JEE Advanced 2021 Paper 2 Question Paper with Solutions

Angular momentum, L = P $\times$ 0.9 = 0.18 kgm²/s

For earth-sun system,

$T_0^2 = {{4{\pi ^2}} \over {G{M_S}}} \times {R^3}$ ..... (i)

For binary system

${T^2} = {{4{\pi ^2}} \over {G[3{M_S} + 6{M_S}]}} \times {(9R)^3}$ .... (ii)

Using Eqs. (i) and (ii), we get

T = 9T₀

So, n = 9

(a) Magnetic field at the plane of the ring is perpendicular to the plane. However, they bend as they move forward.


(b) By symmetry, we can say that B will be same at all the points having the same radial distance. So, B (x, y) will depend on the radial distance $r = \sqrt {{x^2} + {y^2}}$.


(c) ${({B_{net}})_{centre}} = {{{\mu _0}{I_2}} \over {2(2R)}} - {{{\mu _0}{I_1}} \over {2R}}$

$= {{{\mu _0}} \over {4R}}({I_2} - 2{I_1})$, since I₂ > 2I₁, so B_net at the centre will be non-zero in $\otimes$ direction. But at some other point, B_net may be zero.

From the graph, it is clear that B_net = 0 for r$\in$(0, R).

So, option (c) is incorrect.


(d) For the graph, it is clear that B = $-$ ve

In, $\otimes$ direction for r $\in$ (R to 2R), so option (d) is also incorrect.

For lamp,

$p = {{{V^2}} \over R}$

$R = {{100 \times 100} \over {500}} = 20\Omega$

$i = {V \over R} = {{100} \over {20}} = 5A \Rightarrow i = {V \over {\sqrt {{R^2} + X_C^2} }}$

$\Rightarrow 5 = {{200} \over {\sqrt {{{(20)}^2} + {{\left( {{1 \over {2\pi \times 50 \times C}}} \right)}^2}} }}$

$\Rightarrow 400 + {\left( {{1 \over {100\pi C}}} \right)^2} = 1600$

$\Rightarrow {\left( {{1 \over {100\pi C}}} \right)^2} = 1200 \Rightarrow {1 \over {{\pi ^2}{C^2}}} = 1200 \times {10^4}$

$\Rightarrow {(\pi C)^2} = {1 \over {12}} \times {10^{ - 6}}$

$\Rightarrow {C^2} = {1 \over {12{\pi ^2}}} \times {10^{ - 6}}$

$\Rightarrow$ $C = {1 \over {\sqrt {12} \pi}} \times {10^{ - 3}} = 100 \times {10^{ - 6}}F = 100\mu F$

$V = \sqrt {V_C^2 + V_R^2}$

$200 = \sqrt {V_C^2 + {{100}^2}}$

$\Rightarrow {V_C} = 100\sqrt 3$

$\tan \phi = {{{X_C}} \over R} = {{{V_C}} \over {{V_R}}} = {{100\sqrt 3 } \over {100}} = \sqrt 3$

$\phi = 60^\circ$

(p_total) = 0 $\Rightarrow$ (p_total)_f = 0

So, momentum of one nucleus is p in forward direction, then momentum of the other nucleus will be p in backwards direction.

Energy released = ($\Delta$m)c² = $\delta$c² = KE_P + KE_Q

$K{E_P} = {{{p^2}} \over {2{m_P}}},K{E_Q} = {{{p^2}} \over {2{m_Q}}}$

$\Rightarrow K{E_P}:K{E_Q} = {1 \over {{m_P}}}:{1 \over {{m_Q}}} = {m_Q}:{m_P}$

$K{E_P} = \left( {{{{m_Q}} \over {{m_P} + {m_Q}}}} \right)\delta {c^2}$

$K{E_Q} = \left( {{{{m_P}} \over {{m_P} + {m_Q}}}} \right)\delta {c^2}$

$K{E_P} + K{E_Q} = \delta {c^2}$

${{{p^2}} \over {2{m_P}}} + {{{p^2}} \over {2{m_Q}}} = \delta {c^2} \Rightarrow p = c\sqrt {2\left( {{{{m_P}{m_Q}} \over {{m_P} + {m_Q}}}} \right)\delta }$

