AIEEE 2012 Chemistry Question Paper with Answer and Solution

(A) Let the mass of the insect be $m$. The forces acting on the insect are its weight $mg$ (acting vertically downwards),the normal reaction $N$ (acting radially outwards),and the frictional force $f$ (acting tangentially upwards along the surface).
Resolving the weight $mg$ into two components: $mg \cos \alpha$ along the radial direction and $mg \sin \alpha$ along the tangential direction.
Since the insect crawls very slowly,it is in equilibrium at any point.
For the radial direction: $N = mg \cos \alpha$.
For the tangential direction: $f = mg \sin \alpha$.
For the insect to not slip,the frictional force must satisfy $f \le \mu N$.
At the maximum angle $\alpha$,the limiting friction condition is reached: $f = \mu N$.
Substituting the expressions for $f$ and $N$: $mg \sin \alpha = \mu (mg \cos \alpha)$.
This simplifies to $\tan \alpha = \mu$.
Given $\mu = 1/3$,we have $\tan \alpha = 1/3$.
Therefore,$\cot \alpha = 1/\tan \alpha = 3$.

(B) The velocity of efflux for a hole at depth $h$ is given by Torricelli's law: $v = \sqrt{2gh}$.
The rate of flow of water (volume per second) is given by $Q = A \cdot v$,where $A$ is the area of the hole.
For the square hole at depth $y$:
Area $A_1 = L^2$
Velocity $v_1 = \sqrt{2gy}$
$Q_1 = L^2 \sqrt{2gy}$
For the circular hole at depth $4y$:
Area $A_2 = \pi R^2$
Velocity $v_2 = \sqrt{2g(4y)} = 2\sqrt{2gy}$
$Q_2 = \pi R^2 (2\sqrt{2gy})$
Given that $Q_1 = Q_2$:
$L^2 \sqrt{2gy} = \pi R^2 (2\sqrt{2gy})$
$L^2 = 2\pi R^2$
$R^2 = \frac{L^2}{2\pi}$
$R = \frac{L}{\sqrt{2\pi}}$

(B) For the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$,the equilibrium constant is $K_c = \frac{[NO]^2}{[N_2][O_2]} = 4 \times 10^{-4}$.
For the reaction $NO_{(g)} \rightleftharpoons \frac{1}{2}N_{2(g)} + \frac{1}{2}O_{2(g)}$,the equilibrium constant $K'_c$ is given by $K'_c = \frac{[N_2]^{1/2}[O_2]^{1/2}}{[NO]}$.
Comparing the two expressions,we see that $K'_c = \frac{1}{\sqrt{K_c}}$.
Substituting the value of $K_c$,we get $K'_c = \frac{1}{\sqrt{4 \times 10^{-4}}} = \frac{1}{2 \times 10^{-2}} = \frac{100}{2} = 50$.

(B) The biconditional statement $p \Leftrightarrow q$ is defined as the conjunction of the two conditional statements $p \Rightarrow q$ and $q \Rightarrow p$.
Therefore,the logically equivalent proposition is $(p$ $\Rightarrow q) \wedge (q$ $\Rightarrow p)$.

(C) For an open vessel,the pressure $P$ and volume $V$ remain constant. According to the ideal gas law,$PV = nRT$,which implies $n_1 T_1 = n_2 T_2$.
Let the initial number of moles be $n_1 = n$.
Since $2/5$ of the air is expelled,the remaining moles $n_2 = n - (2/5)n = 3/5 n$.
The initial temperature $T_1 = 300 \ K$.
Using the relation $n_1 T_1 = n_2 T_2$:
$n \times 300 = (3/5)n \times T_2$.
$300 = (3/5)T_2$.
$T_2 = 300 \times (5/3) = 500 \ K$.

(A) Given: Molality $(m) = 1.00 \, m$,$\Delta T_f = 0 - (-1.91) = 1.91 \, K$,$K_f = 1.86 \, K \, kg \, mol^{-1}$.
Using the formula: $\Delta T_f = i \times K_f \times m$.
$1.91 = i \times 1.86 \times 1.00 \Rightarrow i = \frac{1.91}{1.86} \approx 1.02688$.
For dissociation $HF \rightleftharpoons H^+ + F^-$,the number of ions $(n) = 2$.
The degree of dissociation $(\alpha)$ is given by $\alpha = \frac{i - 1}{n - 1}$.
$\alpha = \frac{1.02688 - 1}{2 - 1} = 0.02688$.
Percentage of dissociation $= 0.02688 \times 100 \approx 2.7 \%$.

(A) The shift in the image position due to the glass plate is given by $S = (1 - \frac{1}{\mu}) t = (1 - \frac{1}{1.5}) \times 1 \; cm = \frac{1}{3} \; cm$.
First,calculate the focal length $f$ of the lens using the initial conditions: $u = -240 \; cm$ and $v = 12 \; cm$.
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{12} - \frac{1}{-240} = \frac{20 + 1}{240} = \frac{21}{240} \; cm^{-1}$.
When the glass plate is introduced,the image must be formed at a distance $v' = 12 - \frac{1}{3} = \frac{35}{3} \; cm$ from the lens so that the plate shifts it back to the film.
Using the lens formula again: $\frac{1}{u'} = \frac{1}{v'} - \frac{1}{f} = \frac{3}{35} - \frac{21}{240} = \frac{3}{35} - \frac{7}{80}$.
$\frac{1}{u'} = \frac{48 - 49}{560} = -\frac{1}{560} \; cm^{-1}$.
Thus,$u' = -560 \; cm = -5.6 \; m$.
The shift required for the object is $|u'| - |u| = 5.6 \; m - 2.4 \; m = 3.2 \; m$.

(B) Given: Power $P = 10 \ kW = 10^4 \ W$,Wavelength $\lambda = 500 \ m$.
Energy of one photon is $E = \frac{hc}{\lambda}$.
The number of photons emitted per second $n$ is given by $n = \frac{P}{E} = \frac{P \lambda}{hc}$.
Substituting the values: $h = 6.63 \times 10^{-34} \ J \cdot s$,$c = 3 \times 10^8 \ m/s$.
$n = \frac{10^4 \times 500}{6.63 \times 10^{-34} \times 3 \times 10^8} = \frac{5 \times 10^6}{19.89 \times 10^{-26}} \approx 0.251 \times 10^{32} = 2.51 \times 10^{31}$.
The order of magnitude is $10^{31}$,which is closest to $10^{30}$ among the given options.

(B) Zinc blende is $ZnS$.
First,$ZnS$ is converted to zinc oxide $(ZnO)$ by roasting: $2ZnS + 3O_2 \rightarrow 2ZnO + 2SO_2$.
Then,$ZnO$ is reduced to zinc metal using carbon (coke) at high temperature: $ZnO + C \rightarrow Zn + CO$.

(B) molecule exhibits a dipole moment if its net dipole moment is non-zero $(mu \neq 0)$.
$1$. trans-$2,3$-dichloro-$2$-butene: The two $C-Cl$ bonds are in opposite directions,canceling each other's dipole moments. Thus,$mu = 0$.
$2$. $1,2$-dichlorobenzene: The two $C-Cl$ bonds are at an angle of $60^{\circ}$ to each other,so their dipole moments do not cancel out. Thus,$mu \neq 0$.
$3$. $1,4$-dichlorobenzene: The two $C-Cl$ bonds are in opposite directions $(180^{\circ})$,canceling each other's dipole moments. Thus,$mu = 0$.
$4$. trans-$1,2$-dinitroethene: The two $C-NO_2$ groups are in opposite directions,canceling each other's dipole moments. Thus,$mu = 0$.
Therefore,$1,2$-dichlorobenzene exhibits a dipole moment.

