IIT-JEE 2011 Paper 2 Question Paper with Solutions

Let M be the mass of the ring and m that of the ball and let V and $\upsilon$ be their velocity before collision. The initial momentum of the system (ring and ball) in the horizontal direction is

${\overrightarrow p _i} = M\overrightarrow V + m\overrightarrow \upsilon$

$= 2 \times 1 + 0.1 \times ( - 20)$

$= 2 - 2 = 0$

From conservation of momentum, the final momentum of the system ${\overrightarrow p _f} = 0$ in the horizontal direction. Hence ${V_{cm}} = 0$ for the ring, i.e. the ring has pure rotation about its centre of mass. So choice (a) is correct.

The total initial angular momentum of the system about the point of collision is

${L_i} = m\upsilon r - I\omega$

$= m\upsilon r - M{R^2}{V \over R}$

$= m\upsilon r - MRV$

$= 0.1 \times 20 \times 0.75 - 2 \times 0.5 \times 1$

$= 1.5 - 1 = 0.5$ kg m² s^$-$1

From the conservation of angular momentum, the final angular velocity must be anticlockwise. Hence the friction between the ring and the ground is to the left. So the correct choices are (a) and (c).

A particle escapes from the gravitational pull if its total energy (T) i.e., sum of kinetic energy (K) and potential energy (U), is greater than or equal to zero. The condition for just escape T = K + U = 0 i.e.,

K = $-$ U. .............. (1)

In a circular orbit of radius r, gravitational attraction provides the centripetal acceleration, mV²/r = GMm/r², which gives

$r = {{GM} \over {{V^2}}}$ ........ (2)

From equations (1) and (2), the kinetic energy of the particle at the time of injection is given by

$K = - U = - \left( { - {{GMm} \over r}} \right) = {{GMm} \over {GM/{V^2}}} = m{V^2}$.

Here, ${x_1} = A\sin \omega t$

${x_2} = A\sin \left( {\omega t + {{2\pi } \over 3}} \right)$

$\therefore$ ${x_1} + {x_2} = A\sin \omega t + A\sin \left( {\omega t + {{2\pi } \over 3}} \right)$

$= A\sin \omega t + A\left[ {\sin \omega t\cos {{2\pi } \over 3} + \cos \omega t\sin {{2\pi } \over 3}} \right]$

$= A\sin \omega t + A\left[ {\sin \omega t\left( { - {1 \over 2}} \right) + \cos \omega t\left( {{{\sqrt 3 } \over 2}} \right)} \right]$

$= {A \over 2}\sin \omega t + {{\sqrt 3 } \over 2}A\cos \omega t = A\left[ {\sin \omega t\cos {\pi \over 3} + \cos \omega t\sin {\pi \over 3}} \right]$

$= A\sin \left( {\omega t + {\pi \over 3}} \right)$

$\because$ ${x_1} + {x_2} + {x_3} = 0$

$\Rightarrow {x_3} = - ({x_1} + {x_2}) = - A\sin \left( {\omega t + {\pi \over 3}} \right)$

$= A\sin \left( {\omega t + \pi + {\pi \over 3}} \right)$

${x_3} = A\sin \left( {\omega t + {{4\pi } \over 3}} \right)$

$\because$ ${x_3} = B\sin (\omega t + \phi )$

Hence, $B = A,\,\phi = {{4\pi } \over 3}$

Loss of kinetic energy $= {1 \over 2}m{V^2}$

Work done against friction $= \mu mgx$

Gain in potential energy $= {1 \over 2}k{x^2}$

From work-energy principle,

${1 \over 2}m{V^2} = \mu mgx + {1 \over 2}k{x^2}$

$\Rightarrow {1 \over 2} \times 0.18 \times {V^2} = 0.1 \times 0.18 \times 10 \times 0.06 + {1 \over 2} \times 2 \times {(0.06)^2}$

$\Rightarrow V = 0.4 = {4 \over {10}}$ ms^$-$1. Hence, N = 4.

