IIT JEE 2011 Chemistry Question Paper with Answer and Solution

(B) The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at point $(x_1, y_1)$ is given by $\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2$.
Substituting $(x_1, y_1) = (6, 3)$,we get $\frac{a^2x}{6} + \frac{b^2y}{3} = a^2 + b^2$.
Since the normal passes through $(9, 0)$,we substitute $x = 9$ and $y = 0$:
$\frac{a^2(9)}{6} + 0 = a^2 + b^2$
$\frac{3a^2}{2} = a^2 + b^2$
$\frac{1}{2}a^2 = b^2 \Rightarrow a^2 = 2b^2$.
The eccentricity $e$ of the hyperbola is given by $e = \sqrt{1 + \frac{b^2}{a^2}}$.
Substituting $a^2 = 2b^2$:
$e = \sqrt{1 + \frac{b^2}{2b^2}} = \sqrt{1 + \frac{1}{2}} = \sqrt{\frac{3}{2}}$.

(A) The frequency heard by the driver is due to the reflection of sound from the building. The building acts as a stationary source reflecting the sound.
First,the frequency $f_b$ received by the building is given by the Doppler effect formula for a moving source and stationary observer: $f_b = f \left( \frac{v}{v - v_s} \right)$,where $v = 320 \, m/s$,$v_s = 36 \, km/hr = 10 \, m/s$,and $f = 8 \, kHz$.
$f_b = 8 \left( \frac{320}{320 - 10} \right) = 8 \left( \frac{320}{310} \right) \, kHz$.
Now,the building acts as a stationary source reflecting this frequency $f_b$ back to the moving car (observer). The frequency $f'$ heard by the driver is: $f' = f_b \left( \frac{v + v_o}{v} \right)$,where $v_o = 10 \, m/s$.
$f' = \left( 8 \times \frac{320}{310} \right) \times \left( \frac{320 + 10}{320} \right) = 8 \times \frac{330}{310} \approx 8.516 \, kHz$.
Rounding to the nearest provided option,the correct frequency is $8.50 \, kHz$.

(A) For an adiabatic process,the relation between temperature and volume is $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
For helium (a monoatomic gas),the adiabatic index $\gamma = \frac{5}{3}$,so $\gamma - 1 = \frac{2}{3}$.
Given $V_1 = 5.6 \, \text{liter}$ and $V_2 = 0.7 \, \text{liter}$,we have $\frac{V_1}{V_2} = \frac{5.6}{0.7} = 8$.
Substituting these into the relation: $T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma-1} = T_1 (8)^{2/3} = T_1 (2^3)^{2/3} = T_1 (2^2) = 4 T_1$.
At $STP$,the number of moles $n = \frac{5.6 \, \text{liter}}{22.4 \, \text{liter/mol}} = 0.25 = \frac{1}{4} \, \text{mol}$.
The work done in an adiabatic process is $W = \frac{n R (T_1 - T_2)}{\gamma - 1}$.
Substituting the values: $W = \frac{(1/4) R (T_1 - 4 T_1)}{5/3 - 1} = \frac{(1/4) R (-3 T_1)}{2/3} = \frac{-3/4 R T_1}{2/3} = -\frac{9}{8} R T_1$.
The magnitude of work done on the gas is $\frac{9}{8} R T_1$.

(D) Initially,the $2\,\mu F$ capacitor is charged to a potential $V$. The initial energy stored is $U_i = \frac{1}{2} C_1 V^2$,where $C_1 = 2\,\mu F$.
When the switch $S$ is moved to position $2$,the capacitor $C_1$ is connected in parallel with an uncharged capacitor $C_2 = 8\,\mu F$.
The charge redistributes until both capacitors reach a common potential $V' = \frac{C_1 V}{C_1 + C_2} = \frac{2V}{2+8} = 0.2V$.
The final energy stored in the system is $U_f = \frac{1}{2} (C_1 + C_2) (V')^2 = \frac{1}{2} (10\,\mu F) (0.2V)^2 = 0.2 C_1 V^2$.
The energy dissipated as heat is $\Delta U = U_i - U_f = \frac{1}{2} C_1 V^2 - 0.2 C_1 V^2 = 0.3 C_1 V^2$.
The percentage of energy dissipated is $\frac{\Delta U}{U_i} \times 100 = \frac{0.3 C_1 V^2}{0.5 C_1 V^2} \times 100 = 60\%$.
Wait,using the standard formula for energy loss in charge sharing: $\Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2$.
Here $V_1 = V$ and $V_2 = 0$,so $\Delta U = \frac{1}{2} \frac{2 \times 8}{2 + 8} (V - 0)^2 = \frac{1}{2} \times 1.6 \times V^2 = 0.8 \times \frac{1}{2} C_1 V^2$.
Thus,the percentage of energy dissipated is $\frac{\Delta U}{U_i} \times 100 = \frac{C_2}{C_1 + C_2} \times 100 = \frac{8}{2+8} \times 100 = 80\%$.

(D) Initial energy stored in the $2\,\mu F$ capacitor when connected to voltage $V$ is:
$U_{i} = \frac{1}{2} C V^{2} = \frac{1}{2} (2 \times 10^{-6}) V^{2} = 10^{-6} V^{2} \text{ J}$.
When the switch $S$ is turned to position $2$,the $2\,\mu F$ capacitor is connected in parallel with the uncharged $8\,\mu F$ capacitor.
The total charge $Q$ remains conserved:
$Q = C_{1} V = (2 \times 10^{-6}) V$.
The common potential $V_{\text{common}}$ after redistribution is:
$V_{\text{common}} = \frac{Q}{C_{1} + C_{2}} = \frac{2 \times 10^{-6} V}{2 \times 10^{-6} + 8 \times 10^{-6}} = \frac{2V}{10} = \frac{V}{5}$.
The final energy stored in the system is:
$U_{f} = \frac{1}{2} (C_{1} + C_{2}) V_{\text{common}}^{2} = \frac{1}{2} (10 \times 10^{-6}) \left(\frac{V}{5}\right)^{2} = 5 \times 10^{-6} \times \frac{V^{2}}{25} = 0.2 \times 10^{-6} V^{2} \text{ J}$.
The energy dissipated is $\Delta U = U_{i} - U_{f} = 10^{-6} V^{2} - 0.2 \times 10^{-6} V^{2} = 0.8 \times 10^{-6} V^{2} \text{ J}$.
The percentage of energy dissipated is:
$\frac{\Delta U}{U_{i}} \times 100 = \frac{0.8 \times 10^{-6} V^{2}}{10^{-6} V^{2}} \times 100 = 80 \%$.

