AP EAMCET 2006 Chemistry Question Paper with Answer and Solution

(A) According to Heisenberg's uncertainty principle,the relation is $\Delta x \cdot \Delta v \cdot m = \frac{h}{4 \pi}$.
Thus,$\Delta x = \frac{h}{4 \pi m \cdot \Delta v}$.
For particle $A$: $\Delta x_A = \frac{h}{4 \pi m_A \cdot \Delta v_A} = \frac{h}{4 \pi m_A \cdot 0.05}$.
For particle $B$: $\Delta x_B = \frac{h}{4 \pi m_B \cdot \Delta v_B} = \frac{h}{4 \pi (5m_A) \cdot 0.02}$.
Taking the ratio $\frac{\Delta x_A}{\Delta x_B} = \frac{h}{4 \pi m_A \cdot 0.05} \times \frac{4 \pi (5m_A) \cdot 0.02}{h}$.
$\frac{\Delta x_A}{\Delta x_B} = \frac{5 \times 0.02}{0.05} = \frac{0.10}{0.05} = 2$.

(A) According to the ideal gas equation,$PV = nRT$,which can be rearranged as $V = (\frac{nR}{P})T$.
This represents a straight line equation $y = mx$,where the slope $m = \frac{nR}{P}$.
Since the slope is inversely proportional to pressure $(m \propto \frac{1}{P})$,the line with the smallest slope corresponds to the highest pressure.
Looking at the graph,the slope of the line for $p_1$ is the smallest,followed by $p_3$,and then $p_2$.
Therefore,the correct order of pressures is $p_1 > p_3 > p_2$.

(B) Ethyl chloride reacts with sodium ethoxide to form diethyl ether $(A)$ via the Williamson ether synthesis:
$C_2H_5Cl + C_2H_5ONa \rightarrow C_2H_5OC_2H_5 + NaCl$
Diethyl ether is also obtained by the intermolecular dehydration of ethyl alcohol with concentrated $H_2SO_4$ at $140^{\circ}C$:
$2 C_2H_5OH \xrightarrow{H_2SO_4, 140^{\circ}C} C_2H_5OC_2H_5 + H_2O$

(A) When acetone $(CH_3COCH_3)$ is treated with a base like barium hydroxide $(Ba(OH)_2)$,it undergoes an aldol condensation reaction.
Two molecules of acetone react to form $4$-hydroxy-$4$-methylpentan-$2$-one,which is commonly known as diacetone alcohol.
The reaction is: $2CH_3COCH_3 \xrightarrow{Ba(OH)_2} CH_3-C(OH)(CH_3)-CH_2-COCH_3$.

(A) When acetaldehyde $(CH_3CHO)$ is heated with Fehling solution,it undergoes oxidation to form acetate ions,while the $Cu^{2+}$ ions in the Fehling solution are reduced to $Cu^+$ ions.
This results in the formation of a red precipitate of cuprous oxide $(Cu_2O)$.
The chemical equation is:
$CH_3CHO + 2Cu(OH)_2 + NaOH \longrightarrow CH_3COONa + Cu_2O \downarrow (\text{Red}) + 3H_2O$

(D) The reduction of nitrobenzene $(C_6H_5NO_2)$ in an acidic medium using $Zn / HCl$ results in the formation of aniline $(C_6H_5NH_2)$ as the main product.
The chemical reaction is represented as:
$C_6H_5NO_2 + 6[H] \xrightarrow{Zn/HCl} C_6H_5NH_2 + 2H_2O$

(A) The correct matching is as follows:
$1$. Nitrogen molecules produce a band spectrum because molecular spectra consist of bands due to vibrational and rotational energy levels. Thus,$1-C$.
$2$. Incandescent solids emit a continuous spectrum because the atoms are closely packed,leading to overlapping energy levels. Thus,$2-A$.
$3$. Fraunhoffer lines are dark lines observed in the solar spectrum,which are caused by the absorption of light by gases in the solar atmosphere. Thus,$3-B$.
$4$. An electric arc between iron rods produces a line emission spectrum characteristic of the iron atoms. Thus,$4-D$.
Therefore,the correct sequence is $1-C, 2-A, 3-B, 4-D$.

(D) Initially,the oil drop is in equilibrium between the plates. The downward gravitational force $mg$ is balanced by the upward electric force $qE$. Thus,$qE = mg$.
When the polarity of the plates is reversed,the electric force $qE$ now acts downwards,in the same direction as gravity.
The net force on the drop becomes $F_{net} = mg + qE$.
Since $qE = mg$,we have $F_{net} = mg + mg = 2mg$.
The instantaneous acceleration $a$ is given by $a = \frac{F_{net}}{m} = \frac{2mg}{m} = 2g$.
Given $g = 10 ~ms^{-2}$,the acceleration $a = 2 \times 10 = 20 ~ms^{-2}$.

(C) The reaction of diethyl ether $(C_2H_5OC_2H_5)$ with carbon monoxide $(CO)$ in the presence of a Lewis acid catalyst like $BF_3$ at $150^{\circ} C$ and $500 \ atm$ pressure leads to the carbonylation of the ether to form ethyl propionate $(C_2H_5COOC_2H_5)$.
The chemical equation is: $C_2H_5OC_2H_5 + CO \xrightarrow{BF_3, 150^{\circ} C, 500 \ atm} C_2H_5COOC_2H_5$.

(D) Benzene $(C_6H_6)$ consists of a hexagonal ring of six carbon atoms,each bonded to one hydrogen atom.
There are $6$ $C-C$ bonds and $6$ $C-H$ bonds.
In the ring,there are $3$ double bonds and $3$ single bonds.
Each single bond is a $\sigma$ bond,and each double bond consists of one $\sigma$ bond and one $\pi$ bond.
Total $\sigma$ bonds = $6$ ($C-C$ $\sigma$ bonds) + $6$ ($C-H$ $\sigma$ bonds) = $12$ $\sigma$ bonds.
Total $\pi$ bonds = $3$ $\pi$ bonds.
Thus,benzene contains $12$ $\sigma$ bonds and $3$ $\pi$ bonds.

(D) The molecule $C_2H_4$ (ethene) has the structure $CH_2=CH_2$.
Each carbon atom is $sp^2$ hybridized.
$(i)$ There are four $C-H$ sigma bonds formed by the overlap of $sp^2$ orbitals of carbon and $s$ orbitals of hydrogen.
$(ii)$ There is one $C-C$ sigma bond formed by the overlap of $sp^2$ and $sp^2$ orbitals.
$(iii)$ There is one $\pi$ bond formed by the lateral overlap of unhybridized $p_z$ orbitals on each carbon atom.
Thus,the molecule $(X)$ is $C_2H_4$.