Least count of screw gauge

$= {{Pitch} \over {No.\,of\,divisions\,on\,the\,circular\,scale}}$

$= {{0.5\,mm} \over {50}}$ = 0.01 mm

Diameter of ball, $D = MSR + CSR \times LC$

= 2.5 mm + (20)(0.01 mm) = 2.7 mm

Density, $\rho = {{Mass} \over {Volume}} = {M \over {{{4\pi } \over 3}{{\left( {{D \over 2}} \right)}^3}}}$

The relative error in the density is

${{\Delta \rho } \over \rho } = {{\Delta M} \over M} + {{3\Delta D} \over D}$

The relative percentage in the density is

${{\Delta \rho } \over \rho } \times 100 = \left[ {{{\Delta M} \over M} + {{3\Delta D} \over D}} \right] \times 100$

$= 2\% + {{3 \times 0.01} \over {2.7}} \times 100 = 2\% + 1.11\% = 3.11\%$

Gauss's law is valid only if Coulomb's law holds, i.e. if $E \propto {r^{ - 2}}$. Hence, choice (a) is wrong. Gauss's law cannot be used to calculate a non-uniform field distribution around an electric dipole. So choice (b) is also wrong.

Choice (c) is correct because the directions of electric fields are opposite at a point between two similar charges.

Work done ${W_{A \to B}} = q({V_B} - {V_A}) = ({V_B} - {V_A})$ ($\because$ $q = + \,1\,C$)

Hence, choice (d) is correct.

So the correct choices are (c) and (d).

Process A $\to$ B,

It is a isobaric process

P = constant, V $\propto$ T

$\because$ V_B < V_A $\Rightarrow$ T_B < T_A

$\Delta$U = nC_V$\Delta$T = $-$ve

Hence internal energy decreases.

$\Delta$Q = nC_P$\Delta$T = $-$ve

Hence heat is lost.

$\Delta$W = nR$\Delta$T = $-$ve

Hence, work is done on the gas.

Process, B $\to$ C,

It is a isochoric process.

V = constant, P $\propto$ T

$\because$ P_C < P_B $\Rightarrow$ T_C < T_B

$\Delta$U = nC_V$\Delta$T = $-$ve

Hence, internal energy decreases.

$\Delta$W = 0, $\Delta$Q = $\Delta$U = $-$ve

Hence, heat is lost.

Process C $\to$ D,

It is a isobaric process.

P = constant, V $\propto$ T

$\because$ V_D > V_C $\Rightarrow$ T_D > T_C

$\Delta$U = nC_V$\Delta$T = +ve

Hence internal energy increases.

$\Delta$Q = nC_P$\Delta$T = +ve

Hence, heat is gained.

$\Delta$W = nR$\Delta$T = +ve

Hence, work is done by the gas.

Process, D $\to$ A

According to ideal gas equation

${{{P_A}{V_A}} \over {{T_A}}} = {{{P_D}{V_D}} \over {{T_D}}} \Rightarrow {{(3P)(3V)} \over {{T_A}}} = {{(P)(9V)} \over {{T_D}}} \Rightarrow {T_D} = {T_A}$

Hence, it is a isothermal process.

$\therefore$ $\Delta$U = 0

$\because$ V_A < V_D

$\Delta$W = $-$ve, hence work done is done on the gas.

$\Delta$Q = $\Delta$W = $-$ve, hence heat is lost.

The magnetic fields and the induced electric fields form the closed loops.

Let V be the volume of each sphere and let T be the tension in the string.

Buoyant force on sphere A is U_A = d_FVg

Buoyant force on sphere B is U_B = d_FVg

Weight of A is W_A = d_AVg

Weight of B is W_B = d_BVg

The free body diagrams of A and B are as follows. (see Fig. 11.34)

For equilibrium,

U_A = T + W_A and U_B + T = W_B

i.e. d_FVg = T + d_AVg ..... (i)

and d_FVg + T = d_BVg .... (ii)

From Eq. (i) ${d_F} = {T \over {Vg}} + {d_A}$. Hence, d_F > d_A. So choice (a) is correct.

From Eq. (ii) ${d_B} = {T \over {Vg}} + {d_F}$. Hence d_B > d_F. So choice (b) is also correct.

Eliminating T from Eqs. (i) and (ii) we get $2{d_F} = {d_A} + {d_B}$, which is choice (d).

So, the correct choices are (a), (b) and (d).

The silver sphere gets positively charged due to emission of photoelectrons. This positively charged sphere attracts (binds) the emitted photoelectrons. The emitted photoelectrons cannot escape if their kinetic energies (hc/$\lambda$ $-$ $\phi$) are less than or equal to their potential energies $\left( {{1 \over {4\pi {\varepsilon _0}}}{{n{e^2}} \over r}} \right)$. Thus, in limiting case,

${{hc} \over \lambda } - \phi = {1 \over {4\pi {\varepsilon _0}}}{{n{e^2}} \over r}$ ..... (1)

Substitute the values of various parameters in equation (1),

${{1242} \over {200}} - 4.7 = {{n(9 \times {{10}^9})(1.6 \times {{10}^{ - 19}})} \over {{{10}^{ - 2}}}}$,

to get n = 1.04 $\times$ 10⁷. [We have used hc = 1242 eV-nm.]

