IIT JEE 2009 Chemistry Question Paper with Answer and Solution

(C) According to the Hardy-Schulze rule,the coagulating power of an electrolyte depends on the valency of the active ion (ion with charge opposite to that of the sol).
$Sb_2S_3$ sol is a negatively charged sol.
Therefore,the coagulating power depends on the charge of the cation.
The cations present in the given electrolytes are $Na^+$,$Ca^{2+}$,$Al^{3+}$,and $NH_4^+$.
According to the Hardy-Schulze rule,the higher the valency of the coagulating ion,the greater is its coagulating power.
Since $Al^{3+}$ has the highest charge $(+3)$,$Al_2(SO_4)_3$ is the most effective coagulating agent.

(C) Let the height of the ball above the water surface be $x$. The apparent height $x'$ of the ball as seen by the fish is given by $x' = \mu x$, where $\mu = 4/3$ is the refractive index of water.
The speed of the ball as seen by the fish is $v' = \frac{dx'}{dt} = \mu \frac{dx}{dt} = \mu v$, where $v$ is the actual speed of the ball.
First, we calculate the actual speed $v$ of the ball when it is at a height $x = 12.8 \ m$. The ball falls from $H = 20 \ m$. The distance fallen is $h = 20 - 12.8 = 7.2 \ m$.
Using the equation of motion $v^2 = u^2 + 2gh$, with $u = 0$:
$v^2 = 2 \times 10 \times 7.2 = 144$
$v = 12 \ m/s$.
Now, the apparent speed $v'$ seen by the fish is:
$v' = \mu v = (4/3) \times 12 = 16 \ m/s$.

(B) In this compound,the $-CN$ group is the principal functional group,so it is given the highest priority.
Numbering starts from the carbon atom attached to the $-CN$ group.
To assign the lowest locants to the substituents,we number the ring such that the $-Br$ group is at position $2$ and the $-OH$ group is at position $5$.
Thus,the $IUPAC$ name is $2-$bromo$-5-$hydroxybenzonitrile.

(A) The electrostatic force between the charges $+Q$ and $-q$ is a central force,which acts along the line joining the two charges.
Since the force is central,the torque $\vec{\tau} = \vec{r} \times \vec{F}$ about the position of charge $+Q$ is zero.
According to the principle of conservation of angular momentum,if the net torque is zero,the angular momentum $\vec{L}$ remains constant.
In an elliptical orbit,the distance between the charges changes continuously,which causes the magnitude of the electrostatic force to change.
Consequently,the linear speed and the linear momentum of the charge $-q$ are not constant.
Additionally,since the distance from the center changes,the angular velocity is also not constant.

(B) Let the surface charge densities be $\sigma$. Since the surface charge densities on the outer surfaces are equal,we have:
$\sigma = \frac{q_1}{4\pi R^2} = \frac{q_2}{4\pi (2R)^2} = \frac{q_3}{4\pi (3R)^2}$,where $q_1, q_2, q_3$ are the charges on the outer surfaces of the shells.
From the property of induction,the charge on the outer surface of the first shell is $q_1 = Q_1$.
The charge on the outer surface of the second shell is $q_2 = Q_1 + Q_2$.
The charge on the outer surface of the third shell is $q_3 = Q_1 + Q_2 + Q_3$.
Equating the densities: $\frac{Q_1}{R^2} = \frac{Q_1 + Q_2}{4R^2} = \frac{Q_1 + Q_2 + Q_3}{9R^2}$.
From $\frac{Q_1}{1} = \frac{Q_1 + Q_2}{4}$,we get $4Q_1 = Q_1 + Q_2 \Rightarrow Q_2 = 3Q_1$.
From $\frac{Q_1}{1} = \frac{Q_1 + Q_2 + Q_3}{9}$,we get $9Q_1 = Q_1 + (3Q_1) + Q_3 \Rightarrow 9Q_1 = 4Q_1 + Q_3 \Rightarrow Q_3 = 5Q_1$.
Thus,the ratio $Q_1 : Q_2 : Q_3 = Q_1 : 3Q_1 : 5Q_1 = 1 : 3 : 5$.

(C) According to the Hardy-Schulze rule,the coagulating power of an electrolyte depends on the valency of the ion carrying a charge opposite to that of the colloidal particles.
$Sb_2S_3$ sol is a negatively charged sol.
Therefore,the coagulating power depends on the valency of the cation.
The cations present in the given electrolytes are $Na^+$,$Ca^{2+}$,$Al^{3+}$,and $NH_4^+$.
According to the Hardy-Schulze rule,the higher the valency of the coagulating ion,the greater is its coagulating power.
Since $Al^{3+}$ has the highest charge $(+3)$,$Al_2(SO_4)_3$ is the most effective coagulating agent.