(B) In a meter bridge,the balance condition is given by $\frac{X}{R} = \frac{\ell_1 + \alpha}{\ell_2 + \beta}$,where $\alpha$ and $\beta$ are the end corrections at ends $A$ and $B$ respectively.
Given:
Unknown resistance = $X$
Standard resistance $R = 10 \, \Omega$
Null point position $\ell_1 = 52 \, cm$
Length of wire $\ell = 100 \, cm$,so $\ell_2 = 100 - 52 = 48 \, cm$
End correction at $A$,$\alpha = 1 \, cm$
End correction at $B$,$\beta = 2 \, cm$
Substituting the values in the formula:
$\frac{X}{10} = \frac{52 + 1}{48 + 2}$
$\frac{X}{10} = \frac{53}{50}$
$X = \frac{53 \times 10}{50} = \frac{53}{5} = 10.6 \, \Omega$
Therefore,the determined value of '$X$' is $10.6 \, \Omega$.

(A) . According to the postulates of the kinetic theory of gases,collisions between molecules and with the walls of the container are perfectly elastic.
$B$. The momentum transferred to the wall is $\Delta p = 2mu$. Since momentum is proportional to mass $(m)$,heavier molecules transfer more momentum upon collision.
$C$. According to the Maxwell-Boltzmann distribution,the fraction of molecules with very high or very low velocities is extremely small.
$D$. Between two successive collisions,molecules move in straight lines with constant velocity due to the absence of intermolecular forces.

(C) To determine if all atoms lie in one plane,we analyze the hybridization and geometry of each molecule:
$(A)$ $CH_2=CH-CH=CH_2$ ($1,3$-butadiene): Due to rotation around the central $C-C$ single bond,it can exist in $s$-cis and $s$-trans conformations. While the $s$-cis form is planar,the molecule can rotate,and not all conformations are necessarily planar.
$(B)$ $HC\equiv C-CH=CH_2$ (but$-1-$en$-3-$yne): The $HC\equiv C-$ group is linear ($sp$-hybridized). The $-CH=CH_2$ group is planar ($sp^2$-hybridized). Since the linear group is attached to the planar group,the entire molecule remains planar.
$(C)$ $H_2C=C=O$ (ketene): The central carbon is $sp$-hybridized,but the terminal $CH_2$ group is $sp^2$-hybridized. The two $H$-atoms on the $CH_2$ group lie in the same plane as the $C=C=O$ backbone.
$(D)$ $H_2C=C=CH_2$ (allene): The central carbon is $sp$-hybridized. The two terminal $CH_2$ groups lie in perpendicular planes to each other,making the molecule non-planar.
Thus,in compounds $(B)$ and $(C)$,all atoms lie in the same plane.

(B) Step $1$: Analyze the final product. The final product is $2$ moles of acetone $(CH_3COCH_3)$. This implies that the intermediate $Q$ must be $2,3$-dimethylbut-$2$-ene,which upon ozonolysis gives $2$ moles of acetone.
Step $2$: Analyze the conversion of $Q$ to acetone. $Q$ undergoes acid-catalyzed dehydration followed by ozonolysis. $2,3$-dimethylbut-$2$-ene is formed by the dehydration of $2,3$-dimethylbutan-$2$-ol. Thus,$Q$ is $(CH_3)_2CH-CH(OH)-CH_3$ (Option $B$).
Step $3$: Analyze the formation of $Q$ from $P$. $P$ $(C_6H_{10})$ reacts with $dil. H_2SO_4/HgSO_4$ to form a ketone,which is then reduced by $NaBH_4$ to an alcohol $Q$. For $Q$ to be $2,3$-dimethylbutan-$2$-ol,the ketone must be $3,3$-dimethylbutan-$2$-one. This ketone is formed from the hydration of $3,3$-dimethylbut-$1$-yne $(P)$. Thus,$P$ is $(CH_3)_2CH-C \equiv C-H$ (Option $D$).
Therefore,$P$ is $D$ and $Q$ is $B$.

(A) The structure of $Na_2S_4O_6$ is shown below:
$Na^+O^--S(=O)_2-S-S-S(=O)_2-O^-Na^+$
In $Na_2S_4O_6$ (sodium tetrathionate),there are two types of sulphur atoms:
$1$. The two terminal sulphur atoms are bonded to three oxygen atoms (two via double bonds and one via a single bond) and one sulphur atom. Their oxidation number is calculated as $+5$.
$2$. The two central sulphur atoms are bonded only to two other sulphur atoms. Since the electronegativity of identical atoms is the same,their oxidation number is $0$.
The difference in the oxidation numbers of the two types of sulphur atoms is $|5 - 0| = 5$.

(B) The balanced chemical equation for the reaction of bromine with sodium carbonate in aqueous solution is:
$3 Na_2CO_3 + 3 Br_2 \rightarrow 5 NaBr + NaBrO_3 + 3 CO_2$
From the balanced equation,the coefficient of sodium bromide $(NaBr)$ is $5$.

