AIEEE 2008 Chemistry Question Paper with Answer and Solution

(D) We are given the integral $I = \int \frac{5\tan x}{\tan x - 2} dx$.
First,express $\tan x$ as $\frac{\sin x}{\cos x}$:
$I = \int \frac{5(\sin x / \cos x)}{(\sin x / \cos x) - 2} dx = \int \frac{5\sin x}{\sin x - 2\cos x} dx$.
We want to express the numerator in terms of the denominator $(\sin x - 2\cos x)$ and its derivative $(\cos x + 2\sin x)$.
Let $5\sin x = A(\sin x - 2\cos x) + B(\cos x + 2\sin x)$.
Equating coefficients of $\sin x$ and $\cos x$:
$A + 2B = 5$
$-2A + B = 0 \implies B = 2A$.
Substituting $B = 2A$ into the first equation: $A + 2(2A) = 5 \implies 5A = 5 \implies A = 1$.
Then $B = 2(1) = 2$.
So,$5\sin x = 1(\sin x - 2\cos x) + 2(\cos x + 2\sin x)$.
Substituting this into the integral:
$I = \int \frac{(\sin x - 2\cos x) + 2(\cos x + 2\sin x)}{\sin x - 2\cos x} dx$
$I = \int 1 dx + 2 \int \frac{\cos x + 2\sin x}{\sin x - 2\cos x} dx$.
Let $t = \sin x - 2\cos x$,then $dt = (\cos x + 2\sin x) dx$.
$I = x + 2 \int \frac{1}{t} dt = x + 2 \ln |t| + k = x + 2 \ln |\sin x - 2\cos x| + k$.
Comparing this with the given form $x + a \ln |\sin x - 2\cos x| + k$,we get $a = 2$.

(A) If a gas expands at a constant temperature,the average kinetic energy of the molecules remains the same.
The average kinetic energy $(KE)$ of a gas molecule is given by the expression:
$KE = \frac{3}{2} kT$
From this expression,it is clear that $KE \propto T$.
Since the temperature $(T)$ is constant,the kinetic energy $(KE)$ of the molecules must also remain constant.

(A) The speed of sound in an ideal gas is given by the formula: $v = \sqrt{\frac{\gamma RT}{M}}$.
For oxygen $(O_2)$,which is a diatomic gas,the adiabatic index $\gamma_{O_2} = 1.4 = \frac{7}{5}$ and the molar mass $M_{O_2} = 32 \; g/mol$.
Thus,$v_{O_2} = \sqrt{\frac{7RT}{5 \times 32}}$.
For helium $(He)$,which is a monatomic gas,the adiabatic index $\gamma_{He} = 1.67 = \frac{5}{3}$ and the molar mass $M_{He} = 4 \; g/mol$.
Thus,$v_{He} = \sqrt{\frac{5RT}{3 \times 4}}$.
Taking the ratio of the speeds:
$\frac{v_{He}}{v_{O_2}} = \sqrt{\frac{\gamma_{He} / M_{He}}{\gamma_{O_2} / M_{O_2}}} = \sqrt{\frac{5/3}{4} \times \frac{32}{7/5}} = \sqrt{\frac{5}{12} \times \frac{160}{7}} = \sqrt{\frac{800}{84}} = \sqrt{\frac{200}{21}} \approx 3.086$.
Therefore,$v_{He} = 460 \times 3.086 \approx 1420 \; m/s$.

(B) Isoelectronic species are those that possess the same number of electrons.
For $NO^{+}$: $7 + 8 - 1 = 14$ electrons.
For $C_2^{2-}$: $6 \times 2 + 2 = 14$ electrons.
For $CN^{-}$: $6 + 7 + 1 = 14$ electrons.
For $N_2$: $7 \times 2 = 14$ electrons.
Since $NO^{+}, C_2^{2-}, CN^{-}$,and $N_2$ each have $14$ electrons,they are isoelectronic.

(D) The energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{1.312 \times 10^6}{n^2} \ J \ mol^{-1}$.
For $n=1$,$E_1 = -1.312 \times 10^6 \ J \ mol^{-1}$.
For $n=2$,$E_2 = -\frac{1.312 \times 10^6}{2^2} = -\frac{1.312 \times 10^6}{4} = -0.328 \times 10^6 \ J \ mol^{-1}$.
The energy required to excite the electron from $n=1$ to $n=2$ is $\Delta E = E_2 - E_1$.
$\Delta E = (-0.328 \times 10^6) - (-1.312 \times 10^6) \ J \ mol^{-1}$.
$\Delta E = (1.312 - 0.328) \times 10^6 \ J \ mol^{-1} = 0.984 \times 10^6 \ J \ mol^{-1} = 9.84 \times 10^5 \ J \ mol^{-1}$.

