(a) The Kohlrausch law of independent migration of ions state that "limiting molar conductivity of an electrode is the sum of the individual contribution of the cation and the anion of the electrolyte."
(b) The Faraday's first law states that "the amount of chemical reaction occuring at any electrode by passing current is directly proportional to the quantity of electricity passed through the electrolyte.

This is because that sum of ionisation energies (E₁ + E₂) is less in case of Ni than Pt.

When the concentration of electrolyte approaches zero, the molar conductivity is known as limiting molar conductivity \(\left(\Lambda_m^{\circ}\right)\)
\(\Lambda_m^{\circ}=\left(\Lambda_m\right) \text { when } \mathrm{C} \rightarrow 0\)

This reaction involves the preparation of ether by Williamson's synthesis. It takes place via S_N2 attack of an alkoxide ion on primary alkyl halide because here the alkyl halide used is primary. Moreover in method (i) the C-Br bond has some double bond character.

Phenol and methyl iodide

(i) 8MnO₄^-(aq)+3S₂O₃^2-(aq)+H₂O(l) \(\rightarrow \)8MnO₂+6SO₄^2-+2OH^-
(ii) Cr₂O₇^2-(aq) + 6Fe²⁺(aq) +14H⁺(aq) \(\rightarrow \) 2Cr³⁺+6Fe³⁺+7H₂O

(a) Kohlrausch's law of independent migration of ions: According to this law, molar conductivity of an electrolyte, at infinite dilution, can be expressed as the sum of the contributions from its individual ions. If the molar conductivity of the cations is denoted by \(\lambda_{+}^{\infty}\) and that of the anions by \(\lambda_{-}^{\infty},\) then the law of independent migration of ions is
\(\Lambda_{m}^{\infty}=v_{+} \lambda_{+}^{\infty}+v_{-} \lambda_{-}^{\infty}\) 
or \(\Lambda^{\circ}=v_{+} \lambda_{+}^{\circ}+v_{-} \lambda^{\circ}\) 
where v₊ and v_{_} are the number of cations and anions per formula of electrolyte
\(\Lambda_{m}^{\circ} \mathrm{CH}_{3} \mathrm{COOH}=\lambda_{m}^{\circ} \mathrm{CH}_{3} \mathrm{COO}^{-}+\lambda_{m}^{\circ} \mathrm{H}^{+}\) 
The degree of dissociation a can be calculated by the formula \(\alpha=\frac{\left(\Lambda_{m}\right)_{C}}{\left(\Lambda_{m}\right)_{\infty}}\) where (A_m)_c is molar conductivity of a particular cone. an \(\left(\Lambda_{m}\right)_{\infty}\) is the maximum molar conductivity of infinite dilution. It is also denoted by \(\Lambda_{m}^{\circ}\)called limiting molar conductivity.
(b) \(\Lambda_{m}^{\circ}(\mathrm{NaCl})=\lambda^{\circ}\left(\mathrm{Na}^{+}\right)+\lambda^{\circ}(\mathrm{Cl^-})\) 
= 126 S cm² mol^-1         ... (i)
\(\Lambda_{m}^{\circ}(\mathrm{HCl})=\lambda^{\circ}\left(\mathrm{H}^{+}\right)+\lambda^{\circ}\left(\mathrm{Cl}^{-}\right)\) 
= 426 S cm² mol^-1          ... (ii)
\(\Lambda_{m}^{\circ}\left(\mathrm{CH}_{3} \mathrm{COONa}\right)=\lambda^{\circ}\left(\mathrm{Na}^{+}\right)+\lambda^{\circ}\left(\mathrm{CH}_{3} \mathrm{COO}^{-}\right)\) 
= 91 S cm² mol^-1           ...(iii) 
Adding (ii) and (iii) and subtracting (i), we get
\(\Lambda_{m}^{\circ}\left(\mathrm{CH}_{3} \mathrm{COOH}\right)=\lambda^{\circ}\left(\mathrm{H}^{+}\right)+\lambda^{\circ}\left(\mathrm{CH}_{3} \mathrm{COO}^{-}\right)\) 
= \(\Lambda_{m}^{\circ}(\mathrm{HCl})+\Lambda_{m}^{\circ}\left(\mathrm{CH}_{3} \mathrm{COONa}\right) - \Lambda_{m}^{\circ}(\mathrm{NaCl})\) 
= (426 + 91 - 126) = 391 S cm2 mol^-1 