(C) catalyst increases the rate of both forward and backward reactions equally,allowing the system to reach equilibrium faster.
It does not change the position of equilibrium or the value of the equilibrium constant $(K_c)$.
Therefore,the statement in option $(c)$ is incorrect.

(B) Statement $I$ is correct because the modern periodic law states that physical and chemical properties of elements are periodic functions of their atomic numbers,which is equivalent to their electronic configuration.
Statement $II$ is incorrect because fluorine $(F)$ has the highest electronegativity $(4.0)$ among all elements,which is greater than that of chlorine ($Cl$,$3.0$).
Statement $III$ is incorrect because electropositive nature (metallic character) increases from top to bottom in a group due to the increase in atomic size and decrease in ionization enthalpy.
Therefore,only statement $I$ is correct.

(D) Let $m$ be the number of cells in the $12$-cell battery that are wrongly connected. Each wrongly connected cell cancels the emf of one correctly connected cell. Thus,the effective emf of the $12$-cell battery is $(12 - m)E - mE = (12 - 2m)E$.
When the $12$-cell battery and the $2$-cell battery aid each other,the total emf in the circuit is $(12 - 2m)E + 2E = (14 - 2m)E$. The current is given by $i_1 = \frac{(14 - 2m)E}{R} = 3 \text{ A}$ ... $(i)$.
When they oppose each other,the total emf in the circuit is $(12 - 2m)E - 2E = (10 - 2m)E$. The current is given by $i_2 = \frac{(10 - 2m)E}{R} = 2 \text{ A}$ ... (ii).
Dividing equation $(i)$ by equation (ii),we get:
$\frac{3}{2} = \frac{14 - 2m}{10 - 2m}$
$3(10 - 2m) = 2(14 - 2m)$
$30 - 6m = 28 - 4m$
$2 = 2m$
$m = 1$.
Therefore,the number of wrongly connected cells is $1$.

(A) In the steady state,no current flows through the capacitor branch.
Let the common junction point of the resistors and capacitor be $A$ and the common junction of the negative terminals of the batteries be $B$. Let the potential at $B$ be $0 \text{ V}$.
Let the potential at $A$ be $V_A$.
The current $i$ flows through the top and bottom branches:
$i = \frac{(2E - E)}{(r + 2r)} = \frac{E}{3r}$.
The potential at $A$ can be calculated using the top branch:
$V_A - 0 = E - i \cdot r = E - (\frac{E}{3r}) \cdot r = E - \frac{E}{3} = \frac{2E}{3}$.
Since no current flows through the middle branch,the potential drop across the capacitor is equal to the potential difference between $A$ and the positive terminal of the battery $Q$ (which is at potential $E$ relative to $B$).
Potential drop across capacitor $V_C = |E - V_A| = |E - \frac{2E}{3}| = \frac{E}{3}$.

(A) In oxyhaemoglobin,$Fe^{2+}$ is in a low-spin state,which is diamagnetic. Therefore,the statement that $Fe^{2+}$ is paramagnetic in oxyhaemoglobin is incorrect.

(C) Spin only magnetic moments depend upon the number of unpaired electrons; more the number of unpaired electrons,greater will be the spin only magnetic moment.
For $Mn$ $(Z=25)$: $[Ar] 3d^5 4s^2$. Thus,$Mn^{2+} = [Ar] 3d^5$. Number of unpaired electrons = $5$.
For $Cr$ $(Z=24)$: $[Ar] 3d^5 4s^1$. Thus,$Cr^{2+} = [Ar] 3d^4$. Number of unpaired electrons = $4$.
For $V$ $(Z=23)$: $[Ar] 3d^3 4s^2$. Thus,$V^{2+} = [Ar] 3d^3$. Number of unpaired electrons = $3$.
Since the magnetic moment $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons,the order is $Mn^{2+} (5) > Cr^{2+} (4) > V^{2+} (3)$.

(A) When ethyl alcohol $(X)$ is oxidised by acidified potassium dichromate,acetic acid $(Y)$ is obtained as follows:
$3 CH_3CH_2OH(X) + 2 K_2Cr_2O_7 + 8 H_2SO_4 \longrightarrow 3 CH_3COOH(Y) + 2 Cr_2(SO_4)_3 + 2 K_2SO_4 + 11 H_2O$
Carboxylic acids undergo reduction with $LiAlH_4$ to give primary alcohols as follows:
$CH_3COOH(Y) \xrightarrow{LiAlH_4, \text{ether}} CH_3CH_2OH(X)$
Thus,$X$ is $C_2H_5OH$ and $Y$ is $CH_3COOH$.

(C) Given: $i = 96.5 \ A$,$t = 100 \ s$,Atomic weight of $Ag = 108 \ g/mol$.
Charge $Q = i \times t = 96.5 \times 100 = 9650 \ C$.
According to Faraday's first law of electrolysis,$w = \frac{M \times i \times t}{n \times F}$.
For $Ag^+ + e^- \rightarrow Ag$,$n = 1$.
$w = \frac{108 \times 9650}{1 \times 96500} = 10.8 \ g$.
Thus,Assertion $(A)$ is true.
Reason $(R)$ states that mass is inversely proportional to the quantity of electricity,which is incorrect. According to Faraday's law,the mass deposited is directly proportional to the quantity of electricity $(w \propto Q)$.
Therefore,$(A)$ is true but $(R)$ is false.

(B) The reaction for the deposition of silver is: $Ag^+ + e^- \rightarrow Ag(s)$.
Number of moles of $Ag^+$ ions $= \text{Molarity} \times \text{Volume in Liters} = 1 \ M \times 0.125 \ L = 0.125 \ mol$.
Since $1 \ mol$ of $Ag^+$ requires $1 \ mol$ of electrons $(1 \ F)$ for deposition,the total charge $Q$ required is $0.125 \ F$.
$Q = 0.125 \times 96500 \ C = 12062.5 \ C$.
Using the formula $Q = I \times t$,where $I = 241.25 \ A$:
$t = \frac{Q}{I} = \frac{12062.5 \ C}{241.25 \ A} = 50 \ sec$.