E_P = 2E_Q = 2E_R = E(assume)

For metal P and Q,

E₁ - 4 = E_P = E ……..(1)

E₁ - 4.5 = E_A = E/2 ……..(2)

Subtracting equation (2) from (1), we get

E/2 = 0.5

For metal R

E₂ - 5.5 = E_R = E/2 = 0.5

$\Rightarrow$ E₂ = 6

Resistance of potential wire, R₀ = 50$\Omega$

Resistance of 100 m wire = 50$\Omega$

So, resistance of 72 cm wire $= {{50} \over {100}} \times 72 = 36\Omega$

Current, $I = {{{\varepsilon \over 2}} \over {14\Omega + 25\Omega }} = {{{\varepsilon \over 2}} \over {{r_1} + 36\Omega }}$

$\Rightarrow {r_1} = 39 - 36 = 3\Omega$

${n_1} = {n_2} = n = {3 \over 2}$ for minimum deviation.

${r_m} = {A \over 2} = {{60} \over 2} = 30^\circ$ and $i = e = \theta$

$n = {{\sin \left( {{{{\delta _{\min }} + A} \over 2}} \right)} \over {\sin \left( {{A \over 2}} \right)}}$

$\Rightarrow {3 \over 2} = {{\sin \theta } \over {\sin \left( {{{60} \over 2}} \right)}}$

$\Rightarrow \sin \theta = {3 \over 4} \Rightarrow \cos \theta = {{\sqrt 7 } \over 4}$

If ${n_1} = n = {3 \over 2}$ and ${n_2} = n + \Delta n$

$e = \theta + \Delta \theta$

$(1)\sin \theta = n\sin 30^\circ$

$(n + \Delta n)\sin 30^\circ = (1)\sin (\theta + \Delta \theta )$

Solving ${1 \over 2}(\Delta n) = \sin (\theta + \Delta \theta ) - \sin \theta$

${{\Delta n} \over 2} = {{\sin (\theta + \Delta \theta ) - sin\theta } \over {\Delta \theta }} \times \Delta \theta$

${{d(\sin \theta )} \over {d\theta }} = \cos \theta$

${{\Delta n} \over 2} = (\cos \theta )(\Delta \theta ) \Rightarrow {{\Delta n} \over 2} = {{\sqrt 7 } \over 4}(\Delta \theta )$

$\Rightarrow {{\Delta n} \over {\Delta \theta }} = {{\sqrt 7 } \over 2} = {{2.64} \over 2}$

${{\Delta n} \over {\Delta \theta }} = 1.34 > 1$

$\Rightarrow$ $\Delta$n > $\Delta$$\theta$, so option (a) is incorrect.

$\Delta n = (1.34)\Delta \theta$

$\Rightarrow \Delta \theta \propto \Delta n \Rightarrow$ option (b) is correct.

${{\Delta n} \over {\Delta \theta }} = {{\sqrt 7 } \over 2} \Rightarrow {{2.8 \times {{10}^{ - 3}}} \over {\Delta \theta }} = {{\sqrt 7 } \over 2}$

$\Delta \theta = {{5.6 \times {{10}^{ - 3}}} \over {\sqrt 7 }} = \sqrt 7 \times 0.8 \times {10^{ - 3}}$

$\Delta \theta = (2.64 \times 0.8) \times {10^{ - 3}} = 2.11 \times {10^{ - 3}}$ rad

So, option (c) is correct.

Natural frequency of closed pipe,

f = (2n + 1)f₀

f₀ is fundamental frequency

n = 0, 1, 2 .......


Frequency of source received by pipe,

$f' = {f_s}\left[ {{{v - 0} \over {v - u}}} \right]$

For resonance,

$f' = f$

${f_s}\left[ {{v \over {v - u}}} \right] = (2n + 1){f_0}$

If u = 0.8v, f_s = f₀

$f' = {v \over {0.2v}}{f_0} = 5{f_0}$

For n = 2 pipe can be in resonance

Hence, option (a) is correct.