(B) In group $15$ hydrides,the basic character decreases on going down the group due to the increase in the size of the central atom and the dispersion of the lone pair over a larger volume.
Thus,the correct order of basicity is $NH_3 > PH_3 > AsH_3 > SbH_3$.
Therefore,the arrangement $NH_3 < PH_3 < AsH_3 < SbH_3$ is incorrect.

(A) The bond order of a diatomic species depends on the total number of electrons in the molecule or ion.
Species with the same number of electrons are isoelectronic and generally possess the same bond order.
Calculating the total number of electrons for each species:
$CN^{-}$: $6 + 7 + 1 = 14$ electrons.
$NO^{+}$: $7 + 8 - 1 = 14$ electrons.
$CN^{+}$: $6 + 7 - 1 = 12$ electrons.
$O_2^{-}$: $8 + 8 + 1 = 17$ electrons.
Since $CN^{-}$ and $NO^{+}$ both have $14$ electrons,they have the same bond order (which is $3$).

(B) $BF_3$ is a Lewis acid due to the incomplete octet of Boron.
In $BF_3$,the Boron atom has a vacant $2p$-orbital,and the Fluorine atom has lone pairs in its $2p$-orbital.
This allows for $p\pi - p\pi$ back-bonding from the Fluorine to the Boron,which increases the bond order and strengthens the $B-F$ bond.
In $CF_4$,the Carbon atom has a complete octet and no vacant orbitals,so no such back-bonding is possible.
Therefore,the $B-F$ bond in $BF_3$ is stronger than the $C-F$ bond in $CF_4$ due to this back-bonding interaction.

(D) According to $MO$ theory,bond order is inversely proportional to bond length.
$O_2^{2+}$ has a bond order of $3.0$.
$O_2^+$ has a bond order of $2.5$.
$O_2^-$ has a bond order of $1.5$.
$O_2^{2-}$ has a bond order of $1.0$.
Since $O_2^{2+}$ has the highest bond order,it has the shortest bond length.

(B) The total energy change involved in the conversion of $\frac{1}{2} Cl_{2(g)}$ to $Cl^{-}_{(aq)}$ is given by the sum of the enthalpy changes of the individual steps:
$\Delta H = \frac{1}{2} \Delta_{diss} H_{Cl_2}^{\Theta} + \Delta_{eg} H_{Cl}^{\Theta} + \Delta_{Hyd} H_{Cl}^{\Theta}$
Substituting the given values:
$\Delta H = (\frac{1}{2} \times 240) + (-349) + (-381) \ kJ \ mol^{-1}$
$\Delta H = 120 - 349 - 381 \ kJ \ mol^{-1}$
$\Delta H = - 610 \ kJ \ mol^{-1}$

(C) For a reaction to be at equilibrium,$\Delta G = 0$.
Since $\Delta G = \Delta H - T \Delta S$,at equilibrium,$\Delta H = T \Delta S$.
For the reaction $\frac{1}{2} X_2 + \frac{3}{2} Y_2 \to XY_3$,$\Delta H = -30 \ kJ = -30000 \ J$.
Calculating $\Delta S$ for the reaction:
$\Delta S = S^{\circ}(XY_3) - [\frac{1}{2} S^{\circ}(X_2) + \frac{3}{2} S^{\circ}(Y_2)]$
$\Delta S = 50 - [\frac{1}{2} \times 60 + \frac{3}{2} \times 40] \ J \ K^{-1} \ mol^{-1}$
$\Delta S = 50 - [30 + 60] = -40 \ J \ K^{-1} \ mol^{-1}$.
At equilibrium,$T = \frac{\Delta H}{\Delta S} = \frac{-30000 \ J}{-40 \ J \ K^{-1}} = 750 \ K$.

(A) For reaction $X \rightleftharpoons 2Y$:
Initial moles: $a$,$0$
At equilibrium: $a(1 - \alpha)$,$2a\alpha$
Total moles $= a(1 + \alpha)$
$K_{p1} = \frac{(2a\alpha)^2}{a(1 - \alpha)} \times \left(\frac{P_{T1}}{a(1 + \alpha)}\right)^1 = \frac{4\alpha^2 P_{T1}}{1 - \alpha^2}$
For reaction $Z \rightleftharpoons P + Q$:
Initial moles: $b$,$0$,$0$
At equilibrium: $b(1 - \alpha)$,$b\alpha$,$b\alpha$
Total moles $= b(1 + \alpha)$
$K_{p2} = \frac{(b\alpha)(b\alpha)}{b(1 - \alpha)} \times \left(\frac{P_{T2}}{b(1 + \alpha)}\right)^1 = \frac{\alpha^2 P_{T2}}{1 - \alpha^2}$
Given $\frac{K_{p1}}{K_{p2}} = \frac{1}{9}$,we have:
$\frac{4\alpha^2 P_{T1} / (1 - \alpha^2)}{\alpha^2 P_{T2} / (1 - \alpha^2)} = \frac{1}{9}$
$\frac{4 P_{T1}}{P_{T2}} = \frac{1}{9}$
$\frac{P_{T1}}{P_{T2}} = \frac{1}{36}$

