(a) Faraday's first law of electrolysis It states that the amount of chemical reaction occurring at an electrode by passing current is proportional to the quantity of electricity passed through the electrolyte. Charge required for the reduction of 1 mole of Cu²⁺ to Cu = 2F
(b) \(Cu^{ 2+ }Mg\rightarrow Mg^{ 2+ }+Cu\)
Given, E_{^O Cell} = +271 V
By using Nernst equation,
\(E_{ cell }=E^{ 0 }_{ cell }-\frac { 0.059 }{ n } log\frac { Mg^{ 2+ } }{ Cu^{ 2+ } } \)
Here, n = 2 and E⁰ _cell = + 2.71 V
\(\therefore E_{ cell }=2.71-\frac { 0.059 }{ 2 } log\frac { 0.1 }{ 0.01 } \)
\(=2.71-\frac { 0.059 }{ 2 } log10\)
= 2.71- 0.0295
\([\therefore \) log₁₀ = 1]
= 2.68 V

At anode \(Zn(s)\longrightarrow Zn^{ 2+ }(aq)+2e^{ - }\)
At cathode \(Sn^{ 2+ }(aq)+2e^{ - }\longrightarrow sn(s)\)
Overall reaction
\(Zn(s)+Sn^{ 2+ }(aq)\longrightarrow zn^{ 2+ }(aq)+Sn(s)\)
\(E^{ 0 }_{ cell }=E^{ 0 }_{ sin2+/sn }-(-E^{ 0 }_{ zn2+/zn })\)
= -0.14 + 0.76 = 0.62 V
\(\triangle G^{ 0 }=-nFE^{ 0 }\)
= -2 x 96500 x 0.62
\(\triangle G^{ 0 }=-119660Jmol^{ -1 }\) 

(i) The transition elements exhibit variable oxidation states. The variable oxidation states of transition metals are due to the participation of ns and (n - 1) d-electrons. This is because of the very small difference between the energies of (n - 1) d and ns orbitals. For the first five elements, the minimum oxidation state is equal to the number of electrons ---in the 4s orbitals and the other oxidation states are equal to the sum of 4s and some of the 3d-electrons. The highest oxidation state is equal to the sum of 4s and 3d electrons. For the remaining elements, the minimum oxidation state is equal to electrons in 4s-orbitals and the maximum oxidation state is not equal to the sum of 4s and 3d electrons. In general, the oxidation state increases up to the middle and then decreases.

(a) (i) covalent radii: The atomic radii decrease from 5c to Mn because of a number of unpaired electrons increases, therefore, effective nuclear charge increases. The atomic size of Fe, Co, Ni is almost same because the pairing of electrons takes place in d-orbitals causing repulsion and effective nuclear charge does not increase appreciably. Cu and In have bigger size because repulsion between paired electrons increases. Ionic radii of bivalent cations decrease from 5c to Cu due to increase in a number of protons.
(ii) Oxidation states: Transition metals show variable oxidation states due to the tendency of 'd' as well as '5' electrons to take part in bond formation. The highest oxidation state is equal to the total number of electrons in '5' as well as d-orbitals. The maximum oxidation state shown by the elements of first transition series increases from 5c to Mn and then decreases to In. 5c shows maximum + 3 and Mn shows +7, V(+ 5), Cr(+ 6), Fe(+3), Ni(+2), Co(+3), Cu(+2) and In( +2). oxidation state.
(iii) Ionization enthalpies: There is slight and irregular variation in ionization energies of transition metals due to the irregular variation of atomic size. The I.E. of 5d transition series is higher than 3d and 4d transition series because of Lanthanoid contraction, effective nuclear charge increases.
(b) Misch metal is an alloy which contains some of the lanthanoid metals. It contains 45% lanthanoid metals and iron ~ 5% and traces of 5, C, Ca and AI. Misch metal is used in the Mg-based alloy to produce bullets, shell, and lighter flint. Addition of 3% misch metal to magnesium increases its strength and used in making jet engine parts