If u = 0.8v, f_s = 2f₀

$f' = {v \over {0.2v}} \times 2{f_0} = 10{f_0}$

If u = 0.8v, f_s = 0.5f₀

$f' = {v \over {0.2v}} \times 0.5{f_0} = 2.5{f_0}$

Not possible.

If u = 0.5v, f_s = 1.5f₀

$f' = {v \over {0.5v}} \times 1.5{f_0} = 3{f_0}$

For n = 1 f = 3f₀

Pipe can be in resonance.

Hence, option (d) is correct.

For lamp,

$p = {{{V^2}} \over R}$

$R = {{100 \times 100} \over {500}} = 20\Omega$

$i = {V \over R} = {{100} \over {20}} = 5A \Rightarrow i = {V \over {\sqrt {{R^2} + X_C^2} }}$

$\Rightarrow 5 = {{200} \over {\sqrt {{{(20)}^2} + {{\left( {{1 \over {2\pi \times 50 \times C}}} \right)}^2}} }}$

$\Rightarrow 400 + {\left( {{1 \over {100\pi C}}} \right)^2} = 1600$

$\Rightarrow {\left( {{1 \over {100\pi C}}} \right)^2} = 1200 \Rightarrow {1 \over {{\pi ^2}{C^2}}} = 1200 \times {10^4}$

$\Rightarrow {(\pi C)^2} = {1 \over {12}} \times {10^{ - 6}}$

$\Rightarrow {C^2} = {1 \over {12{\pi ^2}}} \times {10^{ - 6}}$

$\Rightarrow$ $C = {1 \over {\sqrt {12} \pi}} \times {10^{ - 3}} = 100 \times {10^{ - 6}}F = 100\mu F$

Free body diagram of rod


$\sum {{F _x} = 0}$

${R_1} = T\cos 45^\circ$

${R_1} = {T \over {\sqrt 2 }}$ ..... (i)

$\sum {{F_y} = 0}$

${R_2} + T\sin 45^\circ = W + \alpha W$

${R_2} + {T \over {\sqrt 2 }} = W(1 + \alpha )$ ..... (ii)

$\sum {{\tau _0} = 0}$

$W{L \over 2} + \alpha WL = {T \over {\sqrt 2 }}L$

$T = \sqrt 2 W\left[ {\alpha + {1 \over 2}} \right]$ ..... (iii)

From Eqs. (ii) and (iii), we get

${R_2} + W\left[ {\alpha + {1 \over 2}} \right] = W(1 + \alpha )$

${R_2} = {W \over 2}$

Hence, option (a) is correct.

From Eqs. (i) and (iii), we get

${R_1} = W\left[ {\alpha + {1 \over 2}} \right]$

$\alpha$ = 0.5, R₁ = W

Hence, option (b) is correct.

From Eq. (iii), if $\alpha$ = 0.5

$T = \sqrt 2 W$

${T_{\max }} = 2\sqrt 2 W$

For rope to break,

$T > 2\sqrt 2 W$

$\sqrt 2 W\left[ {\alpha + {1 \over 2}} \right] > 2\sqrt 2 W \Rightarrow \alpha > {3 \over 2}$

Hence, option (d) is correct.

When the tube just sinks/floats, then average density = density of water

${{Mass} \over {total\,volume}}$ = 1 gm/cc

$\Rightarrow {{5gm} \over {total\,volume}}$ = 1 gm/cc

$\Rightarrow$ Total volume = 5 cc

$\Rightarrow$ Volume of tube + final volume of air in the tube = 5 cc

$\Rightarrow {{5gm} \over {2.5gm/cc}} + {V_f} = 5$

$\Rightarrow$ V_f = 5 $-$ 2 = 3 cc

$\Rightarrow$ $\Delta$V = 0.3 cc

$Q = \Delta U = \Delta {U_1} + \Delta {U_2}$

$\Delta {U_2} = 1 \times 2R \times (\sqrt 2 {T_0} - {T_0}) + 2R{T_0}(\sqrt 2 - 1)$

For $\Delta$U₂ : Using mole conservation of left side.