(B) According to Henry's Law,$P_{N_2} = K_H \times X_{N_2}$.
First,calculate the partial pressure of $N_2$ in air: $P_{N_2} = \text{mole fraction of } N_2 \times \text{Total pressure} = 0.8 \times 5 \ atm = 4 \ atm$.
Now,calculate the mole fraction of $N_2$ in water $(X_{N_2})$: $X_{N_2} = P_{N_2} / K_H = 4 \ atm / (1.0 \times 10^5 \ atm) = 4 \times 10^{-5}$.
Since $X_{N_2} = n_{N_2} / (n_{N_2} + n_{H_2O})$ and $n_{N_2} \ll n_{H_2O}$,we can approximate $X_{N_2} \approx n_{N_2} / n_{H_2O}$.
Given $n_{H_2O} = 10 \ moles$,we have $n_{N_2} = X_{N_2} \times n_{H_2O} = (4 \times 10^{-5}) \times 10 = 4 \times 10^{-4} \ moles$.

(A) According to Gauss's Law,the electric flux $\Phi$ through a closed surface is given by $\Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0}$.
$1$. The disk has a radius $a/4$ and center at $(-a/2, 0, 0)$. The cube extends from $x = -a/2$ to $x = a/2$. The disk lies in the $x-y$ plane. The part of the disk inside the cube is the region from $x = -a/2$ to $x = -a/2 + a/4 = -a/4$. This is exactly half of the disk. Thus,the charge enclosed is $6 \text{ C} / 2 = 3 \text{ C}$.
$2$. The rod has length $a$ and is placed from $x = a/4$ to $x = 5a/4$. The cube extends up to $x = a/2$. The part of the rod inside the cube is from $x = a/4$ to $x = a/2$,which is a length of $a/4$. Since the total length is $a$,the fraction of charge enclosed is $(a/4) / a = 1/4$. Thus,the charge enclosed is $8 \text{ C} \times (1/4) = 2 \text{ C}$.
$3$. The point charge $-7 \text{ C}$ is at $(a/4, -a/4, 0)$,which is inside the cube $(|x| < a/2, |y| < a/2, |z| < a/2)$.
$4$. The point charge $3 \text{ C}$ is at $(-3a/4, 3a/4, 0)$,which is outside the cube.
Total enclosed charge $q_{\text{enclosed}} = 3 \text{ C} + 2 \text{ C} - 7 \text{ C} = -2 \text{ C}$.
Therefore,the electric flux $\Phi = \frac{-2 \text{ C}}{\varepsilon_0}$.

(B) The average atomic mass of an element is calculated by the weighted average of the masses of its isotopes.
Average atomic mass of $Fe = \frac{(5 \times 54) + (90 \times 56) + (5 \times 57)}{100}$
$= \frac{270 + 5040 + 285}{100}$
$= \frac{5595}{100} = 55.95 \ u$

(B) The van der Waals equation for $n$ moles of a real gas is given by:
$(P + \frac{an^2}{V^2})(V - nb) = nRT$
In this equation,the term $\frac{an^2}{V^2}$ is added to the measured pressure $P$ to account for the intermolecular attractive forces between gas molecules.
Thus,$\frac{an^2}{V^2}$ is the correction term for attractive forces.

(B) The correct option is $B$,$2-$Bromo-$5-$hydroxybenzonitrile.
In the given compound,the $-CN$ group is the principal functional group,so the parent chain is named as a derivative of benzonitrile.
The carbon atom attached to the $-CN$ group is assigned position $1$.
Numbering the ring to give the lowest possible locants to the substituents,the $-OH$ group is at position $5$ and the $-Br$ group is at position $2$.
According to alphabetical order,$Bromo$ comes before $hydroxy$.
Therefore,the $IUPAC$ name is $2-$Bromo-$5-$hydroxybenzonitrile.

(A) When sodium metal is heated in excess of air,it primarily forms sodium peroxide $(Na_2O_2)$.
The chemical reaction is: $2Na + O_2 (\text{excess}) \rightarrow Na_2O_2$.
Sodium oxide $(Na_2O)$ is formed when sodium is burned in a limited supply of air.
Therefore,the correct compound formed in excess air is $Na_2O_2$.

(A) The compound $X$ is $hexa-2,4-diene-2,5-diol$. It has two chiral centers at $C_2$ and $C_5$ and one double bond at $C_3=C_4$.
For the $cis$ isomer ($Z$-isomer),the molecule is symmetrical. The number of stereoisomers is given by $2^{(n-1)} + 2^{(n/2 - 1)} = 2^{(2-1)} + 2^{(2/2 - 1)} = 2^1 + 2^0 = 2 + 1 = 3$.
For the $trans$ isomer ($E$-isomer),the molecule is also symmetrical. The number of stereoisomers is $2^{(n-1)} + 2^{(n/2 - 1)} = 2^1 + 2^0 = 3$.
Total stereoisomers = $3 (cis) + 3 (trans) = 6$. Thus,$(A)$ is correct.
For $cis$ isomer: $2$ enantiomers and $1$ meso compound. Thus,$(D)$ is correct.
For $trans$ isomer: $2$ enantiomers and $1$ meso compound. Thus,$(C)$ is incorrect as it states $4$ enantiomers.
Since $(A)$ and $(D)$ are correct,the answer is $(A, D)$.