(1) For a pipe closed at one end, we have

${{{\lambda _f}} \over 4} = L$ or ${\lambda _f} = 4L$

Therefore, the sound waves are longitudinal.

(2) For a pipe open at both ends, we have

${{{\lambda _f}} \over 2} = L$ or ${\lambda _f} = 2L$

Therefore, the sound waves are longitudinal.

(3) For a stretched wire clamped at both ends, we have

${{{\lambda _f}} \over 2} = L$

Vibration on the string is transverse.

(4) For a stretched wire clamped at both ends and at mid-point, we have

${{{\lambda _f}} \over 2} = {L \over 2} \Rightarrow {\lambda _f} = L$

Therefore, the vibration on the string is transverse.

The force exerted on charge +Q by the electric field $\overrightarrow E$ is

$\overrightarrow F = Q\overrightarrow E$

in the direction of $\overrightarrow E$. Since $\overrightarrow F$ is constant, a constant force is added to the applied force. Hence only the mean position will change.

The frequency will be same.

As ${v_0} = {1 \over {2\pi }}\sqrt {{k \over m}}$ does not depend on the constant external force.

Applying Kirchhoff's second law for closed loop CDEFC we get

$- 3 - 2I - I + 6 = 0$

$I = {{6 - 3} \over 3} = 1A$

For the lower path

${V_A} - 3 - 2 \times 1 = {V_B}$

$\therefore$ ${V_A} - {V_B} = 5V$

We can also find the V_AB by considering the upper path

For the upper path,

${V_A} - 6 + 1 \times 1 = {V_B}$

${V_A} - {V_B} = 5V$

The time of flight for the bullet is same as that of the ball and is given by

$t = \sqrt {{{2h} \over g}} = \sqrt {{{2 \times 5} \over {10}}} = 1\,s$

Just after the collision, the velocity of bullet (v₁) is related to its range (R₁) by v₁t = R₁ which gives v₁ = 100 m/s. Similarly, the velocity of the ball is v₂ = 20 m/s. Consider the bullet and the ball together as a system. Along the direction of collision, there is no external force on the system. Hence, linear momentum of the system in the direction of collision is conserved. The linear momentum of the system before and after the collision are

${p_i} = {m_1}V$,

${p_f} = {m_1}{v_1} + {m_2}{v_2}$.

The conservation of linear momentum, ${p_i} = {p_f}$, gives

$V = {{{m_1}{v_1} + {m_2}{v_2}} \over {{m_1}}}$

$= {{(0.01)(100) + (0.2)(20)} \over {0.01}} = 500$ m/s.

We have impedance in the circuit

$Z = \sqrt {{R^2} + {{\left( {{1 \over {\omega C}}} \right)}^2}}$

However, $Z = R\sqrt {1.25}$ and $R\sqrt {1.25} = \sqrt {{R^2} + {{\left( {{1 \over {\omega C}}} \right)}^2}}$

$\Rightarrow 0.25{R^2} = {1 \over {{{(\omega C)}^2}}}$

The time constant is

$RC = \sqrt {{1 \over {0.25 \times {{500}^2}}}} = 4$ ms

We have

${{{n_2}} \over v} - {{{n_1}} \over u} = {{{n_1} - {n_2}} \over R}$

For the first refracting surface (air-oil), we have n₂ = 7/4; n₁ = 1; R = 6 cm. Therefore,

${7 \over {4{v_1}}} - {1 \over {24}} = {{ - (7/4)} \over 6}$

or v₁ = 21 cm and for the second interface (water-oil), we have

${n_1} = {7 \over 4};{n_2} = {4 \over 3}u = {v_1};R = \infty$

Therefore,

${4 \over {3{v_2}}} - {7 \over {4 \times 21}} = 0$

v₂ = 16 cm and v₂ + x = height of water.

