H}}}=\frac{0.987 \mathrm{bar}}{76,480 \mathrm{bar}}=1.29 \times 10^{-5}\)
As 1 litre of water contains 55.5 mol of it, therefore if n represents number of moles of N₂ in solution,
\(x(\text { Nitrogen })=\frac{n \text { mol }}{n \text { mol }+55.5 \text { mol }}=\frac{n}{55.5}=1.29 \times 10^{-5}\)
(n in denominator is neglected as it is < < 55.5)
Thus n = 1.29 × 10^–5 \(\times\) 55.5 mol = 7.16\(\times\) 10^–4 mol
\(=\frac{7.16 \times 10^{-4} \mathrm{~mol} \times 1000 \mathrm{ \ mmol}}{1 \mathrm{~mol}}=0.716 \mathrm{ \ mmol}\)

(i)  Carbylamine Reaction: When primary amne reacts with CHCl₃ and KOH, it forms isocyanide which is an offensive smelling compund.
\(RN{ H }_{ 2 }+CHC{ l }_{ 3 }+3KOH\rightarrow RN\overset { = }{ \rightarrow } C+3KCl+3{ H }_{ 2 }O\)

(ii) Diazotisation: When aniline is reacted with NaNO₂ and conc. HCl at 0-5^o C, benzene diazonium chloride is formed.

(iii)  Hoffman's Bromamide Reaction: When amide react with Br₂ and KOH or NaOH to form lower amines, it is called  Hoffman's Bromamide Reaction.

(iv) Coupling reaction: When benzene diazonium chloride reacts with phenol, the orange azo dye is formed.

(v) Ammonolysis: When alkyl halide with an excess of NH₃ primary amines are formed.
C₂H₅Cl+NH₃ \(\rightarrow\)C₂H₅NH₂+HCl
 (vi)Carbylamine reaction (i) Carbylamine reaction When aliphatic/ aromatic primary amines are heated with chloroform and alc KOH, foul-smelling alkyl isocyanides or carbylamines are obtained. Secondary or tertiary amines do not give this test.
(vii) \(\mathbf{P h}-\mathbf{N H}_{2}+\mathbf{C H C l}_{3}+3 \mathbf{K} \mathbf{O H} \stackrel{\Delta}{\longrightarrow} \mathbf{P h}-\mathbf{N C}+3 \mathbf{K} \mathbf{C l}+\mathbf{3 H}_{2} \mathbf{O}\)
2 ° and 3 ° (aromatic/aliphatic) do not give this reaction.

(i) \(2Fe^{3+}+2I^-\longrightarrow2Fe^{2+}+I_2\)
\(E^0_{cell}=E^0_{I_2/ I^-}-E^0_{Fe^{3+}/Few^{2+}}=0.54 V - (0.77 V) = 0.54 V - 0.11 v = - 0.23 V\)
It is not feasible therefore E⁰_cell is -ve ΔG^o = +ve
(ii) 2Ag⁺(aq) + Cu(s) ⟶ Cu²⁺(aq) + Ag(s)
\(E^0_{cell} = E^0_{Ag^+/Ag} -E^0_{Cu^{2+}/Cu} = +0.80 V - 0.34 V = 0.46 V\)
It is feasible because E⁰_cell is +ve
∴ ΔG⁰ = - ve
(iii) 2Fe³⁺(aq) + 2Br^-(aq) ⟶ 2Fe²⁺(aq) + Br₂'
\(E^0_{cell} = E^0_{Fe^{3+}/Fe^{2+}}- E^0_{Br_2/Br^-} = 0.77 V - 1.09 V = - 0.32 V\)
Since E⁰_cell is -ve, ΔG^o = +ve therefore, reaction is non spontaneous.
(iv) Ag(s) + Fe³⁺(aq) ⟶ Fe²⁺(aq) + Ag⁺(aq)
\(E^0_{cell} = E^0 _{Fe^{3+}/Fe^{2+}}-E^0_{Ag^+/Ag} = 0.77 V - 0.80 V = - 0.03 V\)
since E⁰_cell is -ve, ΔG^o is +ve therefore, reaction is non spontaneous.
(v) Br₂ + 2Fe²⁺(aq) ⟶ 2Br^-(aq) + 2Fe³⁺(aq)
\(E^0_{cell} = E^0_{Br^2/Br^-} E^0_{Fe^{3+}/Fe^{2+}}=1.09 V - 0.77 V = 0.32 V\)
Since E⁰_cell is +ve, ΔG^o = -ve, therefore, reaction will be spontaneous.