(C) Given reactions are:
$(i)$ $CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)}$ ; $K_1$
$(ii)$ $CH_{4(g)} + H_2O_{(g)} \rightleftharpoons CO_{(g)} + 3H_{2(g)}$ ; $K_2$
Adding reaction $(i)$ and $(ii)$:
$(CO_{(g)} + H_2O_{(g)}) + (CH_{4(g)} + H_2O_{(g)}) \rightleftharpoons (CO_{2(g)} + H_{2(g)}) + (CO_{(g)} + 3H_{2(g)})$
Canceling $CO_{(g)}$ from both sides:
$CH_{4(g)} + 2H_2O_{(g)} \rightleftharpoons CO_{2(g)} + 4H_{2(g)}$
This is reaction $(iii)$.
When reactions are added,their equilibrium constants are multiplied.
Therefore,$K_3 = K_1 \cdot K_2$.

(C) The acidic strength depends on the stability of the conjugate base formed after the loss of a proton $(H^{+})$.
$1. \, HSO_3F$ is a superacid,which is the strongest among the given species.
$2. \, H_3O^{+}$ is a strong acid $(pK_a \approx -1.7)$.
$3. \, HSO_4^-$ is a moderately strong acid $(pK_a \approx 1.99)$.
$4. \, HCO_3^-$ is a weak acid $(pK_a \approx 10.3)$.
Therefore,the correct order of acidic strength is:
$HCO_3^- < HSO_4^- < H_3O^{+} < HSO_3F$
Which corresponds to the sequence: $i < iii < ii < iv$.

(C) The salt $BA$ is formed from a weak acid $(HA)$ and a weak base $(BOH)$.
In aqueous solution,the salt undergoes hydrolysis as follows:
$BA + H_2O \rightleftharpoons BOH + HA$
The $pH$ of a salt of a weak acid and a weak base is given by the formula:
$pH = \frac{1}{2} pK_w + \frac{1}{2} pK_a - \frac{1}{2} pK_b$
Given $pK_w = 14.00$,$pK_a = 4.80$,and $pK_b = 4.78$.
Substituting these values into the formula:
$pH = \frac{1}{2} (14.00 + 4.80 - 4.78)$
$pH = \frac{1}{2} (14.02)$
$pH = 7.01$

(D) On the industrial scale,hydrogen is prepared from water gas using the water-gas shift reaction:
$CO + H_2O(g) \xrightarrow{\text{catalyst}} CO_2 + H_2$
The resulting $CO_2$ is removed by scrubbing the mixture with an alkali solution (e.g.,$NaOH$ or $K_2CO_3$):
$CO_2 + 2NaOH \longrightarrow Na_2CO_3 + H_2O$
Thus,$CO$ is oxidized to $CO_2$ using steam in the presence of a catalyst,and the $CO_2$ is subsequently removed by absorption in an alkali.

(B) To determine the absolute configuration,we assign priorities to the groups attached to each chiral center using the Cahn-Ingold-Prelog $(CIP)$ rules.
For the first chiral center (left): The groups are $-OH$ $(1)$,$-COOH$ $(2)$,$-CH(OH)COOH$ $(3)$,and $-H$ $(4)$. Since $-H$ is on a dashed bond (pointing away),the sequence $1$ $\rightarrow 2$ $\rightarrow 3$ is clockwise,which corresponds to the $R$ configuration.
For the second chiral center (right): The groups are $-OH$ $(1)$,$-COOH$ $(2)$,$-CH(OH)COOH$ $(3)$,and $-H$ $(4)$. Since $-H$ is on a dashed bond (pointing away),the sequence $1$ $\rightarrow 2$ $\rightarrow 3$ is clockwise,which corresponds to the $R$ configuration.
Thus,the absolute configuration is $R, R$.

(B) The stability of the $\sigma-$complex (arenium ion) depends on the substituents present on the benzene ring.
Nitrobenzene contains a $-NO_2$ group,which is a strong electron-withdrawing group ($-I$ and $-M$ effect). This group destabilizes the positively charged $\sigma-$complex by withdrawing electron density.
In contrast,benzene has no such electron-withdrawing substituent.
Therefore,the $\sigma-$complex formed from benzene is more stable and has lower energy compared to the $\sigma-$complexes formed from nitrobenzene (at ortho,meta,or para positions).

