AIPMT 2003 Chemistry Question Paper with Answer and Solution

(A) If two vectors are perpendicular,their dot product must be equal to zero.
According to the problem:
$(\overrightarrow{A} + \overrightarrow{B}) \cdot (\overrightarrow{A} - \overrightarrow{B}) = 0$
$\Rightarrow \overrightarrow{A} \cdot \overrightarrow{A} - \overrightarrow{A} \cdot \overrightarrow{B} + \overrightarrow{B} \cdot \overrightarrow{A} - \overrightarrow{B} \cdot \overrightarrow{B} = 0$
Since the dot product is commutative,$\overrightarrow{A} \cdot \overrightarrow{B} = \overrightarrow{B} \cdot \overrightarrow{A}$,so the middle terms cancel out:
$\Rightarrow A^2 - B^2 = 0$
$\Rightarrow A^2 = B^2$
$\therefore A = B$
This means the two forces are equal to each other in magnitude.

(A) Let the two force vectors be $\overrightarrow{F_1}$ and $\overrightarrow{F_2}$.
Given that their vector sum $(\overrightarrow{F_1} + \overrightarrow{F_2})$ is perpendicular to their vector difference $(\overrightarrow{F_1} - \overrightarrow{F_2})$.
Two vectors are perpendicular if their dot product is zero.
Therefore,$(\overrightarrow{F_1} + \overrightarrow{F_2}) \cdot (\overrightarrow{F_1} - \overrightarrow{F_2}) = 0$.
Expanding the dot product:
$\overrightarrow{F_1} \cdot \overrightarrow{F_1} - \overrightarrow{F_1} \cdot \overrightarrow{F_2} + \overrightarrow{F_2} \cdot \overrightarrow{F_1} - \overrightarrow{F_2} \cdot \overrightarrow{F_2} = 0$.
Since the dot product is commutative,$\overrightarrow{F_1} \cdot \overrightarrow{F_2} = \overrightarrow{F_2} \cdot \overrightarrow{F_1}$,so these terms cancel out.
We are left with $|\overrightarrow{F_1}|^2 - |\overrightarrow{F_2}|^2 = 0$.
This implies $|\overrightarrow{F_1}|^2 = |\overrightarrow{F_2}|^2$,or $|\overrightarrow{F_1}| = |\overrightarrow{F_2}|$.
Thus,the forces are equal in magnitude.

(A) The speed of sound is $v$. The speed of the observer is $v_0 = v/5$. The source is stationary,so $v_s = 0$.
The apparent frequency $f'$ is given by the Doppler effect formula: $f' = f \left( \frac{v + v_0}{v} \right)$.
Substituting the values: $f' = f \left( \frac{v + v/5}{v} \right) = f \left( \frac{6v/5}{v} \right) = 1.2f$.
Since the source is stationary,the wavelength $\lambda$ of the sound waves in the medium does not change for the observer. Therefore,the apparent wavelength is $\lambda' = \lambda$.
Thus,the apparent frequency is $1.2f$ and the apparent wavelength is $\lambda$.

(C) The intensity $I$ of light from a point source follows the inverse square law,$I \propto \frac{1}{d^2}$.
Since the number of photoelectrons emitted per second is directly proportional to the intensity of incident light,we have $n \propto I \propto \frac{1}{d^2}$.
Given initial distance $d_1 = 1\;m$ and final distance $d_2 = 2\;m$.
Therefore,the ratio of the number of electrons is $\frac{n_2}{n_1} = \left( \frac{d_1}{d_2} \right)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.
Thus,$n_2 = \frac{n_1}{4}$.
The energy of each emitted electron depends on the frequency of the incident light,not on the intensity,so the energy remains unchanged.

(C) The relationship between the heat supplied $(q)$,molar heat capacity at constant pressure $(C_p)$,and the temperature change $(\Delta T)$ is given by the formula: $q = n \cdot C_p \cdot \Delta T$
Given:
$q = 1.0 \, kJ = 1000 \, J$
$C_p = 75 \, J \, K^{-1} \, mol^{-1}$
Mass of water = $100 \, g$
Molar mass of water $(H_2O)$ = $18 \, g \, mol^{-1}$
Calculate the number of moles $(n)$:
$n = \frac{\text{mass}}{\text{molar mass}} = \frac{100 \, g}{18 \, g \, mol^{-1}} = 5.55 \, mol$
Substitute the values into the formula:
$1000 = (\frac{100}{18}) \times 75 \times \Delta T$
$1000 = 5.55 \times 75 \times \Delta T$
$1000 = 416.67 \times \Delta T$
$\Delta T = \frac{1000}{416.67} \approx 2.4 \, K$
Therefore,the increase in temperature is $2.4 \, K$.

(D) The relationship between wavelength $(\lambda)$,velocity of light $(c)$,and frequency $(\nu)$ is given by $\lambda = \frac{c}{\nu}$.
Substituting the given values: $\lambda = \frac{3.0 \times 10^8 \ m \ s^{-1}}{8 \times 10^{15} \ s^{-1}} = 0.375 \times 10^{-7} \ m = 3.75 \times 10^{-8} \ m$.
To convert the wavelength into nanometres $(nm)$,we multiply by $10^9 \ nm/m$: $\lambda = 3.75 \times 10^{-8} \ m \times 10^9 \ nm/m = 37.5 \ nm$.
The value closest to $37.5 \ nm$ is $4 \times 10^1 \ nm$.