(A) $B_2$: $\sigma 1s^2 \sigma^{\star} 1s^2 \sigma 2s^2 \sigma^{\star} 2s^2 \pi 2p_y^1 \pi 2p_z^1$. It is paramagnetic. Bond order $= (6-4)/2 = 1$. Mixing of $s$ and $p$ orbitals occurs. $B_2$ undergoes oxidation and reduction.
$(B)$ $N_2$: $\sigma 1s^2 \sigma^{\star} 1s^2 \sigma 2s^2 \sigma^{\star} 2s^2 \sigma 2p_x^2 \pi 2p_y^2 \pi 2p_z^2$. It is diamagnetic. Bond order $= (10-4)/2 = 3$. Mixing of $s$ and $p$ orbitals occurs. $N_2$ undergoes oxidation and reduction.
$(C)$ $O_2^{-}$: Paramagnetic with bond order $= 1.5$. $O_2^{-}$ undergoes oxidation and reduction.
$(D)$ $O_2$: $\sigma 1s^2 \sigma^{\star} 1s^2 \sigma 2s^2 \sigma^{\star} 2s^2 \sigma 2p_x^2 \pi 2p_y^2 \pi 2p_z^2 \pi^{\star} 2p_y^1 \pi^{\star} 2p_z^1$. Paramagnetic with bond order $= 2$. $O_2$ undergoes oxidation and reduction.
Correct mapping: $(A)$ $\rightarrow p, q, r, t; (B)$ $\rightarrow q, r, s, t; (C)$ $\rightarrow p, q, r; (D)$ $\rightarrow p, q, r, s$.

(A) Any point $Q$ on the line $\vec{r}$ can be represented as $Q(1 - 3\mu, -1 + \mu, 2 + 5\mu)$.
The vector $\overrightarrow{PQ}$ is given by $\overrightarrow{PQ} = (1 - 3\mu - 3)\hat{i} + (-1 + \mu - 2)\hat{j} + (2 + 5\mu - 6)\hat{k} = (-3\mu - 2)\hat{i} + (\mu - 3)\hat{j} + (5\mu - 4)\hat{k}$.
Since $\overrightarrow{PQ}$ is parallel to the plane $x - 4y + 3z = 1$,it must be perpendicular to the normal vector $\vec{n} = \hat{i} - 4\hat{j} + 3\hat{k}$ of the plane.
Therefore,$\overrightarrow{PQ} \cdot \vec{n} = 0$.
$1(-3\mu - 2) - 4(\mu - 3) + 3(5\mu - 4) = 0$.
$-3\mu - 2 - 4\mu + 12 + 15\mu - 12 = 0$.
$8\mu - 2 = 0$.
$8\mu = 2$,which gives $\mu = \frac{1}{4}$.

(A) Given the equation $z\bar{z}^3+\bar{z}z^3=350$.
We can factor this as $z\bar{z}(z^2+\bar{z}^2)=350$.
Let $z=x+iy$,then $z\bar{z}=x^2+y^2$ and $z^2+\bar{z}^2=(x+iy)^2+(x-iy)^2=2(x^2-y^2)$.
Substituting these into the equation: $(x^2+y^2)(2(x^2-y^2))=350$.
$(x^2+y^2)(x^2-y^2)=175$.
Since $175 = 25 \times 7$,we set $x^2+y^2=25$ and $x^2-y^2=7$.
Adding these equations gives $2x^2=32$,so $x^2=16$,which means $x=\pm 4$.
Subtracting gives $2y^2=18$,so $y^2=9$,which means $y=\pm 3$.
The vertices of the rectangle are $(4,3), (4,-3), (-4,-3),$ and $(-4,3)$.
The length of the rectangle is $|4 - (-4)| = 8$ and the breadth is $|3 - (-3)| = 6$.
Area $= 8 \times 6 = 48$ square units.

(D) Given $z = \cos \theta + i \sin \theta$,by De Moivre's Theorem,$z^n = \cos(n\theta) + i \sin(n\theta)$.
Thus,$\operatorname{Im}(z^{2m-1}) = \sin((2m-1)\theta)$.
The sum is $S = \sum_{m=1}^{15} \sin((2m-1)\theta) = \sin \theta + \sin 3\theta + \sin 5\theta + \dots + \sin 29\theta$.
This is an arithmetic progression of angles with $n = 15$ terms,first term $a = \theta$,and common difference $d = 2\theta$.
The sum of the series $\sum_{k=0}^{n-1} \sin(a + kd) = \frac{\sin(nd/2)}{\sin(d/2)} \sin(a + (n-1)d/2)$.
Substituting $n=15, a=\theta, d=2\theta$:
$S = \frac{\sin(15 \times 2\theta / 2)}{\sin(2\theta / 2)} \sin(\theta + (14 \times 2\theta / 2)) = \frac{\sin(15\theta)}{\sin \theta} \sin(\theta + 14\theta) = \frac{\sin^2(15\theta)}{\sin \theta}$.
At $\theta = 2^{\circ}$,$15\theta = 30^{\circ}$.
$S = \frac{\sin^2(30^{\circ})}{\sin 2^{\circ}} = \frac{(1/2)^2}{\sin 2^{\circ}} = \frac{1}{4 \sin 2^{\circ}}$.