(A) In the $IUPAC$ system of nomenclature,the priority order for functional groups is determined by specific rules.
According to the $IUPAC$ priority table,the carboxylic acid group $(-COOH)$ has the highest priority among the given groups.
The sulfonic acid group $(-SO_3H)$ follows,then the amide group $(-CONH_2)$,and finally the aldehyde group $(-CHO)$.
Therefore,the correct decreasing order is: $-COOH > -SO_3H > -CONH_2 > -CHO$.

(D) The reaction is an ozonolysis of but$-2-$ene $(CH_3-CH=CH-CH_3)$.
Step $1$: Reaction with $O_3$ forms an ozonide intermediate $(A)$.
Step $2$: Reductive cleavage of the ozonide with $Zn$ and $H_2O$ breaks the $C=C$ bond.
$CH_3-CH=CH-CH_3 + O_3$ $\rightarrow \text{Ozonide}$ $\xrightarrow{Zn, H_2O} 2CH_3CHO$
Thus,the product $B$ is acetaldehyde $(CH_3CHO)$.

(B) Terminal alkynes contain an acidic hydrogen atom attached to an $sp$-hybridized carbon atom.
These acidic protons can be removed by strong bases like sodium metal in liquid ammonia ($NaNH_2$ is formed in situ or sodium acts as a reducing agent for the proton) to form sodium acetylides.
Among the given options,$CH_3CH_2C \equiv CH$ is a terminal alkyne,whereas the others are internal alkynes.
Therefore,$CH_3CH_2C \equiv CH$ reacts with sodium in liquid ammonia.

(D) $CH_3MgX$ is a Grignard reagent,which acts as a strong base.
Terminal alkynes like $CH_3-C \equiv CH$ contain an acidic hydrogen atom attached to the $sp$-hybridized carbon.
The Grignard reagent abstracts this acidic proton to form an alkane.
The reaction is: $CH_3MgX + CH_3-C \equiv CH \rightarrow CH_4 + CH_3-C \equiv C-MgX$.
Therefore,the product formed is methane $(CH_4)$.

(C) The protective power of a colloid is inversely proportional to its gold number.
Given gold numbers are:
$A = 0.50$
$B = 0.01$
$C = 0.10$
$D = 0.005$
Arranging the gold numbers in increasing order: $0.005 (D) < 0.01 (B) < 0.10 (C) < 0.50 (A)$.
Since protective power is inversely proportional to the gold number,the order of protective power is $A < C < B < D$.

(B) For a general reaction $aA \rightarrow bB$,the rate of reaction is given by:
Rate $= -\frac{1}{a} \frac{d[A]}{dt} = \frac{1}{b} \frac{d[B]}{dt}$
Given the reaction $\frac{1}{2} A \rightarrow 2B$,we have $a = \frac{1}{2}$ and $b = 2$.
Substituting these values into the rate expression:
$-\frac{1}{1/2} \frac{d[A]}{dt} = \frac{1}{2} \frac{d[B]}{dt}$
$-2 \frac{d[A]}{dt} = \frac{1}{2} \frac{d[B]}{dt}$
Dividing both sides by $2$:
$-\frac{d[A]}{dt} = \frac{1}{4} \frac{d[B]}{dt}$

(C) Boric acid is a Lewis acid,not a protonic acid.
Beryllium typically exhibits a coordination number of $4$.
$B_2H_6 \cdot 2NH_3$ is known as 'inorganic benzene' is incorrect; borazine $(B_3N_3H_6)$ is known as inorganic benzene.
Both $BeCl_2$ and $AlCl_3$ exist as polymeric structures with bridged chloride atoms in the solid state.

(D) In the complex $[E(en)_2(C_2O_4)]NO_2$,$en$ (ethylenediamine) and $C_2O_4^{2-}$ (oxalate) are both bidentate ligands.
Coordination number = $(2 \times 2) + (1 \times 2) = 6$.
Let the oxidation state of $E$ be $x$.
The complex ion is $[E(en)_2(C_2O_4)]^+$,as the counter ion $NO_2^-$ has a $-1$ charge.
$x + 2(0) + 1(-2) = +1$.
$x - 2 = +1 \Rightarrow x = +3$.
Thus,the coordination number is $6$ and the oxidation state is $+3$.

(C) The ozone layer primarily absorbs ultraviolet $(UV)$ radiation from the sun,preventing it from reaching the earth's surface. It does not block infrared radiation. Therefore,statement $C$ is incorrect. All other statements are scientifically accurate.