(C) The balanced chemical equation for the Haber process is: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$.
Initially,we have $30 \ L$ of $N_2$ and $30 \ L$ of $H_2$.
According to the stoichiometry,$1 \ L$ of $N_2$ reacts with $3 \ L$ of $H_2$ to produce $2 \ L$ of $NH_3$.
If all $30 \ L$ of $H_2$ were consumed,it would require $10 \ L$ of $N_2$ and produce $20 \ L$ of $NH_3$ (theoretical yield).
Since the reaction yielded only $50\%$ of the expected product,the actual amount of $NH_3$ produced is $50\% \times 20 \ L = 10 \ L$.
For $10 \ L$ of $NH_3$ to be produced,$5 \ L$ of $N_2$ and $15 \ L$ of $H_2$ must have reacted.
Remaining $N_2 = 30 \ L - 5 \ L = 25 \ L$.
Remaining $H_2 = 30 \ L - 15 \ L = 15 \ L$.
Thus,the final composition is $10 \ L$ ammonia,$25 \ L$ nitrogen,and $15 \ L$ hydrogen.

(B) protonic acid is a substance that can donate a proton ($H^+$ ion) in an aqueous solution.
$SO_2(OH)_2$ (Sulfuric acid),$PO(OH)_3$ (Phosphoric acid),and $SO(OH)_2$ (Sulfurous acid) all contain ionizable $H$ atoms attached to oxygen,which can be released as $H^+$ ions.
$B(OH)_3$ (Boric acid) acts as a Lewis acid by accepting an $OH^-$ ion from water to form $[B(OH)_4]^-$,releasing $H^+$ from the water molecule,not from itself. Therefore,it is not a protonic acid.

(C) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by the equation: $\Delta H = \Delta E + \Delta n_g RT$,which implies $\Delta H - \Delta E = \Delta n_g RT$.
For the reaction $C_3H_{8(g)} + 5O_{2(g)} \to 3CO_{2(g)} + 4H_2O_{(l)}$,the change in the number of gaseous moles $(\Delta n_g)$ is calculated as: $\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$.
$\Delta n_g = 3 - (1 + 5) = 3 - 6 = -3$.
Substituting this value into the equation: $\Delta H - \Delta E = -3RT$.

(C) The molar mass of water $(H_2O)$ is $18 \ g \ mol^{-1}$.
Given,molar heat capacity $(C_p)$ = $75 \ J \ K^{-1} \ mol^{-1}$.
Specific heat capacity $(c)$ = $\frac{C_p}{\text{Molar mass}} = \frac{75}{18} \approx 4.17 \ J \ g^{-1} \ K^{-1}$.
Heat supplied $(Q)$ = $1.0 \ kJ = 1000 \ J$.
Mass of water $(m)$ = $100 \ g$.
Using the formula $Q = m \cdot c \cdot \Delta T$:
$1000 = 100 \times 4.17 \times \Delta T$.
$\Delta T = \frac{1000}{417} \approx 2.4 \ K$.

(D) The standard enthalpy of formation,$\Delta H_f^o$,is defined as the enthalpy change when $1 \ mol$ of a compound is formed from its constituent elements in their standard states.
For option $D$,$Xe_{(g)}$ and $F_{2(g)}$ are the standard states of the elements at $298 \ K$ and $1 \ bar$ pressure.
Since $1 \ mol$ of $XeF_{4(s)}$ is formed from its elements in their standard states,$\Delta H_{react}^o = \Delta H_f^o$ for $XeF_{4(s)}$.

(B) The transformation is $C_{\text{graphite}} \rightarrow C_{\text{diamond}}$.
For this process, $\Delta G = \Delta G^o + \int_{P_1}^{P_2} \Delta V \, dP = 0$ at equilibrium.
$\Delta G^o = -\Delta V \times P$ (assuming $\Delta V$ is constant).
$\Delta V = V_{\text{diamond}} - V_{\text{graphite}} = \frac{M}{\rho_{\text{diamond}}} - \frac{M}{\rho_{\text{graphite}}}$.
Given $M = 12 \, g \, mol^{-1}$, $\rho_{\text{diamond}} = 3.31 \, g \, cm^{-3}$, $\rho_{\text{graphite}} = 2.25 \, g \, cm^{-3}$.
$\Delta V = 12 \times (\frac{1}{3.31} - \frac{1}{2.25}) \, cm^3 \, mol^{-1} = 12 \times (0.3021 - 0.4444) = -1.7076 \, cm^3 \, mol^{-1} = -1.7076 \times 10^{-6} \, m^3 \, mol^{-1}$.
$P = \frac{-\Delta G^o}{\Delta V} = \frac{-1895 \, J \, mol^{-1}}{-1.7076 \times 10^{-6} \, m^3 \, mol^{-1}} \approx 1.11 \times 10^9 \, Pa$.
Given the options, the closest value is $9.92 \times 10^8 \, Pa$.

(C) The ionic radius $(r)$ is inversely proportional to the effective nuclear charge $(Z_{eff})$.
According to the relationship $r \propto \frac{1}{Z_{eff}}$,as the effective nuclear charge increases,the electrons are pulled more strongly towards the nucleus,resulting in a decrease in the ionic radius.
Therefore,the correct option is $(C)$.

(B) $1$. Identify the longest carbon chain. The longest chain contains $8$ carbon atoms,so the parent alkane is octane.
$2$. Number the chain from the end that gives the lowest locants to the substituents. Numbering from left to right gives substituents at positions $3$ and $4$.
$3$. At position $3$,there is a methyl group $(-CH_3)$,and at position $4$,there is an ethyl group $(-CH_2CH_3)$.
$4$. According to $IUPAC$ rules,substituents are listed in alphabetical order: ethyl comes before methyl.
$5$. Therefore,the correct name is $4-$ethyl$-3-$methyloctane.