${{{p_0}{V_0}} \over {R{T_0}}} = {{2\sqrt 2 {p_0}{{3{V_0}} \over 2}} \over {R{T_L}}} \Rightarrow {T_L} = 3\sqrt 2 {T_0}$

$\Delta {U_1} = 1 \times 2R \times (3\sqrt 2 {T_0} - {T_0}) = 2R{T_0}(3\sqrt 2 - 1)$

$Q = 2R{T_0}(\sqrt 2 - 1) + 2R{T_0}(3\sqrt 2 - 1) = 2R{T_0}(4\sqrt 2 - 2)$

${{\Delta Q} \over {R{T_0}}} = 2(4\sqrt 2 - 1)$

When the tube just sinks/floats, then average density = density of water

${{Mass} \over {total\,volume}}$ = 1 gm/cc

$\Rightarrow {{5gm} \over {total\,volume}}$ = 1 gm/cc

$\Rightarrow$ Total volume = 5 cc

$\Rightarrow$ Volume of tube + final volume of air in the tube = 5 cc

$\Rightarrow {{5gm} \over {2.5gm/cc}} + {V_f} = 5$

$\Rightarrow$ V_f = 5 $-$ 2 = 3 cc

$\Rightarrow$ $\Delta$V = 0.3 cc

For isothermal process,

${p_i}{V_i} = {p_f}{V_f} \Rightarrow {p_f} = {10^5} \times {{3.3} \over 3}$

p_f = 1.1 $\times$ 10⁵

p_f $-$ p_i = 1.1 $\times$ 10⁵ $-$ 10⁵

= 0.1 $\times$ 10⁵ = 10 $\times$ 10³ Pa $\Rightarrow$ Y = 10

Angular momentum, L = P $\times$ 0.9 = 0.18 kgm²/s


Just after string becomes taut; there will be no velocity along the string.

$\therefore$ ${V_ \bot } = {{P\cos \theta } \over m} = {{0.2 \times 0.9} \over {1 \times 0.1}} = 1.8$ m/s

$\therefore$ $K = {1 \over 2}mv_ \bot ^2 = {1 \over 2} \times 0.1 \times {1.8^2} = 0.162$ J

Dimension of electric field $[E] = [M{A^{ - 1}}L{T^{ - 3}}]$

Dimensions of magnetic field [B]

$[B] = [M{A^{ - 1}}L{T^{ - 2}}]$

Dimensions of magnetic permebility

$[{\mu _0}] = [M{A^{ - 2}}{T^{ - 2}}L]$

$[S] = {{EB} \over {{\mu _0}}} = {{[M{A^{ - 1}}L{T^{ - 3}}][M{A^{ - 1}}{T^{ - 2}}]} \over {[M{A^{ - 2}}{T^{ - 2}}L]}} = [M{T^{ - 3}}]$

(a) H₃PO₃ gives H₃PO₄ and PH₃ on heating. This reaction is known as disproportionation reaction.

$4{H_3}P{O_3}\buildrel \Delta \over \longrightarrow 3{H_3}P{O_4} + P{H_3}$

(b) P in H₃PO₄ is in its highest oxidation state i.e. + 5 hence, it cannot act as reducing agent. But P in H₃PO₃ is in oxidation state, + 3 hence, it can act as reducing agent.

(c) H₃PO₃ contains two $-$OH groups, hence it is a dibasic acid.


(d) Hydrogen attached to P, does not ionise in water. Therefore, options (a), (b) and (d) are correct.