(B) The hydrolysis of substituted chlorosilanes leads to the formation of silicones.
$1$. $R_3SiCl$ gives a dimer on hydrolysis,which terminates the chain.
$2$. $R_2SiCl_2$ gives linear silicone polymers.
$3$. $RSiCl_3$ has three chlorine atoms attached to the silicon atom. Upon hydrolysis,it forms $RSi(OH)_3$,which has three hydroxyl groups available for condensation. This allows the polymer to grow in three dimensions,resulting in a cross-linked silicone polymer.
Therefore,$RSiCl_3$ is the correct answer.

(B) The reaction is $\frac{1}{2} X_2 + \frac{3}{2} Y_2 \to XY_3$.
First,calculate the change in entropy $\Delta S$ for the reaction:
$\Delta S = S^{\circ}(XY_3) - [\frac{1}{2} S^{\circ}(X_2) + \frac{3}{2} S^{\circ}(Y_2)]$
$\Delta S = 50 - [\frac{1}{2} \times 60 + \frac{3}{2} \times 40] = 50 - [30 + 60] = 50 - 90 = -40 \, J \, K^{-1} \, mol^{-1}$.
At equilibrium,the Gibbs free energy change $\Delta G = 0$.
Since $\Delta G = \Delta H - T \Delta S$,at equilibrium,$\Delta H = T \Delta S$.
Given $\Delta H = -30 \, kJ = -30000 \, J$.
$T = \frac{\Delta H}{\Delta S} = \frac{-30000 \, J}{-40 \, J \, K^{-1}} = 750 \, K$.

(D) The formula for the mutual inductance $M$ of two coaxial solenoids is given by $M = \frac{\mu_0 N_1 N_2 A}{\ell}$.
Given values:
$N_1 = 300$
$N_2 = 400$
$A = 10 \, cm^2 = 10 \times 10^{-4} \, m^2 = 10^{-3} \, m^2$
$\ell = 20 \, cm = 0.2 \, m = 2 \times 10^{-1} \, m$
$\mu_0 = 4\pi \times 10^{-7} \, T \cdot m \cdot A^{-1}$
Substituting these values into the formula:
$M = \frac{(4\pi \times 10^{-7}) \times 300 \times 400 \times 10^{-3}}{0.2}$
$M = \frac{4\pi \times 10^{-7} \times 120000 \times 10^{-3}}{0.2}$
$M = \frac{4\pi \times 10^{-7} \times 120}{0.2}$
$M = 4\pi \times 10^{-7} \times 600$
$M = 2400\pi \times 10^{-7} \, H$
$M = 2.4\pi \times 10^{-4} \, H$.

(A) The terminal speed $(v_t)$ is reached when the net force acting on the ball is zero.
The forces acting on the ball are:
$1$. Weight of the ball acting downwards: $W = V \rho_1 g$
$2$. Buoyant force acting upwards: $F_B = V \rho_2 g$
$3$. Viscous force acting upwards: $F_v = kv_t^2$
At terminal speed,the forces are balanced:
$W = F_B + F_v$
$V \rho_1 g = V \rho_2 g + kv_t^2$
Rearranging the equation to solve for $v_t$:
$kv_t^2 = V \rho_1 g - V \rho_2 g$
$kv_t^2 = Vg(\rho_1 - \rho_2)$
$v_t^2 = \frac{Vg(\rho_1 - \rho_2)}{k}$
$v_t = \sqrt{\frac{Vg(\rho_1 - \rho_2)}{k}}$

(A) The pitch of the screw gauge is the distance moved by the spindle in one full rotation. Since $2$ full turns cover $1\,mm$,the pitch is $\text{pitch} = \frac{1\,mm}{2} = 0.5\,mm$.
The least count $(LC)$ is given by $\frac{\text{pitch}}{\text{total divisions}} = \frac{0.5\,mm}{50} = 0.01\,mm$.
The observed diameter is calculated as $\text{Main Scale Reading} + (\text{Circular Scale Division} \times LC)$.
$\text{Observed diameter} = 3\,mm + (35 \times 0.01\,mm) = 3.35\,mm$.
The actual diameter is given by $\text{Actual diameter} = \text{Observed diameter} - \text{Zero error}$.
$\text{Actual diameter} = 3.35\,mm - (+0.03\,mm) = 3.32\,mm$.

