Much larger third ionisation energy of Mn (where the required change is d⁵ to d⁴) is mainly responsible for this. This also explains why the +3 state of Mn is of little importance.

The cell can be written as Mg | Mg²⁺(0.130M) |  Ag⁺(0.0001M) | Ag
\(E_{(\text {cell })}=E_{\text {(cell) }}^{\ominus}-\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\left[\mathrm{Mg}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}}\)
\(=3.17 \mathrm{~V}-\frac{0.059 \mathrm{~V}}{2} \log \frac{0.130}{(0.0001)^{2}}\) 
= 3.17 V – 0.21V = 2.96 V.

- 0.465°C

In an aqueous solution, KOH almost completely ionises to give OH^– ions. OH^– ions is a strong nucleophile, which leads the alkyl chloride to undergo a substitution reaction to form alcohol.

On the other hand, an alcoholic soln of KOH contains alkoxide (RO^–) ion, which is a strong base. Thus, it can abstract a hydrogen from the β ^– carbon of the alkyl chloride and form an alkene by eliminating a molecule of HCl.

OH^– ion is a much weaker base than RO– ion. Also, OH^– ion is highly solvated in an aqueous solution and as a result the basic character of OH’ ion decreases. Therefore, it cannot abstract a hydrogen from β - carbon.

(i) Mg
(ii) Fe
(iii) Co
(iv) Pt

Threshold energy is the minimum energy which must be possessed by reacting molecules in order to undergo effective collision which leads to formation of product molecule.

Nitro group at ortho-position in chlorobenzene withdraws the electron density from the benzene ring and thus facilitates the attack of nucleophile on haloarene. -NO, at ortho-position

(a) Lanthanoids, mostly show +3 oxidation state but some of them show +2 and +4 oxidation states also due to the stability of electronic configuration (4f⁰, 4f ⁷ and 4f¹⁴ ).
(b) (i) They show variable oxidation states.
(ii) They form coloured ions.
(c) Zn may not be regarded as transition metal because neither Zn nor Zn²⁺ have incompletely filled d-orbital.

The structures of different dihalogen derivatives of propane are :

[Cr(NH₃)₆]³⁺ is paramagnetic due to presence of 3 unpaired electrons. Cr (Z = 24) has electronic configuration [Ar]3d ⁵4s ¹ .
Cr³⁺ has electronic configuration [Ar]3d ³ .
It undergoes d ² sp ³ hybridisation to form six hybrid orbitals which are occupied by electron pairs donated by six ammonia ligands.[Ni(CN)₄]^2- is diamagnetic as all electrons are paired. Ni (Z = 28) has electronic configuration [Ar]3d ⁸ 4s ². Ni ²⁺ has electronic configuration [Ar]3d ⁸. It undergoes dsp ² hybridisation to form 4 hybrid orbitals which are occupied by electron pairs donated by four cyanide ligands.

Ag⁺ ion is stronger Lewis acid than Na⁺. Ag⁺ has 18-electron shell configuration (known as pseudo inert gas configuration) and causes greater polarization than Na⁺, an ion with 8-electron shell configuration. This is due to the fact that in the case of 18-electrons shell ion there are 10 d-electrons in addition to 88-and p-electrons. The d-electrons do not shield the nuclear charge effectively and therefore, they have increased the effective nuclear charge. Hence they cause greater polarization and therefore, Ag⁺ is a stronger Lewis acid than Na⁺ ion.

\(K=1.29\times { 10 }^{ -2 }{ ohm }^{ -1 }{ cm }^{ -1 }\)
\(K=\frac { 1 }{ R } \times \frac { 1 }{ a } \Rightarrow \frac { 1 }{ a } =K\times R=1.29\times { 10 }^{ -2 }\times 85=109.65\times { 10 }^{ -2 }=1.0965{ cm }^{ -1 }\)
\({ \wedge }_{ m }=\frac { 1000K }{ M } =\frac { 1000 }{ M } \times \frac { 1 }{ R } \times \frac { 1 }{ a } \)
\({ \wedge }_{ m }=\frac { 1000\times 1\times 1.0965 }{ 0.052\times 96 } =\frac { 1096.50 }{ 4.992 } =219.65 \ S \ { cm }^{ 2 }{ mol }^{ -1 }\)