(C) For a shell with principal quantum number $n$,the total number of orbitals is $n^2$.
For $n=3$,the total number of orbitals is $3^2 = 9$.
Each orbital can hold a maximum of $2$ electrons,one with $m_s = +1/2$ and one with $m_s = -1/2$.
Therefore,for $n=3$,there are $9$ orbitals,and each can accommodate exactly one electron with $m_s = -1/2$.
Thus,the maximum number of electrons with $n=3$ and $m_s = -1/2$ is $9$.

(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Using $h = 6.626 \times 10^{-34} \ J \cdot s$,$c = 3 \times 10^8 \ m/s$,and $\lambda = 300 \times 10^{-9} \ m$:
$E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9}} \ J = 6.626 \times 10^{-19} \ J$.
Converting to $eV$ by dividing by $1.602 \times 10^{-19} \ J/eV$:
$E = \frac{6.626 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 4.136 \ eV$.
Photoelectric effect occurs if the work function $\phi < E$. Thus,metals with $\phi < 4.136 \ eV$ will show the effect.
Comparing the given values: $Li \ (2.4)$,$Na \ (2.3)$,$K \ (2.2)$,and $Mg \ (3.7)$ are all less than $4.136 \ eV$.
Therefore,there are $4$ such metals.

(B) The total pressure inside the vessel is equal to the external pressure,which is $P_{total} = 1 \ atm$.
The unknown compound exerts a vapour pressure of $0.68 \ atm$. Since it is in equilibrium with its vapour,the partial pressure of the unknown compound is $P_{unknown} = 0.68 \ atm$.
The partial pressure of $He$ is $P_{He} = P_{total} - P_{unknown} = 1 \ atm - 0.68 \ atm = 0.32 \ atm$.
Using the ideal gas law for $He$: $PV = nRT$.
$0.32 \ atm \times V = 0.1 \ mol \times 0.0821 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1} \times 273 \ K$.
$0.32 \times V = 2.24133$.
$V = \frac{2.24133}{0.32} \approx 7.004 \ L$.
Thus,the total volume is close to $7 \ L$.

(A) $3$-Bromo-$3$-cyclopentylhexane has three different types of $\beta$-hydrogens (not $\alpha$-hydrogens as $\alpha$ is the carbon attached to $Br$). These are marked as $(a)$,$(b)$,and $(c)$ in the structure.
Dehydrobromination leads to the formation of three structurally isomeric alkenes:
$1$. From $\beta$-hydrogen $(a)$: $CH_3CH_2CH=C(cyclopentyl)CH_2CH_3$ (exists as $E$ and $Z$ isomers).
$2$. From $\beta$-hydrogen $(b)$: $CH_3CH_2CH_2C(cyclopentyl)=CHCH_3$ (exists as $E$ and $Z$ isomers).
$3$. From $\beta$-hydrogen $(c)$: $CH_3CH_2CH_2C(cyclopentyl)=CHCH_2CH_3$ (this is the same as the product from $(b)$).
Wait,re-evaluating the structure: The $\beta$-carbons are the two $CH_2$ groups of the hexane chain and the $CH$ group of the cyclopentyl ring.
- Removal of $H$ from $CH_2$ (position $a$): $CH_3CH_2CH=C(cyclopentyl)CH_2CH_3$ ($E/Z$ isomers).
- Removal of $H$ from $CH_2$ (position $b$): $CH_3CH_2CH_2C(cyclopentyl)=CHCH_3$ ($E/Z$ isomers).
- Removal of $H$ from $CH$ (position $c$): $CH_3CH_2CH_2C(cyclopentyl)=C(CH_2CH_3)_2$ (this is a different alkene,no $E/Z$ isomerism).
Total alkenes = $2 (E/Z) + 2 (E/Z) + 1 = 5$.

(C) Let $\vec{v} = \lambda \vec{a} + \mu \vec{b}$.
Substituting the vectors $\vec{a}$ and $\vec{b}$,we get $\vec{v} = \lambda(\hat{i} + \hat{j} + \hat{k}) + \mu(\hat{i} - \hat{j} + \hat{k}) = (\lambda + \mu)\hat{i} + (\lambda - \mu)\hat{j} + (\lambda + \mu)\hat{k}$.
The projection of $\vec{v}$ on $\vec{c}$ is given by $\frac{\vec{v} \cdot \vec{c}}{|\vec{c}|} = \frac{1}{\sqrt{3}}$.
Given $\vec{c} = \hat{i} - \hat{j} - \hat{k}$,we have $|\vec{c}| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{3}$.
Thus,$\frac{(\lambda + \mu)(1) + (\lambda - \mu)(-1) + (\lambda + \mu)(-1)}{\sqrt{3}} = \frac{1}{\sqrt{3}}$.
$\Rightarrow \lambda + \mu - \lambda + \mu - \lambda - \mu = 1$.
$\Rightarrow \mu - \lambda = 1$,or $\mu = \lambda + 1$.
Substituting $\mu$ back into $\vec{v}$,we get $\vec{v} = (2\lambda + 1)\hat{i} - \hat{j} + (2\lambda + 1)\hat{k}$.
For $\lambda = 1$,we get $\vec{v} = 3\hat{i} - \hat{j} + 3\hat{k}$,which matches option $C$.

(D) For set $P$,we have $\sin \theta - \cos \theta = \sqrt{2} \cos \theta$.
This simplifies to $\sin \theta = (\sqrt{2} + 1) \cos \theta$,which implies $\tan \theta = \sqrt{2} + 1$.
For set $Q$,we have $\sin \theta + \cos \theta = \sqrt{2} \sin \theta$.
This simplifies to $\cos \theta = (\sqrt{2} - 1) \sin \theta$,which implies $\tan \theta = \frac{1}{\sqrt{2} - 1}$.
Rationalizing the denominator,$\tan \theta = \frac{1(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{\sqrt{2} + 1}{2 - 1} = \sqrt{2} + 1$.
Since both sets $P$ and $Q$ represent the same condition $\tan \theta = \sqrt{2} + 1$,it follows that $P = Q$.
Therefore,option $(D)$ is the correct answer.