(A) At terminal speed,the net force acting on the ball is zero.
The forces acting on the ball are:
$1$. Weight of the ball acting downwards: $W = V \rho_1 g$
$2$. Buoyant force (upthrust) acting upwards: $F_B = V \rho_2 g$
$3$. Viscous force acting upwards: $F_{viscous} = kv_t^2$
For equilibrium at terminal velocity $v_t$:
$W - F_B = F_{viscous}$
$V \rho_1 g - V \rho_2 g = kv_t^2$
$Vg(\rho_1 - \rho_2) = kv_t^2$
$v_t^2 = \frac{Vg(\rho_1 - \rho_2)}{k}$
$v_t = \sqrt{\frac{Vg(\rho_1 - \rho_2)}{k}}$

(A) In $BF_3$,the boron atom is electron-deficient and has an empty $2p$ orbital.
Fluorine atoms have lone pairs in their $2p$ orbitals.
This allows for $p\pi-p\pi$ back-bonding from $F$ to $B$,which gives the $B-F$ bond partial double bond character.
This additional bonding interaction increases the bond dissociation energy.
In $CF_4$,carbon has a complete octet and cannot accept electrons via back-bonding,so no such $p\pi-p\pi$ interaction exists.

(A) $1$. When the multimeter is connected between $P$ and $Q$,no conduction is observed. This implies that $P$ and $Q$ are of the same type (both $n$-type or both $p$-type). Since the transistor is working,$P$ and $Q$ must be the collector and emitter,and $R$ must be the base.
$2$. When the negative terminal of the multimeter is connected to $R$ (base) and the positive terminal is connected to $P$ or $Q$ (collector/emitter),conduction is observed. This means the base-collector and base-emitter junctions are forward-biased when the base is at a lower potential than the collector/emitter.
$3$. In an $NPN$ transistor,the base is $p$-type and the collector/emitter are $n$-type. Forward biasing occurs when the $p$-type base is connected to the positive terminal and $n$-type regions to the negative terminal. However,the problem states the negative terminal is at $R$. This indicates the base is $n$-type and the collector/emitter are $p$-type,which corresponds to a $PNP$ transistor.
$4$. Re-evaluating the standard multimeter test: If the negative terminal is at $R$ and conduction occurs,$R$ must be $p$-type. Thus,$R$ is the base of a $PNP$ transistor.

(A) The forces acting on the ball are the gravitational force,the buoyant force,and the viscous force. When the ball reaches its terminal speed,it is in dynamic equilibrium. Let the terminal speed of the ball be $v_T$.
The forces acting downwards are the weight of the ball: $W = V\rho_1 g$.
The forces acting upwards are the buoyant force: $F_B = V\rho_2 g$ and the viscous force: $F_v = kv_T^2$.
In dynamic equilibrium,the net force is zero:
$V\rho_1 g = V\rho_2 g + kv_T^2$
$kv_T^2 = Vg(\rho_1 - \rho_2)$
$v_T^2 = \frac{Vg(\rho_1 - \rho_2)}{k}$
$v_T = \sqrt{\frac{Vg(\rho_1 - \rho_2)}{k}}$

(B) Given: $m_1 = 0.50 \ kg$,$u_1 = 2.00 \ ms^{-1}$,$m_2 = 1.00 \ kg$,$u_2 = 0 \ ms^{-1}$.
Since the collision is perfectly inelastic,the bodies move together after the collision.
The loss in kinetic energy during a perfectly inelastic collision is given by the formula:
$\Delta K = \frac{m_1 m_2}{2(m_1 + m_2)} (u_1 - u_2)^2$
Substituting the values:
$\Delta K = \frac{0.50 \times 1.00}{2(0.50 + 1.00)} (2.00 - 0)^2$
$\Delta K = \frac{0.50}{2(1.50)} (4)$
$\Delta K = \frac{0.50}{3.00} \times 4 = \frac{2}{3} \approx 0.67 \ J$.

(A) At terminal speed $(v_t)$,the net force acting on the ball is zero. The forces acting on the ball are its weight $(W)$ acting downwards,and the buoyant force $(B)$ and viscous force $(F_v)$ acting upwards.
The condition for terminal speed is:
$W = B + F_v$
Substituting the expressions for the forces:
$V \rho_1 g = V \rho_2 g + k v_t^2$
Rearranging the equation to solve for $v_t^2$:
$k v_t^2 = V \rho_1 g - V \rho_2 g$
$k v_t^2 = Vg(\rho_1 - \rho_2)$
Solving for $v_t$:
$v_t^2 = \frac{Vg(\rho_1 - \rho_2)}{k}$
$v_t = \sqrt{\frac{Vg(\rho_1 - \rho_2)}{k}}$
Therefore,the terminal speed of the ball is $\sqrt{\frac{Vg(\rho_1 - \rho_2)}{k}}$.

(D) The reaction is an ozonolysis of but$-2-$ene.
$1$. The alkene $CH_3-CH=CH-CH_3$ reacts with $O_3$ to form an ozonide intermediate $(A)$.
$2$. The ozonide undergoes reductive cleavage with $Zn/H_2O$ to yield two molecules of acetaldehyde $(CH_3CHO)$.
Therefore,the compound $B$ is $CH_3CHO$.