(D) The carbocation is at $C-3$. $A$ hydride shift from $C-2$ to $C-3$ occurs because it results in a more stable carbocation at $C-2$.
The positive charge at $C-2$ is stabilized by resonance (conjugation) from the lone pair of electrons on the oxygen atom of the $-OH$ group.
Therefore,the $H$ atom at $C-2$ is the most likely to migrate to the positively charged carbon at $C-3$.

(B) To determine the stability of resonance structures,we follow these rules:
$1$. Structures with more covalent bonds are more stable (complete octets).
$2$. Structures with fewer formal charges are more stable.
$3$. Negative charges should be on more electronegative atoms,and positive charges on less electronegative atoms.
Analyzing the structures:
$(I)$ $H_2C=N^+=N^-$: Has $4$ bonds,octets are complete for all atoms. It is the most stable.
$(III)$ $H_2C^--N^+=N$: Has $4$ bonds,but the negative charge is on the carbon atom (less electronegative than nitrogen). It is the next most stable.
$(II)$ $H_2C^+-N=N^-$: Has $3$ bonds,carbon has an incomplete octet. Less stable.
$(IV)$ $H_2C^--N=N^+$: Has $3$ bonds,nitrogen has an incomplete octet. Least stable.
Thus,the correct order is $I > III > II > IV$.

(B) The correct option is $B$.
Explanation:
$1$. $A$ state function is a property whose value depends only on the current state of the system and not on the path taken to reach that state.
$2$. Internal energy $(U)$ and molar enthalpy $(H_m)$ are state functions because they depend only on the state variables of the system.
$3$. Work $(w)$ and heat $(q)$ are path functions,meaning their values depend on the specific process or path taken to change the state of the system.
$4$. Therefore,both $A$ (Internal energy) and $D$ (Molar enthalpy) are state functions.

(A) Small amines such as $NH_3$,$CH_3NH_2$,and $(CH_3)_2NH$ undergo unsymmetrical cleavage of diborane to form ionic products of the type $[BH_2(X)_2]^+ [BH_4]^-$.
The reaction is: $B_2H_6 + 2X \rightarrow [BH_2(X)_2]^+ [BH_4]^-$.
Large amines,such as $(CH_3)_3N$,undergo symmetrical cleavage of diborane to form borane-amine adducts: $B_2H_6 + 2N(CH_3)_3 \rightarrow 2H_3B \leftarrow N(CH_3)_3$.
Therefore,$X$ can be $NH_3$,$CH_3NH_2$,or $(CH_3)_2NH$.

(A) The reactions are as follows:
$(A)$ $3Cu + 8HNO_3 \text{ (dil)} \rightarrow 3Cu(NO_3)_2 + 2NO + 4H_2O$. Products are $NO$ $(p)$ and $Cu(NO_3)_2$ $(s)$.
$(B)$ $Cu + 4HNO_3 \text{ (conc)} \rightarrow Cu(NO_3)_2 + 2NO_2 + 2H_2O$. Products are $NO_2$ $(q)$ and $Cu(NO_3)_2$ $(s)$.
$(C)$ $4Zn + 10HNO_3 \text{ (dil)} \rightarrow 4Zn(NO_3)_2 + N_2O + 5H_2O$. Products are $N_2O$ $(r)$ and $Zn(NO_3)_2$ $(t)$.
$(D)$ $Zn + 4HNO_3 \text{ (conc)} \rightarrow Zn(NO_3)_2 + 2NO_2 + 2H_2O$. Products are $NO_2$ $(q)$ and $Zn(NO_3)_2$ $(t)$.
Thus,the correct matching is $A-p, s; B-q, s; C-r, t; D-q, t$.

(A) The heat released during combustion is absorbed by the calorimeter: $q = C_V \times \Delta T$.
Given $C_V = 2.5 \ kJ \ K^{-1}$ and $\Delta T = 298.45 \ K - 298.0 \ K = 0.45 \ K$.
Heat released for $3.5 \ g$ of gas $= 2.5 \ kJ \ K^{-1} \times 0.45 \ K = 1.125 \ kJ$.
Moles of gas $= \frac{\text{mass}}{\text{molecular weight}} = \frac{3.5 \ g}{28 \ g \ mol^{-1}} = 0.125 \ mol$.
Enthalpy of combustion per mole $= \frac{\text{Heat released}}{\text{moles}} = \frac{1.125 \ kJ}{0.125 \ mol} = 9 \ kJ \ mol^{-1}$.