Elevation in boiling point required \((\triangle { T }_{ b })=100-99.{ 63 }^{ \circ }=0.{ 37 }^{ \circ }\)
Mass of solvent (water), \({ w }_{ 1 }=500\quad g\)
Molar mass of solvent, \({ M }_{ 1 }=18 \ g \ { mol }^{ -1 }\)
Molar mass of solute,\({ C }_{ 12 }{ H }_{ 22 }{ O }_{ 11 }=342 \ g{ \ mol }^{ -1 }\)
Applying the formula, \({ M }_{ 2 }=\frac { 1000\triangle { T }_{ b }{ w }_{ 2 } }{ { w }_{ 1 }\times \triangle { T }_{ b } } \)
or  \({ w }_{ 2 }=\frac { { M }_{ 2 }\times { w }_{ 1 }\triangle { T }_{ b } }{ 1000\times { K }_{ b } } =\frac { 342 \ g{ \ mol }^{ -1 }\times 500 \ g\times 0.37 \ K }{ 1000 \ g{ \ kg }^{ -1 }\times 0.52 \ K \ kg \ { \ mol }^{ -1 } } =121.7 \ g\)

For first order reaction, \(t=\frac { 2.303 }{ k } \log { \frac { a }{ a-x } } \)
99% completion means that x = 99% of a = 0.99 a
\({ t }_{ 99 }\%\)\(=\frac { 2.303 }{ k } \log { \frac { a }{ a-0.99a } } =\frac { 2.303 }{ k } \log { { 10 }^{ 2 } } =2\times \frac { 2.303 }{ k } \)
90% completion means that x = 90% of a = 0.90 a
\(\therefore \quad { t }_{ 90 }\%\)\(=\frac { 2.303 }{ k } \log { \frac { a }{ a-0.99a } } =\frac { 2.303 }{ k } \log { { 10 } } =2\times \frac { 2.303 }{ k } \)
\(\therefore \quad \frac { { t }_{ 99 } }{ { t }_{ 90 } } ={ \left( \frac { 2\times 2.303 }{ k } \right) }/{ \left( \frac { 2.303 }{ k } \right) }=2\quad \)\({ t }_{ 99 }\%\)\(=2\times { t }_{ 90 }\%\)

(i) Haloalkanes react with Mg in dry ether to form Grignard reagents (RMgX), which react with water to form hydrocarbons.


(iii) Haloalkanes react with Na in dry ether to give hydrocarbons containing double number of C-atoms.

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.

(i) \([CoF_6]^{3-}\)
\( [Co(H_2O)_6]^{2+}\)
\([Co(CN)_6]^{3-}\)
(ii) \([FeF_6]^{3-}\)
\(\ [Fe(H_2O)_6]^{2+}\)
\([Fe(CN)_6]^{4-}\)
It is diamagnetic due to the absence of unpaired electrons.

\(t_1={2.303\over k_1}log{a\over 0.10a}=t_1={2.303\over k_2}log{a\over a-0.25a}\)
As t₁ = t₂ \({2.303\over k_1}log{a\over 0.90a}={2.303\over k_2}log{a\over 0.7a};\ {k_2\over k_1}={log(100/75)\over log(100/90)}=2.73\)
But \(log{k_2\over k_1}={E_a\over 2.303R}\left(T_2-T_1\over T_1T_2\right)\)
Putting k₂/k₁) = 2.73, R = 8.314 J K^-1 mol^-1, T₁ = 298 K, T₂ = 308 K, we get E_a = 76.6 kJ mol^-1
Further, k - Ae^{-E_a / RT} or log k = log A -\({E_a\over 2.303RT}\)
Putting A = 3.56 x 10⁹s^-1, R = 8.314 x 10^-3 kJ K^-1 mol^-1, E_a = 76.6 kJ mol^-1, T = 318 K, we get
k = 9.3 x 10^-4 S^-1.