(B) Enantiomers are stereoisomers that are non-superimposable mirror images of each other.
In option $(B)$,the two structures are mirror images of each other and are non-superimposable,which defines them as enantiomers.
Other options represent either identical compounds,diastereomers,or constitutional isomers.

(D) The reaction of an alkene with $NaIO_4$ in the presence of $KMnO_4$ (Lemieux-von Rudloff reagent) results in the oxidative cleavage of the $C=C$ double bond.
In the compound $CH_3-C(CH_3)=CH-CH_3$,the double bond is cleaved.
The substituted carbon $CH_3-C(CH_3)=$ is oxidized to a ketone,$CH_3COCH_3$ (acetone).
The $=CH-CH_3$ part is oxidized to an aldehyde,which is further oxidized to a carboxylic acid,$CH_3COOH$ (acetic acid),under these conditions.
Therefore,the final products are $CH_3COCH_3 + CH_3COOH$.

(A) The reactivity towards electrophilic aromatic substitution depends on the electron density of the benzene ring.
$1$. The $-NH_2$ group in aniline $(I)$ is a strong electron-donating group due to its $+M$ effect,which significantly increases the electron density on the ring,making it highly reactive.
$2$. Benzene $(II)$ has no substituents,serving as the reference point.
$3$. The $-NO_2$ group in nitrobenzene $(III)$ is a strong electron-withdrawing group due to its $-M$ and $-I$ effects,which significantly decreases the electron density on the ring,making it the least reactive.
Therefore,the correct order of reactivity is $I > II > III$.

(C) rocket propellant consists of a fuel and an oxidizer.
Liquid hydrogen $(H_2)$ acts as the fuel,and liquid oxygen $(O_2)$ acts as the oxidizer.
This combination is highly efficient and is commonly used in space launch vehicles because it provides a high specific impulse.

(C) $DNA$ (deoxyribonucleic acid) is the molecule that carries the genetic information of all living beings.
The structure consists of two polynucleotide strands that wind around each other to form a double helix.
Each strand has a backbone made of alternating sugar (deoxyribose) and phosphate molecules.
The two strands are held together by hydrogen bonding between the nitrogenous bases.
Adenine $(A)$ pairs with Thymine $(T)$ via two hydrogen bonds,and Cytosine $(C)$ pairs with Guanine $(G)$ via three hydrogen bonds.
Therefore,the double-helical structure of $DNA$ is stabilized by hydrogen bonding.

(C) Pattern formation is the developmental process by which cells in a developing embryo acquire identities that lead to the formation of a spatially organized and differentiated structure. This process involves the establishment of polarity along the anterior/posterior,dorsal/ventral,and medial/lateral axes,ensuring that organs and tissues develop in their correct anatomical positions.

(B) The correct answer is $B$. The establishment of polarity in a developing embryo along different axes (anterior/posterior,dorsal/ventral,or medial/lateral) is induced by the organizer phenomenon. This process involves specific regions of the embryo that secrete signaling molecules to direct the differentiation and spatial organization of surrounding tissues.

(B) $Zinc$ is an essential micronutrient required by plants for various physiological processes.
Specifically,$Zinc$ is required for the synthesis of the plant hormone $Auxin$ (specifically $Indole-3-acetic$ $acid$ or $IAA$).
Therefore,a deficiency of $Zinc$ leads to a reduction in the biosynthesis of $Auxin$,which subsequently affects plant growth and development.

(A) During the process of transcription in prokaryotes and eukaryotes,the $RNA$ polymerase enzyme recognizes and binds to a specific sequence on the $DNA$ template strand known as the promoter.
This binding event is the first step in initiating transcription.
The promoter is located towards the $5'$ end (upstream) of the structural gene.

(A) $Crotalaria juncea$ (Sunn hemp) and $Alhagi camelorum$ (Camel thorn) are widely recognized for their role in soil enrichment.
$Crotalaria juncea$ is a leguminous plant that fixes atmospheric nitrogen,making it an excellent green manure crop.
$Alhagi camelorum$ is often used in sandy soils to improve soil fertility and structure.
Therefore,these plants are specifically utilized in agricultural practices to enhance soil nutrient content.

(A) Green manure refers to plants that are grown and then incorporated into the soil to improve its fertility and structure.
$Crotalaria juncea$ (Sunn hemp) is a well-known leguminous plant used extensively as green manure because it fixes atmospheric nitrogen.
$Alhagi camelorum$ (Camel thorn) is also known to be used in sandy soils for soil improvement and as a green manure crop.
Therefore,the correct pair is $Crotalaria juncea$ and $Alhagi camelorum$.

(C) For a body rolling without slipping down an inclined plane,the velocity $v$ at the bottom is given by the conservation of energy: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
Since the body is a solid cylinder,its moment of inertia about the central axis is $I = \frac{1}{2}MR^2$ and the condition for rolling without slipping is $v = R\omega$.
Substituting these into the energy equation:
$Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}(\frac{1}{2}MR^2)(\frac{v}{R})^2$
$Mgh = \frac{1}{2}Mv^2 + \frac{1}{4}Mv^2$
$Mgh = \frac{3}{4}Mv^2$
$v^2 = \frac{4}{3}gh$
$v = \sqrt{\frac{4}{3}gh}$.