(B) The formula for root mean square speed is $V_{rms} = \sqrt{\frac{3RT}{M_X}}$.
The formula for most probable speed is $V_{mp} = \sqrt{\frac{2RT}{M_Y}}$.
Given that $V_{rms} (X) = V_{mp} (Y)$,we have:
$\sqrt{\frac{3R(400)}{40}} = \sqrt{\frac{2R(60)}{M_Y}}$.
Squaring both sides:
$\frac{3 \times 400}{40} = \frac{2 \times 60}{M_Y}$.
$30 = \frac{120}{M_Y}$.
$M_Y = \frac{120}{30} = 4$.
Thus,the molecular weight of gas $Y$ is $4$. Hence,the correct answer is option $B$.

(C) The sodium salt of a weak acid and a strong base undergoes hydrolysis. The $pH$ of such a salt solution is calculated using the formula:
$pH = \frac{1}{2} (pK_w + pK_a + \log C)$
Given:
$K_a = 1.0 \times 10^{-4} \implies pK_a = -\log(1.0 \times 10^{-4}) = 4$
$K_w = 1.0 \times 10^{-14} \implies pK_w = 14$
$C = 0.01 \ M = 10^{-2} \ M \implies \log C = -2$
Substituting these values:
$pH = \frac{1}{2} (14 + 4 + (-2)) = \frac{1}{2} (16) = 8$

(B) In the crystalline state of $AlCl_3$,the $Cl^{-}$ ions form a close-packed lattice structure.
$Al^{3+}$ ions occupy the octahedral voids within this lattice.
Since each $Al^{3+}$ ion is surrounded by $6$ $Cl^{-}$ ions in an octahedral geometry,the coordination number of $Al$ is $6$.

(A) The molecular formula $C_5H_{10}$ corresponds to a degree of unsaturation $(DU)$ of $1$. This means the compound can either be cyclic or contain one double bond.
For cyclic isomers,we consider the following structures:
$1$. Cyclopentane
$2$. Methylcyclobutane
$3$. Ethylcyclopropane
$4$. $1,1$-Dimethylcyclopropane
$5$. cis-$1,2$-Dimethylcyclopropane (optically inactive,meso)
$6$. trans-$1,2$-Dimethylcyclopropane (optically active,$d$-isomer)
$7$. trans-$1,2$-Dimethylcyclopropane (optically active,$l$-isomer)
Counting these,we have $1$ (cyclopentane) + $1$ (methylcyclobutane) + $1$ (ethylcyclopropane) + $1$ ($1,1$-dimethylcyclopropane) + $1$ (cis-$1,2$-dimethylcyclopropane) + $2$ (enantiomers of trans-$1,2$-dimethylcyclopropane) = $7$ isomers.
Thus,the total number of cyclic structural and stereoisomers is $7$.

(C) Since the line makes equal angles with the coordinate axes,its direction cosines are equal. Let the direction cosines be $l, m, n$. Since $l=m=n$ and $l^2+m^2+n^2=1$,we have $3l^2=1$,so $l=m=n=\frac{1}{\sqrt{3}}$ (given positive direction cosines).
The equation of the line passing through $P(2,-1,2)$ with direction cosines $(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$ is $\frac{x-2}{1/\sqrt{3}} = \frac{y+1}{1/\sqrt{3}} = \frac{z-2}{1/\sqrt{3}} = t$.
Any point on this line is given by $(2+\frac{t}{\sqrt{3}}, -1+\frac{t}{\sqrt{3}}, 2+\frac{t}{\sqrt{3}})$.
Since this point lies on the plane $2x+y+z=9$,we substitute the coordinates into the plane equation:
$2(2+\frac{t}{\sqrt{3}}) + (-1+\frac{t}{\sqrt{3}}) + (2+\frac{t}{\sqrt{3}}) = 9$
$4 + \frac{2t}{\sqrt{3}} - 1 + \frac{t}{\sqrt{3}} + 2 + \frac{t}{\sqrt{3}} = 9$
$5 + \frac{4t}{\sqrt{3}} = 9$
$\frac{4t}{\sqrt{3}} = 4$
$t = \sqrt{3}$.
Since $t$ represents the distance $PQ$,the length of the line segment $PQ$ is $\sqrt{3}$.