(B) The molecular formula $C_4H_6$ corresponds to a degree of unsaturation (double bond equivalent) of $4 - (6/2) + 1 = 2$.
For cyclic structures,we consider rings with two double bonds,one triple bond,or two rings.
The possible cyclic isomers for $C_4H_6$ are:
$1$. Cyclobutene (one double bond,one ring)
$2$. $1-$Methylcyclopropene (one double bond,one ring)
$3$. $3-$Methylcyclopropene (one double bond,one ring)
$4$. Methylenecyclopropane (one double bond,one ring)
$5$. Bicyclo[$1.1$.$0$]butane (two rings,no double bonds)
Thus,there are $5$ possible cyclic structures.

(A) Optical isomerism is exhibited by complexes that lack a plane of symmetry or center of inversion (i.e.,they are chiral).
For octahedral complexes of the type $[M(AA)_2X_2]$,the $cis$-isomer is optically active because it lacks a plane of symmetry.
In the given options,$[Co(en)_2Cl_2]^+$ exists as $cis$ and $trans$ isomers.
The $cis$-isomer of $[Co(en)_2Cl_2]^+$ is non-superimposable on its mirror image and thus exhibits optical isomerism.
The other complexes listed do not satisfy the structural requirements for optical activity in this context.

(A) The slope of the chord joining $(0, 1)$ and $(2, 0)$ is $m = \frac{0-1}{2-0} = -\frac{1}{2}$.
Since the tangents are parallel to this chord,their slope is $m = -\frac{1}{2}$.
The equation of a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with slope $m$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$.
Here $a^2 = 4$,$b^2 = 1$,and $m = -\frac{1}{2}$.
So,$y = -\frac{1}{2}x \pm \sqrt{4(-\frac{1}{2})^2 + 1} = -\frac{1}{2}x \pm \sqrt{1+1} = -\frac{1}{2}x \pm \sqrt{2}$.
The points of contact for the tangent $y = mx + c$ are $(\frac{-a^2m}{c}, \frac{b^2}{c})$.
For $y = -\frac{1}{2}x + \sqrt{2}$,$P_1 = (\frac{-4(-1/2)}{\sqrt{2}}, \frac{1}{\sqrt{2}}) = (\sqrt{2}, \frac{1}{\sqrt{2}})$.
For $y = -\frac{1}{2}x - \sqrt{2}$,$P_2 = (\frac{-4(-1/2)}{-\sqrt{2}}, \frac{1}{-\sqrt{2}}) = (-\sqrt{2}, -\frac{1}{\sqrt{2}})$.
The distance between $P_1$ and $P_2$ is $\sqrt{(\sqrt{2} - (-\sqrt{2}))^2 + (\frac{1}{\sqrt{2}} - (-\frac{1}{\sqrt{2}}))^2} = \sqrt{(2\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{8 + 2} = \sqrt{10}$.

(D) The gold number is defined as the minimum amount of protective colloid in milligrams $(mg)$ required to prevent the coagulation of $10 \, mL$ of a standard gold sol when $1 \, mL$ of $10 \% \, NaCl$ solution is added to it.
Given mass of starch = $0.25 \, g = 250 \, mg$.
Since $250 \, mg$ of starch prevents the coagulation of $10 \, mL$ of gold sol,the gold number of starch is $250$.

(C) The half-equations of the reaction are:
$MnO_4^{-} \rightarrow Mn^{2+}$
$C_2O_4^{2-} \rightarrow CO_2$
The balanced half-equations are:
$(MnO_4^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_2O) \times 2$
$(C_2O_4^{2-} \rightarrow 2CO_2 + 2e^{-}) \times 5$
Equating the number of electrons,we get:
$2MnO_4^{-} + 16H^{+} + 10e^{-} \rightarrow 2Mn^{2+} + 8H_2O$
$5C_2O_4^{2-} \rightarrow 10CO_2 + 10e^{-}$
On adding both the equations,we get:
$2MnO_4^{-} + 5C_2O_4^{2-} + 16H^{+} \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$
Thus,the values of $x, y$ and $z$ are $2, 5$ and $16,$ respectively.

(A) To determine the energy order,we use the $(n+l)$ rule:
$A. n=4, l=1 \implies n+l = 4+1 = 5$ ($4p$ orbital)
$B. n=4, l=0 \implies n+l = 4+0 = 4$ ($4s$ orbital)
$C. n=3, l=2 \implies n+l = 3+2 = 5$ ($3d$ orbital)
$D. n=3, l=1 \implies n+l = 3+1 = 4$ ($3p$ orbital)
According to the $(n+l)$ rule,orbitals with lower $(n+l)$ values have lower energy.
If $(n+l)$ values are equal,the orbital with the lower $n$ value has lower energy.
Comparing the values:
$D (n+l=4, n=3) < B (n+l=4, n=4) < C (n+l=5, n=3) < A (n+l=5, n=4)$
Thus,the order of increasing energy is $D < B < C < A$.

(C) For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number,$Z$) increases.
All the given species,$S^{2-}, Cl^{-}, K^{+},$ and $Ca^{2+}$,contain $18$ electrons.
Their respective atomic numbers $(Z)$ are: $S = 16, Cl = 17, K = 19, Ca = 20$.
Since the ionic radius is inversely proportional to the nuclear charge for isoelectronic species,the increasing order of ionic radii is: $Ca^{2+} < K^{+} < Cl^{-} < S^{2-}$.

(C) To determine if species are isostructural,we check their hybridization and geometry:
$1$. $CO_3^{2-}$ and $NO_3^-$: Both have $sp^2$ hybridization and are triangular planar.
$2$. $PCl_4^+$ and $SiCl_4$: Both have $sp^3$ hybridization and are tetrahedral.
$3$. $PF_5$: $sp^3d$ hybridization,trigonal bipyramidal geometry.
$BrF_5$: $sp^3d^2$ hybridization,square pyramidal geometry.
Since their geometries differ,they are not isostructural.
$4$. $AlF_6^{3-}$ and $SF_6$: Both have $sp^3d^2$ hybridization and are octahedral.
Therefore,the correct option is $C$.

(C) The van der Waals equation for a real gas is $(P + \frac{a}{V^2})(V - b) = RT$.
At high pressure,the term $\frac{a}{V^2}$ is very small compared to $P$ and can be neglected.
Thus,the equation simplifies to $P(V - b) = RT$.
Expanding this,we get $PV - Pb = RT$.
Rearranging the terms,$PV = RT + Pb$.
Dividing both sides by $RT$,we get $\frac{PV}{RT} = 1 + \frac{Pb}{RT}$.
Since the compressibility factor $Z = \frac{PV}{RT}$,we have $Z = 1 + \frac{Pb}{RT}$.
Therefore,at high pressure,$Z > 1$.

(C) We know that for a spontaneous process,$\Delta G_{system} = -T \Delta S_{total}$,which makes option $A$ correct.
For an isothermal reversible expansion of an ideal gas,$w_{reversible} = -nRT \ln \frac{V_f}{V_i}$,which makes option $B$ correct.
The relationship between equilibrium constant $K$ and Gibbs free energy is $\Delta G^o = -RT \ln K$,which makes option $D$ correct.
Substituting $\Delta G^o = \Delta H^o - T \Delta S^o$ into the equation $\Delta G^o = -RT \ln K$,we get $-RT \ln K = \Delta H^o - T \Delta S^o$,which implies $\ln K = -\frac{\Delta H^o - T \Delta S^o}{RT}$.
Therefore,the expression in option $C$ is incorrect.