(A) $1$. Nitration of $Toluene$ $(C_6H_5CH_3)$ with $HNO_3/H_2SO_4$ gives a mixture of $o-$ and $p-$ nitrotoluenes.
$2$. Reduction of these with $Sn/HCl$ converts the $-NO_2$ group into $-NH_2$ group,yielding $o-$ and $p-$ toluidines.
$3$. Diazotisation of these amines with $NaNO_2/HCl$ at $0-5 \ ^{\circ}C$ forms the corresponding diazonium salts.
$4$. Heating these diazonium salts with cuprous bromide $(CuBr)$ (Sandmeyer reaction) replaces the diazonium group with a bromine atom.
$5$. Thus,the final product is a mixture of $o-$ bromotoluene and $p-$ bromotoluene.

(A) Let the number of atoms of element $Y$ in the $ccp$ lattice be $n = 4$.
The number of tetrahedral voids is $2n = 2 \times 4 = 8$.
Element $X$ occupies $2/3$ of the tetrahedral voids,so the number of $X$ atoms $= 8 \times \frac{2}{3} = \frac{16}{3}$.
The ratio of $X:Y$ is $\frac{16}{3} : 4$.
Multiplying by $3$,we get the ratio $16:12$,which simplifies to $4:3$.
Therefore,the formula of the compound is $X_4Y_3$.

(A) The relative lowering of vapour pressure is given by Raoult's Law: $\frac{P^o - P_S}{P^o} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$ for dilute solutions.
$1.$ Calculate moles of glucose $(n_2)$: $n_2 = \frac{18 \, g}{180 \, g/mol} = 0.1 \, mol$.
$2.$ Calculate moles of water $(n_1)$: $n_1 = \frac{178.2 \, g}{18 \, g/mol} = 9.9 \, mol$.
$3.$ Using the formula $\frac{P^o - P_S}{P^o} = \frac{n_2}{n_1 + n_2}$:
$\frac{17.5 - P_S}{17.5} = \frac{0.1}{9.9 + 0.1} = \frac{0.1}{10} = 0.01$.
$4.$ Solve for $P_S$:
$17.5 - P_S = 17.5 \times 0.01 = 0.175$.
$P_S = 17.5 - 0.175 = 17.325 \, mm \, Hg$.

(D) The cell reaction is: $2Cr(s) + 3Fe^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + 3Fe(s)$.
$n = 6$ (number of electrons transferred).
$E^o_{cell} = E^o_{cathode} - E^o_{anode} = -0.42 - (-0.72) = 0.30 \, V$.
Using the Nernst equation: $E_{cell} = E^o_{cell} - \frac{0.059}{n} \log \frac{[Cr^{3+}]^2}{[Fe^{2+}]^3}$.
$E_{cell} = 0.30 - \frac{0.059}{6} \log \frac{(0.1)^2}{(0.01)^3}$.
$E_{cell} = 0.30 - \frac{0.059}{6} \log \frac{10^{-2}}{10^{-6}} = 0.30 - \frac{0.059}{6} \log(10^4)$.
$E_{cell} = 0.30 - \frac{0.059}{6} \times 4 = 0.30 - 0.0393 = 0.2607 \, V \approx 0.26 \, V$.

(D) The reduction of metal sulphides by carbon is generally not spontaneous because the $\Delta G$ value for the reaction $MS + C \rightarrow M + CS_2$ is positive.
In contrast,the reduction of metal oxides by carbon is spontaneous because the $\Delta G$ value for $MO + C \rightarrow M + CO_2$ is negative.
This difference arises because $CO_2$ is thermodynamically much more stable than $CS_2$.
Therefore,the stability of the metal sulphide compared to the oxide and the relative stability of $CO_2$ versus $CS_2$ are significant factors.
The volatility of $CO_2$ compared to $CS_2$ is not a thermodynamic factor that determines the feasibility of the reduction process.

(B) The main reason for actinoids exhibiting a larger number of oxidation states compared to lanthanoids is the smaller energy difference between the $5f$ and $6d$ orbitals compared to the energy difference between the $4f$ and $5d$ orbitals.
Because the $5f$,$6d$,and $7s$ orbitals in actinoids have comparable energies,electrons from these orbitals can participate in bonding,leading to a wider range of oxidation states.

(A) The magnitude of crystal field splitting energy $(\Delta_o)$ depends on the strength of the ligand field.
According to the spectrochemical series,the strength of ligands increases in the order: $I^- < Br^- < Cl^- < F^- < OH^- < C_2O_4^{2-} < H_2O < NH_3 < en < NO_2^- < CN^-$.
Among the given ligands,$CN^-$ is the strongest field ligand.
Therefore,the complex containing $CN^-$ will have the highest crystal field splitting energy.
The order of $\Delta_o$ for the given complexes is: $[Co(H_2O)_6]^{3+} < [Co(C_2O_4)_3]^{3-} < [Co(NH_3)_6]^{3+} < [Co(CN)_6]^{3-}$.
Thus,the correct option is $A$.