(B) Let the slope of line $L$ be $m$. The slope of the given line $\sqrt{3}x + y = 1$ is $m_1 = -\sqrt{3}$.
Given the angle between the lines is $60^{\circ}$,we have $\tan 60^{\circ} = \left| \frac{m - m_1}{1 + m \cdot m_1} \right|$.
$\sqrt{3} = \left| \frac{m - (-\sqrt{3})}{1 + m(-\sqrt{3})} \right| = \left| \frac{m + \sqrt{3}}{1 - \sqrt{3}m} \right|$.
Case $1$: $\frac{m + \sqrt{3}}{1 - \sqrt{3}m} = \sqrt{3}$ $\Rightarrow m + \sqrt{3} = \sqrt{3} - 3m$ $\Rightarrow 4m = 0$ $\Rightarrow m = 0$.
The equation of line $L$ is $y - (-2) = 0(x - 3) \Rightarrow y + 2 = 0$.
Case $2$: $\frac{m + \sqrt{3}}{1 - \sqrt{3}m} = -\sqrt{3}$ $\Rightarrow m + \sqrt{3} = -\sqrt{3} + 3m$ $\Rightarrow 2m = 2\sqrt{3}$ $\Rightarrow m = \sqrt{3}$.
The equation of line $L$ is $y - (-2) = \sqrt{3}(x - 3)$ $\Rightarrow y + 2 = \sqrt{3}x - 3\sqrt{3}$ $\Rightarrow y - \sqrt{3}x + 2 + 3\sqrt{3} = 0$.

(A) In neutral medium:
$MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$
(Involves $3$ electrons)
In alkaline medium:
$MnO_4^- + e^- \rightarrow MnO_4^{2-}$
$MnO_4^{2-} + 2H_2O + 2e^- \rightarrow MnO_2 + 4OH^-$
Overall: $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$
(Involves $3$ electrons)
In acidic medium:
$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$
(Involves $5$ electrons)
Comparing with the options:
$A$: $3$ electrons in neutral medium (Correct)
$B$: $5$ electrons in neutral medium (Incorrect)
$C$: $3$ electrons in alkaline medium (Correct)
$D$: $5$ electrons in acidic medium (Correct)
Thus,the correct combination is $A, C, D$.

(A) The equilibrium $2 Cu^{+} \rightleftharpoons Cu + Cu^{2+}$ shifts to the left if the concentration of $Cu^{+}$ is increased or the concentration of $Cu^{2+}$ is decreased,or if $Cu^{+}$ is removed from the solution by precipitation or complexation.
$Cu(I)$ forms insoluble compounds such as $CuCl$,$CuCN$,and $CuSCN$.
When $Cl^{-}$,$CN^{-}$,or $SCN^{-}$ are added,they react with $Cu^{+}$ to form these insoluble precipitates,effectively removing $Cu^{+}$ from the equilibrium mixture.
According to Le Chatelier's principle,removing a product (or in this case,a reactant $Cu^{+}$) shifts the equilibrium to the left (backward direction) to counteract the change.
Therefore,the presence of $Cl^{-}$,$SCN^{-}$,and $CN^{-}$ shifts the equilibrium to the left.

(A) This is a case of simultaneous solubility of salts with a common ion $(Cl^{-})$.
Since $K_{sp}(CuCl) \gg K_{sp}(AgCl)$,the concentration of $Cl^{-}$ ions in the solution is primarily determined by the dissolution of $CuCl$.
For $CuCl \rightleftharpoons Cu^{+} + Cl^{-}$,let $s$ be the solubility.
$K_{sp}(CuCl) = s^2 = 1.0 \times 10^{-6} \Rightarrow s = 10^{-3} \ M$.
Thus,$[Cl^{-}] \approx 10^{-3} \ M$.
Now,for $AgCl \rightleftharpoons Ag^{+} + Cl^{-}$,the solubility product expression is:
$K_{sp}(AgCl) = [Ag^{+}][Cl^{-}] = 1.6 \times 10^{-10}$.
Substituting the value of $[Cl^{-}]$:
$[Ag^{+}] \times 10^{-3} = 1.6 \times 10^{-10}$.
$[Ag^{+}] = 1.6 \times 10^{-7} \ M$.
Comparing this with $1.6 \times 10^{-x}$,we get $x = 7$.

(B) The given compound is $3$-methylpentane,$CH_3CH_2CH(CH_3)CH_2CH_3$.
Monochlorination can occur at different positions:
$1$. Substitution at the terminal $CH_3$ group (equivalent to $CH_2CH_2Cl$): This creates a chiral center at the $C-3$ position. It forms a pair of enantiomers ($2$ isomers).
$2$. Substitution at the $CH_2$ group: This creates two chiral centers (at $C-2$ and $C-3$). This results in $2^2 = 4$ stereoisomers (two pairs of enantiomers).
$3$. Substitution at the $CH$ group: The resulting product is $3$-chloro-$3$-methylpentane,which is achiral ($1$ isomer).
$4$. Substitution at the $CH_3$ group attached to the $CH$ (i.e.,$CH_2Cl$): This creates a chiral center at the $C-3$ position ($1$ isomer).
Total isomers = $2 + 4 + 1 + 1 = 8$.

(C) The carbon atom directly attached to the positively charged carbon is called the $\alpha$-carbon,and the hydrogen atoms attached to these $\alpha$-carbons are called $\alpha$-hydrogens.
In the given carbocation,the central carbocation is attached to:
$1$. $A$ methyl group $(-CH_3)$,which has $3$ $\alpha$-hydrogens.
$2$. An ethyl group $(-CH_2CH_3)$,which has $2$ $\alpha$-hydrogens.
$3$. $A$ cyclohexyl group,which has $1$ $\alpha$-hydrogen (at the bridgehead position attached to the carbocation).
Total number of $\alpha$-hydrogens = $3 + 2 + 1 = 6$.
The number of hyperconjugative structures is equal to the number of $\alpha$-hydrogens,which is $6$.