(C) For a body rolling without slipping,the total kinetic energy $E_{\text{total}}$ is the sum of translational kinetic energy $E_t$ and rotational kinetic energy $E_r$.
$E_t = \frac{1}{2}mv^2$
$E_r = \frac{1}{2}I\omega^2$
Since $I = mK^2$ and $\omega = \frac{v}{R}$,we have $E_r = \frac{1}{2}(mK^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2(\frac{K^2}{R^2})$.
$E_{\text{total}} = E_t + E_r = \frac{1}{2}mv^2 + \frac{1}{2}mv^2(\frac{K^2}{R^2}) = \frac{1}{2}mv^2(1 + \frac{K^2}{R^2})$.
The fraction of total energy associated with rotational kinetic energy is $\frac{E_r}{E_{\text{total}}}$.
$\frac{E_r}{E_{\text{total}}} = \frac{\frac{1}{2}mv^2(\frac{K^2}{R^2})}{\frac{1}{2}mv^2(1 + \frac{K^2}{R^2})} = \frac{\frac{K^2}{R^2}}{\frac{R^2 + K^2}{R^2}} = \frac{K^2}{K^2 + R^2}$.

(B) According to the principle of conservation of angular momentum,since no external torque acts on the system,the initial angular momentum equals the final angular momentum.
$L_1 = L_2$
$I_1 \omega_1 = I_2 \omega_2$
Initially,the moment of inertia of the ring is $I_1 = MR^2$ and angular velocity is $\omega$.
After placing four masses $m$ at the rim,the new moment of inertia is $I_2 = MR^2 + 4(mR^2) = (M + 4m)R^2$.
Substituting these into the conservation equation:
$MR^2 \omega = (M + 4m)R^2 \omega_2$
$\omega_2 = \frac{MR^2 \omega}{(M + 4m)R^2} = \frac{M\omega}{M + 4m}$.

(D) Vitamin $B_{12}$,also known as cyanocobalamin,contains a central cobalt ion in the $+3$ oxidation state,i.e.,$Co(III)$,coordinated within a corrin ring system.

(D) The direction of a reaction can be predicted by comparing the reaction quotient $Q$ with the equilibrium constant $K_c$.
If $Q > K_c$,the concentration of products is higher than at equilibrium,so the reaction proceeds in the reverse direction (right to left) to reach equilibrium.
If $Q < K_c$,the reaction proceeds in the forward direction (left to right).
If $Q = K_c$,the reaction is at equilibrium.
Therefore,the correct condition for the reaction to proceed from right to left is $Q > K_c$.

(C) The dissociation of $AgI$ is given by: $AgI(s) \rightleftharpoons Ag^+(aq) + I^-(aq)$.
Given $K_{sp} = 1.0 \times 10^{-16}$.
In a $10^{-4}\,\text{N}$ solution of $KI$,the concentration of $I^-$ ions is $[I^-] = 10^{-4}\,\text{M}$ (since $KI$ is a strong electrolyte).
Let $S$ be the solubility of $AgI$ in the presence of $KI$.
Then $[Ag^+] = S$ and $[I^-] = 10^{-4} + S \approx 10^{-4}$ (since $S$ is very small).
$K_{sp} = [Ag^+][I^-] = S \times 10^{-4} = 1.0 \times 10^{-16}$.
$S = \frac{1.0 \times 10^{-16}}{10^{-4}} = 1.0 \times 10^{-12}\,\text{mol L}^{-1}$.

(D) Internal energy $(U)$ is a state function.
$A$ state function depends only on the initial and final states of the system,not on the path taken.
For any cyclic process,where the system returns to its initial state,the total change in the state function is always zero.
Since the system starts at state $A$,goes to state $B$,and returns to state $A$,the net change in internal energy $\Delta U_{net} = U_{final} - U_{initial} = U_A - U_A = 0$.

(A) The entropy change $(\Delta S)$ for a phase transition is given by the formula $\Delta S = \frac{\Delta H_{fus}}{T}$.
Given,$\Delta H_{fus} = 6.0\, kJ\, mol^{-1} = 6000\, J\, mol^{-1}$.
The temperature $T = 0^{\circ}C = 273\, K$.
Substituting the values: $\Delta S = \frac{6000\, J\, mol^{-1}}{273\, K} = 21.98\, J\, K^{-1}\, mol^{-1}$.
Thus,the correct option is $A$.

(B) To find the oxidation state of sulphur $(S)$ in each anion,we use the rule that the sum of oxidation states equals the charge of the ion.
$1$. For $S_2O_4^{2-}$: $2x + 4(-2) = -2 \implies 2x - 8 = -2 \implies 2x = +6 \implies x = +3$.
$2$. For $SO_3^{2-}$: $x + 3(-2) = -2 \implies x - 6 = -2 \implies x = +4$.
$3$. For $S_2O_6^{2-}$: $2x + 6(-2) = -2 \implies 2x - 12 = -2 \implies 2x = +10 \implies x = +5$.
Comparing the values: $+3 < +4 < +5$.
Therefore,the order is $S_2O_4^{2-} < SO_3^{2-} < S_2O_6^{2-}$.

(B) The reaction $CH_4 + Cl_2 \xrightarrow{h\nu} CH_3Cl + HCl$ is a free-radical substitution reaction.
In this reaction,the $Cl-Cl$ bond undergoes homolytic fission in the presence of light $(h\nu)$ to generate chlorine free radicals $(Cl^{\bullet})$.
These radicals then attack the methane molecule to substitute a hydrogen atom,which is a characteristic step of free-radical substitution.