Therefore,

$x = 18 - 16 = 2$

If I is the intensity of the incident light then R + T = I. Since the incident ray is travelling in a denser medium (glass), it will be totally reflected at a certain critical angle $\theta$_c. For $\theta$ < $\theta$_c a part of the incident intensity is reflected and the remaining part is transmitted. But for $\theta$ > $\theta$_c, there is no refracted ray. Hence for $\theta$ > $\theta$_c, the value of R is 100%. Hence the only correct option is (c).

In case A,

The capacitive reactance is

$X_C^A = {1 \over {\omega C}}$

Impedance of the circuit is

${Z_A} = \sqrt {{{(R)}^2} + {{\left( {{1 \over {\omega C}}} \right)}^2}}$

$I_R^A = {V \over {\sqrt {{{(R)}^2} + {{\left( {{1 \over {\omega C}}} \right)}^2}} }}$ ...... (i)

$V_C^A = {{I_R^A} \over {\omega C}} = {V \over {\sqrt {{{(R\omega C)}^2} + 1} }}$ ...... (ii)

In case B,

The capacitive reactance is

$X_C^B = {1 \over {\omega (4C)}} = {1 \over {4\omega C}}$

Impedance of the circuit is

${Z_B} = \sqrt {{R^2} + {{\left( {{1 \over {4\omega C}}} \right)}^2}}$

$I_R^B = {V \over {\sqrt {{R^2} + {{\left( {{1 \over {4\omega C}}} \right)}^2}} }}$ ..... (iii)

$V_C^B = {V \over {\sqrt {{{(4R\omega C)}^2} + 1} }}$ ..... (iv)

From (i) and (iii), we conclude that

$I_R^A < I_R^B$

From (ii) and (iv), we conclude that

$V_C^A > V_C^B$

u = 10 ms^$-$1, $\theta$ = 60$^\circ$

Time of flight is

$t = {{2u\sin \theta } \over g} = {{2 \times 10 \times \sin 60^\circ } \over {10}} = \sqrt 3 \,s$

Let v be the velocity of the train. The horizontal velocity of ball at the instant it is thrown $= (v + {u_x}) = (v + u\cos \theta )$. Therefore, the horizontal range of the ball with respect to the ground is

$R = (v + u\cos \theta )t$, where $t = \sqrt 3 \,s$

It is clear that

Distance travelled by ball in time $t + 1.15 = R$

i.e. $vt + {1 \over 2}a{t^2} + 1.15 = (v + u\cos \theta )t$

$\Rightarrow {1 \over 2}a{t^2} + 1.15 = (u\cos \theta )t$

$\Rightarrow {1 \over 2}a \times {(\sqrt 3 )^2} + 1.15 = (10\cos 60^\circ ) \times \sqrt 3$

$\Rightarrow a = 5$ ms^$-$2

Magnetic field at the centre of a circular loop of radius r and carrying a current $I = {{{\mu _0}I} \over {2r}}$. The direction of the field is along z-direction if the current is anticlockwise.

Consider a small element of width dr. The current through the element is

$dI = {{total\,current\,in\,spiral} \over {total\,width\,of\,spiral}} \times width\,of\,element$

$= {{Idr} \over {(b - a)}}$

$\therefore$ $B = \int\limits_a^b {{{{\mu _0}NdI} \over {2r}} = \int\limits_a^b {{{{\mu _0}NI} \over {2(b - a)}}{{dr} \over r}} }$

$= {{{\mu _0}NI} \over {2(b - a)}}\int\limits_a^b {{{dr} \over r} = {{{\mu _0}NI} \over {2(b - a)}}\ln \left( {{b \over a}} \right)}$

To find the cell potential under non-standard conditions, we can use the Nernst equation, which is given as:

$E = E^o - \frac{RT}{nF} \ln(Q)$

Where:

• $E^o$ is the standard cell potential (1.67 V).
• $R$ is the gas constant (8.314 J/mol·K).
• $T$ is the temperature in Kelvin (298 K for 25°C).
• $n$ is the number of moles of electrons transferred per mole of reaction (4 in this case).
• $F$ is the Faraday constant (96485 C/mol).
• $Q$ is the reaction quotient.