(A) . $CO_{2(s)} \rightarrow CO_{2(g)}$: This is a phase transition (sublimation). It is endothermic $(\Delta H > 0)$ and the entropy increases $(\Delta S > 0)$. Thus,$A \rightarrow p, r, s$.
$B$. $CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)}$: This is a chemical decomposition. It is endothermic $(\Delta H > 0)$ and the entropy increases $(\Delta S > 0)$ due to the formation of gas. Thus,$B \rightarrow r, s$.
$C$. $2H_{(g)} \rightarrow H_{2(g)}$: $A$ chemical bond is formed,which is exothermic $(\Delta H < 0)$. The system becomes more ordered,so entropy decreases $(\Delta S < 0)$. Thus,$C \rightarrow t$.
$D$. $P_{(\text{white, solid})} \rightarrow P_{(\text{red, solid})}$: This is a phase transition and an allotropic change. Red phosphorus is more stable and ordered than white phosphorus,so entropy decreases $(\Delta S < 0)$. Thus,$D \rightarrow p, q, t$.

(A) Reaction $(A)$ is an intramolecular aldol condensation. It involves the formation of a carbanion $(t)$,followed by nucleophilic addition $(s)$ to the carbonyl group,and finally dehydration $(r)$ to form the conjugated enone.
Reaction $(B)$ involves the reaction of a Grignard reagent with a ketone,which is a nucleophilic addition $(s)$,followed by an intramolecular nucleophilic substitution $(p)$ by the oxygen atom on the carbon bearing the chlorine atom.
Reaction $(C)$ is an acid-catalyzed intramolecular cyclization (acetal/ether formation),which involves nucleophilic addition $(s)$ of the hydroxyl group to the carbonyl,followed by dehydration $(r)$.
Reaction $(D)$ is an acid-catalyzed cyclization involving the formation of a carbocation,followed by an intramolecular electrophilic substitution $(q)$ on the benzene ring,and dehydration $(r)$ to restore aromaticity.
Thus,the correct matches are: $A \rightarrow r, s, t$; $B \rightarrow p, s$; $C \rightarrow r, s$; $D \rightarrow q, r$. The correct option is $A$.

(D) For set $P$: $\sin \theta - \cos \theta = \sqrt{2} \cos \theta$
$\Rightarrow \sin \theta = (\sqrt{2} + 1) \cos \theta$
$\Rightarrow \tan \theta = \sqrt{2} + 1$
For set $Q$: $\sin \theta + \cos \theta = \sqrt{2} \sin \theta$
$\Rightarrow \cos \theta = (\sqrt{2} - 1) \sin \theta$
$\Rightarrow \frac{1}{\sqrt{2} - 1} = \tan \theta$
$\Rightarrow \tan \theta = \frac{\sqrt{2} + 1}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \sqrt{2} + 1$
Since both sets $P$ and $Q$ represent the same set of values for $\theta$ satisfying $\tan \theta = \sqrt{2} + 1$,we have $P = Q$.

(C) Given $\overline{a}=\hat{i}+\hat{j}+\hat{k}$,$\overline{b}=\hat{i}-\hat{j}+\hat{k}$ and $\overline{c}=\hat{i}-\hat{j}-\hat{k}$.
Since $\overline{v}$ lies in the plane of $\overline{a}$ and $\overline{b}$,we can write $\overline{v} = m\overline{a} + n\overline{b}$.
$\overline{v} = m(\hat{i}+\hat{j}+\hat{k}) + n(\hat{i}-\hat{j}+\hat{k}) = (m+n)\hat{i} + (m-n)\hat{j} + (m+n)\hat{k} \quad \dots(i)$
The projection of $\overline{v}$ on $\overline{c}$ is given by $\frac{\overline{v} \cdot \overline{c}}{|\overline{c}|} = \frac{1}{\sqrt{3}}$.
$|\overline{c}| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{3}$.
So,$\frac{(m+n)(1) + (m-n)(-1) + (m+n)(-1)}{\sqrt{3}} = \frac{1}{\sqrt{3}}$.
$m+n - m+n - m-n = 1 \implies n-m = 1 \implies n = m+1$.
Substituting $n = m+1$ into $(i)$:
$\overline{v} = (m+m+1)\hat{i} + (m-(m+1))\hat{j} + (m+m+1)\hat{k} = (2m+1)\hat{i} - \hat{j} + (2m+1)\hat{k}$.
For $m=1$,$\overline{v} = 3\hat{i} - \hat{j} + 3\hat{k}$,which matches option $C$.

(D) Very pure nitrogen can be obtained by the thermal decomposition of sodium or barium azide.
$2NaN_3 \xrightarrow{573 \ K} 2Na + 3N_2$
Azide salt of barium can be obtained in the purest form as well as the decomposition product contains solid $Ba$ as a by-product along with gaseous nitrogen,hence no additional step of separation is required.
$Ba(N_3)_2 \xrightarrow{\text{heat}} Ba + 3N_2$
Hence,option $D$ is correct.

(B) $Ni^{2+}$ has a $3d^8$ electronic configuration.
$1$. For $[NiCl_4]^{2-}$,$Cl^{-}$ is a weak field ligand,so it does not cause pairing of electrons. The hybridization is $sp^3$,resulting in a tetrahedral geometry.
$2$. For $[Ni(CN)_4]^{2-}$,$CN^{-}$ is a strong field ligand,which causes pairing of $3d$ electrons. The hybridization is $dsp^2$,resulting in a square planar geometry.
$3$. For $[Ni(H_2O)_6]^{2+}$,$H_2O$ is a ligand that forms an octahedral complex with $Ni^{2+}$ involving $sp^3d^2$ hybridization.
Thus,the shapes are tetrahedral,square planar,and octahedral,respectively. Therefore,option $B$ is correct.