(B) Given standard reduction potentials:
$E^{\circ}_{Zn^{2+}|Zn} = -0.76 \ V$
$E^{\circ}_{Cu^{2+}|Cu} = 0.34 \ V$
$E^{\circ}_{Ag^{+}|Ag} = 0.8 \ V$
For cell $(1)$: $Zn|Zn^{2+}||Cu^{2+}|Cu$
$E^{\circ}_{cell(1)} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.34 - (-0.76) = 1.10 \ V$
For cell $(2)$: $Zn|Zn^{2+}||Ag^{+}|Ag$
$E^{\circ}_{cell(2)} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.8 - (-0.76) = 1.56 \ V$
For cell $(3)$: $Cu|Cu^{2+}||Ag^{+}|Ag$
$E^{\circ}_{cell(3)} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.8 - 0.34 = 0.46 \ V$
Comparing the values: $1.56 \ V (2) > 1.10 \ V (1) > 0.46 \ V (3)$.
Therefore,the correct order is $2 > 1 > 3$.

(C) The magnetic field $B$ at a distance $r$ from an infinitely long wire is given by $B = \frac{\mu_0 i}{2 \pi r}$.
The loop has two sides of length $b$ parallel to the wire at distances $x$ and $x+l$ from the wire.
The motional emf induced in a conductor moving in a magnetic field is $e = B v L$.
The emf induced in the side at distance $x$ is $e_1 = B_1 v b = \left(\frac{\mu_0 i}{2 \pi x}\right) v b$.
The emf induced in the side at distance $x+l$ is $e_2 = B_2 v b = \left(\frac{\mu_0 i}{2 \pi (x+l)}\right) v b$.
Since the loop moves away,these emfs oppose each other. The net magnitude of the emf is $|e| = |e_1 - e_2|$.
$|e| = \frac{\mu_0 i v b}{2 \pi} \left( \frac{1}{x} - \frac{1}{x+l} \right) = \frac{\mu_0 i v b}{2 \pi} \left( \frac{x+l-x}{x(x+l)} \right) = \frac{\mu_0 i l b v}{2 \pi x(x+l)}$.

(D) The magnetic field $B$ produced by a square loop of side $L$ carrying current $I$ at its centre is given by the sum of fields due to four wires of length $L$ at a distance $d = L/2$ from the centre.
For one wire,the field is $B_{wire} = \frac{\mu_0 I}{4 \pi d} (\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 I}{4 \pi (L/2)} (2 \times \frac{1}{\sqrt{2}}) = \frac{\mu_0 I}{\sqrt{2} \pi L}$.
Since there are four such wires,the total magnetic field at the centre is $B = 4 \times \frac{\mu_0 I}{\sqrt{2} \pi L} = \frac{2 \sqrt{2} \mu_0 I}{\pi L}$.
Since $L \gg l$,we can assume the magnetic field $B$ is uniform over the area of the smaller loop $S_2 = l^2$.
The magnetic flux linked with the smaller loop is $\phi_2 = B \times S_2 = \frac{2 \sqrt{2} \mu_0 I}{\pi L} \times l^2$.
The mutual inductance $M$ is defined as $M = \frac{\phi_2}{I} = \frac{2 \sqrt{2} \mu_0}{\pi} \frac{l^2}{L}$.
Therefore,$M \propto \frac{l^2}{L}$.

(B) The charges are located at $x=0$ $(q/2)$,$x=a$ $(-q)$,and $x=2a$ $(q/2)$.
The point $P$ is at a distance $r$ from the charge $-q$ at $x=a$. Since $P$ is on the $x$-axis to the right of the charges,its coordinate is $x_P = a + r$.
The distances of point $P$ from the three charges are:
$1$. From $q/2$ at $x=0$: $d_1 = (a+r) - 0 = r+a$
$2$. From $-q$ at $x=a$: $d_2 = (a+r) - a = r$
$3$. From $q/2$ at $x=2a$: $d_3 = (a+r) - 2a = r-a$
The total electric potential $V$ at point $P$ is the sum of potentials due to individual charges:
$V = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q/2}{r+a} - \frac{q}{r} + \frac{q/2}{r-a} \right]$
$V = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{1}{r+a} - \frac{2}{r} + \frac{1}{r-a} \right]$
$V = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{r(r-a) - 2(r^2-a^2) + r(r+a)}{r(r^2-a^2)} \right]$
$V = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{r^2 - ar - 2r^2 + 2a^2 + r^2 + ar}{r(r^2-a^2)} \right]$
$V = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{2a^2}{r(r^2-a^2)} \right]$
Since $r \gg a$,we can approximate $r^2 - a^2 \approx r^2$:
$V \approx \frac{q}{8 \pi \varepsilon_0} \cdot \frac{2a^2}{r^3} = \frac{q a^2}{4 \pi \varepsilon_0 r^3}$

(B) The electric potential $V$ at a point due to a system of charges is the algebraic sum of the potentials due to individual charges.
The positions of the charges are $x_1 = 0$,$x_2 = a$,and $x_3 = 2a$. The point $P$ is at $x = a+r$.
The distances of point $P$ from the charges are:
$r_1 = (a+r) - 0 = r+a$
$r_2 = (a+r) - a = r$
$r_3 = (a+r) - 2a = r-a$
The total potential $V_P$ is:
$V_P = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q/2}{r+a} - \frac{q}{r} + \frac{q/2}{r-a} \right]$
$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{1}{2(r+a)} - \frac{1}{r} + \frac{1}{2(r-a)} \right]$
$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{r(r-a) - 2(r^2-a^2) + r(r+a)}{2r(r^2-a^2)} \right]$
$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{r^2 - ar - 2r^2 + 2a^2 + r^2 + ar}{2r(r^2-a^2)} \right]$
$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{2a^2}{2r(r^2-a^2)} \right] = \frac{q a^2}{4 \pi \varepsilon_0 r(r^2-a^2)}$
Since $a \ll r$,we have $r^2 - a^2 \approx r^2$.
Therefore,$V_P = \frac{q a^2}{4 \pi \varepsilon_0 r^3}$.

(D) Chlorofluorocarbons $(CFCs)$ are responsible for the depletion of the ozone layer in the stratosphere.
$CFCl_3$ decomposes in the presence of ultraviolet radiation to produce chlorine free radicals $(Cl^{\bullet})$.
These chlorine free radicals react with ozone $(O_3)$ to form chlorine monoxide $(ClO^{\bullet})$ and oxygen $(O_2)$:
$Cl^{\bullet} + O_3 \rightarrow ClO^{\bullet} + O_2$
Thus,the chlorine free radical $(Cl^{\bullet})$ is the species that reacts with ozone.