The reaction quotient, $Q$, can be calculated based on the given conditions and the reaction:

$Q = \frac{[\text{Fe}^{2+}]^2}{[\text{H}^+]^4 \cdot P(O_2)}$

Substituting the given values:

• $[\text{Fe}^{2+}] = 10^{-3} \text{ M}$
• $[\text{H}^+] = 10^{-\text{pH}} = 10^{-3} \text{ M}$
• $P(O_2) = 0.1 \text{ atm}$

$Q = \frac{(10^{-3})^2}{(10^{-3})^4 \cdot 0.1}$

$Q = \frac{10^{-6}}{10^{-12} \cdot 0.1}$

$Q = \frac{10^{-6}}{10^{-13}}$

$Q = 10^{7}$

Now, substituting the values into the Nernst equation:

$E = 1.67 \text{ V} - \frac{8.314 \times 298}{4 \times 96485} \ln(10^{7})$

Calculating the term $\frac{RT}{nF}$;

$\frac{RT}{nF} = \frac{8.314 \times 298}{4 \times 96485}$

$\approx \frac{2476}{385940}$

$\approx 0.0064 \text{ V}$

The natural logarithm of $10^{7}$ is about 16.1 (since $\ln(10) \approx 2.303$, so $\ln(10^{7}) = 7 \times \ln(10) \approx 16.1$).

Thus,

$E = 1.67 \text{ V} - 0.0064 \times 16.1$

$E = 1.67 \text{ V} - 0.103 \text{ V}$

$E = 1.567 \text{ V}$

Therefore, the cell potential under the given conditions is approximately 1.57 V, so option D (1.57 V) is the correct answer.

To find the freezing point of the solution, we will use the formula for freezing point depression, which is given by:

$\Delta T_f = i \cdot K_f \cdot m$

where:

• $\Delta T_f$ is the freezing point depression.


• $i$ is the van't Hoff factor (number of particles the solute splits into or forms in solution).


• $K_f$ is the cryoscopic constant (freezing point depression constant), for water it's 1.86 K kg mol^-1.


• $m$ is the molality of the solution.

Given:

• Solute is $K_3[Fe(CN)_6]$, and molecular weight (M) = 329 g/mol.


• Mass of solute = 0.1 g.


• Mass of solvent (water) = 100 g.

First, we calculate the molality $m$ of the solution:

$m = \frac{{\text{Moles of solute}}}{{\text{Kilograms of solvent}}}$

Moles of solute:

$\text{Moles of solute} = \frac{{\text{Mass of solute}}}{{\text{Molecular weight}}} = \frac{0.1}{329} \text{mol}$

$\text{Moles of solute} = 3.04 \times 10^{-4} \text{ mol}$

Kilograms of solvent:

$\text{Kilograms of solvent} = 100 \text{ g} = 0.1 \text{ kg}$

Now molality:

$m = \frac{3.04 \times 10^{-4} \text{ mol}}{0.1 \text{ kg}} = 0.00304 \text{ mol/kg}$

Next, calculate the van't Hoff factor $i$:

• The solute $K_3[Fe(CN)_6]$ disassociates into 4 ions: 3 $K^+$ and 1 $[Fe(CN)_6]^{3-}$. Therefore, $i = 4$.

Substituting these into the freezing point depression formula:

$\Delta T_f = i \cdot K_f \cdot m = 4 \cdot 1.86 \cdot 0.00304$

$\Delta T_f = 0.02259 \text{ K}$

Convert to degrees Celsius and remember that this will lower the freezing point, so it should be negative (since the freezing point depression means the solution freezes at a lower temperature than pure solvent):

$\Delta T_f = -0.02259 \text{ °C}$

So, $\Delta T_f = -2.259 \times 10^{-2} \text{ °C}$

From the given options, the closest to this calculated value of $\Delta T_f$ is:

Option A: -2.3 $\times$ 10^-2oC. Hence, Option A is the correct answer.

Analysis of each transformation:

(A) CO₂(s) $\to$ CO₂(g)

This transformation represents the sublimation of solid carbon dioxide (dry ice) into gaseous carbon dioxide. This process is:

• Phase transition (p): The substance changes from solid to gas phase.
• $\Delta H$ is positive (r): Sublimation is an endothermic process, meaning it absorbs heat from the surroundings.
• $\Delta S$ is positive (s): The entropy increases as the substance moves from a more ordered solid state to a less ordered gaseous state.

(B) CaCO₃(s) $\to$ CaO(s) + CO₂(g)

This transformation is the thermal decomposition of calcium carbonate to form calcium oxide and carbon dioxide. This process involves:

• $\Delta H$ is positive (r): Decomposition reactions are typically endothermic, requiring energy to break chemical bonds.
• $\Delta S$ is positive (s): The process results in an increase in entropy, primarily because a gas (CO₂) is produced from a solid compound.