For process 1, entropy is constant thus, q is constant

$\therefore$ ${W_I} = \Delta U = n{C_{V,m}}\Delta T$

$= - (2250 - 450)R = - 1800R$ .... (i)

$\Delta U = n{C_{V,m}}\Delta T$

$- 1800R = 1 \times {{5R} \over 2} \times \Delta T$

$\Rightarrow \Delta T = - 720K$

${T_2} - {T_1} = - 720K$

${T_2} = - 720K + 900K = 180K$

For process II,

${W_{II}} = - nR{T_2}\ln \left( {{{{V_3}} \over {{V_2}}}} \right)$

$= - 1 \times R \times 180\ln \left( {{{{V_3}} \over {{V_2}}}} \right)$ .... (ii)

As, both work done for process I and II are equal,

Therefore, W_I = W_II

$- 1800R = - R \times 180 \times \ln \left( {{{{V_3}} \over {{V_2}}}} \right)$

$\ln \left( {{{{V_3}} \over {{V_2}}}} \right) = {{1800} \over {180}} \Rightarrow \ln \left( {{{{V_3}} \over {{V_2}}}} \right) = 10$

Given,

At anode : $X(s) \to \mathop {{X^{2 + }}(aq)}\limits_{(0.001\,M)} + 2{e^ - }$

At cathode : $\mathop {{Y^{2 + }}(aq)}\limits_{(0.1\,M)} + 2{e^ - } \to Y(s)$

Overall reaction $X(s) + {Y^{2 + }}(aq) \to {X^{2 + }}(aq) + Y(s)$

Nernst equation,

${E_{cell}} = E_{cell}^o - {{0.06} \over 2}\log {{[{X^{2 + }}]} \over {[{Y^{2 + }}]}}$

${E_{cell}} = E_{cell}^o - 0.03\log {{0.001} \over {0.1}}$

${E_{cell}} = E_{cell}^o + 0.06$

E_cell should be positive for a reaction to be spontaneous.

(a) $E_{C{d^{2 + }}/Cd}^o = - 0.40V,E_{N{i^{2 + }}/Ni}^o = - 0.24V$

$E_{cell}^o = E_{cathode}^o - E_{anode}^o = - 0.24 - ( - 0.40)$

$E_{cell}^o = 0.16$

$E_{cell}^{} = E_{cell}^o + 0.06 = 0.16 + 0.06 = 0.22V$

Reaction is spontaneous.

(b) $E_{C{d^{2 + }}Cd}^o = - 0.40V$ $E_{F{e^{2 + }}/Fe}^o = - 0.44V$

$E_{cell}^o = - 0.44 + 0.40 = - 0.04V$

$E_{cell}^{} = - 0.04 + 0.06 = 0.02V$

Reaction is spontaneous.

(c) $E_{N{i^{2 + }}/Ni}^o = - 0.24V$

$E_{P{b^{2 + }}/Pb}^o = - 0.13V$

$E_{cell}^o = - 0.13 + 0.24 = 0.11V$

$E_{cell}^{} = 0.11 + 0.06 = 0.17V$

Reaction is spontaneous.

(d) $E_{N{i^{2 + }}/Ni}^o = - 0.24V$

$E_{F{e^{2 + }}/Fe}^o = - 0.44V$

$E_{cell}^o = - 0.44 + 0.24 = - 0.20V$

$E_{cell}^o = - 0.20 + 0.06 = - 0.14V$

Reaction is non-spontaneous.

Therefore, the correct combinations of X and Y are (a), (b) and (c).

Degree of dissociation = $\alpha$

Limiting molar conductivity, $\Lambda _m^o$ = 4 $\times$ 10² S cm² mol^$-$1

Molar conductivity, $\Lambda _m$ = y $\times$ 10² S cm² mol^$-$1

Molar conductivity of dilution, $\Lambda _m^o$ = 3y $\times$ 10² S cm² mol^$-$1

Concentration before dilution = C

Concentration after dilution = ${C \over {20}}$

Using relation, $\alpha = {{\Lambda _m^{}} \over {\Lambda _m^o}}$ .... (i)