(A) The chemical formula of carnallite is $KCl \cdot MgCl_2 \cdot 6 H_2 O$.
Therefore,the metal ions present in carnallite are $K^{+}$ and $Mg^{2+}$.

(B) Chloroform $(CHCl_3)$ undergoes aerial oxidation in the presence of light to form a highly poisonous gas called phosgene $(COCl_2)$.
$2CHCl_3 + O_2 \xrightarrow{light} 2COCl_2 + 2HCl$
To prevent this,a small amount of ethyl alcohol $(C_2H_5OH)$ is added to chloroform.
Ethyl alcohol acts as a negative catalyst and converts any phosgene formed back into harmless diethyl carbonate.

(C) Only alcoholic $KOH$ undergoes dehydrohalogenation reaction with alkyl halides. When ethyl chloride $(CH_3CH_2Cl)$ is heated with alcoholic $KOH$,it eliminates a molecule of $HCl$ to form ethylene $(CH_2=CH_2)$.
The reaction is as follows:
$CH_3-CH_2Cl + KOH (\text{alc.}) \rightarrow CH_2=CH_2 + KCl + H_2O$

(D) Friedel-Crafts acylation involves the reaction of benzene with an acyl halide (such as $CH_3COCl$) or an acid anhydride in the presence of a Lewis acid catalyst like anhydrous aluminum chloride $(AlCl_3)$.
The reaction is as follows:
$C_6H_6 + CH_3COCl \xrightarrow{AlCl_3} C_6H_5COCH_3 + HCl$
Thus,benzene reacts with acetyl chloride $(CH_3COCl)$ to form acetophenone.

(C) $30 \%$ solution of hydrogen peroxide can be obtained by the electrolysis of $50 \%$ sulphuric acid followed by vacuum distillation.
The first product of electrolysis is perdisulphuric acid $(H_2S_2O_8)$,which reacts with water during distillation to form $H_2O_2$.
$2H_2SO_4 \longrightarrow 2H^{+} + 2HSO_4^-$
$2HSO_4^- \longrightarrow H_2S_2O_8 + 2e^-$ (At anode)
$H_2S_2O_8 + 2H_2O \longrightarrow 2H_2SO_4 + H_2O_2$
Here,'$X$' is $H_2SO_4$ and '$Y$' is $H_2S_2O_8$.
$H_2SO_4$ (sulphuric acid) contains zero peroxy bonds.
$H_2S_2O_8$ (Marshall's acid) contains one peroxy bond $(-O-O-)$.
Therefore,the number of peroxy bonds in $X$ and $Y$ are zero and $1$ respectively.

(B) When the block is lowered slowly,the equilibrium position is reached when the spring force equals the gravitational force: $k d = m g$,which implies $d = \frac{m g}{k}$.
When the block is allowed to fall suddenly from the unstretched position,the block undergoes simple harmonic motion. At the maximum extension $x$,the change in gravitational potential energy equals the elastic potential energy stored in the spring:
$m g x = \frac{1}{2} k x^2$
Solving for $x$ (where $x \neq 0$):
$x = \frac{2 m g}{k}$
Since $d = \frac{m g}{k}$,we substitute this into the equation:
$x = 2 d$.

(A) Let $M_1$ be the magnetic moment of magnet $AB$ and $M_2$ be the magnetic moment of magnet $CD$. Point $P(2,2)$ lies on the axial line of magnet $AB$ at a distance $r_1 = 2$ from its center $(0,2)$,and on the equatorial line of magnet $CD$ at a distance $r_2 = 2$ from its center $(2,0)$.
The magnetic field at $P$ due to $AB$ (axial) is $B_1 = \frac{\mu_0}{4\pi} \frac{2M_1}{r_1^3} = 10^{-7} \times \frac{2M_1}{2^3} = 10^{-7} \times \frac{M_1}{4}$.
The magnetic field at $P$ due to $CD$ (equatorial) is $B_2 = \frac{\mu_0}{4\pi} \frac{M_2}{r_2^3} = 10^{-7} \times \frac{M_2}{2^3} = 10^{-7} \times \frac{M_2}{8}$.
Given the resultant field $B = B_1 + B_2 = 100 \times 10^{-7} \ T$:
$10^{-7} (\frac{M_1}{4} + \frac{M_2}{8}) = 100 \times 10^{-7} \Rightarrow 2M_1 + M_2 = 800$ (Eq. $i$).
When the poles of $CD$ are reversed,the direction of $B_2$ is reversed,so $B' = B_1 - B_2 = 50 \times 10^{-7} \ T$:
$10^{-7} (\frac{M_1}{4} - \frac{M_2}{8}) = 50 \times 10^{-7} \Rightarrow 2M_1 - M_2 = 400$ (Eq. $ii$).
Adding (Eq. $i$) and (Eq. $ii$): $4M_1 = 1200 \Rightarrow M_1 = 300 \ Am^2$.
Substituting $M_1$ in (Eq. $i$): $2(300) + M_2 = 800 \Rightarrow M_2 = 200 \ Am^2$.

(B) The radius of a circular path of a charged particle in a magnetic field is given by $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$.
Since kinetic energy $K$ and magnetic field $B$ are the same for all particles, we have $r \propto \frac{\sqrt{m}}{q}$.
For a proton $(p)$: mass $m_p = m$, charge $q_p = q$. So, $r_p \propto \frac{\sqrt{m}}{q}$.
For a deuteron $(d)$: mass $m_d = 2m$, charge $q_d = q$. So, $r_d \propto \frac{\sqrt{2m}}{q}$.
For an $\alpha$-particle $(\alpha)$: mass $m_\alpha = 4m$, charge $q_\alpha = 2q$. So, $r_\alpha \propto \frac{\sqrt{4m}}{2q} = \frac{2\sqrt{m}}{2q} = \frac{\sqrt{m}}{q}$.
Thus, the ratio $r_p : r_d : r_\alpha = \frac{\sqrt{m}}{q} : \frac{\sqrt{2m}}{q} : \frac{\sqrt{m}}{q} = 1 : \sqrt{2} : 1$.

(D) The magnetic force on a charged particle is given by $\vec{F} = q(\vec{v} \times \vec{B})$.
$1$. If the velocity $\vec{v}$ is parallel or anti-parallel to the magnetic field $\vec{B}$,then $\vec{v} \times \vec{B} = 0$,so $\vec{F} = 0$. The particle continues in a straight line.
$2$. If the velocity $\vec{v}$ is perpendicular to the magnetic field $\vec{B}$,the force acts as a centripetal force,causing the particle to move in a circle.
$3$. If the velocity $\vec{v}$ makes an angle $\theta$ (where $\theta \neq 0^\circ, 90^\circ, 180^\circ$) with the magnetic field $\vec{B}$,the velocity component parallel to the field remains constant,while the perpendicular component causes circular motion. This results in a helical trajectory.
Therefore,all three trajectories are possible depending on the angle between $\vec{v}$ and $\vec{B}$.