Conductivity The inverse of resistivity is called conductivity. It is denoted by K (kappa). It is also known as specific conductance. k = 1/p SI unit of conductivity is Sm^-1 or ohm^-1 m^-1
Molar conductivity It is defined as the conductance of the solution which contains one mole of the electrolyte such that entire solution is in between the two electrodes kept one centimeter apart, and large enough to contain all the electrolytes.
Molar conductivity, \(\Lambda _{ m }=\frac { K }{ C } \) 
Variation of conductivity and molar conductivity with concentration.

Conductivity and molar conductivity change with change in concentration of electrolyte ..Conductivity always decreases with decrease in concentration for both weak as well as strong electrolytes.
But molar conductivity increases with decrease in concentration. For strong electrolytes, \(\Lambda _{ m }\) increases slowly with dilution but for weak electrolyte, \(\Lambda _{ m}\) increases steeply on dilution, especially near lower concentrations.

(a) (i) \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+3 \mathrm{H}_{2} \mathrm{~S}(g)+8 \mathrm{H}^{+}(a q) \) \(\longrightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{~S}+7 \mathrm{H}_{2} \mathrm{O}\)
(ii) \( 2 \mathrm{Cu}^{2+}(a q)+2 \mathrm{I}^{-}(a q) \longrightarrow 2 \mathrm{Cu}^{+}(a q)+\mathrm{I}_{2}(s)\)
(b) (i) It is because V in lower oxidation the state is less stable than Cr which is less stable than Mn. That is why MnO₄ is a best oxidizing agent and VO²⁺ is least.
(ii) M n (2 5) has electronic configuration [Ar]4s23ds, electronic configuration of Mn²⁺ is [Ar]4so3ds. After losing 2 electrons, it has half filled d-orbital, which is more stable that is why Mn²⁺ has exceptionally high third ionization energy, i.e. the energy required to remove
the third electron is very high.
(iii) It is because in Cr³⁺, d3 (half filled t2g orbitals) is more stable in aqueous solution than Fe³⁺ i.e. Cr³⁺ is more stable than Fe³⁺.

(i)

(ii)
 
(iii)
 
(iv)

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Advantages of Fuel cell
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Z = 27
\(\Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^{7} 4 \mathrm{~s}^{2}\) 
\(\therefore \mathrm{M}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{7}\) 

i.e., 3 unpaired electrons
\(\therefore n=3\) 
\(\Rightarrow \sqrt{n(n+2)}=\mu\) 
\(\Rightarrow \sqrt{3(3+2)}=\mu\) 
\(\Rightarrow \sqrt{15}=\mu\) 
\(\mu \approx 4 \mathrm{BM}\)

(i) Kolbes reaction.
It is a decarboxylative dimerisation of two carboxylic acids (or carboxylate ions)  to form carbon carbon bond of alkane.

(ii) Reimer-Tiemann reaction.
It involves the ortho-formylation of phenols. Phenol is converted to salicylaldehyde.