(A) The titration of oxalic acid with $KMnO_4$ in the presence of $HCl$ gives an unsatisfactory result because $KMnO_4$ is a strong oxidizing agent and it oxidizes $HCl$ to $Cl_2$ gas along with the oxidation of oxalic acid.
This leads to an error in the titration as the volume of $KMnO_4$ consumed is higher than that required for the oxidation of oxalic acid alone.

(D) Nucleophilic substitution bimolecular $S_{N}2$ reaction prefers a less sterically hindered site for the nucleophilic attack.
Lesser the steric hindrance,faster is the $S_{N}2$ reaction.
The order of reactivity for $S_{N}2$ reactions is: $\text{Methyl halide} > 1^{\circ} > 2^{\circ} > 3^{\circ}$.
$S_{N}2$ reactions involve complete inversion of configuration (Walden inversion).
Among the given options,$CH_3Cl$ is a methyl halide,which is the least sterically hindered and thus undergoes $S_{N}2$ reaction with complete stereochemical inversion.

(B) When phenol reacts with concentrated $H_2SO_4$,it undergoes sulphonation to form phenol-$2, 4-$disulphonic acid.
Subsequently,when this product reacts with concentrated $HNO_3$,the sulphonic acid groups are replaced by nitro groups to yield $2, 4, 6-$trinitrophenol,which is commonly known as picric acid.
Therefore,option $B$ is correct.

(D) Bakelite is a thermosetting polymer formed by the condensation polymerization of phenol and formaldehyde $(HCHO)$.
In the presence of an acid or base catalyst,phenol reacts with formaldehyde to form $o$- and $p$-hydroxymethylphenol intermediates.
These intermediates undergo further polymerization to form a linear polymer called Novolac.
Upon heating with more formaldehyde,Novolac undergoes cross-linking to form the infusible,hard,and cross-linked polymer known as Bakelite.

(C) $\alpha-D-(+)$-glucose and $\beta-D-(+)$-glucose differ in configuration only at the $C-1$ carbon atom,which is the anomeric carbon.
Such diastereomers that differ in configuration specifically at the $C-1$ atom are known as anomers.

(B) The correct order for basic strength of group $15$ hydrides is $NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3$.
As the size of the central atom increases,the electron density on the central atom decreases,making it less available for donation,thus decreasing basic strength.
Therefore,the sequence $NH_3 < PH_3 < AsH_3 < SbH_3$ is incorrect as it represents increasing basic strength,whereas it should be decreasing.

(A) The protective power of a colloid is inversely proportional to its gold number.
Lower gold number indicates higher protective power.
Given gold numbers: $A = 0.50$,$B = 0.01$,$C = 0.10$,$D = 0.005$.
Arranging in increasing order of gold number: $D (0.005) < B (0.01) < C (0.10) < A (0.50)$.
Therefore,the order of protective power is $A < C < B < D$.
Thus,the correct option is $A$ or $C$ (as both represent the same sequence).

(D) In the given complex,there are two $en$ ligands and one $C_2O_4$ ligand. Both are bidentate ligands.
Coordination number = $(2 \times 2) + (1 \times 2) = 4 + 2 = 6$.
Let the oxidation state of $E$ be $x$.
The complex is $[E(en)_2(C_2O_4)]NO_2$. Since $NO_2$ is a counter ion with a charge of $-1$,the complex cation $[E(en)_2(C_2O_4)]^+$ has a charge of $+1$.
$x + 2(0) + 1(-2) = +1$
$x - 2 = +1$
$x = +3$.
Therefore,the coordination number is $6$ and the oxidation state is $+3$. Thus,option $(d)$ is correct.

(B) For a general reaction $aA \rightarrow bB$,the rate of reaction is given by:
Rate $= -\frac{1}{a} \frac{d[A]}{dt} = \frac{1}{b} \frac{d[B]}{dt}$
Given the reaction $\frac{1}{2} A \rightarrow 2 B$,we have $a = \frac{1}{2}$ and $b = 2$.
Substituting these values:
$-\frac{1}{1/2} \frac{d[A]}{dt} = \frac{1}{2} \frac{d[B]}{dt}$
$-2 \frac{d[A]}{dt} = \frac{1}{2} \frac{d[B]}{dt}$
Multiplying both sides by $\frac{1}{2}$:
$-\frac{d[A]}{dt} = \frac{1}{4} \frac{d[B]}{dt}$