(A) The reaction for path $(i)$ is: ${ }_{13}^{27} Al +{ }_2^4 \alpha \rightarrow{ }_{14}^{30} Si +{ }_1^1 p (X)$.
Here,$X$ is a proton $({ }_1^1 p)$ because the mass number and atomic number are conserved ($27+4 = 30+1$ and $13+2 = 14+1$).
The reaction for path $(ii)$ is: ${ }_{13}^{27} Al +{ }_2^4 \alpha \rightarrow{ }_{15}^{30} P +{ }_0^1 n (Y)$.
Here,$Y$ is a neutron $({ }_0^1 n)$ because the mass number and atomic number are conserved ($27+4 = 30+1$ and $13+2 = 15+0$).
The subsequent decay of phosphorus is: ${ }_{15}^{30} P \rightarrow{ }_{14}^{30} Si +{ }_{+1}^{0} e (Z)$.
Here,$Z$ is a positron $({ }_{+1}^{0} e)$ because the atomic number decreases by $1$.
Therefore,$X$ is proton,$Y$ is neutron,and $Z$ is positron.
Thus,the correct option is $A$.

(C) Molarity is defined as the number of moles of solute dissolved in per litre of solution.
Molarity,$M = \frac{n}{V}$
$n =$ number of moles of solute = mass $/$ molar mass
$V =$ volume of solution
Mass of urea = $120 \ g$
Molar mass of urea = $60 \ g / mol$
Mass of water = $1000 \ g$
Density of solution,$\rho = 1.15 \ g / mL$
$n = \frac{120}{60} = 2 \ mol$
Total mass of solution = $m_{solute} + m_{solvent} = 120 \ g + 1000 \ g = 1120 \ g$
Volume of solution,$V = \frac{\text{mass}}{\text{density}} = \frac{1120 \ g}{1.15 \ g / mL} \approx 973.91 \ mL = 0.97391 \ L$
Molarity,$M = \frac{n}{V} = \frac{2 \ mol}{0.97391 \ L} \approx 2.05 \ M$
Hence,the correct option is $C$.

(A) The reaction between $AgNO_3$ and $KCl$ is: $Ag^+(aq) + NO_3^-(aq) + K^+(aq) + Cl^-(aq) \rightarrow AgCl(s) + K^+(aq) + NO_3^-(aq)$.
Initially,the solution contains $K^+$ and $Cl^-$ ions. As $AgNO_3$ is added,$Ag^+$ ions react with $Cl^-$ ions to form $AgCl$ precipitate.
Since $AgCl$ is insoluble,the $Cl^-$ ions are replaced by $NO_3^-$ ions in the solution.
The ionic mobility of $Cl^-$ is higher than that of $NO_3^-$,so the conductance of the solution decreases until the equivalence point is reached.
After the equivalence point,adding more $AgNO_3$ increases the concentration of $Ag^+$ and $NO_3^-$ ions in the solution,which leads to an increase in the conductance.
Therefore,the graph shows a decrease followed by an increase,which corresponds to plot $(P)$.

(C) The acidity of carboxylic acids is significantly influenced by substituents on the benzene ring.
$o$-hydroxybenzoic acid (salicylic acid) exhibits the ortho-effect,where the proximity of the hydroxyl group to the carboxyl group stabilizes the conjugate base through intramolecular hydrogen bonding and steric factors,making it more acidic than the other options provided.

(A) The reaction shown is the first step of the Gabriel phthalimide synthesis.
Phthalimide reacts with $KOH$ to form potassium phthalimide,which acts as a nucleophile.
This nucleophile then undergoes an $S_N2$ reaction with the alkyl halide,$4-bromobenzyl chloride$ $(Br-C_6H_4-CH_2Cl)$.
In $S_N2$ reactions,the nucleophile attacks the carbon atom attached to the leaving group.
Since the $CH_2Cl$ group is more reactive towards $S_N2$ than the aryl bromide ($C-Br$ bond),the nitrogen of the phthalimide attacks the $CH_2$ carbon,displacing the $Cl^-$ ion.
Therefore,the product is $N-(4-bromobenzyl)phthalimide$.

(A) The ore cassiterite is $SnO_{2}$.
$1$. The extraction involves the reduction of the oxide ore $SnO_{2}$ using carbon (smelting): $SnO_{2} + 2C \rightarrow Sn + 2CO$.
$2$. Cassiterite often contains iron as an impurity in the form of $FeWO_{4}$ (ferberite) or other iron oxides. This is removed by magnetic separation or by converting it into slag during the smelting process.
Therefore,the process involves carbon reduction of an oxide ore and the removal of iron impurity.

(B) Adsorption is always exothermic because it involves the release of energy as surface forces are satisfied.
$(B)$ Physisorption is reversible and weak,but at higher temperatures,it can provide enough activation energy for chemical bond formation,transforming into chemisorption.
$(C)$ This statement is incorrect. Physisorption decreases with an increase in temperature. Chemisorption initially increases with temperature (due to activation energy) and then decreases at very high temperatures.
$(D)$ Chemisorption involves chemical bond formation,which releases more energy than the weak van der Waals forces in physisorption. It is slow because it requires a significant activation energy.