(i) Molar mass of CH₂Cl₂ = 12 × 1 + 1 × 2 + 35.5 × 2 = 85 g mol^–1
Molar mass of CHCl₃ = 12 × 1 + 1 × 1 + 35.5 × 3 = 119.5 g mol^–1
\(\text {Moles of } \mathrm{CH}_{2} \mathrm{Cl}_{2} \quad=\frac{40 \mathrm{~g}}{85 \mathrm{~g} \mathrm{~mol}^{-1}}=0.47 \mathrm{~mol}\)
\(\text {Moles of } \mathrm{CHCl}_{3} \quad=\frac{25.5 \mathrm{~g}}{119.5 \mathrm{~g} \mathrm{~mol}^{-1}}=0.213 \mathrm{~mol}\)
\(x_{\mathrm{CH}_{2} \mathrm{Cl}_{2}}=\frac{0.47 \mathrm{~mol}}{0.683 \mathrm{~mol}}=0.688\)
x_CHCl3 = 1.00 – 0.688 = 0.312
\(p_{\text {total }}=p_{1}^{0}+\left(p_{2}^{0}-p_{1}^{0}\right) x_{2}=200+(415-200) \times 0.688\)
= 200 + 147.9 = 347.9 mm Hg
(ii) y₁ = p₁/p_total, we can calculate the mole fraction of the components in gas phase (y₁).
\(p_{\mathrm{CH}_{2} \mathrm{Cl}_{2}}=0.688 \times 415 \mathrm{~mm} \mathrm{Hg}=285.5 \mathrm{~mm} \mathrm{Hg}\)
\(p_{\mathrm{CHCl}_{3}}=0.312 \times 200 \mathrm{~mm} \mathrm{Hg}=62.4 \mathrm{~mm} \mathrm{Hg}\)
\(y_{\mathrm{CH}_{2} \mathrm{Cl}_{2}}=285.5 \mathrm{~mm} \mathrm{Hg} / 347.9 \mathrm{~mm} \mathrm{Hg}=0.82\)
\(y_{\mathrm{CHCl}_{3}}=62.4 \mathrm{~mm} \mathrm{Hg} / 347.9 \mathrm{~mm} \mathrm{Hg}=0.18\)

\({ N }_{ 1 }{ V }_{ 1 }+{ N }_{ 2 }{ V }_{ 2 }+{ N }_{ 3 }{ V }_{ 3 }={ N }_{ 4 }{ V }_{ 4 }={ N }_{ 4 }({ V }_{ 1 }+{ V }_{ 2 }+{ V }_{ 3 })\)

E) If assertion is false but reason is true.
Explanation:
The nature of the cathode can affect the order of discharge of ions.

(a) Both (A) and (R) are correct, (R) is the correct explanation of (A).
Here, both (A) and (R) are true but (R) is not the correct explanation of (A).
Alkyl halides are insoluble in water because they are unable to form hydrogen bonds with water or break pre-existing hydrogen bonds.

(b): In amalgam, solute and solvent are uniformly distributed.

(b): In these complexes, the metal and ligand form a bond that involves the rt-electrons of the ligand and so it is a ㅠ -bonded organometallic compound.

(a) Cis-platin
(b) Coordination number of Co is 6 because 'en' (ethane 1, 2-diammine) is bidentate ligand.
(c) \(\left[\mathrm{An}(\mathrm{CN})_{2}\right]^{\ominus}\) Dicyanido aurate(1)
(d) EDTA, it is hexadentate ligand.
(e) It is done by reacting with sodium thio sulphate
AgBr + 2Na₂S₂O₃ \(\rightarrow\) Na₃[Ag(SP₃)₂] + NaBr
                                 Sodium dithio sulphato argentate (I)          

(a) \(\mathrm{Pb}+\mathrm{SO}_{4}^{2-} \longrightarrow \mathrm{PbSO}_{4}+2 \mathrm{e}^{-}\) (At anode)
(b) \(\mathrm{PbO}_{2}+4 \mathrm{H}^{+}+2 \mathrm{e}^{-}+\mathrm{SO}_{4}^{2-} \longrightarrow \mathrm{PbSO}_{4}+2 \mathrm{H}_{2} \mathrm{O}\)
2 Faraday is required.
(c) \(\mathrm{M}=\frac{\text { percentage by mass } \times d \times 10}{\text { Molar mass }}\)
\(=\frac{38 \times 1.30 \times 10}{98}=\frac{494}{98}=5.041 \mathrm{M}\)
(d) Mass of solution after discharge = 3500 mL x 1.14 g mL^-1 = 3990 g
Mass of H₂ SO₄ present in solution (20%) = \(\frac{20}{100} \times 3990 \mathrm{~g}=798 \mathrm{~g}\)
(e) \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{dil}) \longrightarrow 2 \mathrm{H}^{+}+\mathrm{SO}_{4}^{2-}\)
H₂O \(\rightarrow\) H⁺ + OH^-
At cathode: 2H⁺ + 2e^- \(\rightarrow\) H₂(g)
At anode: \(2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 4 \mathrm{H}^{+}+4 \mathrm{e}^{-}+\mathrm{O}_{2}(g)\)
H₂ gas is liberated at cathode and O₂ gas is formed at anode.