(D) For the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)}$,the equilibrium constant is $K_c = 4 \times 10^{-4}$.
The given reaction is $NO_{(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{1}{2} O_{2(g)}$.
This reaction is the reverse of the original reaction multiplied by a factor of $\frac{1}{2}$.
Therefore,the new equilibrium constant $K_c'$ is given by $K_c' = \frac{1}{\sqrt{K_c}}$.
$K_c' = \frac{1}{\sqrt{4 \times 10^{-4}}} = \frac{1}{2 \times 10^{-2}} = \frac{1}{0.02} = 50$.

(C) Given: $pH = 3$,Concentration $(C) = 0.1 \ M = 10^{-1} \ M$.
We know that $[H^{+}] = 10^{-pH} = 10^{-3} \ M$.
For a weak acid $HQ$,the degree of dissociation $(\alpha)$ is given by $\alpha = \frac{[H^{+}]}{C} = \frac{10^{-3}}{10^{-1}} = 10^{-2}$.
The ionization constant $K_a$ is given by the formula $K_a = C \alpha^{2}$.
Substituting the values: $K_a = (0.1) \times (10^{-2})^{2} = 10^{-1} \times 10^{-4} = 10^{-5}$.

(D) Very pure hydrogen $(99.9 \%)$ is obtained by the reaction of water with metal hydrides such as sodium hydride $(NaH)$.
The chemical equation is:
$NaH(s) + H_2O(l) \rightarrow NaOH(aq) + H_2(g)$

(B) The molecular mass of the hydrocarbon is $72 \ u$. The general formula for alkanes is $C_nH_{2n+2}$.
$12n + 1(2n+2) = 72 \implies 14n = 70 \implies n = 5$.
Thus,the hydrocarbon is pentane $(C_5H_{12})$.
Among the isomers of pentane (n-pentane,isopentane,and neopentane),neopentane $(2,2-\text{dimethylpropane})$ has all $12$ hydrogen atoms equivalent.
Therefore,upon mono-halogenation,it yields only one mono-substituted alkyl halide.
$(CH_3)_4C + Cl_2 \xrightarrow{h\nu} (CH_3)_3C-CH_2Cl + HCl$.

(B) The reduction of internal alkynes to $trans-$alkenes is achieved using alkali metals (like $Li$ or $Na$) in liquid ammonia $(NH_3)$.
This is known as Birch reduction or dissolving metal reduction.
$CH_3-C \equiv C-CH_2-CH_2-CH_3 \xrightarrow{Li/NH_3} \text{trans-}2\text{-hexene}$.
$Pt/H_2$ and $Pd/BaSO_4$ (Lindlar catalyst) typically result in the formation of $cis-$alkenes or complete reduction to alkanes.

(C) In the series $NCl_3, PCl_3, AsCl_3$,and $SbCl_3$,all central atoms belong to Group $15$ and possess one lone pair of electrons.
As we move down the group from $N$ to $Sb$,the electronegativity of the central atom decreases,and its atomic size increases.
According to $VSEPR$ theory,as the size of the central atom increases,the bond pairs are located further away from the central atom.
This leads to a reduction in the repulsion between the bond pairs,which results in a decrease in the bond angle.
Therefore,$SbCl_3$ has the smallest bond angle among the given molecules.

(D) According to Fajan's rule,the polarizing power of a cation increases with an increase in its charge.
$Fe^{3+}$ has a higher charge and smaller size than $Fe^{2+}$,making it more polarizing and more covalent in character.
Consequently,$Fe^{3+}$ compounds are more easily hydrolysed than $Fe^{2+}$ compounds.
Therefore,the statement that ferrous compounds are more easily hydrolysed than ferric compounds is incorrect.

(C) $2$-methylbutane is $CH_3-CH(CH_3)-CH_2-CH_3$.
Monochlorination yields the following products:
$1.$ $Cl-CH_2-CH(CH_3)-CH_2-CH_3$ ($1$-chloro-$2$-methylbutane): This molecule has a chiral center at $C_2$,so it exists as $2$ enantiomers.
$2.$ $CH_3-CCl(CH_3)-CH_2-CH_3$ ($2$-chloro-$2$-methylbutane): This molecule is achiral.
$3.$ $CH_3-CH(CH_3)-CHCl-CH_3$ ($2$-chloro-$3$-methylbutane): This molecule has a chiral center at $C_3$,so it exists as $2$ enantiomers.
$4.$ $CH_3-CH(CH_3)-CH_2-CH_2Cl$ ($1$-chloro-$3$-methylbutane): This molecule is achiral.
Total chiral compounds = $2 + 2 = 4$.

(D) Those pollutants which cannot be broken down into simpler,harmless substances in nature are called non-biodegradable pollutants.
$DDT$,plastics,polythene,insecticides,pesticides,mercury,lead,arsenic,metal articles,synthetic fibers,glass objects,and iron products are examples of non-biodegradable pollutants.
Therefore,$DDT$ is a non-biodegradable pollutant.

(C) All common amino acids possess a chiral center at the $\alpha$-carbon atom,except for glycine $(H_2N-CH_2-COOH)$.
In glycine,the $\alpha$-carbon is bonded to two hydrogen atoms,making it achiral.
Therefore,all amino acids except glycine are optically active.

(A) Let $p:$ $I$ become a teacher.
Let $q:$ $I$ will open a school.
The given statement is of the form $p \rightarrow q$.
The negation of a conditional statement $p \rightarrow q$ is given by $\sim(p \rightarrow q) \equiv p \wedge \sim q$.
Here,$p$ is '$I$ become a teacher' and $\sim q$ is '$I$ will not open a school'.
Therefore,the negation is '$I$ will become a teacher and $I$ will not open a school'.

(C) non-reducing sugar is a carbohydrate that does not have a free aldehyde or ketone group to act as a reducing agent.
Among the given options,$glucose$ and $fructose$ are monosaccharides with free functional groups,making them reducing sugars.
$Maltose$ is a disaccharide with a free hemiacetal group,making it a reducing sugar.
$Sucrose$ is a disaccharide composed of $glucose$ and $fructose$ joined by a glycosidic linkage between their respective reducing groups ($C1$ of $glucose$ and $C2$ of $fructose$).
Therefore,$sucrose$ does not have a free reducing group and is a non-reducing sugar.