(a) It is due to irregular variations in sum of first and second ionisation energies and sublimation energies. It is also due to stability of electronic configuration.
(b) (i) Cr₂O₇^2- + 3Sn²⁺ + 14H⁺➝ 2Cr³⁺ + 3Sn⁴⁺ + 7H₂O
(ii) MnO₄^- + 5Fe²⁺ + 8H⁺ ➝ SFe³⁺ + Mn²⁺ + 4H₂O

(a) CH₃OC₂H₅ undergoes cleavage by S_N2 mechanism to give CH₃I and C₂H₅OH while (CH₃)₃COCH₃ undergoes cleavage by S_N1 mechanism to give (CH₃)₃CI and CH₃OH.

Carbohydrates are the source of energy in plants. Their two main functions are growth and metabolism.

Since the coordination number of Mn²⁺ ion in the complex ion is 4, it will be either tetrahedral (sp³ hybridisation) or square planar (dsp² hybridisation). But the fact that the magnetic moment of the complex ion is 5.9 BM, it should be tetrahedral in shape rather than square planar because of the presence of five unpaired electrons in the d orbitals.

o-Nitrophenol is steam-volatile due to chelation (intramolecular H-bonding) and hence can be separated by steam distillation from p-nitrophenol which is not steam volatile because of intermolecular H-bonding. 

The rate of disinetegration of a radioactive substance is directly proportional to amount of the substance present. Hence, it is a reaction of first order.

(i) (d) : Density of solution = 1.202 g/mL
Volume of solution = \(\frac{100 \mathrm{~g}}{1.202 \mathrm{~g} / \mathrm{mL}}=83.2 \mathrm{~mL}\)
Molarity = \(\frac{n_{\mathrm{KI}}}{\text { Volume of solution in } \mathrm{L}}\)
\(=\frac{0.120 \mathrm{~mol}}{0.0832 \mathrm{~L}}=1.4423 \mathrm{~mol} \mathrm{~L}^{-1}\)
(ii) (a): \(x_{2}=\frac{n_{2}}{n_{1}+n_{2}} ; x_{1}=\frac{n_{1}}{n_{1}+n_{2}} ; \frac{x_{2}}{x_{1}}=\frac{n_{2}}{n_{1}}\)
\(\frac{x_{2}}{x_{1}}=\frac{m_{2} / M_{2}}{m_{1} / M_{1}}=\frac{m_{2}}{m_{1}} \times \frac{M_{1}}{M_{2}}\)          ...(i)
Molality = \(\frac{n_{2}}{m_{1}}=\frac{m_{2}}{M_{2} \times m_{1}}\)              ...(ii)
From(i) and (ii), m = \(\frac{x_{2}}{x_{1}} \times \frac{1}{M_{1}} ; x_{1}=1-x_{2}\)
Hence. x₂ = \(\frac{m M_{1}}{1+m M_{1}}\)
(iii) (a) : Mass does not depend on temperature while volume does. Hence, molarity depends on temperature.
(iv) (b): 1M solution contains 1 mole of solute in less than 1000 g of the solvent whereas 1 m solution has 1 mole of the solute in 1000 g of the solvent.

(i) (a): Spin only magnetic moment, \(\mu=\sqrt{n(n+2)} \text { B.M. }\) where n = number of unpaired electrons.
As the number of unpaired electrons in Cr²⁺ ([Ar ]3d⁴) and Fe²⁺([Ar]3d⁶) are same, hence [Cr(H₂O)₆]²⁺ and [Fe(H₂O)₆]²⁺ will have same magnetic moment.
(ii) (a): Fe⁺ : [Ar] 3d⁶ 4s¹
When the weak field ligand H₂O and strong field ligand NO⁺ attack, the configuration changes as follows: Fe+ : [Ar] 3d⁷ 4s°

ஃ Fe⁺ has 3 unpaired electrons.

In presence of strong ethylenediamine ligand the electrons get paired.

Thus inner orbital complex with no unpaired electrons.
(iv) (a)