(C) 2H $\to$ H₂(g)

This is the formation of hydrogen gas (H₂) from atomic hydrogen (H). This transformation:

• $\Delta S$ is negative (t): Two individual hydrogen atoms combine into a diatomic molecule, decreasing the system's entropy since fewer particles are free to move independently.

(D) P_{(white, solid)} $\to$ P_{(red, solid)}

This transformation represents the allotropy change from white phosphorus to red phosphorus:

• Allotropic change (q): Both forms are the same element, phosphorus, showing in different structural forms.
• $\Delta S$ is negative (t): Red phosphorus is more thermodynamically stable and less reactive than white phosphorus, implying a decrease in entropy.

Option Evaluation:

Upon reviewing each transformation and comparing the mappings:

• Option A agrees completely with the above analysis with proper mapping for each statement.

Thus, Option A is correct:

A $\to$ p,r,s; B $\to$ r,s; C $\to$ t; D $\to$ p,q,t

The given chemical equation represents a first-order reaction:

$2N_2O_5 (g) \to 4NO_2 (g) + O_2 (g)$

In first-order reactions, the rate of reaction is directly proportional to the concentration of the reactant. The rate law can be expressed as:

$Rate = k[N_2O_5]$

where k is the rate constant.

The integrated rate equation for a first-order reaction is:

$\ln [N_2O_5] = -kt + \ln [N_2O_5]_0$

where [N_2O_5]_0 is the initial concentration and [N_2O_5] is the concentration at time t.

Rearranging gives:

$[N_2O_5] = [N_2O_5]_0 e^{-kt}$

Now, let's analyze each option:

Option A: "the concentration of the reactant decreases exponentially with time".

This matches the expression $[N_2O_5] = [N_2O_5]_0 e^{-kt}$ which shows an exponential decline in the concentration of the reactant. Therefore, option A is correct.

Option B: "the half-life of the reaction decreases with increasing temperature".

In first-order reactions, the half-life is determined by the equation $t_{1/2} = \frac{0.693}{k}.$ Increases in temperature typically increase the rate constant k, thus decreasing the half-life. However, it is not directly related to the half-life itself but rather to the rate constant via the Arrhenius equation. Nonetheless, the essence of the statement is correct as increasing temperature generally decreases the half-life.

Option C: "the half-life of the reaction depends on the initial concentration of the reactant".

The half-life for a first-order reaction given by $t_{1/2} = \frac{0.693}{k}$ is independent of the initial concentration [N_2O_5]_0. Therefore, option C is incorrect.

Option D: "the reaction proceeds to 99.6% completion in eight half-life durations".

Completion to a certain level can be found using the formula for decay over multiple half-lives, given as $[N_2O_5] = [N_2O_5]_0 \times \left(\frac{1}{2}\right)^n$ where n is the number of half-lives. For eight half-lives, we have:

$Final\, Concentration = [N_2O_5]_0 \times \left(\frac{1}{2}\right)^8 = [N_2O_5]_0 \times \frac{1}{256}$

The reaction proceeds to $\frac{255}{256} \approx 99.6\%$ completion. Thus, option D is correct.

In summary, options A, B, and D are correct, while option C is incorrect.

(i) The solubility of $\mathrm{AgCl}\left(s\right)$ in saturated solution is expressed as :

$H_{2}O+{\mathrm{AgCl}}_{\left(𝑠\right)}\to \underset{𝑥}{{\mathrm{Ag}}^{+}}\left(𝑎𝑞\right)+\underset{𝑥}{{\mathrm{Cl}}^{-}}\left(𝑎𝑞\right)$

Though $\mathrm{AgCl}\left(s\right)$ has low solubility with $K_{𝑠𝑝}=1.6$ $\times {10}^{-10}$ still some silver $\left({\mathrm{Ag}}^{+}\right)$and chloride $\left({\mathrm{Cl}}^{-}\right)$ are dissolved in solution. Let the concentration of these ions be $𝑥{\mathrm{mol}}^{-1}$ (in solution).

$\begin{array}{cc}[\mathrm{Ag}{]}^{+}& =\left[{\mathrm{Cl}}^{-}\right]=𝑥\\ & =1.6\times {10}^{-10}\\ \mathrm{CuCl}\left(𝑠\right)+H_{2}O\left(𝑙\right)& \rightleftharpoons {\mathrm{Cu}}^{+}\left(𝑎𝑞\right)+{\mathrm{Cl}}^{-}\left(𝑎𝑞\right)\end{array}$

(ii) Similarly though CuCl has low solubility in aqueous solution $[{𝐾}_{𝑠𝑝}=1.0\times {10}^{-6}]$, still some copper $\left({\mathrm{Cu}}^{+}\right)$and chloride $\left({\mathrm{Cl}}^{-}\right)$are dissolved in the solution. Let the concentration of these ions be $𝑦\mathrm{mol}{L}^{-1}$ (in solution).