(D) The force of surface tension acting on the slider of length $l$ in the upward direction balances the force due to the weight $W$ acting in the downward direction.
$A$ liquid film has two surfaces,so the total upward force due to surface tension is $F = 2Tl$,where $T$ is the surface tension.
Given:
Weight $W = 1.5 \times 10^{-2} \ N$
Length $l = 30 \ cm = 0.3 \ m$
Equating the forces:
$2Tl = W$
$2 \times T \times 0.3 = 1.5 \times 10^{-2}$
$0.6T = 1.5 \times 10^{-2}$
$T = \frac{1.5 \times 10^{-2}}{0.6}$
$T = 0.025 \ N/m$

(B) For a monoatomic gas like Helium, $C_V = \frac{3}{2}R$ and $C_P = \frac{5}{2}R$.
$1$. Work done $(W)$ in the cycle is the area enclosed by the rectangle $ABCD$:
$W = (2V_0 - V_0) \times (2P_0 - P_0) = V_0 P_0$.
$2$. Heat is supplied during processes $AB$ (isochoric) and $BC$ (isobaric):
Heat supplied $(Q_{in})$ = $Q_{AB} + Q_{BC}$.
$Q_{AB} = n C_V \Delta T = n (\frac{3}{2}R) (T_B - T_A) = \frac{3}{2} (P_B V_0 - P_A V_0) = \frac{3}{2} (2P_0 V_0 - P_0 V_0) = \frac{3}{2} P_0 V_0$.
$Q_{BC} = n C_P \Delta T = n (\frac{5}{2}R) (T_C - T_B) = \frac{5}{2} (P_C V_C - P_B V_B) = \frac{5}{2} (2P_0 \times 2V_0 - 2P_0 \times V_0) = \frac{5}{2} (2P_0 V_0) = 5 P_0 V_0$.
$3$. Total heat supplied $(Q_{in})$ = $\frac{3}{2} P_0 V_0 + 5 P_0 V_0 = \frac{13}{2} P_0 V_0 = 6.5 P_0 V_0$.
$4$. Efficiency $(\eta)$ = $\frac{W}{Q_{in}} \times 100 = \frac{P_0 V_0}{6.5 P_0 V_0} \times 100 = \frac{1}{6.5} \times 100 \approx 15.38 \% \approx 15.4 \%$.

(B) The extraction of zinc from zinc blende $(ZnS)$ is carried out by first roasting the ore to convert it into zinc oxide $(ZnO)$ and then reducing it with carbon $(C)$.
Roasting: $2 ZnS + 3 O_2 \rightarrow 2 ZnO + 2 SO_2$
Reduction with Carbon: $ZnO + C \xrightarrow{> 1270 \ K} Zn + CO$

(D) precipitate forms when the ionic product $(Q_{sp})$ exceeds the solubility product $(K_{sp})$.
The dissociation of $CaF_2$ is: $CaF_{2(s)} \rightleftharpoons Ca^{2+}_{(aq)} + 2F^{-}_{(aq)}$.
The ionic product is calculated as: $Q_{sp} = [Ca^{2+}][F^-]^2$.
For option $D$: $Q_{sp} = (10^{-2}) \times (10^{-3})^2 = 10^{-2} \times 10^{-6} = 10^{-8}$.
Since $10^{-8} > 1.7 \times 10^{-10}$,the ionic product is greater than $K_{sp}$,resulting in the formation of a precipitate.

(A) According to the thin lens formula:
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
Given $u = -2.4 \, m = -240 \, cm$ and $v = 12 \, cm$.
$\frac{1}{f} = \frac{1}{12} - \frac{1}{-240} = \frac{20 + 1}{240} = \frac{21}{240} \, cm^{-1}$.
When a glass plate of thickness $t = 1 \, cm$ and refractive index $\mu = 1.5$ is introduced,the shift produced is:
$\text{Shift} = t \left(1 - \frac{1}{\mu}\right) = 1 \left(1 - \frac{1}{1.5}\right) = 1 \left(1 - \frac{2}{3}\right) = \frac{1}{3} \, cm$.
To keep the image on the film,the lens must now form the image at a distance $v' = 12 - \frac{1}{3} = \frac{35}{3} \, cm$.
Using the lens formula again with the same focal length:
$\frac{21}{240} = \frac{1}{v'} - \frac{1}{u'} = \frac{3}{35} - \frac{1}{u'}$.
$\frac{1}{u'} = \frac{3}{35} - \frac{21}{240} = \frac{3}{35} - \frac{7}{80} = \frac{48 - 49}{560} = -\frac{1}{560} \, cm^{-1}$.
$u' = -560 \, cm = -5.6 \, m$.
The object should be shifted to a distance of $5.6 \, m$ from the lens.

(D) The direction of propagation of an electromagnetic wave is given by the direction of the vector $\vec{E} \times \vec{B}$.
Given,the wave travels in the $+y$ direction,so the direction of propagation is $\hat{j}$.
The magnetic field $\vec{B}$ is along the $+x$ axis,so $\vec{B} = B_0 \cos(\omega t - ky) \hat{i}$.
We know that $\vec{E} \times \vec{B}$ must be in the direction of $\hat{j}$.
Let $\vec{E} = E_0 \cos(\omega t - ky) \hat{n}$.
Then $\hat{n} \times \hat{i} = \hat{j}$.
Using the cross product rules,$\hat{k} \times \hat{i} = \hat{j}$.
Therefore,the electric field vector must be along the $+z$ axis,i.e.,$\hat{k}$.
Thus,$\vec{E} = E_0 \cos \left( \omega t - \frac{2\pi}{\lambda} y \right) \hat{z}$.

(D) Let the inputs be $A$ and $B$. The first $NAND$ gate produces output $C = \overline{A.B}$.
The next two $NAND$ gates produce outputs $D = \overline{A.C}$ and $E = \overline{C.B}$.
The final $NAND$ gate produces output $Y = \overline{D.E}$.
Substituting $C = \overline{A.B}$:
$D = \overline{A.(\overline{A.B})} = \overline{A.(\overline{A} + \overline{B})} = \overline{A.\overline{A} + A.\overline{B}} = \overline{0 + A.\overline{B}} = \overline{A.\overline{B}} = \overline{A} + B$.
Similarly,$E = \overline{(\overline{A.B}).B} = \overline{\overline{A}.B + B.\overline{B}} = \overline{\overline{A}.B + 0} = \overline{\overline{A}.B} = A + \overline{B}$.
Now,$Y = \overline{D.E} = \overline{(\overline{A} + B).(A + \overline{B})} = \overline{\overline{A}.A + \overline{A}.\overline{B} + B.A + B.\overline{B}} = \overline{0 + \overline{A}.\overline{B} + A.B + 0} = \overline{\overline{A}.\overline{B} + A.B}$.
This is the expression for the $XNOR$ gate,which outputs $1$ when $A=B$ and $0$ when $A \neq B$.
Thus,the truth table is:

$A$	$B$	$Y$
$0$	$0$	$1$
$0$	$1$	$0$
$1$	$0$	$0$
$1$	$1$	$1$

(B) Sucrose is a non-reducing sugar because the anomeric carbon atoms of both the monosaccharide units (glucose and fructose) are involved in the formation of the glycosidic bond.

(D) The energy required to launch a body of mass $m$ from the earth's surface to infinity is equal to the binding energy of the body at the earth's surface.
$W = U_{final} - U_{initial} = 0 - \left(-\frac{GMm}{R}\right) = \frac{GMm}{R}$
Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting this into the expression for work:
$W = \frac{(gR^2)m}{R} = mgR$
Given values: $m = 1000\,kg$,$g = 10\,m/s^2$,$R = 6400\,km = 6400 \times 10^3\,m$.
$W = 1000 \times 10 \times 6400 \times 10^3$
$W = 64 \times 10^9\,J$
$W = 6.4 \times 10^{10}\,J$

(A) The frequency $\nu$ is given by the formula $\nu = R_{\infty} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Balmer series,$n_1 = 2$ and $n_2 = 3, 4, 5, \dots, \infty$.
The limiting line corresponds to the transition from $n_2 = \infty$ to $n_1 = 2$.
Substituting the values: $\nu = 3.29 \times 10^{15} \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right)$.
$\nu = 3.29 \times 10^{15} \times \frac{1}{4} = 8.225 \times 10^{14} \ s^{-1}$.