(B) When a magnetic needle is placed in a uniform magnetic field $B$,the north pole experiences a force $F = mB$ and the south pole experiences a force $F = -mB$,where $m$ is the pole strength.
Since the forces are equal in magnitude and opposite in direction,the net force $F_{net} = F + (-F) = 0$.
However,because these forces act at different points,they create a torque $\tau = p_m \times B$,which tends to rotate the needle to align it with the magnetic field.
Therefore,torque is present but there is no net force.

(C) Given expression: $\sqrt{12-\sqrt{68+48 \sqrt{2}}}$
We simplify the inner radical $\sqrt{68+48 \sqrt{2}}$:
$68+48 \sqrt{2} = 68+2 \times 24 \times \sqrt{2} = 68+2 \times 6 \times 4 \sqrt{2}$
$= (6)^2 + (4 \sqrt{2})^2 + 2 \times 6 \times 4 \sqrt{2} = (6+4 \sqrt{2})^2$
So,$\sqrt{68+48 \sqrt{2}} = 6+4 \sqrt{2}$
Now substitute this back into the expression:
$\sqrt{12-(6+4 \sqrt{2})} = \sqrt{6-4 \sqrt{2}}$
$= \sqrt{6-2 \times 2 \times 2 \sqrt{2}} = \sqrt{6-2 \times 2 \times \sqrt{8}}$
$= \sqrt{(2)^2 + (\sqrt{2})^2 - 2 \times 2 \times \sqrt{2}} = \sqrt{(2-\sqrt{2})^2}$
$= 2-\sqrt{2}$

(B) Given that $\alpha, \beta, \gamma$ are the roots of the equation $x^3-6x^2+11x+6=0$.
From the relation between roots and coefficients:
$\alpha+\beta+\gamma = 6$
$\alpha \beta+\beta \gamma+\gamma \alpha = 11$
$\alpha \beta \gamma = -6$
We need to find the value of $\Sigma \alpha^2 \beta+\Sigma \alpha \beta^2$.
$\Sigma \alpha^2 \beta+\Sigma \alpha \beta^2 = \alpha^2 \beta+\beta^2 \gamma+\gamma^2 \alpha+\alpha \beta^2+\beta \gamma^2+\gamma \alpha^2$
$= \alpha \beta(\alpha+\beta)+\beta \gamma(\beta+\gamma)+\gamma \alpha(\gamma+\alpha)$
$= \alpha \beta(6-\gamma)+\beta \gamma(6-\alpha)+\gamma \alpha(6-\beta)$
$= 6(\alpha \beta+\beta \gamma+\gamma \alpha)-3 \alpha \beta \gamma$
$= 6(11)-3(-6)$
$= 66+18 = 84$.

(A) Given roots are $\alpha = \sin ^2 18^{\circ}$ and $\beta = \cos ^2 36^{\circ}$.
We know that $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$.
Sum of roots $= \sin ^2 18^{\circ} + \cos ^2 36^{\circ} = \left(\frac{\sqrt{5}-1}{4}\right)^2 + \left(\frac{\sqrt{5}+1}{4}\right)^2$.
$= \frac{5+1-2\sqrt{5}}{16} + \frac{5+1+2\sqrt{5}}{16} = \frac{12}{16} = \frac{3}{4}$.
Product of roots $= \sin ^2 18^{\circ} \cdot \cos ^2 36^{\circ} = \left(\frac{\sqrt{5}-1}{4}\right)^2 \cdot \left(\frac{\sqrt{5}+1}{4}\right)^2$.
$= \left(\frac{(\sqrt{5}-1)(\sqrt{5}+1)}{16}\right)^2 = \left(\frac{5-1}{16}\right)^2 = \left(\frac{4}{16}\right)^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16}$.
The quadratic equation is given by $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
$x^2 - \frac{3}{4}x + \frac{1}{16} = 0$.
Multiplying by $16$,we get $16x^2 - 12x + 1 = 0$.

(C) Let $z = 1+i\sqrt{3}$. The roots are given by $z_k = r^{1/2n} e^{i(\theta + 2k\pi)/2n}$ for $k = 0, 1, \dots, 2n-1$,where $r = |z| = 2$ and $\theta = \pi/3$.
The product of the $m$ roots of $z^m = A$ is given by $(-1)^{m-1} (-A)$.
Here,$m = 2n$ and $A = 1+i\sqrt{3}$.
Product $= (-1)^{2n-1} (-(1+i\sqrt{3}))$.
Since $2n-1$ is odd,$(-1)^{2n-1} = -1$.
Product $= (-1) \times (-(1+i\sqrt{3})) = 1+i\sqrt{3}$.

(C) The terminal velocity $v$ of an air bubble rising in a liquid is given by the formula: $v = \frac{2}{9} \frac{r^2 \rho g}{\eta}$,where $\rho$ is the density of the liquid,$r$ is the radius,and $\eta$ is the coefficient of viscosity.
Rearranging for $\eta$: $\eta = \frac{2}{9} \frac{r^2 \rho g}{v}$.
Given values: $r = 1 \ cm = 10^{-2} \ m$,$\rho = 1.5 \ g/cc = 1.5 \times 10^3 \ kg/m^3$,$v = 0.25 \ cm/s = 0.25 \times 10^{-2} \ m/s$,and $g = 9.8 \ m/s^2$.
Substituting the values: $\eta = \frac{2}{9} \cdot \frac{(10^{-2})^2 \cdot (1.5 \times 10^3) \cdot 9.8}{0.25 \times 10^{-2}}$.
$\eta = \frac{2}{9} \cdot \frac{10^{-4} \cdot 1500 \cdot 9.8}{0.0025} = \frac{2}{9} \cdot \frac{0.15 \cdot 9.8}{0.0025} = \frac{2}{9} \cdot \frac{1.47}{0.0025} = \frac{2}{9} \cdot 588 \approx 130.6 \ Pa \cdot s$.
Thus,the coefficient of viscosity is approximately $130 \ Pa \cdot s$.