$\left[{\mathrm{Cu}}^{+}\right]=\left[{\mathrm{Cl}}^{-}\right]=𝑦$

(iii) The salts $\mathrm{AgCl}\left(s\right)$ and $\mathrm{CuCl}\left(s\right)$ are present in equilibrium with their ions as follows :

$\text{AgCl}\left(𝑠\right)+{\text{H}}_2\text{O}\rightleftharpoons \underset{𝑥}{{\text{Ag}}^{+}\left(𝑎𝑞\right)}+\underset{𝑥+𝑦}{{\text{Cl}}^{-}\left(𝑎𝑞\right)}$

${𝐾}_{𝑠𝑝}\left(\text{AgCl}\right)=1.6\times {10}^{-10}=𝑥(𝑥+𝑦)\dots \text{(i)}$

${\text{CuCl}}_{\left(𝑠\right)}+{\text{H}}_2\text{O}\rightleftharpoons \underset{𝑥}{{\text{Cu}}^{+}\left(𝑎𝑞\right)}+\underset{𝑦+𝑥}{{\text{Cl}}^{-}\left(𝑎𝑞\right)}$

${𝐾}_{𝑠𝑝}\left(\text{CuCl}\right)=1.0\times {10}^{-6}=𝑦\times (𝑥+𝑦)\dots \text{(ii)}$

Dividing equation (ii) by (i) :

$\frac{1.0\times {10}^{-6}}{1.6\times {10}^{-10}}=\frac{𝑦}{𝑥}$

$\frac{𝑦}{𝑥}=\frac{1.0}{1.6}\times {10}^4$ …(iii)

No. of moles of CuCl = $0.1\text{ mol}$

Volume of solution = $1\text{ L}$

$\begin{array}{cc}& \text{ Concentration of }\mathrm{CuCl}=\frac{\text{ No. of moles of }\mathrm{CuCl}}{\text{ Volume of solution }}\\ \\ & \text{ Concentration of }\mathrm{CuCl}=\frac{0.1\mathrm{mol}}{1L}=0.1\end{array}$

$\begin{array}{cc}{K}_{𝑠𝑝}& =1.0\times {10}^{-6}=𝑦\times 𝑦\\ \\ 𝑦& =\sqrt{\surd {10}^{-6}}={10}^{-3}\mathrm{mol}{L}^{-1}\end{array}$

Substituting value of $𝑦$ in equation (iii),

$\begin{array}{cc}\frac{\left[{\mathrm{Cu}}^{+}\right]}{\left[{\mathrm{Ag}}^{+}\right]}=\frac{𝑦}{𝑥}& =\frac{{10}^{-3}\mathrm{mol}{L}^{-1}}{𝑥}={10}^4\frac{1.0}{1.6}\\ 𝑥& =1.6\times {10}^{-7}\end{array}$

The concentration of silver ion, $\left[{\mathrm{Ag}}^{+}\right]=𝑥$ $=1.6\times {10}^{-7}\mathrm{mol}$ L.

Hence, the value of $𝑥$ in 1.6 $\times {10}^{-𝑥}$ is 7 .

(i) The shape of a regular octahedron :

When the octahedron is truncated from all edges.

(ii) Each face of truncated octahedron is hexagonal.

Since, there are 8 faces, there are eight hexagons.

The molecular formula of the mineral haematite is ${\mathrm{Fe}}_2{O}_3$ and that of magnetite is ${\mathrm{Fe}}_3{O}_4$ respective. Both contain iron ( Fe ) but in different oxidation states.

Oxidation state of iron ( Fe ) in :

(a) Haematite $\left({\mathrm{Fe}}_2{O}_3\right)$

Let the oxidation state of iron ( Fe ) be $𝑥$

$\begin{array}{cc}2𝑥+3\times (-2)& =0\\ \\ 2𝑥& =6\\ \\ 𝑥& =+3\end{array}$

Iron exist as Fe (III).