(A) $1.$ When $Cu$ rod $(M)$ is dipped in $AgNO_3$ $(N)$,$Cu$ displaces $Ag^+$ ions: $Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)$. The formation of $Cu^{2+}$ turns the solution light blue.
$2.$ Addition of $NaCl$ to the solution results in the precipitation of $AgCl$ $(O)$: $Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)$ (white precipitate).
$3.$ Addition of excess $NH_3$ dissolves the $AgCl$ precipitate by forming a soluble complex: $AgCl(s) + 2NH_3(aq) \rightarrow [Ag(NH_3)_2]^+(aq) + Cl^-(aq)$. Simultaneously,$Cu^{2+}$ ions react with $NH_3$ to form an intense blue complex: $Cu^{2+}(aq) + 4NH_3(aq) \rightarrow [Cu(NH_3)_4]^{2+}(aq)$.
Thus,$M = Cu$,$N = AgNO_3$,and the final solution contains $[Ag(NH_3)_2]^+$ and $[Cu(NH_3)_4]^{2+}$. The correct option is $(A)$.

(C) Molecular weight of decapeptide $= 796 \ g/mol$.
$A$ decapeptide contains $10$ amino acid units,so it has $(10-1) = 9$ peptide bonds.
Upon complete hydrolysis,$9$ molecules of water are added.
Total weight of water added $= 9 \times 18 \ g/mol = 162 \ g/mol$.
Total weight of hydrolysis products $= 796 + 162 = 958 \ g$.
Total weight of glycine in the product $= \frac{47.0}{100} \times 958 \ g = 450.26 \ g \approx 450 \ g$.
Molecular weight of glycine $= 75 \ g/mol$.
Number of glycine units $= \frac{450 \ g}{75 \ g/mol} = 6$.

(D) Haematite is $Fe_2O_3$,where the oxidation state of iron is $+3$.
Magnetite is $Fe_3O_4$,which is a mixed oxide represented as $FeO \cdot Fe_2O_3$.
In $FeO$,the oxidation state of iron is $+2$,and in $Fe_2O_3$,it is $+3$.
Therefore,the oxidation states of iron in haematite and magnetite are $III$ and $II, III$ respectively.

(C) $K \Rightarrow K_3[Fe(CN)_6], Fe^{3+} = 3d^5$ (has $5$ unpaired electrons). It is paramagnetic.
$L \Rightarrow [Co(NH_3)_6]Cl_3, Co^{3+} = 3d^6$. $NH_3$ is a strong field ligand,causing pairing. It is diamagnetic.
$M \Rightarrow Na_3[Co(ox)_3], Co^{3+} = 3d^6$. Oxalate $(ox^{2-})$ acts as a strong field ligand with $Co^{3+}$,causing pairing. It is diamagnetic.
$N \Rightarrow [Ni(H_2O)_6]Cl_2, Ni^{2+} = 3d^8$ (has $2$ unpaired electrons). It is paramagnetic.
$O \Rightarrow K_2[Pt(CN)_4], Pt^{2+} = 5d^8$. $CN^{-}$ is a strong field ligand and $5d$ orbitals have large splitting. It is diamagnetic.
$P \Rightarrow [Zn(H_2O)_6](NO_3)_2, Zn^{2+} = 3d^{10}$. All electrons are paired. It is diamagnetic.
Therefore,the diamagnetic complexes are $L, M, O, P$.

(A) In an acidified aqueous solution,the concentration of $S^{2-}$ ions is very low due to the common ion effect of $H^+$ ions from the acid.
Only the sulphides of Group $II$ cations,such as $Cu^{2+}$ and $Hg^{2+}$,have a sufficiently low solubility product $(K_{sp})$ to precipitate under these conditions.
The sulphides of $Mn^{2+}$ and $Ni^{2+}$ (Group $III$ and $IV$ cations) have higher $K_{sp}$ values and remain in the solution.
Therefore,$CuS$ and $HgS$ are precipitated.
Hence,the correct option is $A$.

(D) The Nernst equation for the given cell reaction is:
$E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log Q$
Here,$n = 4$ (number of electrons transferred).
The reaction quotient $Q$ is given by:
$Q = \frac{[Fe^{2+}]^2}{P(O_2) \times [H^+]^4}$
Given $[Fe^{2+}] = 10^{-3} \ M$,$P(O_2) = 0.1 \ atm$,and $pH = 3$,so $[H^+] = 10^{-3} \ M$.
$Q = \frac{(10^{-3})^2}{0.1 \times (10^{-3})^4} = \frac{10^{-6}}{0.1 \times 10^{-12}} = \frac{10^{-6}}{10^{-13}} = 10^7$
Now,substitute the values into the Nernst equation:
$E_{cell} = 1.67 - \frac{0.0591}{4} \log(10^7)$
$E_{cell} = 1.67 - \frac{0.0591 \times 7}{4}$
$E_{cell} = 1.67 - 0.1034 \approx 1.57 \ V$

(A) The dissociation of $K_3[Fe(CN)_6]$ is given by: $K_3[Fe(CN)_6] \rightarrow 3K^{+} + [Fe(CN)_6]^{3-}$.
Since $1 \ mol$ of $K_3[Fe(CN)_6]$ produces $4 \ mol$ of ions,the van't Hoff factor $i = 4$.
The formula for depression in freezing point is $\Delta T_f = i \times K_f \times m$.
Here,molality $m = \frac{w_2 \times 1000}{M_2 \times w_1} = \frac{0.1 \times 1000}{329 \times 100} = \frac{1}{329} \ mol \ kg^{-1}$.
$\Delta T_f = 4 \times 1.86 \times \frac{1}{329} \approx 0.0226 \ K \approx 0.023 \ K$.
Since the freezing point of pure water is $0 \ {}^{\circ}C$,the freezing point of the solution is $T_f = 0 - \Delta T_f = -0.023 \ {}^{\circ}C = -2.3 \times 10^{-2} \ {}^{\circ}C$.