(D) Given: Radius $r = \frac{20}{\pi} \; m$,final velocity $v = 80 \; m/s$,initial velocity $u = 0$,and number of revolutions $n = 2$.
The total angular displacement $\theta = 2\pi \times n = 2\pi \times 2 = 4\pi \; rad$.
The final angular velocity $\omega = \frac{v}{r} = \frac{80}{20/\pi} = 4\pi \; rad/s$.
Using the rotational kinematic equation $\omega^2 = \omega_0^2 + 2\alpha\theta$,where $\omega_0 = 0$:
$(4\pi)^2 = 0 + 2 \times \alpha \times 4\pi$
$16\pi^2 = 8\pi\alpha$
$\alpha = \frac{16\pi^2}{8\pi} = 2\pi \; rad/s^2$.
The tangential acceleration $a_t$ is given by $a_t = r\alpha$.
$a_t = \left(\frac{20}{\pi}\right) \times 2\pi = 40 \; m/s^2$.

(A) The initial angular momentum of the ring is $L = I\omega = Mr^2\omega$.
When four particles of mass $m$ are placed on the ring,the new moment of inertia of the system becomes $I' = Mr^2 + 4(mr^2) = (M + 4m)r^2$.
Since there is no external torque acting on the system,the angular momentum remains conserved.
Therefore,$L_{initial} = L_{final}$.
$Mr^2\omega = (M + 4m)r^2\omega'$.
Solving for the new angular velocity $\omega'$,we get $\omega' = \frac{Mr^2\omega}{(M + 4m)r^2} = \frac{M\omega}{M + 4m}$.

(A) Given reactions:
$(1) \ N_2 + 3H_2 \rightleftharpoons 2NH_3 \quad K_1$
$(2) \ N_2 + O_2 \rightleftharpoons 2NO \quad K_2$
$(3) \ H_2 + \frac{1}{2} O_2 \rightleftharpoons H_2O \quad K_3$
Target reaction:
$(4) \ 2NH_3 + \frac{5}{2} O_2 \rightleftharpoons 2NO + 3H_2O \quad K$
To obtain equation $(4)$,we perform the operation: $(2) + 3 \times (3) - (1)$.
Applying the rules of equilibrium constants:
$K = \frac{K_2 \times (K_3)^3}{K_1} = \frac{K_2 K_3^3}{K_1}$

(D) The atomic number of Palladium $(Pd)$ is $46$.
The electronic configuration of $Pd$ in its ground state is $[Kr] 4d^{10} 5s^0$.
The value $l = 2$ corresponds to the $d$-subshell.
In the $4d^{10}$ configuration,there are $10$ electrons in the $d$-subshell.
Therefore,the number of electrons with $l = 2$ is $10$.

(C) The reaction of an alkene with $KMnO_4$ (a strong oxidizing agent) leads to the oxidative cleavage of the double bond.
For the compound $CH_3-C(CH_3)=CH-CH_3$ ($2$-methylbut$-2-$ene):
$1$. The double bond breaks between the $C_2$ and $C_3$ carbons.
$2$. The $C_2$ carbon is bonded to two methyl groups,so it forms acetone $(CH_3COCH_3)$.
$3$. The $C_3$ carbon is bonded to one hydrogen and one methyl group,so it forms acetic acid $(CH_3COOH)$ upon further oxidation.
Thus,the products are $CH_3COCH_3 + CH_3COOH$.

(D) The method of zone refining is based on the principle that the impurities are more soluble in the molten state of the metal than in the solid state.
$A$ circular mobile heater is fixed at one end of a rod of the impure metal.
The molten zone moves along with the heater as it is moved forward.
As the heater moves forward,the pure metal crystallises out of the melt,while the impurities remain in the molten zone and are carried forward.

(A) Boron is an essential micronutrient for plants. Its primary physiological roles include:
$1$. Boron is required for the uptake and utilization of $Ca^{2+}$.
$2$. It is essential for pollen germination and tube growth.
$3$. It plays a critical role in cell differentiation and carbohydrate translocation.
$4$. Specifically,boron facilitates the transport of sugars across the cell membrane by forming borate-sugar complexes,which are more easily transported than free sugars.
Therefore,the correct option is $A$.

(A) Enantiomers are non-superimposable mirror images of each other.
In option $A$, the two structures are non-superimposable mirror images.
Structure $1$ is $(2S, 3S)$-$3$-bromo-$2$-chlorobutane and structure $2$ is $(2R, 3R)$-$3$-bromo-$2$-chlorobutane.
Since they are non-superimposable mirror images, they are enantiomers.

(C) For $SO_3^{2-}$: $x + 3(-2) = -2 \implies x = +4$.
For $S_2O_4^{2-}$: $2x + 4(-2) = -2 \implies 2x = +6 \implies x = +3$.
For $S_2O_6^{2-}$: $2x + 6(-2) = -2 \implies 2x = +10 \implies x = +5$.
Comparing the oxidation states: $+3 < +4 < +5$.
Thus,the increasing order is $S_2O_4^{2-} < SO_3^{2-} < S_2O_6^{2-}$.

(A) According to the rate law,the rate of reaction is given by the expression:
$\text{Rate} = k[\text{conc.}]^{x}$
Where $x$ is the overall order of the reaction.
To make the rate of reaction equal to the rate constant $(k)$,the term $[\text{conc.}]^{x}$ must be equal to $1$.
This is possible only when the order of the reaction $(x)$ is $0$,because $[\text{conc.}]^{0} = 1$.
Therefore,if the rate of reaction is equal to the rate constant,the order of the reaction is $0$.