Dissociation constant, ${K_a} = {{C{\alpha ^2}} \over {(1 - \alpha )}}$

Putting Eq. (i),

${K_a} = {{C\Lambda _m^2} \over {\Lambda _m^{o2}\left( {1 - {{\Lambda _m^{}} \over {\Lambda _m^o}}} \right)}} = {{C\Lambda _m^2} \over {\Lambda _m^o(\Lambda _m^o - \Lambda _m^{})}}$

Dissociation constant before dillution,

${K_a} = {{C{{(y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - y \times {{10}^2})}}$ ..... (ii)

Dissociation constant after dilution,

${K_a} = {{{C \over {20}}{{(3y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - 3y \times {{10}^2})}}$ ..... (iii)

Comparing Eqs. (ii) and (iii),

${{C{{(y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - y \times {{10}^2})}} = {{C{{(3y \times {{10}^2})}^2}} \over {20(4 \times {{10}^2})(4 \times {{10}^2} - 3y \times {{10}^2})}}$

${{{y^2} \times {{10}^4}} \over {{{10}^2}(4 - y)}} = {{9{y^2} \times {{10}^4}} \over {{{10}^2} \times 20(4 - 3y)}}$

${1 \over {(4 - y)}} = {9 \over {20(4 - 3y)}}$

$80 - 60y = 36 - 9y$

$80 - 36 = 60y - 9y$

$44 = 51y$

$\Rightarrow y = {{44} \over {51}}$ .... (iv)

Putting in Eq. (i),

$\alpha = {{\Lambda _m^{}} \over {\Lambda _m^o}} = {{y \times {{10}^2}} \over {4 \times {{10}^2}}}$

$\alpha = {{44} \over {51 \times 4}} = {{11} \over {51}}$

$\alpha = 0.22$

The value of $\alpha$ is 0.22.

(a) ${[FeC{l_4}]^ - }$

Atomic number of Fe = 26

Oxidation number of Fe = x + 4 ($-$1) = $-$1

$\Rightarrow$ x = 3

Electronic configuration of Fe = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s²

Electronic configuration of Fe³⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s⁰

Cl^- is a weak field ligand, so no pairing of electrons occurs.


Hybridisation is sp³

Geometry is tetrahedral.

${[Fe{(CO)_4}]^{2 - }}$ Oxidation number of Fe = x + 4(0) = - 2 $\Rightarrow$ x = - 2

Electronic configuration of Fe^2- = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s²



CO is a strong field ligand and cause electrons of 3d orbitals to pair up and shift electrons of 4s to 3d orbital.



One 4s and three 4p orbitals undergoes hybridisation to form four sp³ hybrid orbitals which arrange into a tetrahedral geometry.



Each of the four sp³ hybrid orbitals accept a pair of electrons from carbonyl ligand.



Geometry is tetrahedral.

(b) ${[Co{(CO)_4}]^ - }$ Atomic number of Co = 27

Oxidation number of Co = x + 4(0) = $-$1 $\Rightarrow$ x = $-$1

Electronic configuration of Co = 1s²2s²2p⁶3s²3p⁶3d⁷4s²

Electronic configuration of Co^- = 1s²2s²2p⁶3s²3p⁶3d⁸4s²

CO is a strong field ligand, so pairing of electrons occurs.


Hybridisation is sp³

Geometry is tetrahedral.

${[CoC{l_4}]^{2 - }}$

Oxidation number of Co = x + 4($-$1) = $-$2

$\Rightarrow$ x = + 2

Electronic configuration of Co²⁺ = 1s²2s²2p⁶3s²3p⁶3d⁷4s⁰

Cl^- is a weak field ligand, so pairing of electrons do not occur.


Hybridisation is sp³

Geometry is tetrahedral

(c) $[Ni{(CO)_4}]$ Oxidation number of Ni = x + 4(0) = 0 $\Rightarrow$ x = 0

Atomic number of Ni = 28

Electronic configuration of Ni = 1s²2s²2p⁶3s²3p⁶3d⁸4s²

CO is a strong field ligand, so pairing of electrons occur.


Hybridisation is sp³

Geometry is tetrahedral.