(C) The reaction described is the azo coupling reaction,which is characteristic of primary aromatic amines.
Primary aromatic amines react with $NaNO_2$ and $HCl$ at $0-5 \ ^{\circ}C$ to form stable diazonium salts.
These diazonium salts then undergo electrophilic substitution (coupling) with electron-rich aromatic compounds like $\beta$-naphthol in an alkaline medium to form intensely colored azo dyes.
Among the given options:
$A$ is $N,N$-dimethylaniline (tertiary aromatic amine).
$B$ is $N$-methylaniline (secondary aromatic amine).
$C$ is $p$-toluidine (primary aromatic amine).
$D$ is benzylamine (primary aliphatic amine,which forms an unstable diazonium salt that decomposes to form alcohols).
Thus,$p$-toluidine $(C)$ is the correct compound that forms a stable diazonium salt and subsequently an azo dye.

(B) The reaction involves the acid-catalyzed addition of an alcohol $(RCH_2OH)$ to the double bond of $3,4-dihydro-2H-pyran$.
This reaction proceeds through the formation of a resonance-stabilized oxocarbenium ion intermediate.
The alcohol acts as a nucleophile and attacks the electrophilic carbon of the oxocarbenium ion.
After the loss of a proton $(H^+)$,the final product formed is a tetrahydropyranyl ether.
Since the product contains an alkoxy group attached to a carbon atom that is also bonded to another oxygen atom (part of the ring),it is classified as an acetal (specifically,a cyclic acetal).

(B) The given structure is a six-membered pyranose ring.
At the anomeric carbon $(C1)$,the $-OH$ group is in the equatorial (up) position,which identifies it as the $\beta$-anomer.
Since the substituent at the anomeric carbon is a hydrogen atom $(-H)$ rather than a hydroxymethyl group $(-CH_2OH)$,it is an aldohexose (specifically $\beta$-$D$-glucopyranose).
Therefore,the correct option is $B$.

(A) For a first-order reaction,the concentration at time $t$ is given by $C_t = C_0 e^{-kt}$,which shows exponential decay. Thus,statement $A$ is correct.
The half-life is $t_{1/2} = \frac{\ln 2}{k}$. Since the rate constant $k$ increases with temperature according to the Arrhenius equation,$t_{1/2}$ decreases as temperature increases. Thus,statement $B$ is correct.
For a first-order reaction,$t_{1/2} = \frac{0.693}{k}$,which is independent of the initial concentration $C_0$. Thus,statement $C$ is incorrect.
After $n$ half-lives,the remaining reactant is $C_t = \frac{C_0}{2^n}$. For $n = 8$,$C_t = \frac{C_0}{2^8} = \frac{C_0}{256}$.
Percentage completion = $\frac{C_0 - C_t}{C_0} \times 100 = \frac{C_0 - \frac{C_0}{256}}{C_0} \times 100 = (1 - \frac{1}{256}) \times 100 \approx 99.6 \%$. Thus,statement $D$ is correct.
Therefore,statements $A, B,$ and $D$ are correct.

(A) To form a condensation polymer with adipic acid $(HOOC-(CH_2)_4-COOH)$,we need a diamine $(NH_2-(CH_2)_n-NH_2)$.
$1$. If $X = CONH_2$,the reaction with $Br_2 / NaOH$ (Hofmann bromamide degradation) gives $NH_2-(CH_2)_4-NH_2$. This is a diamine,which reacts with adipic acid to form a polyamide (Nylon-$6$,$4$).
$2$. If $X = CN$,the reaction with $H_2 / Ni / \text{heat}$ (catalytic hydrogenation) gives $NH_2-(CH_2)_6-NH_2$. This is a diamine,which reacts with adipic acid to form a polyamide (Nylon-$6$,$6$).
Both $(C)$ and $(D)$ result in the formation of a diamine,which can undergo condensation polymerization with the given dicarboxylic acid. Thus,the correct pairs are $(C)$ and $(D)$.

(A) $PCl_5$ reacts with compounds containing oxygen to form $POCl_3$ under specific conditions. Let us analyze each compound:
$1$. $O_2$: Does not react with $PCl_5$.
$2$. $CO_2$: Does not react with $PCl_5$.
$3$. $SO_2$: $PCl_5 + SO_2 \rightarrow POCl_3 + SOCl_2$. (Reacts)
$4$. $H_2O$: $PCl_5 + H_2O \rightarrow POCl_3 + 2HCl$. (Reacts with limited water)
$5$. $H_2SO_4$: $2PCl_5 + H_2SO_4 \rightarrow 2POCl_3 + SO_2Cl_2 + 2HCl$. (Reacts)
$6$. $P_4O_{10}$: $6PCl_5 + P_4O_{10} \rightarrow 10POCl_3$. (Reacts)
Thus,the compounds that react to give $POCl_3$ are $SO_2, H_2O, H_2SO_4$,and $P_4O_{10}$. The total number of such compounds is $4$.

(A) The complex $[Cr(H_2O)_5Cl]Cl_2$ dissociates in water to release $2$ moles of ionizable $Cl^-$ ions per mole of the complex.
Number of millimoles of $[Cr(H_2O)_5Cl]Cl_2 = 30 \ mL \times 0.01 \ M = 0.3 \ mmol$.
Since each mole of complex provides $2$ moles of $Cl^-$, the total millimoles of $Cl^- = 0.3 \times 2 = 0.6 \ mmol$.
For complete precipitation, $Ag^+$ reacts with $Cl^-$ in a $1:1$ molar ratio $(Ag^+ + Cl^- \rightarrow AgCl(s))$.
Therefore, millimoles of $AgNO_3$ required $= 0.6 \ mmol$.
Using the formula $M = \frac{n}{V(in \ L)}$ or $M = \frac{mmol}{V(in \ mL)}$, we have $0.1 = \frac{0.6}{V}$.
$V = \frac{0.6}{0.1} = 6 \ mL$.

(B) truncated octahedron is an Archimedean solid.
It is formed by cutting off the corners of a regular octahedron.
It consists of $6$ square faces and $8$ hexagonal faces.
Therefore,the number of hexagonal faces present in a truncated octahedron is $8$.