(a) (i) Mole fraction (x): It is the ratio of number of moles of a particular component to the total number of moles of all the components. For example, mole fraction of component A, \(X_A={n_A\over n_A+n_B}\), nA is the number of moles of
component 'A' and n_B is the number of moles of component 'B'.
(ii) Van't Hoff Factor (i): The ratio of experimental value of a colligative property to the theoretical value (calculated on the basis of normal behaviour of solute) is known as van't Hoff factor. Experimental determined value
\(i={Experimental\ determined\ value\ of\ the\ colligative\ property\over Calculated\ value\ of\ the\ same\ from\ the\ formula}\)  
Since colligative properties are inversely proportional to the molecular mass of solute, therefore, in terms of molecular mass, \(i={Theoretical\ molecular\ mass\ (from\ the\ formula)\over Molecular\ mass\ determined\ experimentally}\)
\(i={Normal\ molecular\ mass\over Observed\ molecular\ mass}\)
(b) Here, π = 13.3 mm Hg = \({13.3\over 760}atm\)
[760 .; mm Hg = 1 atm]
\(V = 10 mL ={10\over 1000}=0.01L\)
R = 0.0821 L atm mol^-1 K^-l,
T = 25 + 273 = 298 K
W_B = 100 mg = \({100\over 1000}=0.01g\)
πV = nRT,
\(\Rightarrow\ {13.3\over 760}\times0.01={W_B\over M_B}\times0.0821\times298\)
\(\Rightarrow\ {13.3\over 760}\times0.01={0.1\over M_B}\times0.0821\times298\)
\(M_B={0.lg \times 0.0821\ L\ atm\ mol^{-1} K^{-1}\times 298K \times 760\over13.3\ atm\times 0.01L}\)
\(={1859.401\over 0.133}g\ mol^{-1}\)
= 13980.45 9 mol^-1or 1.4 x 10⁴ g mol^-1

Raoult's law states that, for a solution of volatile liquids, the partial vapour pressure of each component in solution is equal to the product of the vapour pressure of the pure component and its mole fraction. For a binary solution of two components A and B
\({ p }_{ A }={ p }_{ A }^{ \circ }X{ x }_{ A }\\ { p }_{ B }={ p }_{ B }^{ \circ }X{ x }_{ B }\)
Differences between ideal and non-ideal solutions

Conductivity of an electrolyte solution decreases with dilution because the number of ions per unit volume furnished by an electrolyte decreases with dilution.

Mn³⁺ = 3d⁴ = 4 unpaired electron, Cr³⁺ = 3d³ = 3 unpaired electrons, V³⁺ = 3d² = 2 unpaired electrons, Ti³⁺= 3d ¹ = 1 unpaired electron. Cr³⁺ is most stable out of these in aqueous solution because it has half filled t_2g level (i.e., t³_2g) .

\({ \triangle G }^{ o }={ -nE }^{ o }F,-{ \triangle G }^{ o }={ W }_{ max' }\) decrease in free energy is measure of maximum work obtainable from the cell.

When oxalic acid is added to acidic solution of KMnO₄, its colour disappear due to the reduction of MnO₄^-  ion to Mn ²⁺. Chemical reaction occurring during this neutralisation reaction is as follows:
\( 5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-}+\underset{\text { (Coloured) }}{2 \mathrm{MnO}_{4}^{-}}+16 \mathrm{H}^{+} \longrightarrow \underset{\text { (Colourless) }}{2 \mathrm{Mn}^{2+}}+8 \mathrm{H}_{2} \mathrm{O} +10 \mathrm{CO}_{2} \)

Camphor has a very high cryoscopic constant (=39.7^o). Hence, It gives a large depression in melting point when an organic solute is dissolved in it.

(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.

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) Electronic configuration
Lanthanoids = [Xe] 4f^0-14 5d^0-1 6s²
Actinoids = [Rn] 5f^0-14 6d^0-1 7s²
(ii) Oxidation states In lanthanoids, +3 oxidation state is most common along with + 2 and + 4. While in actinoids, there is a greater range of oxidation states because 5f, 6d and 7s levels are of comparable energies.They show + 2, + 3, + 4, + 5, + 6 and + 7 oxidation states. Common oxidation state in actinoids is + 3.
(iii) Chemical reactivity Lanthanoids are less reactive than actinoids. Actually, earlier members of lanthanoids are quite reactive similar to calcium but with increasing atomic number, they behave more like aluminium. Lanthanoids react with dilute acids to liberate H₂ gas while actinoids react with boiling water and gives a mixture of oxide and hydride.