(D) The mass of a radioactive sample at time $t$ is given by the decay law: $M = M_0 e^{-\lambda t}$.
Here,$M_0 = 10 \, g$ is the initial mass.
The mean life $\tau$ is defined as $\tau = \frac{1}{\lambda}$,so $\lambda = \frac{1}{\tau}$.
We are asked to find the mass after two mean lives,i.e.,$t = 2\tau = \frac{2}{\lambda}$.
Substituting the value of $t$ in the decay equation:
$M = 10 e^{-\lambda (2/\lambda)} = 10 e^{-2} = \frac{10}{e^2}$.
Given that $e \approx 2.718$,then $e^2 \approx 7.389$.
$M = \frac{10}{7.389} \approx 1.35 \, g$.

(D) Zone refining is based on the principle that impurities are more soluble in the molten state of the metal than in the solid state.
As the heater moves,the pure metal crystallizes out of the melt,while the impurities remain in the molten zone and are carried to the end of the rod.
This method is commonly used for purifying semiconductors like $Si$,$Ge$,and $Ga$.

(A) $1$. $F$ is more electronegative than $Cl$,making the $H-F$ bond stronger than the $H-Cl$ bond. Thus,$HF$ does not donate a proton easily,making it a weaker acid than $HCl$. Therefore,statement $A$ is incorrect.
$2$. Iodide $(I^-)$ has the largest size and lowest ionization energy among halides,making it the strongest reducing agent.
$3$. Fluorine is the most electronegative element and only shows an oxidation state of $-1$ (except in $F_2$ where it is $0$).
$4$. Since $Cl$ is more electronegative than $Br$,the $O-H$ bond in $HOCl$ is more polarized than in $HOBr$,making $HOCl$ a stronger acid.

(C) The potential energy of the solid cylinder at height $h$ is $U = Mgh$.
When the cylinder reaches the bottom,its total kinetic energy $(K.E.)$ is the sum of translational and rotational kinetic energy:
$K.E. = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2$
Since the cylinder rolls without slipping,$\omega = \frac{v}{R}$. For a solid cylinder,the moment of inertia $I = \frac{1}{2} M R^2$.
Substituting these into the kinetic energy equation:
$K.E. = \frac{1}{2} M v^2 + \frac{1}{2} (\frac{1}{2} M R^2) (\frac{v}{R})^2$
$K.E. = \frac{1}{2} M v^2 + \frac{1}{4} M v^2 = \frac{3}{4} M v^2$
By the law of conservation of energy,the potential energy at the top equals the kinetic energy at the bottom:
$Mgh = \frac{3}{4} M v^2$
Solving for $v$:
$v^2 = \frac{4}{3} gh$
$v = \sqrt{\frac{4}{3} gh}$

(B) For $SO_3^{2-}$: $x + 3(-2) = -2 \Rightarrow x = +4$
For $S_2O_4^{2-}$: $2x + 4(-2) = -2$ $\Rightarrow 2x = +6$ $\Rightarrow x = +3$
For $S_2O_6^{2-}$: $2x + 6(-2) = -2$ $\Rightarrow 2x = +10$ $\Rightarrow x = +5$
The oxidation states are $+3, +4, +5$ for $S_2O_4^{2-}, SO_3^{2-}, S_2O_6^{2-}$ respectively.
Thus,the correct order is $S_2O_4^{2-} < SO_3^{2-} < S_2O_6^{2-}$.

(A) . The statement "$HF$ is a stronger acid than $HCl$" is false.
$HF$ is a weak acid in aqueous solution due to strong intermolecular hydrogen bonding and high bond dissociation enthalpy of the $H-F$ bond,whereas $HCl$ is a strong acid.

(B) The enthalpy of solution formation is the sum of the enthalpy changes of the individual steps:
$\Delta H_{soln} = \Delta H_1 + \Delta H_2 + \Delta H_3$
For an ideal solution,the net enthalpy change of mixing is zero,meaning $\Delta H_{mix} = 0$.
However,the question asks for the expression of $\Delta H_{soln}$ based on the given steps.
By Hess's Law,the total enthalpy change is the sum of the enthalpy changes of the intermediate steps:
$\Delta H_{soln} = \Delta H_1 + \Delta H_2 + \Delta H_3$

(A) The fraction of unoccupied sites (vacancies) is given by the formula: $\text{Fraction} = \frac{\rho_{\text{X-ray}} - \rho_{\text{pyknometric}}}{\rho_{\text{X-ray}}}$
Given: $\rho_{\text{X-ray}} = 2.178 \times 10^3 \ kg \ m^{-3}$ and $\rho_{\text{pyknometric}} = 2.165 \times 10^3 \ kg \ m^{-3}$
Difference $= (2.178 - 2.165) \times 10^3 = 0.013 \times 10^3 \ kg \ m^{-3}$
Fraction $= \frac{0.013 \times 10^3}{2.178 \times 10^3} = \frac{0.013}{2.178} \approx 5.96 \times 10^{-3}$

(B) The rate law for a reaction is given by $Rate = k[A]^n$,where $n$ is the order of the reaction.
If the rate of the reaction is equal to the rate constant $(Rate = k)$,then $[A]^n$ must be equal to $1$.
This is only possible when $n = 0$.
Therefore,the reaction is of zero order.

(B) For a first-order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
In the first case,initial concentration $[A]_0 = 0.8 \ mol$ and remaining concentration $[A]_t = 0.8 - 0.6 = 0.2 \ mol$. However,the problem states $0.8 \ mol$ of $A$ produces $0.6 \ mol$ of $B$,implying the ratio of reactant remaining is $\frac{0.8 - 0.6}{0.8} = \frac{0.2}{0.8} = \frac{1}{4}$.
Wait,the standard first-order equation is $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
For the first case: $k = \frac{2.303}{1} \log \frac{0.8}{0.8 - 0.6} = 2.303 \log \frac{0.8}{0.2} = 2.303 \log 4$.
For the second case: $t = \frac{2.303}{k} \log \frac{0.9}{0.9 - 0.675} = \frac{2.303}{k} \log \frac{0.9}{0.225} = \frac{2.303}{k} \log 4$.
Since $k$ is the same,$t = 1 \ hr$.