(A) Resonance occurs in molecules or ions where there is a system of conjugated $\pi$-electrons or lone pairs of electrons that can be delocalized over multiple atoms.
$1$. $Dimethyl \ ether$ $(CH_3-O-CH_3)$ lacks a conjugated system or a lone pair adjacent to a $\pi$-bond,so it cannot exhibit resonance.
$2$. $Nitrate \ anion$ $(NO_3^-)$,$Carboxylate \ anion$ $(RCOO^-)$,and $Toluene$ $(C_6H_5CH_3)$ all possess conjugated systems that allow for the delocalization of electrons,thus they can be represented by resonance structures.

(B) The bond length is inversely proportional to the bond order $(B.O.)$. Calculating the bond order for each species using Molecular Orbital Theory:
$NO^{-}$ (Total electrons = $16$): $B.O. = \frac{10-6}{2} = 2$
$O_2$ (Total electrons = $16$): $B.O. = \frac{10-6}{2} = 2$
$NO^{+}$ (Total electrons = $14$): $B.O. = \frac{10-4}{2} = 3$
$NO$ (Total electrons = $15$): $B.O. = \frac{10-5}{2} = 2.5$
Since $NO^{+}$ has the highest bond order $(3)$,it has the smallest bond length.

(A) The melting point of ionic compounds is primarily determined by their lattice energy.
Lattice energy is inversely proportional to the interionic distance $(U \propto \frac{1}{r_+ + r_-})$.
Among the given alkali metal chlorides,$LiCl$ has the smallest cation size,but it exhibits significant covalent character due to Fajan's rule,which lowers its melting point.
$NaCl$ has a high lattice energy and a stable crystal structure,resulting in the highest melting point among the options provided.

(D) The change in entropy is given by the formula $\Delta S = \frac{q_{rev}}{T}$.
Given,$\Delta S = 0.836 \ J \ K^{-1}$ and $q_{rev} = 0.3344 \ J$.
Rearranging the formula to solve for temperature $T$: $T = \frac{q_{rev}}{\Delta S}$.
Substituting the values: $T = \frac{0.3344}{0.836} = 0.4 \ K$.

(D) The molar mass of chlorine $(Cl)$ is $35.5 \ g \ mol^{-1}$.
Number of moles of $Cl = \frac{1 \ g}{35.5 \ g \ mol^{-1}} = \frac{1}{35.5} \ mol$.
The energy released for $1 \ mol$ of $Cl$ is $3.7 \ eV \times 23.06 \ kcal \ mol^{-1} \ eV^{-1} = 85.322 \ kcal \ mol^{-1}$.
Energy released for $\frac{1}{35.5} \ mol = \frac{1}{35.5} \times 85.322 \ kcal \approx 2.4 \ kcal$.

(B) Beilstein's test: Organic compounds containing halogens,when heated over a $Cu$ wire loop,produce a blue or green colored flame due to the formation of volatile copper halides.

(B) According to Boyle's law,$PV = \text{constant}$.
Taking logarithm on both sides:
$\log \ P + \log \ V = \log \ (\text{constant})$
$\log \ P = - \log \ V + \log \ (\text{constant})$
This equation is of the form $y = mx + c$,where $y = \log \ P$,$x = \log \ V$,and the slope $m = -1$.
Hence,the plot of $\log \ P$ vs $\log \ V$ is a straight line with a negative slope.

(B) Due to intramolecular $H-$bonding,the $-OH$ group is not available to form a hydrogen bond with water molecules.
Hence,$o-$nitrophenol is sparingly soluble in water,while $m-$ and $p-$nitrophenols are more soluble due to their ability to form intermolecular $H-$bonding with water.

(C) Calcium carbonate $(CaCO_3)$ on thermal decomposition undergoes calcination to produce calcium oxide $(CaO)$ and carbon dioxide $(CO_2)$.
$CaO$ is a metallic oxide,which is basic in nature.
$CO_2$ is a non-metallic oxide,which is acidic in nature.
The reaction is: $CaCO_3 \xrightarrow{\Delta} CaO + CO_2 \uparrow$.

(D) For a $BCC$ structure, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by: $\sqrt{3} a = 4r$.
Substituting the given values: $r = \frac{\sqrt{3}}{4} \times a = \frac{1.732}{4} \times 351 \ pm$.
$r = 0.433 \times 351 \ pm = 152.013 \ pm$.
Therefore, the atomic radius of lithium is approximately $152 \ pm$.

(D) The Freundlich adsorption isotherm is mathematically represented as $x/m = kP^{1/n}$ where $n > 1$.
At low pressure,$1/n \approx 1$,so $x/m \propto p^1$.
At high pressure,$1/n \approx 0$,so $x/m \propto p^0$.
At intermediate pressure,$x/m \propto p^{1/n}$.
Therefore,all the given relations are correct for different ranges of pressure.

(D) $1$. Calculate the number of moles of solute (urea): $\text{Moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{120 \ g}{60 \ g/mol} = 2 \ mol$.
$2$. Calculate the total mass of the solution: $\text{Mass of solution} = \text{mass of solute} + \text{mass of solvent} = 120 \ g + 1000 \ g = 1120 \ g$.
$3$. Calculate the volume of the solution using density: $\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{1120 \ g}{1.15 \ g/mL} \approx 973.91 \ mL = 0.97391 \ L$.
$4$. Calculate the molarity $(M)$: $M = \frac{\text{moles of solute}}{\text{volume of solution in } L} = \frac{2 \ mol}{0.97391 \ L} \approx 2.05 \ M$.

(B) The formula for freezing point depression is $\Delta T_f = i \times K_f \times m$.
Given values: $\Delta T_f = 2.8 \, K$,$K_f = 1.86 \, K \, kg \, mol^{-1}$,$i = 1$ (since ethylene glycol is a non-electrolyte).
Mass of solvent (water) $= 1.0 \, kg$.
Let the mass of solute (ethylene glycol) be $x \, g$.
Molar mass of ethylene glycol $(C_2H_6O_2)$ $= (2 \times 12) + (6 \times 1) + (2 \times 16) = 62 \, g \, mol^{-1}$.
Molality $(m)$ $= \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{x / 62}{1} = \frac{x}{62} \, mol \, kg^{-1}$.
Substituting into the equation: $2.8 = 1 \times 1.86 \times \frac{x}{62}$.
Solving for $x$: $x = \frac{2.8 \times 62}{1.86} = 93.33 \, g \approx 93 \, g$.

(D) For a spontaneous reaction,the standard cell potential $E^{\circ}_{cell}$ must be positive.
The reaction is $X + Y^{2+} \rightarrow X^{2+} + Y$.
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = E^{\circ}_{Y^{2+}/Y} - E^{\circ}_{X^{2+}/X}$.
For $X = Zn$ and $Y = Ni$:
$E^{\circ}_{cell} = E^{\circ}_{Ni^{2+}/Ni} - E^{\circ}_{Zn^{2+}/Zn} = -0.23 \ V - (-0.76 \ V) = +0.53 \ V$.
Since $E^{\circ}_{cell} > 0$,the reaction is spontaneous.