(A) Any point $Q$ on the line $\vec{r}$ can be represented as $Q(1 - 3\mu, -1 + \mu, 2 + 5\mu)$.
Given $P(3, 2, 6)$,the vector $\vec{PQ}$ is calculated as:
$\vec{PQ} = (1 - 3\mu - 3)\hat{i} + (-1 + \mu - 2)\hat{j} + (2 + 5\mu - 6)\hat{k} = (-3\mu - 2)\hat{i} + (\mu - 3)\hat{j} + (5\mu - 4)\hat{k}$.
Since $\vec{PQ}$ is parallel to the plane $x - 4y + 3z = 1$,it must be perpendicular to the normal vector $\vec{n} = \hat{i} - 4\hat{j} + 3\hat{k}$ of the plane.
Therefore,$\vec{PQ} \cdot \vec{n} = 0$.
$1(-3\mu - 2) - 4(\mu - 3) + 3(5\mu - 4) = 0$.
$-3\mu - 2 - 4\mu + 12 + 15\mu - 12 = 0$.
$8\mu - 2 = 0$.
$8\mu = 2 \Rightarrow \mu = \frac{1}{4}$.

(C) According to the Hardy-Schulze rule,the coagulating power of an electrolyte depends on the valency of the active ion (the ion carrying a charge opposite to that of the colloidal particles).
$Sb_2S_3$ sol is a negatively charged sol. Therefore,it is coagulated by the positively charged ions of the added electrolytes.
The positively charged ions in the given electrolytes are:
$Na^+$ (from $Na_2SO_4$),
$Ca^{2+}$ (from $CaCl_2$),
$Al^{3+}$ (from $Al_2(SO_4)_3$),
$NH_4^+$ (from $NH_4Cl$).
According to the Hardy-Schulze rule,the coagulating power increases with the increase in the magnitude of the charge on the ion. The order of coagulating power is $Al^{3+} > Ca^{2+} > Na^+ = NH_4^+$.
Since $Al^{3+}$ has the highest charge $(+3)$,$Al_2(SO_4)_3$ is the most effective coagulating agent for $Sb_2S_3$ sol.
Hence,option $C$ is correct.

(A) Total pressure $P_T = 5 \ atm$. The mole fraction of nitrogen in air is $0.8$.
Partial pressure of nitrogen $P_{N_2} = P_T \times X_{air} = 5 \times 0.8 = 4 \ atm$.
According to Henry's law,$P_{N_2} = K_H \cdot X_{sol}$,where $X_{sol}$ is the mole fraction of $N_2$ in the solution.
$X_{sol} = \frac{P_{N_2}}{K_H} = \frac{4}{1.0 \times 10^5} = 4 \times 10^{-5}$.
Since $X_{sol} = \frac{n_{N_2}}{n_{N_2} + n_{H_2O}} \approx \frac{n_{N_2}}{n_{H_2O}}$ because $n_{N_2} \ll n_{H_2O}$.
Given $n_{H_2O} = 10 \ moles$,we have $4 \times 10^{-5} = \frac{n_{N_2}}{10}$.
Therefore,$n_{N_2} = 4 \times 10^{-4} \ moles$.

(B) Phosphorus trioxide $(P_4O_6)$ is prepared by burning white phosphorus in a limited supply of oxygen.
To ensure the reaction stops at $P_4O_6$ and does not proceed to $P_4O_{10}$,the reaction is carried out in an atmosphere of nitrogen $(N_2)$ which acts as an inert diluent.
The chemical equation is:
$P_4 + 3O_2 \xrightarrow{N_2} P_4O_6$
Thus,$X$ is a mixture of $O_2$ and $N_2$.

(A) The compounds are: $(I)$ Phenol,$(II)$ $4$-chlorophenol,$(III)$ Benzoic acid,$(IV)$ $4$-methylbenzoic acid.
$1$. Carboxylic acids are significantly more acidic than phenols due to the resonance stabilization of the carboxylate anion compared to the phenoxide ion.
$2$. Comparing $(III)$ and $(IV)$: $(III)$ is benzoic acid. In $(IV)$,the $-CH_3$ group is an electron-donating group (via inductive and hyperconjugation effects),which destabilizes the carboxylate anion,making $(IV)$ less acidic than $(III)$. Thus,$III > IV$.
$3$. Comparing $(I)$ and $(II)$: $(II)$ is $4$-chlorophenol. The $-Cl$ atom is an electron-withdrawing group (via inductive effect),which stabilizes the phenoxide ion,making $(II)$ more acidic than $(I)$. Thus,$II > I$.
$4$. Combining these,the overall order of acidity is $III > IV > II > I$.
Therefore,option $A$ is correct.

(D) The strength of intermolecular forces in polymers follows the order: $\text{Fibers} > \text{Thermoplastics} > \text{Elastomers}$.
$1$. $\text{Nylon}$ is a fiber and has strong hydrogen bonding.
$2$. $\text{Cellulose}$ is a natural polymer with strong hydrogen bonding.
$3$. $\text{Poly(vinyl chloride)}$ is a thermoplastic with dipole-dipole interactions.
$4$. $\text{Natural rubber}$ is an elastomer held together by weak van der Waals forces.
Therefore,$\text{natural rubber}$ has the weakest intermolecular forces.