(A) The stress in the wire is given by $\text{Stress} = \frac{\text{Tension}}{\text{Area of cross-section}}$.
To avoid breaking,the stress in the wire must not exceed the breaking stress.
Let the tension in the wire be $T$ and the acceleration of the system be $a$.
The equations of motion for the two blocks are:
For the $1 ~kg$ block: $T - 1(10) = 1a \implies T - 10 = a$
For the $2 ~kg$ block: $2(10) - T = 2a \implies 20 - T = 2a$
Adding the two equations: $(T - 10) + (20 - T) = a + 2a \implies 10 = 3a \implies a = \frac{10}{3} ~m/s^2$.
Substituting $a$ into the first equation: $T = 10 + \frac{10}{3} = \frac{40}{3} ~N$.
The area of cross-section of the wire is $A = \pi r^2$.
The stress is $\frac{T}{A} = \frac{40/3}{\pi r^2}$.
Equating this to the breaking stress: $\frac{40/3}{\pi r^2} = 2 \times 10^9$.
$r^2 = \frac{40}{3 \times \pi \times 2 \times 10^9} = \frac{20}{3 \pi \times 10^9} \approx \frac{20}{9.4247 \times 10^9} \approx 2.122 \times 10^{-9} ~m^2$.
$r = \sqrt{2.122 \times 10^{-9}} \approx 4.606 \times 10^{-5} ~m$.

(B) The formula for the horizontal range $R$ is given by $R = \frac{u^2 \sin 2\theta}{g}$.
Given $R = 20 \ m$ and $\theta = 30^{\circ}$,we have $20 = \frac{u^2 \sin(60^{\circ})}{g}$.
Thus,$\frac{u^2}{g} = \frac{20}{\sin 60^{\circ}} = \frac{20}{\sqrt{3}/2} = \frac{40}{\sqrt{3}}$.
The formula for maximum height $H$ is $H = \frac{u^2 \sin^2 \theta}{2g}$.
Substituting the values,$H = \frac{1}{2} \times \left(\frac{u^2}{g}\right) \times \sin^2 30^{\circ}$.
$H = \frac{1}{2} \times \frac{40}{\sqrt{3}} \times \left(\frac{1}{2}\right)^2$.
$H = \frac{20}{\sqrt{3}} \times \frac{1}{4} = \frac{5}{\sqrt{3}} \ m$.

(A) Let $\vec{v}_{r,g}$ be the velocity of rain with respect to the ground,$\vec{v}_{m,g}$ be the velocity of the man with respect to the ground,and $\vec{v}_{r,m}$ be the velocity of rain with respect to the man.
When the man is at rest,the rain appears to fall at $30^{\circ}$ with the vertical,which is the direction of $\vec{v}_{r,g}$.
When the man runs at $10 ~km/h$,the rain hits him vertically,meaning the horizontal component of $\vec{v}_{r,m}$ is zero.
The relative velocity is given by $\vec{v}_{r,m} = \vec{v}_{r,g} - \vec{v}_{m,g}$.
For $\vec{v}_{r,m}$ to be vertical,the horizontal component of $\vec{v}_{r,g}$ must be equal to the horizontal component of $\vec{v}_{m,g}$.
Thus,$v_{r,g} \sin 30^{\circ} = v_{m,g} = 10 ~km/h$.
$v_{r,g} = \frac{10}{\sin 30^{\circ}} = \frac{10}{0.5} = 20 ~km/h$.

(C) In this system,the fixed bob is located directly above the oscillating bob.
Since the string is vertical,the electrostatic force of repulsion between the two identical charges $q$ acts along the line of the string.
This electrostatic force acts in the same direction as the tension in the string and does not have any component perpendicular to the string.
Therefore,the restoring force for the small oscillations of the pendulum remains solely dependent on the component of gravity,$mg \sin \theta$.
Since the electrostatic force does not affect the restoring torque or the effective acceleration due to gravity for the oscillating bob,the time period remains unchanged.
Thus,the time period is $T = 2 \pi \sqrt{\frac{l}{g}}$.

(A) Silica,represented as $SiO_2$,is widely used in the manufacturing of optical instruments due to its high transparency and thermal stability.

(C) The hydrolysis of nitrogen trichloride $(NCl_3)$ involves the reaction with water to produce ammonia $(NH_3)$ and hypochlorous acid $(HOCl)$.
The balanced chemical equation is:
$NCl_3 + 3H_2O \longrightarrow NH_3 + 3HOCl$
Therefore,$X$ is $HOCl$.

(A) The formation of ozone from oxygen is an endothermic reaction,not an exothermic reaction.
$3 O_2 \underset{\text{silent electric discharge}}{\rightleftharpoons} 2 O_3 \quad \Delta H = +284.5 \ kJ \ \text{mol}^{-1}$.
Therefore,the statement $3 O_2 \underset{\text{silent electric discharge}}{\rightleftharpoons} 2 O_3; \Delta H = -284.5 \ kJ$ is incorrect because the enthalpy change is positive,not negative.

(D) $I$. Bleaching powder $(CaOCl_2)$ reacts with ethanol or acetone to produce chloroform $(CHCl_3)$. This is a standard laboratory preparation method.
$II$. Bleaching powder decomposes in the presence of cobalt chloride $(CoCl_2)$ catalyst to release oxygen gas: $2CaOCl_2 \xrightarrow{CoCl_2} 2CaCl_2 + O_2$.
$III$. Fluorine is prepared by the electrolysis of a fused mixture of potassium hydrogen fluoride $(KHF_2)$ and anhydrous hydrogen fluoride $(HF)$. Aqueous $KHF_2$ cannot be used because water would be oxidized to oxygen instead of fluoride ions being oxidized to fluorine gas.
Therefore,statements $I$ and $II$ are correct.

(B) The abundance of noble gases in the atmosphere by volume/weight is as follows:
$Ar \approx 0.93\%$,$Ne \approx 0.0018\%$,$Kr \approx 0.00011\%$.
Therefore,the correct order of occurrence is $Ar > Ne > Kr$.

(A) In step-$1$,$CH_3 CH_2 OH$ is converted to $CH_3 CHO$ by $Cl_2$. This is an oxidation reaction where the primary alcohol group $(-CH_2 OH)$ is oxidized to an aldehyde group $(-CHO)$.
In step-$2$,$CH_3 CHO$ reacts with $3 Cl_2$ to form $Cl_3 C CHO$. This is a chlorination reaction where the hydrogen atoms of the methyl group are substituted by chlorine atoms.