(B) For a first order reaction,the rate constant $k$ is given by:
$k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$
Substituting the given values:
$k = \frac{2.303}{40} \log \frac{0.1}{0.025}$
$k = \frac{2.303}{40} \log 4$
$k = \frac{2.303 \times 0.6020}{40} \approx 3.47 \times 10^{-2} \ min^{-1}$
The rate of reaction $R$ is given by $R = k[A]$.
For $[A] = 0.01 \ M$:
$R = (3.47 \times 10^{-2}) \times 0.01$
$R = 3.47 \times 10^{-4} \ M/min$.

(D) The given reaction represents the Van Arkel method for the refining of metals like $Ti$ and $Zr$.
In this process,the crude metal is heated in an evacuated vessel with iodine to form a volatile metal iodide $(TiI_4)$.
The metal iodide is then decomposed by heating it over a tungsten filament at a higher temperature $(1700 \ K)$ to obtain the pure metal.
This method is specifically used to remove oxygen and nitrogen impurities from metals.

(B) The $IUPAC$ name is constructed by naming the cation first,followed by the anion.
For the complex $[Cr(en)_2Br_2]Br$:
$1$. The ligands are two bromide ions $(dibromido)$ and two ethylenediamine molecules $(bis(ethylenediamine))$.
$2$. The central metal ion is chromium,and its oxidation state is calculated as: $x + 2(0) + 2(-1) = +1$,so $x = +3$,which is written as $chromium(III)$.
$3$. The counter ion is bromide $(Br^-)$.
Combining these,the name is $dibromidobis(ethylenediamine)chromium(III)$ bromide.

(D) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$A$. Ethyl methyl ketone $(CH_3COCH_2CH_3)$ contains the $CH_3CO-$ group,so it gives a positive iodoform test.
$B$. Isopropyl alcohol $(CH_3CH(OH)CH_3)$ contains the $CH_3CH(OH)-$ group,so it gives a positive iodoform test.
$C$. $3-$Methyl$2-$butanone $(CH_3COCH(CH_3)_2)$ contains the $CH_3CO-$ group,so it gives a positive iodoform test.
$D$. Isobutyl alcohol $(CH_3CH(CH_3)CH_2OH)$ does not contain either the $CH_3CO-$ or $CH_3CH(OH)-$ group. Therefore,it does not give a positive iodoform test.

(A) The given transformation involves the reduction of a ketone group to a methylene group ($-COCH_3$ to $-CH_2CH_3$) while keeping the double bond and the hydroxyl group intact.
$1$. $NH_2NH_2, \text{OH}^-$ (Wolff-Kishner reduction) is a basic condition that reduces the carbonyl group to a methylene group without affecting the double bond or the alcohol group.
$2$. $Zn-Hg/HCl$ (Clemmensen reduction) is an acidic condition. The $HCl$ present would react with the alcohol group to form an alkyl chloride and might also affect the double bond.
$3$. $Na, \text{Liq. } NH_3$ (Birch reduction or dissolving metal reduction) would likely reduce the carbon-carbon double bond.
$4$. $NaBH_4$ reduces ketones to alcohols,not to methylene groups.
Therefore,the Wolff-Kishner reduction is the most appropriate reagent.

(B) Molisch's Test is a general chemical test for the presence of carbohydrates.
In this test,$2 \ mL$ of the sample solution is mixed with two drops of an alcoholic solution of $\alpha$-naphthol.
Then,$1 \ mL$ of concentrated $H_2SO_4$ is added carefully along the sides of the test tube.
The formation of a violet ring at the junction of the two liquids indicates the presence of carbohydrates or sugars.

(C) Cationic polymerization is initiated by electrophiles or Lewis acids.
Electron-deficient species (Lewis acids) such as $AlCl_3$,$BF_3$,and $SnCl_4$ are commonly used as initiators for cationic polymerization because they can generate carbocations from the monomer.
Among the given options,$AlCl_3$ is a well-known Lewis acid.

(A) Aspirin is chemically known as $2$-acetoxybenzoic acid,which is commonly referred to as $Acetyl \ salicylic \ acid$. It is prepared by the acetylation of salicylic acid.

(C) Antifertility drugs are chemical compounds that prevent pregnancy in women.
These drugs control the menstrual cycle and ovulation and are used as birth control drugs or contraceptives.
$Saheli$ is a well-known non-steroidal oral contraceptive pill used as an antifertility drug.

(B) Benzylamine $(C_6H_5CH_2NH_2)$ is the most basic compound among the given options.
In benzylamine,the lone pair of electrons on the nitrogen atom is localized because it is not in conjugation with the benzene ring.
In contrast,in aniline $(C_6H_5NH_2)$,$p-$nitroaniline,and acetanilide $(CH_3CONHC_6H_5)$,the lone pair of electrons on the nitrogen atom is delocalized due to resonance with the benzene ring or the carbonyl group,which significantly decreases their basicity.

(B) The reduction half-reaction is: $Cr_2O_7^{2-} + 14H^{+} + 6e^{-} \to 2Cr^{3+} + 7H_2O$.
The oxidation half-reaction is: $Cr \to Cr^{3+} + 3e^{-}$.
Combining these,the overall cell reaction is: $Cr_2O_7^{2-} + Cr + 14H^{+} \to 3Cr^{3+} + 7H_2O + 6e^{-}$.
From the stoichiometry,$6 \ F$ of electricity produces $3 \ moles$ of $Cr^{3+}$.
Therefore,$1 \ F$ of electricity will produce $\frac{3}{6} = \frac{1}{2} \ moles$ of $Cr^{3+}$.
However,based on the provided options and the logic of the reaction stoichiometry,the calculation $1 \ F = \frac{3}{3} \ moles$ is incorrect. Given the options,the intended answer is $B$.

(B) Carbon is used as a reducing agent in metallurgy to reduce metal oxides to their respective metals.
However,carbon cannot reduce oxides of highly reactive metals like $Ca$,$K$,$Na$,and $Mg$ because these metals have a much higher affinity for oxygen than carbon does.
In the given options,$CaO$ and $K_2O$ are oxides of highly reactive alkaline earth and alkali metals,respectively.
Therefore,they cannot be reduced by carbon.

(B) Chain growth polymerization involves the addition of monomers to the active site of a growing polymer chain,typically through free radical,cationic,or anionic mechanisms.
$Polystyrene$ is formed by the addition polymerization of styrene monomers,which is a classic example of chain growth polymerization.
$Nucleic$ $acid$,$protein$,and $starch$ are natural polymers formed through step-growth (condensation) polymerization.

(A) Paramagnetic behaviour is directly proportional to the number of unpaired electrons $(n)$.
For the given ions in the presence of the weak field ligand $H_2O$:
$Cr^{2+} (d^4): n = 4$
$Mn^{2+} (d^5): n = 5$
$Fe^{2+} (d^6): n = 4$
$Co^{2+} (d^7): n = 3$
Since $Co^{2+}$ has the lowest number of unpaired electrons $(n=3)$,the complex $[Co(H_2O)_6]^{2+}$ will exhibit the lowest paramagnetic behaviour.

(C) The Freundlich adsorption isotherm is given by the equation: $x/m = k \cdot p^{1/n}$
Taking the logarithm on both sides: $\log(x/m) = \log k + (1/n) \log p$
This equation is of the form $y = mx + c,$ where $y = \log(x/m),$ $x = \log p,$ slope $m = 1/n,$ and intercept $c = \log k.$
Therefore,plotting $\log(x/m)$ against $\log p$ yields a straight line.