Mass of solute, \({ \omega }_{ B }=5g\)
Mass of water, \({ \omega }_{ A }=100g\)
Molecular mass of water, \({ M }_{ A }=18\)
Vapour pressure of water, \({ p }_{ A }^{ 0 }=23.755\quad torr\)
Vapour pressure of solution, \({ p }_{ A }=23.402\quad torr\)
Lowering in vapour pressure = \({ p }_{ A }^{ 0 }-{ p }_{ A }=23.755 -23.402=0.353\)
\(Now,\frac { { p }_{ A }^{ 0 }-{ p }_{ A } }{ { p }_{ A }^{ 0 } } =\frac { { \omega }_{ B }{ M }_{ A } }{ { \omega }_{ A }{ M }_{ B } }\)
\( \frac { 0.353 }{ 23.755 } =\frac { 5\times 18 }{ 100\times { M }_{ B } }\)
\( \\ or \ { M }_{ B }=\frac { 5\times 18 }{ 100\times 0.353 } \times 23.755=60.56\)

The cell constant is given by the equation:
Cell constant = G* = conductivity × resistance
= 1.29 S/m × 100 \(\Omega\) = 129 m^–1 = 1.29 cm^–1
= \(\frac{G^{*}}{R}=\frac{129 \mathrm{~m}^{-1}}{520 \Omega}\) = 0.248 S m^–1
Concentration = 0.02 mol L^–1
= 1000 × 0.02 mol m–3 = 20 mol m^–3
Molar conductivity = \(A_{m}=\frac{\kappa}{c}\) 
= \(\frac{248 \times 10^{-3} \mathrm{Sm}^{-1}}{20 \mathrm{~mol} \mathrm{~m}^{-3}}\) = 124 × 10^–4 S m²mol^–1
Alternatively, \(\kappa=\frac{1.29 \mathrm{~cm}^{-1}}{520 \Omega}\) = 0.248 × 10^–2 S cm^–1
and Λ _m = κ × 1000 cm3 L^–1 molarity^–1
= \(\frac{0.248 \times 10^{-2} \mathrm{~S} \mathrm{~cm}^{-1} \times 1000 \mathrm{~cm}^{3} \mathrm{~L}^{-1}}{0.02 \mathrm{~mol} \mathrm{~L}^{-1}}\) 
= 124 S cm² mol^–1

(a) No
(b) Yes
(c) No

1.56 V

(a) It is because neither they nor their ions have incompletely filled d-orbitals.
(b) Scandium (Sc) and Zinc (Zn).
(c) It is because they show variable oxidation state, can form intermediate complexes and have large surface area for adsorption of gases.
(d) It is due to strong interatomic forces of attraction due to presence of unpaired electrons.
(e) It is because Cu^{2 +} has one unpaired electron and undergoes d-d transition by absorbing light from visible region and radiate blue colour, where as Zn^{2 +} is colourless due to absence of unpaired electron.

(i) (b) : \(0.059=\frac{+0.059}{2} \log \frac{0.001}{\left[M^{2+}\right]}\) 
\(\log \frac{0.001}{\left[M^{2+}\right]}=2 \text { or }\left[M^{2+}\right]=10^{-5}\) 
Let solubility of salt be S mol/litre
\(\begin{array}{cc} \text {Thus,} \ M X_{2} & \rightarrow M^{2+}+2 X^- \\ S & S& 2 S \end{array}\) 
\(\therefore\) K_sp = 4S³ = 4 x (10^-5 )³ = 4 x 10^-15
(ii) (d) : \(\Delta G=-n F E=-2 \times 96500 \times 0.059\) 
= -11387 J mol^-1 = -11.4 kJ mol^-1
(iii) (a) : \(E_{\mathrm{cell}}^{\circ}=\frac{0.059}{1} \log K_{\mathrm{C}}\) 
\(E_{\mathrm{cell}}^{\circ}=E_{\mathrm{Fe}^{2+} / \mathrm{Fe}^{3+}}^{\circ}+E_{\mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}}^{\circ}\) 
= -0.68 + 1.44 = 0.76 V
\(\log _{10} K_{C}=\frac{0.76}{0.059}=12.88\) 
K_C = 7.6 x 10¹²
(iv) (a)