(A) and $C$ are correct statements.
$A$. Incorrect: Frenkel defect is favoured by a large difference in the sizes of cation and anion.
$B$. Correct: Frenkel defect is a dislocation defect where an ion moves from its lattice site to an interstitial site.
$C$. Correct: Trapping of an electron in an anionic vacancy leads to the formation of $F$-centers,which impart color to the crystal.
$D$. Incorrect: Schottky defects decrease the density of the solid,thus affecting its physical properties.

(B) Geometrical isomerism in coordination compounds occurs when ligands can be arranged in different spatial orientations around the central metal atom.
$A$. $[Pt(en)Cl_2]$ is a square planar complex of the type $[M(AA)X_2]$. It does not show geometrical isomerism because the chelating ligand $(en)$ occupies adjacent positions,and the two $Cl$ ligands are also adjacent.
$B$. $[Pt(en)_2]Cl_2$ is a square planar complex of the type $[M(AA)_2]$. It does not show geometrical isomerism as all possible arrangements are identical.
$C$. $[Pt(en)_2Cl_2]Cl_2$ is an octahedral complex of the type $[M(AA)_2X_2]$. It exhibits geometrical isomerism,existing as $cis$ and $trans$ isomers.
$D$. $[Pt(NH_3)_2Cl_2]$ is a square planar complex of the type $[MA_2X_2]$. It exhibits geometrical isomerism,existing as $cis$ and $trans$ isomers.
Thus,compounds $C$ and $D$ exhibit geometrical isomerism.

(A) The reaction described is the Lauth's violet or Methylene Blue test for sulfides.
$1.$ $X$ is $Na_2S$ because it provides $S^{2-}$ ions which react with $p$-amino-$N,N$-dimethylaniline in the presence of an oxidizing agent $(Y)$ to form methylene blue.
$2.$ $Y$ is $FeCl_3$ (an oxidizing agent). $FeCl_3$ reacts with potassium hexacyanoferrate$(II)$ $(K_4[Fe(CN)_6])$ to form Prussian blue $(Fe_4[Fe(CN)_6]_3)$,which is an intense blue precipitate. This precipitate dissolves in excess reagent to form a soluble complex.
$3.$ $FeCl_3$ reacts with potassium hexacyanoferrate$(III)$ $(K_3[Fe(CN)_6])$ to form $Fe[Fe(CN)_6]$ (iron$(III)$ hexacyanoferrate$(III)$),which gives a brown coloration.
Therefore,$X = Na_2S$,$Y = FeCl_3$,and $Z = Fe[Fe(CN)_6]$.
The correct option is $(D, C, B)$.

(B) $1$. Compound $P$ must be $4-$phenylbut$-3-$en$-2-$one (structure $B$ in the first set) because it contains a methyl ketone group (gives positive iodoform test) and has the correct carbon skeleton to form the subsequent products.
$2$. Reaction of $P$ with $MeMgBr$ followed by dehydration gives $Q$ ($1$,$1$-dimethyl-1H-indene,structure $A$ in the second set). Ozonolysis of $Q$ gives $R$ ($2$-($2$-acetylphenyl)$-2-$methylpropanal,structure $B$ in the second set).
$3$. Intramolecular aldol condensation of $R$ gives $S$ ($3$,$3$-dimethyl$-3,4-$dihydronaphthalen-$1$(2H)-one,structure $B$ in the third set).
Thus,the correct sequence is $P=B, Q=A, R=B, S=B$. The answer is $(B, A, B)$.

(A) . $CH_3CH_2CH_2CN$ (nitrile) undergoes reduction with $Pd-C/H_2$ $(p)$,$SnCl_2/HCl$ (Stephen reduction,$q$),and $DIBAL-H$ $(s)$.
$B$. $CH_3CH_2OCOCH_3$ (ester) undergoes alkaline hydrolysis $(t)$.
$C$. $CH_3-CH=CH-CH_2OH$ (allylic alcohol) undergoes reduction of the double bond with $Pd-C/H_2$ $(p)$ and can be reduced with $DIBAL-H$ $(s)$.
$D$. $CH_3CH_2CH_2CH_2NH_2$ (primary amine) gives the carbylamine test (foul smell) with $CHCl_3/alc. KOH$ $(r)$.

(D) The Arrhenius equation is given by $k = Ae^{-E_a/RT}$.
Taking the logarithm on both sides,we get $\log k = \log A - \frac{E_a}{2.303RT}$.
Comparing this with the given equation $\log k = -(2000) \frac{1}{T} + 6.0$:
For the pre-exponential factor $A$,$\log A = 6.0$,which gives $A = 10^6 \ s^{-1}$.
For the activation energy $E_a$,$\frac{E_a}{2.303R} = 2000$.
$E_a = 2000 \times 2.303 \times 8.314 \ J \ mol^{-1} \approx 38314 \ J \ mol^{-1} = 38.3 \ kJ \ mol^{-1}$.
Thus,the correct option is $D$.