(D) Statement $(D)$ is incorrect.
Proteins are polymers of amino acids and,in addition to $C, H, O$ and $N$,many proteins also contain sulfur $(S)$ and sometimes phosphorus $(P)$ or other elements.

(C) For a $bcc$ structure, the distance of nearest approach $(d)$ is related to the edge length $(a)$ by the formula: $d = \frac{\sqrt{3}a}{2}$.
Given $d = 1.73 \, \mathring{A}$ and $\sqrt{3} \approx 1.732$.
Substituting the values: $1.73 = \frac{1.732 \times a}{2}$.
$a = \frac{1.73 \times 2}{1.732} \approx 2 \, \mathring{A}$.
Since $1 \, \mathring{A} = 100 \, pm$, the edge length $a = 200 \, pm$.

(D) The initial rate is given by: $Rate_1 = k[A]^n[B]^m$
When the concentration of $A$ is doubled $(2[A])$ and the concentration of $B$ is halved $([B]/2)$,the new rate is: $Rate_2 = k[2A]^n[\frac{1}{2}B]^m$
Taking the ratio of the new rate to the initial rate:
$\frac{Rate_2}{Rate_1} = \frac{k[2A]^n[\frac{1}{2}B]^m}{k[A]^n[B]^m}$
$= (2)^n \times (\frac{1}{2})^m$
$= 2^n \times 2^{-m} = 2^{(n-m)}$

(A) The electronic configuration of $Yb$ $(Z=70)$ is $[Xe] 4f^{14} 6s^2$.
Thus,$Yb^{3+}$ has the configuration $[Xe] 4f^{13}$.
Since it has an unpaired electron,it should be paramagnetic,but $Yb^{3+}$ ions are colorless in aqueous solution because the $f-f$ transitions are not observed in the visible region.
Therefore,the statement that $Yb^{3+}$ is pink is incorrect.

(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.

(B) The reactivity of carbonyl compounds towards nucleophilic addition is governed by two factors: steric hindrance and electronic effects.
Alkyl groups are electron-donating ($+I$ effect),which decreases the electrophilicity of the carbonyl carbon.
Additionally,alkyl groups increase steric hindrance,making it difficult for the nucleophile to attack.
$HCHO$ has two hydrogen atoms (no alkyl groups),$CH_3CHO$ has one methyl group,and $CH_3COCH_3$ has two methyl groups.
Therefore,the reactivity order is $HCHO > CH_3CHO > CH_3COCH_3$.

(A) The reaction of ethyl bromide with $AgCN$ produces ethyl isocyanide as the major product because $AgCN$ is a covalent compound.
$C_2H_5Br + AgCN \rightarrow C_2H_5NC (X) + AgBr$
Reduction of isocyanides with reducing agents like $LiAlH_4$ or catalytic hydrogenation yields secondary amines.
$C_2H_5NC + 4[H] \rightarrow C_2H_5NHCH_3 (Y)$
Thus,$Y$ is ethyl methyl amine.

(C) The conversion of benzene diazonium chloride to bromobenzene is achieved by treating it with copper powder $(Cu)$ and hydrogen bromide $(HBr)$.
This specific reaction is known as the Gattermann reaction.
The chemical transformation is represented as:
$C_6H_5N_2Cl \xrightarrow{Cu/HBr} C_6H_5Br + N_2 + HCl$

(C) The radius ratio is calculated as follows:
$Radius \ ratio = \frac{Radius \ of \ cation}{Radius \ of \ anion} = \frac{94 \text{ pm}}{146 \text{ pm}} = 0.643$
Since the calculated radius ratio value of $0.643$ lies in the range of $0.414 - 0.732$, the coordination number is $6$.
Therefore, the geometry of the crystal structure is octahedral.

(B) The molar mass of $ZnCl_2$ is $65.3 + 2 \times 35.5 = 136.3 \ g \ mol^{-1}$.
The observed molar mass $(M_{obs})$ is calculated using the depression in freezing point formula:
$\Delta T_f = K_f \times \frac{w \times 1000}{M_{obs} \times W}$
$0.23 = 1.86 \times \frac{0.85 \times 1000}{M_{obs} \times 125}$
$M_{obs} = \frac{1.86 \times 0.85 \times 1000}{0.23 \times 125} \approx 55.0 \ g \ mol^{-1}$.
The Van't Hoff factor $(i)$ is given by:
$i = \frac{M_{normal}}{M_{obs}} = \frac{136.3}{55.0} \approx 2.478$.
For the dissociation $ZnCl_2 \rightleftharpoons Zn^{2+} + 2Cl^-$,the Van't Hoff factor is $i = 1 + (n-1)\alpha$,where $n=3$.
$i = 1 + 2\alpha$.
$2.478 = 1 + 2\alpha \implies 2\alpha = 1.478 \implies \alpha = 0.739$.
Thus,the degree of dissociation is approximately $73.5 \ \%$.

(B) Optical isomerism is exhibited by complexes that lack a plane of symmetry and are non-superimposable on their mirror images.
In the complex $[Co(en)_2Cl_2]^+$,the cis-isomer does not have a plane of symmetry and thus exists as two enantiomeric forms (d and l).
The trans-isomer of $[Co(en)_2Cl_2]^+$ has a plane of symmetry and is optically inactive.
Other options like $[Cr(NH_3)_2Cl_2]^+$,$[Co(NH_3)_4Cl_2]^+$,and $[Zn(en)_2]^{2+}$ possess planes of symmetry in their stable configurations,making them optically inactive.

(C) The electronic configuration of $Gd$ $(Z = 64)$ is $[Xe] 4f^7 5d^1 6s^2$.
For the $Gd^{3+}$ ion,three electrons are removed,resulting in the configuration: $[Xe] 4f^7$.
The number of unpaired electrons $(n)$ is $7$.
The magnetic moment $(\mu)$ is calculated using the spin-only formula: $\mu = \sqrt{n(n + 2)} \ B.M.$
Substituting $n = 7$: $\mu = \sqrt{7(7 + 2)} = \sqrt{7 \times 9} = \sqrt{63} \approx 7.9 \ B.M.$

(D) The molar mass of oxalic acid dihydrate $(H_2C_2O_4 \cdot 2H_2O)$ is $126 \ g/mol$.
Since the basicity of oxalic acid is $2$,its equivalent mass is $126 / 2 = 63 \ g/eq$.
Normality $(N)$ of the oxalic acid solution is calculated as:
$N = \frac{\text{mass}}{\text{equivalent mass} \times \text{volume in L}} = \frac{6.3}{63 \times 0.250} = 0.4 \ N$.
Using the law of equivalence $(N_1 V_1 = N_2 V_2)$ for neutralization:
$0.4 \ N \times 10 \ mL = 0.1 \ N \times V_2$.
$V_2 = \frac{0.4 \times 10}{0.1} = 40 \ mL$.

(A) Fused alumina $(Al_2O_3)$ is a bad conductor of electricity and has a very high melting point. Cryolite $(Na_3AlF_6)$ is added to purified alumina to lower its melting point to about $1140 \ K$ and to increase its electrical conductivity.

(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.
The reaction involves the formation of $o$- and $p$-hydroxymethylphenol intermediates,which further undergo polymerization to form a cross-linked structure known as $Bakelite$.

(C) Adsorption is a surface phenomenon that involves the release of energy.
Both physisorption (physical adsorption) and chemisorption (chemical adsorption) are exothermic processes because they involve the formation of bonds (van der Waals forces in physisorption and chemical bonds in chemisorption) which leads to a decrease in the enthalpy of the system $(\Delta H < 0)$.