(B) The correct matches are as follows:
$A$. Grignard reagent is $CH_3MgX$ $(3)$.
$B$. Clemmensen reduction uses $Zn-Hg / conc. HCl$ $(4)$.
$C$. Rosenmund reduction uses $H_2 / Pd-BaSO_4$ $(1)$.
$D$. Wolff-Kishner reduction uses $N_2H_4 / KOH$ in ethylene glycol $(2)$.
Thus,the correct sequence is $A-3, B-4, C-1, D-2$.

(C) For a $1^{\text{st}}$ order reaction,the integrated rate equation is given by:
$t = \frac{2.303}{k} \log \left(\frac{a}{a-x}\right)$
For the $\left(\frac{3}{4}\right)^{\text{th}}$ life,$x = \frac{3}{4}a$.
Substituting this value into the equation:
$t_{3/4} = \frac{2.303}{k} \log \left(\frac{a}{a - \frac{3}{4}a}\right)$
$t_{3/4} = \frac{2.303}{k} \log \left(\frac{a}{\frac{1}{4}a}\right)$
$t_{3/4} = \frac{2.303}{k} \log (4)$

(A) In step-$1$: $CH_3 CH_2 OH + Cl_2 \rightarrow CH_3 CHO + 2HCl$. Here,$Cl_2$ acts as an oxidizing agent,converting the primary alcohol $(-CH_2OH)$ into an aldehyde $(-CHO)$.
In step-$2$: $CH_3 CHO + 3Cl_2 \rightarrow CCl_3 CHO + 3HCl$. Here,$Cl_2$ replaces the hydrogen atoms of the methyl group with chlorine atoms,which is a chlorination reaction (specifically,a substitution reaction).

(B) The reaction of ethyl chloride $(C_2H_5Cl)$ with sodium ethoxide $(C_2H_5ONa)$ is a Williamson ether synthesis,which produces diethyl ether $(C_2H_5-O-C_2H_5)$ as compound $A$.
The reaction is: $C_2H_5Cl + C_2H_5ONa \rightarrow C_2H_5-O-C_2H_5 + NaCl$.
Diethyl ether is also obtained by the intermolecular dehydration of ethyl alcohol $(C_2H_5OH)$ in the presence of concentrated $H_2SO_4$ at $140^{\circ}C$ $(413 \ K)$:
$2C_2H_5OH \xrightarrow{conc. H_2SO_4, 140^{\circ}C} C_2H_5-O-C_2H_5 + H_2O$.

(B) The correct matches are:
$A$. Grignard reagent is $CH_3MgX$ $(3)$.
$B$. Clemmensen reduction uses $Zn-Hg / \text{conc. } HCl$ $(4)$.
$C$. Rosenmund reduction uses $H_2 / Pd-BaSO_4$ $(1)$.
$D$. Wolff-Kishner reduction uses $N_2H_4 / KOH / (CH_2OH)_2$ $(2)$.
Thus,the correct matching is $A-3, B-4, C-1, D-2$.

(A) When acetone $(CH_3COCH_3)$ is treated with a base like barium hydroxide $(Ba(OH)_2)$,it undergoes an aldol condensation reaction.
Two molecules of acetone react to form $4$-hydroxy-$4$-methylpentan-$2$-one,which is commonly known as diacetone alcohol.
The reaction is as follows:
$2CH_3COCH_3 \xrightarrow{Ba(OH)_2} CH_3-CO-CH_2-C(OH)(CH_3)_2$
Therefore,the correct product is represented by option $A$.

(A) When acetaldehyde $(CH_3CHO)$ is heated with Fehling solution,it undergoes oxidation to form acetate ions,while the $Cu^{2+}$ ions in the Fehling solution are reduced to $Cu^+$ ions.
This results in the formation of a red precipitate of cuprous oxide $(Cu_2O)$.
The chemical reaction is:
$CH_3CHO + 2Cu(OH)_2 + NaOH \longrightarrow CH_3COONa + Cu_2O \downarrow (\text{Red}) + 3H_2O$

(D) The reduction of nitrobenzene $(C_6H_5NO_2)$ in an acidic medium using $Zn / HCl$ results in the formation of aniline $(C_6H_5NH_2)$ as the main product.
The chemical reaction is:
$C_6H_5NO_2 + 6[H] \xrightarrow{Zn/HCl} C_6H_5NH_2 + 2H_2O$

(C) The reaction of ethers with carbon monoxide $(CO)$ in the presence of a Lewis acid catalyst like $BF_3$ at high temperature $(150^{\circ} C)$ and high pressure $(500 \ atm)$ is a carbonylation reaction.
Diethyl ether $(C_2H_5OC_2H_5)$ reacts with $CO$ to form ethyl propionate $(C_2H_5COOC_2H_5)$ according to the following equation:
$C_2H_5OC_2H_5 + CO \xrightarrow{BF_3, 150^{\circ} C, 500 \ atm} C_2H_5COOC_2H_5$.

(A) In oxyhaemoglobin,$Fe^{2+}$ is in a low-spin state,which makes it diamagnetic. Therefore,the statement that $Fe^{2+}$ is paramagnetic in oxyhaemoglobin is incorrect.

(C) Spin only magnetic moments depend upon the number of unpaired electrons; the greater the number of unpaired electrons,the greater will be the spin only magnetic moment.
$Mn$ $(Z=25)$: $[Ar] 3d^5 4s^2$
$Mn^{2+}$: $[Ar] 3d^5$,number of unpaired electrons = $5$
$Cr$ $(Z=24)$: $[Ar] 3d^5 4s^1$
$Cr^{2+}$: $[Ar] 3d^4$,number of unpaired electrons = $4$
$V$ $(Z=23)$: $[Ar] 3d^3 4s^2$
$V^{2+}$: $[Ar] 3d^3$,number of unpaired electrons = $3$
Since the magnetic moment $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons,the order of magnetic moment is $Mn^{2+} > Cr^{2+} > V^{2+}$.

(A) When ethyl alcohol $(X)$ is oxidised by acidified potassium dichromate,acetic acid $(Y)$ is formed:
$3 CH_3 CH_2 OH + 2 K_2 Cr_2 O_7 + 8 H_2 SO_4 \longrightarrow 3 CH_3 COOH + 2 Cr_2(SO_4)_3 + 2 K_2 SO_4 + 11 H_2 O$
Carboxylic acids undergo reduction with $LiAlH_4$ to give primary alcohols:
$CH_3 COOH \xrightarrow{LiAlH_4} CH_3 CH_2 OH$
Thus,$X$ is $C_2 H_5 OH$ and $Y$ is $CH_3 COOH$.