(A) In $Cr(CO)_6$, the central metal atom $Cr$ is in the $0$ oxidation state.
The ground state electronic configuration of $Cr$ is $[Ar] 3d^5 4s^1$.
$CO$ is a strong field ligand, which causes the pairing of all $6$ electrons ($5$ from $3d$ and $1$ from $4s$) into the $3d$ orbitals.
As a result, the number of unpaired electrons $(n)$ is $0$.
The spin-only magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n = 0$, we get $\mu = \sqrt{0(0+2)} = 0 \ BM$.

(A, B, D) The reduction potential of $NO_3^{-}$ is $E^0 = +0.96 \ V$.
$A$ metal is oxidized by $NO_3^{-}$ if its standard oxidation potential is greater than the reduction potential of $NO_3^{-}$,or equivalently,if its standard reduction potential $E^0_{red}$ is less than $E^0_{red}(NO_3^{-}) = +0.96 \ V$.
Comparing the given values:
$V: E^0 = -1.19 \ V < +0.96 \ V$ (Oxidized)
$Fe: E^0 = -0.04 \ V < +0.96 \ V$ (Oxidized)
$Hg: E^0 = +0.86 \ V < +0.96 \ V$ (Oxidized)
$Au: E^0 = +1.40 \ V > +0.96 \ V$ (Not oxidized)
Thus,$V, Fe,$ and $Hg$ are oxidized by $NO_3^{-}$.
The pairs containing these metals are $(A) V$ and $Hg$,$(B) Hg$ and $Fe$,and $(D) Fe$ and $V$.

(A) $N_2O$ has the structure $N \equiv N^+ - O^-$,which contains an $N-N$ bond.
$N_2O_3$ has the structure $O=N-N(=O)-O$,which contains an $N-N$ bond.
$N_2O_4$ has the structure $O_2N-NO_2$,which contains an $N-N$ bond.
$N_2O_5$ has the structure $O_2N-O-NO_2$,which does not contain an $N-N$ bond.
Therefore,the oxides containing $N-N$ bonds are $N_2O, N_2O_3,$ and $N_2O_4$.

(A) In sugar $X$ (Sucrose),the glycosidic linkage is between $C1$ of $\alpha$-$D$-glucose and $C2$ of $\beta$-$D$-fructose. Both anomeric carbons are involved in the linkage,making it a non-reducing sugar.
In sugar $Y$ (Maltose),the glycosidic linkage is between $C1$ of $\alpha$-$D$-glucose and $C4$ of another glucose unit. One anomeric carbon remains free (hemiacetal),making it a reducing sugar.
Regarding the linkages: $X$ has an $\alpha$-linkage (from glucose) and $Y$ has an $\alpha$-linkage (maltose is $\alpha(1\to4)$). However,based on the provided options and standard structural analysis,$X$ is non-reducing and $Y$ is reducing (Statement $B$). The linkage in $X$ is $\alpha, \beta$ and in $Y$ is $\alpha$. Thus,$(B)$ and $(C)$ are the correct statements.

(A) Undergoes nucleophilic substitution of $Br^-$. Undergoes elimination of $HBr$. Does not undergo nucleophilic addition. Does not esterify with acetic anhydride,but can be dehydrogenated. Thus,$(A-p, q, t)$.
$(B)$ Undergoes nucleophilic substitution with $SOCl_2, PCl_5$ etc. Does not undergo elimination. Does not undergo nucleophilic addition. Undergoes esterification with acetic anhydride. Undergoes dehydrogenation to give $C_6H_5CHO$. Thus,$(B-p, s, t)$.
$(C)$ Does not undergo nucleophilic substitution (no leaving group). Does not undergo elimination. Undergoes nucleophilic addition at the carbonyl carbon of $-CHO$. Undergoes esterification with acetic anhydride (at $-OH$ group). Does not undergo dehydrogenation. Thus,$(C-r, s)$.
$(D)$ Undergoes aromatic nucleophilic substitution $(S_NAr)$. Does not undergo elimination,nucleophilic addition,esterification,or dehydrogenation. Thus,$(D-p)$.

(A) The reaction for the alkaline oxidative fusion of manganese dioxide is:
$2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$
The product formed is potassium manganate,$K_2MnO_4$,which contains the manganate ion,${MnO_4}^{2-}$.
To find the oxidation state of $Mn$ in ${MnO_4}^{2-}$:
Let the oxidation state of $Mn$ be $x$.
$x + 4 \times (-2) = -2$
$x - 8 = -2$
$x = +6$
Thus,the oxidation number of $Mn$ in the product is $6$.

(C) In $CuSO_4 \cdot 5 H_2 O$,the copper ion $(Cu^{2+})$ is six-coordinated.
Four water molecules are directly coordinated to the $Cu^{2+}$ ion.
The remaining fifth water molecule is held by hydrogen bonding between the coordinated water molecules and the sulphate $(SO_4^{2-})$ ions.
Therefore,the number of water molecules directly bonded to the metal centre is $4$.