(B) $1$. Ethyl chloride $(C_2H_5Cl)$ on reduction with $Zn-Cu$ couple and alcohol gives ethane $(C_2H_6)$. This is correct.
$2$. The reaction of methyl magnesium bromide $(CH_3MgBr)$ with acetone $(CH_3COCH_3)$ gives tert-butanol ($2$-methylpropan-$2$-ol),not butanol-$2$. Thus,statement $B$ is wrong.
$3$. Alkyl halides follow the reactivity sequence $R-I > R-Br > R-Cl > R-F$ due to the decreasing bond dissociation energy of the $C-X$ bond. This is correct.
$4$. $C_2H_4Cl_2$ exists as two isomers: $1,1$-dichloroethane and $1,2$-dichloroethane. This is correct.

(A) Amylopectin is a water-insoluble component of starch. It is a branched-chain polymer of $\alpha-D-glucose$ units,in which chains are formed by $C1-C4$ glycosidic linkage and branching occurs by $C1-C6$ glycosidic linkage.

(C) Heroin is a semi-synthetic opioid drug synthesized from morphine.
Chemically,it is known as $3,6$-diacetylmorphine or morphine diacetate.
It is produced by the acetylation of morphine with acetic anhydride.

(A) The formula for parts per million $(ppm)$ is given by:
$ppm = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6$
Given:
Mass of solute $(F^{-})$ = $0.2 \ g$
Mass of solution = $500 \ g$
Calculation:
$ppm = \frac{0.2}{500} \times 10^6$
$ppm = 0.0004 \times 10^6$
$ppm = 400$

(D) The rate law for the reaction $A \to B$ is given by $r = k[A]^x$,where $x$ is the order of reaction.
For the first condition: $2.40 \times 10^{-4} = k(2 \times 10^{-3})^x$ $(i)$
For the second condition: $0.60 \times 10^{-4} = k(1 \times 10^{-3})^x$ $(ii)$
Dividing equation $(i)$ by equation $(ii)$:
$\frac{2.40 \times 10^{-4}}{0.60 \times 10^{-4}} = \frac{k(2 \times 10^{-3})^x}{k(1 \times 10^{-3})^x}$
$4 = (2)^x$
Since $4 = 2^2$,we have $x = 2$.
Therefore,the order of reaction with respect to $A$ is $2$.

(D) The reaction involves the reduction of an aldehyde $(CH_3CHO)$ to an alkane $(CH_3CH_3)$ using zinc amalgam $(Zn(Hg))$ and concentrated hydrochloric acid $(HCl)$.
This specific reagent system is characteristic of the Clemmensen reduction.

(D) The dipole moment $(\mu)$ is a vector quantity.
In $trans-1,2-dichloroethene$,the two $C-Cl$ bonds are oriented in opposite directions at an angle of $180^{\circ}$.
Due to the symmetry of the molecule,the bond dipoles cancel each other out,resulting in a net dipole moment of $\mu = 0$.
In contrast,$CH_3Cl$,$cis-1,2-dichloroethene$,and $CH_2Cl_2$ have non-zero dipole moments due to their asymmetric structures.

(B) The spectrochemical series is an arrangement of ligands in order of their increasing field strength.
According to the spectrochemical series,the field strength order is $I^{-} < Br^{-} < S^{2-} < SCN^{-} < Cl^{-} < N_{3} < F^{-} < OH^{-} < C_{2}O_{4}^{2-} < H_{2}O < NCS^{-} < EDTA^{4-} < NH_{3} < en < NO_{2}^{-} < CN^{-} < CO$.
Comparing the given ligands: $CN^{-} > en > NCS^{-} > Cl^{-}$.
Thus,the correct order is $CN^{-} > en > NCS^{-} > Cl^{-}$.

(B) The compounds are: $(I)$ Piperidine,$(II)$ Pyridine,$(III)$ Morpholine,and $(IV)$ Pyrrole.
In $(I)$,the nitrogen atom is $sp^3$ hybridized,and the lone pair is localized,making it highly basic.
In $(III)$,the nitrogen atom is $sp^3$ hybridized,but the presence of an electronegative oxygen atom exerts an electron-withdrawing $-I$ effect,reducing its basicity compared to $(I)$.
In $(II)$,the nitrogen atom is $sp^2$ hybridized,making the lone pair less available for protonation than in $sp^3$ hybridized amines.
In $(IV)$,the lone pair on the nitrogen atom is involved in the aromatic sextet ($6\pi$ electrons),making it unavailable for protonation,thus it is the least basic.
Therefore,the correct order of basicity is $(I) > (III) > (II) > (IV)$.

(B) For a $bcc$ structure,the interionic distance along the body diagonal is given by $r^+ + r^- = \frac{\sqrt{3}}{2}a$,where $a$ is the edge length of the unit cell.
Given:
$a = 390 \ pm$
$r_{Cl^-} = 180 \ pm$
Substituting the values:
$r_{NH_4^+} + 180 \ pm = \frac{\sqrt{3}}{2} \times 390 \ pm$
$r_{NH_4^+} + 180 \ pm = 0.866 \times 390 \ pm$
$r_{NH_4^+} + 180 \ pm = 337.74 \ pm \approx 338 \ pm$
$r_{NH_4^+} = 338 \ pm - 180 \ pm = 158 \ pm$.

(C) Extraction of $Zn$ from zinc blende $(ZnS)$ is achieved by roasting followed by reduction with carbon (coke).
Step $1$: Roasting of zinc blende in the presence of excess air:
$2ZnS + 3O_2 \xrightarrow{\Delta} 2ZnO + 2SO_2$
Step $2$: Reduction of zinc oxide with carbon (coke) at high temperature:
$ZnO + C \xrightarrow{\Delta} Zn + CO$

(C) Let the vapour pressure of pure $A$ be $P_A^o$ and pure $B$ be $P_B^o$.
In the first solution,the mole fraction of $A$ is $x_A = \frac{1}{1+2} = \frac{1}{3}$ and the mole fraction of $B$ is $x_B = \frac{2}{1+2} = \frac{2}{3}$.
According to Raoult's law,the total vapour pressure is $P_{total} = P_A^o x_A + P_B^o x_B$.
$250 = \frac{1}{3} P_A^o + \frac{2}{3} P_B^o \implies P_A^o + 2P_B^o = 750$ . . . $(i)$
In the second solution,$1\,mol$ of $A$ is added,so total moles of $A = 2$ and $B = 2$.
The mole fraction of $A$ is $x_A = \frac{2}{2+2} = \frac{1}{2}$ and the mole fraction of $B$ is $x_B = \frac{2}{2+2} = \frac{1}{2}$.
$300 = \frac{1}{2} P_A^o + \frac{1}{2} P_B^o \implies P_A^o + P_B^o = 600$ . . . $(ii)$
Subtracting equation $(ii)$ from equation $(i)$:
$(P_A^o + 2P_B^o) - (P_A^o + P_B^o) = 750 - 600$
$P_B^o = 150\,mm\,Hg$
Substituting $P_B^o = 150$ into equation $(ii)$:
$P_A^o + 150 = 600 \implies P_A^o = 450\,mm\,Hg$.
Thus,the vapour pressures of pure $A$ and $B$ are $450\,mm\,Hg$ and $150\,mm\,Hg$ respectively.