${[Ni{(CN)_4}]^{2 - }}$

Oxidation number of Ni = x + 4($-$1) = $-$2 $\Rightarrow$ x = + 2

Electronic configuration of Ni²⁺ = 1s²2s²2p⁶3s²3p⁶3d⁸4s⁰

CN is a strong field ligand, so pairing of electrons occur.


Hybridisation is dsp².

Geometry is square planar.

(d) ${[Cu{(py)_4}]^ + }$

Oxidation number of Cu = x + 4(0) = + 1 $\Rightarrow$ x = + 1

Atomic number of Cu = 29

Electronic configuration of Cu = 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹

Electronic configuration of Cu⁺ = 1s²2s²2p⁶3s²3p⁶3d¹⁰4s⁰


Hybridisation is sp³

Geometry is tetrahedral.

${[Cu{(CN)_4}]^{3 - }}$

Oxidation number of Cu = x + 4($-$1) = $-$ 3 $\Rightarrow$ x = + 1


Hybridisation is sp³

Geometry is tetrahedral.

$\mathop {2Cl{O_2}}\limits_{Chlorine\,dioxide} + 2{O_3}\buildrel {} \over \longrightarrow \mathop {C{l_2}{O_6}}\limits_{Chloral\,perchlorate} + 2{O_2}$

Oxidation state of Cl in Cl₂O₆ = 2x + 6($-$2) = 0

2x $-$ 12 = 0

2x = 12 $\Rightarrow$ x = + 6

Average oxidation state of Cl in Cl₂O₆ is 6.

(a) If KNO₂ (Q)-on reacting with PhCH₂Br(P) gives PhCH₂ONO which on reduction with LiAlH₄ does not give PhCH₂NH₂.

(b) If product R formed is benzylamine not benzene amine. KCN reacts with PhCH₂ Br to give PhCH₂CN which does not have foul smelling.

(c) If AgNO₂ (Q) reacts with PhCH₂ Br to give PhCH₂NO₂ which on reduciton give PhCH₂NH₂ (Phenyl methanamine or benzyl amine).

(d) If amide reacts with LiAlH 4 (W) to give amine. PhCH₂ Br reacts with AgCN to give PhCH₂NO₂ .

As the concentration of reactant becomes half at t = 50 s. So, half-time of reaction is 50 s.

Given,

${{dP} \over {dt}} = k{[X]^1}$

$\Rightarrow - {1 \over 2}{{dX} \over {dt}} = {{dP} \over {dt}} = k{[X]^1}$

$- {{dX} \over {dt}} = 2k{[X]^1}$

${{ - dX} \over {[X]}} = 2k\,dt$

$- \int_{{X_0}}^X {{{dX} \over X} = 2k\,\int_0^t {dt} }$

$- [\ln X]_{{X_0}}^X = 2k[t]_0^t$

$- \ln X + \ln {X_0} = 2kt$

$\ln {{{X_0}} \over X} = 2kt$

At t_1/2, $X = {{{X_0}} \over 2}$

$\ln {{2{X_0}} \over {{X_0}}} = 2k{t_{1/2}}$

$k = {{\ln 2} \over {2{t_{1/2}}}} = {{0.693} \over {100}} = 6.93 \times {10^{ - 3}}{s^{ - 1}}$

At t = 50 s, $- {{dx} \over {dt}} = 2k{[X]^1};[X] = 1$ mol

So, ${{ - dX} \over {dt}} = 2 \times 6.93 \times {10^{ - 3}} \times 1 = 13.86 \times {10^{ - 3}}$ mol L^-1 s^-1

At t = 100 s, $- {1 \over 2}{{dX} \over {dt}} = {{ - dY} \over {dt}} \Rightarrow {{ - dY} \over {dt}} = k{[Y]^1};[Y] = {1 \over 2}$

$- {{dY} \over {dt}} = 6.93 \times {10^{ - 3}} \times {1 \over 2} = 3.46 \times {10^{ - 3}}$ mol L^-1 s^-1

So, options (b), (c) and (d) are correct.