(D) The relationship between the activation energy of the forward reaction $(E_{a,f})$ and the reverse reaction $(E_{a,r})$ is given by the equation: $E_{a,f} - E_{a,r} = \Delta H$,where $\Delta H$ is the enthalpy change of the reaction.
For an exothermic reaction,$\Delta H < 0$,which implies $E_{a,r} > E_{a,f}$.
For an endothermic reaction,$\Delta H > 0$,which implies $E_{a,r} < E_{a,f}$.
Therefore,the activation energy for the reverse reaction can be less than or more than ${E_a}$ depending on the nature of the reaction.

(A) The Arrhenius equation is given by $k = A e^{-E^*/RT}$.
Taking the logarithm on both sides,we get $\log \, k = \log \, A - \frac{E^*}{2.303 R} \cdot \frac{1}{T}$.
This equation follows the linear form $y = mx + c$,where $y = \log \, k$ and $x = \frac{1}{T}$.
Plotting $\log \, k$ versus $\frac{1}{T}$ yields a straight line with a slope equal to $-\frac{E^*}{2.303 R}$.
Thus,the activation energy $E^*$ can be determined from the slope of this plot.

(B) The reaction for the electrolysis of $Al_2O_3$ is the reverse of the formation reaction: $\frac{2}{3} Al_2O_3 \to \frac{4}{3} Al + O_2$.
For this reaction,$\Delta G = +827 \ kJ \ mol^{-1} = 827000 \ J \ mol^{-1}$.
The number of electrons involved in the reaction is calculated from the oxidation state change: $Al^{3+} + 3e^- \to Al$. Since there are $\frac{4}{3}$ moles of $Al$,the total electrons $n = \frac{4}{3} \times 3 = 4$.
Using the formula $\Delta G = -nFE$,the minimum emf $E$ required is $E = -\frac{\Delta G}{nF}$.
$E = -\frac{827000 \ J \ mol^{-1}}{4 \times 96500 \ C \ mol^{-1}} = -2.14 \ V$.
Since we are looking for the minimum external potential to drive the non-spontaneous reaction,the magnitude is $2.14 \ V$.

(B) The cell reaction is $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$.
Using the $Nernst$ equation: $E = E^o - \frac{0.0591}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$.
For ${E_1}$: $[Zn^{2+}] = 0.01 \ M$ and $[Cu^{2+}] = 1.0 \ M$,so ${E_1} = E^o - \frac{0.0591}{2} \log \frac{0.01}{1} = E^o + 0.0591$.
For ${E_2}$: $[Zn^{2+}] = 1.0 \ M$ and $[Cu^{2+}] = 0.01 \ M$,so ${E_2} = E^o - \frac{0.0591}{2} \log \frac{1}{0.01} = E^o - 0.0591$.
Therefore,${E_1} > {E_2}$.

(A) According to the adsorption theory of catalysis,the catalyst provides a surface where reactant molecules get adsorbed.
This adsorption increases the concentration of reactants at the active sites and,more importantly,it provides an alternative pathway with lower activation energy,thereby increasing the rate of the reaction.

(A) The catalytic activity of transition metals is primarily due to their ability to adopt $Variable \ oxidation \ states$ and provide a large surface area with free valencies.
These properties allow them to form unstable intermediate compounds with reactants,providing an alternative reaction path with lower activation energy.
Therefore,the correct option is $(A)$.

(A) This method is based on the principle that the impurities are more soluble in the melt than in the solid state of the metal.
$A$ circular mobile heater is fixed at one end of a rod of the impure metal.
The molten zone moves along with the heater which is moved forward.
As the heater moves forward,the pure metal crystallises out of the melt and the impurities pass on into the adjacent molten zone.

(A) The basic character of metal oxides depends on the oxidation state and the metallic nature of the element.
For transition metal monoxides $(MO)$,the metal is in the $+2$ oxidation state.
As we move from left to right across the $3d$ series $(Ti$ $\rightarrow V$ $\rightarrow Cr$ $\rightarrow Fe)$,the electronegativity of the metal increases and its metallic character decreases.
Consequently,the ionic character of the $M-O$ bond decreases,leading to a decrease in the basic character of the oxides.
Therefore,the order of basic character is $TiO > VO > CrO > FeO$.

(A) The chemical formula for sodium nitroprusside is $Na_2[Fe(CN)_5(NO)] \cdot 2H_2O$.
In this complex,the ligand $NO$ is treated as $NO^+$ (nitrosonium ion).
The oxidation state of $Fe$ is calculated as: $2(+1) + x + 5(-1) + (+1) = 0$,which gives $x = +2$.
Thus,the central metal is $Fe(II)$.
The ligands are five cyanide ions (pentacyanido) and one nitrosyl group (nitrosyl).
According to $IUPAC$ rules,the name is sodium pentacyanidonitrosylferrate$(II)$.

(A) Geometric isomerism in octahedral complexes occurs when there are different spatial arrangements of ligands around the central metal atom.
For complexes of the type $[MA_5B]$,all positions are equivalent relative to the unique ligand $B$,meaning no geometric isomers can be formed.
Complexes of the type $[MA_4B_2]$ exhibit $cis$ and $trans$ isomers.
Complexes of the type $[MA_3B_3]$ exhibit $fac$ (facial) and $mer$ (meridional) isomers.
Complexes of the type $[MA_2B_4]$ exhibit $cis$ and $trans$ isomers.
Therefore,$[MA_5B]$ is the correct answer.