(iii) Williamson ether synthesis.
Sodium alkoxide reacts with alkyl halide to form an ether.
\(\mathbf{C}_{2} \mathrm{H}_{5} \mathrm{ONa}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl} \rightarrow \mathrm{C}_{2} \mathrm{H}_{5}-\mathrm{O}-\mathrm{C}_{2} \mathrm{H}_{5}+\mathrm{NaCl}\)
(iv) Unsymmetrical ether.
Two different alkyl groups are attached to O atom. The general formula is R−O−R′. An example is ethyl methyl ether 
\(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{3}\)

(i)From the electronic configuration of the 3d transition series, we can see that, as we go from left to right, the number of unpaired electrons increases from Sc to Cr and decreases thereafter. So the paramagnetic nature of the 3d transition series increaes till Cr and then decreases regularly.
(ii)Transition metals have vacant d-orbitals and because of this their valency is variable so they can be used as oxidising agent as well as reducing agent. also they have large surface area so they can act as a good catalyst.
(iii)The transition metals form interstitial compounds because there are vacant spaces in the lattice of transition metals which can be filled by small atoms like H,C,N etc.

(i) Sum of enthalphy of atomisation and ionisation is greater than enthalphy of hydration. Thus, \(E_{(M^{2+}/Mn)}^0\) is positive for copper.
(ii) Ionisation enthalpy is lower than hydration enthalphy due to stable 3d⁵ configuration. Thus, \(E_{(M^{2+}/Mn)}^0\) is more negative.
(iii) Cr²⁺ is a stronger reducing agent because it can lose electron to form Cr³⁺ which has stable 3d³ configuration (half-filled t_2g³). Further is more negative than Fe²⁺

\({ Cr }^{ 2+ }\) is a stronger reducing agent than \({ Fe }^{ 2+ }\). This is because the configuration of \({ Cr }^{ 2+ }\) changes from \({ d }^{ 4 }\) to \({ d }^{ 3 }\) configuration is stable \(({ t }_{ 2g }^{ 3 })\) being half-filled \({ t }_{ 2g }\) level.

(i) Propane-1, 2, 3-triol
(ii) 2-Methylphenol
(iii) 4-Methylphenol.

(i) (c):


It has 9 C-atoms.
(iii) (b): Sodium phenoxide when heated with CO₂ at 400 K under a pressure of 4-7 atm followed by acidification gives 2-hydroxybenzoic acid (salicylic acid) as the main product along with a small amount of 4-hydroxybenzoic acid. This reaction is called Kolbe's reaction.

Salol is used as an intestinal antiseptic.

(i) (b) : Equivalent conductance of CaSO₄ :
\(\Lambda_{\mathrm{CaSO}_{4}}^{\infty}=\lambda_{\mathrm{Ca}^{2+}}^{\infty}+\lambda_{\mathrm{SO}_{4}^{2-}}^{\infty}\) 
\(\lambda_{\mathrm{Ca}^{2+}}^{\infty}=\left(\mu_{\mathrm{Ca}^{2+}}\right) F ; \lambda_{\mathrm{SO}_{4}^{2-}}^{\infty}=\left(\mu_{\mathrm{SO}_{4}^{2-}}\right) F\) 
\(\mu_{\mathrm{Ca}^{2+}}\) and \(\mu_{\mathrm{SO}_{4}^{2-}}\) - are ionic mobilities. 
\(\Lambda_{\mathrm{CaSO}_{4}}^{\infty}=F(12.33+16.58) \times 10^{-4}\) 
= 96500 x 10^{- 4} x 28.91 = 279
(ii) (c) : \(\alpha=\frac{\Lambda_{C}}{\Lambda^{\infty}} \Rightarrow 0.1=\frac{\Lambda_{C}}{279} \Rightarrow \Lambda_{C}=27.9\) 
(iii) (a)
(iv) (c) : \(\Lambda_{m\left(\mathrm{CaCl}_{2}\right)}^{\circ}=\lambda_{\mathrm{Ca}^{2+}}^{\circ}+2 \lambda_{\mathrm{Cl}^{-}}^{\circ}\) 
 = (118.88 x 10^{- 4} ) + 2(77.33 x 10^{- 4} ) 
= 273.54 x 10^-4 m² mho mol^-1