(C) Given: $i = 96.5 \ A$,$t = 100 \ s$,Atomic weight of $Ag = 108 \ g/mol$.
Using Faraday's law: $Q = i \times t = 96.5 \ A \times 100 \ s = 9650 \ C$.
According to the reaction $Ag^+ + e^- \rightarrow Ag$,$1 \ mole$ of $e^-$ $(96500 \ C)$ deposits $108 \ g$ of $Ag$.
Mass of $Ag$ deposited $= \frac{108 \times 9650}{96500} = 10.8 \ g$.
Thus,statement $(A)$ is true.
Faraday's first law of electrolysis states that the mass of a substance deposited is directly proportional to the quantity of electricity passed $(w = Z \times Q)$.
Therefore,statement $(R)$ is false.

(B) The reaction for the deposition of silver is: $Ag^+ + e^- \rightarrow Ag(s)$.
Number of moles of $Ag^+$ in $125 \ mL$ of $1 \ M \ AgNO_3$ solution is: $n = M \times V(L) = 1 \times 0.125 = 0.125 \ mol$.
Since $1 \ mol$ of $Ag^+$ requires $1 \ mol$ of electrons $(1 \ F)$ to deposit,$0.125 \ mol$ of $Ag^+$ requires $0.125 \ F$.
Total charge $Q = 0.125 \times 96500 \ C = 12062.5 \ C$.
Using the formula $Q = I \times t$,where $I = 241.25 \ A$:
$t = \frac{Q}{I} = \frac{12062.5}{241.25} = 50 \ sec$.

(B) Given standard reduction potentials:
$E^{\circ}_{Zn^{2+}|Zn} = -0.76 \ V$
$E^{\circ}_{Cu^{2+}|Cu} = 0.34 \ V$
$E^{\circ}_{Ag^{+}|Ag} = 0.80 \ V$
For cell $(1)$: $Zn|Zn^{2+}||Cu^{2+}|Cu$
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.34 - (-0.76) = 1.10 \ V$
For cell $(2)$: $Zn|Zn^{2+}||Ag^{+}|Ag$
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.80 - (-0.76) = 1.56 \ V$
For cell $(3)$: $Cu|Cu^{2+}||Ag^{+}|Ag$
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.80 - 0.34 = 0.46 \ V$
Comparing the values: $1.56 \ V (2) > 1.10 \ V (1) > 0.46 \ V (3)$.
Therefore,the correct order is $2 > 1 > 3$.

(B) Chloroform $(CHCl_3)$ undergoes aerial oxidation in the presence of light to form a highly poisonous gas called phosgene $(COCl_2)$.
$2CHCl_3 + O_2 \xrightarrow{light} 2COCl_2 + 2HCl$
To prevent this,$1\%$ of ethyl alcohol $(C_2H_5OH)$ is added to chloroform.
Ethyl alcohol acts as a negative catalyst and converts any phosgene formed back into harmless diethyl carbonate.

(C) Only alcoholic $KOH$ undergoes a dehydrohalogenation reaction with alkyl halides. When ethyl chloride $(CH_3CH_2Cl)$ is heated with alcoholic $KOH$,it undergoes elimination to form ethylene $(CH_2=CH_2)$.
The reaction is as follows:
$CH_3CH_2Cl + KOH (alc.) \rightarrow CH_2=CH_2 + KCl + H_2O$

(C) The hydrolysis of nitrogen trichloride $(NCl_3)$ involves the reaction with water to produce ammonia $(NH_3)$ and hypochlorous acid $(HOCl)$.
The balanced chemical equation is:
$NCl_3 + 3H_2O \longrightarrow NH_3 + 3HOCl$
Therefore,$X$ is $HOCl$.

(A) The formation of ozone from oxygen is an endothermic reaction,not an exothermic reaction.
The correct reaction is: $3 O_2 \underset{\text{silent electric discharge}}{\rightleftharpoons} 2 O_3 ; \Delta H = +284.5 \ kJ$.
Therefore,the statement $3 O_2 \underset{\text{silent electric discharge}}{\rightleftharpoons} 2 O_3 ; \Delta H = -284.5 \ kJ$ is incorrect.

(D) $I$. Bleaching powder $(CaOCl_2)$ reacts with ethanol to produce chloroform $(CHCl_3)$. This is a standard laboratory preparation method.
$II$. Bleaching powder decomposes in the presence of cobalt chloride $(CoCl_2)$ catalyst to release oxygen gas: $2CaOCl_2 \xrightarrow{CoCl_2} 2CaCl_2 + O_2$.
$III$. Fluorine is prepared by the electrolysis of a fused mixture of potassium hydrogen fluoride $(KHF_2)$ and anhydrous hydrogen fluoride $(HF)$. Aqueous $KHF_2$ cannot be used because water would be oxidized to oxygen instead of fluoride ions being oxidized to fluorine.
Thus,statements $I$ and $II$ are correct.

(B) The abundance of noble gases in the atmosphere by volume (which is proportional to weight for these gases) is as follows:
$Ar$ $(0.934\%)$ > $Ne$ $(0.0018\%)$ > $Kr$ $(0.00011\%)$.
Therefore,the correct order is $Ar > Ne > Kr$.

(C) According to the Clausius-Clapeyron equation,the relationship between vapour pressure $(P)$ and temperature $(T)$ is given by: $\log_{10} P = -\frac{\Delta H_{vap}}{2.303 R} \cdot \frac{1}{T} + C$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log_{10} P$ and $x = \frac{1}{T}$,the slope $m = -\frac{\Delta H_{vap}}{2.303 R}$ is negative.
Therefore,plotting $\log_{10} P$ on the $y$-axis and $\frac{1}{T}$ on the $x$-axis yields a straight line with a negative slope.

(B) Isotones are species having the same number of neutrons.
In $_{20}Ca^{40}$,the number of neutrons is $40 - 20 = 20$.
In $_{18}Ar^{40}$,the number of neutrons is $40 - 18 = 22$.
Since the number of neutrons is different,$_{20}Ca^{40}$ and $_{18}Ar^{40}$ are not isotones.
Therefore,the statement in option $B$ is incorrect.

(C) The colloidal solution of gold is prepared by dispersing solid gold particles in a liquid medium (usually water).
Since gold is a metal,it does not have an affinity for the dispersion medium.
Therefore,it cannot be prepared by simple mixing and requires special methods.
Such colloids are classified as lyophobic (solvent-hating) colloids.