(b) Magnetite $\left({\mathrm{Fe}}_3{O}_4\right)$

The mineral magnetite is made by two iron oxides $\mathrm{FeO}.{\mathrm{Fe}}_2{O}_3$.

The oxidation state of iron in ${\mathrm{Fe}}_2{O}_3$ is +3 and oxidation state of iron in FeO can be calculated as :

$\begin{array}{cc}𝑥+2\times (-2)& =0\\ \\ 𝑥& =+2\end{array}$

Iron exist as Fe (II).

Hence, the oxidation state of iron in haematite is III and that in magnetite is II, III respectively.

In K₃Fe(CN)₆ :

CN^$-$ being a strong field ligand, it causes pairing of electrons, so only one unpaired electron. The complex is paramagnetic.

In Co[(NH₃)₆]Cl₃ :

NH₃ being a strong field ligand it causes pairing of spins. There is no unpaired electron, hence the complex is diamagnetic.

In Na₃[Co(oxalate)₃] :

Oxalate being a strong field ligand it causes pairing of electron spins. There is no unpaired electron, hence the complex is diamagnetic.

In [Ni(H₂O)₆]Cl₂ :

Since H₂O is a weak field ligand, no pairing takes place. There are two unpaired electrons, so the complex is paramagnetic.

In K₂[Pt(CN)₄] :

CN^$-$ being a strong field ligand, it causes pairing of electrons spins, so no unpaired electron. Hence, the complex is diamagnetic.

In [Zn(H₂O)₆(NO₃)₂] : Zn²⁺ has 3d¹⁰ configuration. Therefore, it is a diamagnetic complex.

In presence of mineral acid $H^{+}$(with no water) positively charged centre is generated at second position on the ring.

Since, oxygen is electronegative, it shifts pi electron towards itself generating a carbocation which attack one lone pair of electron on oxygen of alcohol.

Loss of proton $\left(H^{+}\right)$from oxygen generates acetal.

Cu²⁺ and Hg²⁺ ions belong to Group II of inorganic salt analysis and their sulphides are very less soluble in aqueous medium.

Hence, CuS and HgS will be precipitated out even in acidic medium by passing H₂S gas.

Azo dye test is given by aromatic primary amines.

(i) The cyclic structure of pyranose ring

This is 3D structure of glucose formed as a result of cyclisation of an aldohexose.

(ii) There are 2 types of groups in cyclic structure; groups that lie perpendicular to the plane of pyranose ring are axial and other are axial groups.

The fischer form of pyranose ring has groups above the ring on left and groups below the ring on right.

The cyclic form of pyranose ring is obtained as a result of reaction of hydroxyl group (–OH) at fifth carbon to the electropositive carbon of aldehydic group. The reaction happens as follows :

For the equilibrium reaction in aqueous medium :

$2{\mathrm{Cu}}^{+}\rightleftharpoons {\mathrm{Cu}}^0+{\mathrm{Cu}}^{2+}$

(i) If concentration of copper (I) is reduced, then according to Le-chatelier's principle the equilibrium reaction will shift backward.

(ii) The copper (II) ion reacts with chloride $\left({\mathrm{Cl}}^{-}\right)$ion forming copper (III) chloride, which reacts with copper to produce the precipitate of CuCl .

$\begin{array}{cc}& {\mathrm{Cu}}^{2+}+2{\mathrm{Cl}}^{-}\to {\mathrm{CuCl}}_2\\ \\ & {\mathrm{CuCl}}_2+\mathrm{Cu}\to 2\mathrm{CuCl}\end{array}$

Since, the end product involves consumption of copper (I) ion to form CuCl , hence, reactive shifts backward.

(iii) The copper (II) ion also reacts with cyanide $\left({\mathrm{CN}}^{-}\right)$ion forming copper (II) cyanide which on further reactions form ${\left[\mathrm{Cu}\right(\mathrm{CN}{)}_4]}^{3-}$ with copper in (I) oxidation state shifting reaction backward.

$\begin{array}{cc}{\mathrm{Cu}}^{2+}+2{\mathrm{CN}}^{-}& \to \mathrm{Cu}(\mathrm{CO}{)}_2\\ \\ 2\mathrm{Cu}(\mathrm{CN}{)}_2& \to 2\mathrm{CuCN}+(\mathrm{CN}{)}_2\\ \\ \mathrm{CuCN}+3{\mathrm{CN}}^{-}& \to {\left[\mathrm{Cu}\right(\mathrm{CN}{)}_4]}^{3-}\end{array}$