(D) The atomic number of $Co$ is $27$. The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$.
In the complex ion $[CoF_6]^{3-}$,let the oxidation state of $Co$ be $x$. Then $x + 6(-1) = -3$,which gives $x = +3$.
The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
Since $F^-$ is a weak field ligand,it does not cause pairing of electrons in the $3d$ orbitals.
According to crystal field theory,for $d^6$ configuration in an octahedral field with weak ligands,the distribution is $t_{2g}^4 e_g^2$.
The number of unpaired electrons is $4$.

(A) $(CH_3)_4Sn$ (Tetramethyltin) contains only $\sigma$-bonds between the metal and the carbon atoms of the alkyl groups.
$K[PtCl_3(\eta^2 - C_2H_4)]$ (Zeise's salt) contains a $\pi$-bonded ethylene ligand.
$Fe(\eta^5 - C_5H_5)_2$ (Ferrocene) is a sandwich compound with $\pi$-bonded cyclopentadienyl rings.
$Cr(\eta^6 - C_6H_6)_2$ (Dibenzenechromium) is a sandwich compound with $\pi$-bonded benzene rings.
Therefore,$(CH_3)_4Sn$ is the only $\sigma$-bonded organometallic compound among the given options.

(B) The reaction of $m$-chlorobenzaldehyde with concentrated $KOH$ $(50\%)$ is a classic example of the Cannizzaro reaction.
Aldehydes that do not contain $\alpha$-hydrogen atoms undergo self-oxidation and reduction (disproportionation) in the presence of concentrated alkali.
$m$-chlorobenzaldehyde lacks $\alpha$-hydrogens,so it undergoes the Cannizzaro reaction to form $m$-chlorobenzoate and $m$-chlorobenzyl alcohol.

(A) The reaction of a ketone $(RR'C=O)$ with $HCN$ in the presence of $KCN$ (catalyst) leads to the formation of a cyanohydrin,where $A = RR'C(OH)(CN)$.
Next,the reduction of the cyanohydrin $(RR'C(OH)(CN))$ using a strong reducing agent like $LiAlH_4$ reduces the nitrile group $(-CN)$ to a primary amine group $(-CH_2NH_2)$.
Therefore,$A = RR'C(OH)(CN)$ and $B = LiAlH_4$.

(C) The acid strength of a compound is determined by the stability of its conjugate base. The more stable the conjugate base,the stronger the acid.
$1$. $RCOOH$ (Carboxylic acid) is the strongest acid among the given options because its conjugate base,the carboxylate ion $(RCOO^-)$,is resonance-stabilized.
$2$. $HOH$ $(H_2O)$ is a stronger acid than $ROH$ (alcohol) because the alkyl group in $ROH$ is electron-donating ($+I$ effect),which destabilizes the alkoxide ion $(RO^-)$.
$3$. $HC \equiv CH$ (terminal alkyne) is the weakest acid among these because the $sp$-hybridized carbon is less electronegative than oxygen,and the negative charge on the carbon atom is less stable compared to the negative charge on oxygen.
Therefore,the correct order of acid strength is: $RCOOH > HOH > ROH > HC \equiv CH$.

(A) The reaction sequence is as follows:
$1.$ Acetic acid reacts with $PCl_5$ to form acetyl chloride $(A)$:
$CH_3COOH + PCl_5 \to CH_3COCl (A) + POCl_3 + HCl$
$2.$ Acetyl chloride reacts with benzene in the presence of anhydrous $AlCl_3$ (Friedel-Crafts acylation) to form acetophenone $(B)$:
$CH_3COCl + C_6H_6 \xrightarrow{\text{anh. } AlCl_3} C_6H_5COCH_3 (B) + HCl$
$3.$ Acetophenone reacts with ethylmagnesium bromide followed by hydrolysis to form $2$-phenylbutan-$2$-ol $(C)$:
$C_6H_5COCH_3 + C_2H_5MgBr$ $\xrightarrow{\text{ether}} C_6H_5-C(OMgBr)(CH_3)-C_2H_5$ $\xrightarrow{H_2O} C_6H_5-C(OH)(CH_3)-C_2H_5 (C)$

(A) The reaction sequence is as follows:
$1$. $p$-Toluidine reacts with acetic anhydride $(Ac_2O)$ to form $N$-($4$-methylphenyl)acetamide (acetanilide derivative),which is compound $A$.
$2$. Compound $A$ undergoes electrophilic aromatic substitution with $Br_2$ in $CH_3COOH$. The $-NHCOCH_3$ group is strongly activating and ortho/para directing. Since the para position is occupied by the $-CH_3$ group,bromine enters the ortho position relative to the $-NHCOCH_3$ group,forming $2$-bromo-$4$-methylacetanilide,which is compound $B$.
$3$. Finally,acid-catalyzed hydrolysis of compound $B$ removes the acetyl group to regenerate the amine,yielding $2$-bromo-$4$-methylaniline as the final product $C$.

(D) Neoprene is a synthetic rubber formed by the free-radical polymerization of chloroprene ($2$-chloro$-1,3-$butadiene).
The chemical reaction is:
$n(CH_2 = C(Cl) - CH = CH_2) \rightarrow [-CH_2 - C(Cl) = CH - CH_2 -]_n$
Thus,the monomer is $CH_2 = C(Cl) - CH = CH_2$.

(B) . Acrilan is the trade name for polyacrylonitrile $(PAN)$.
It is a synthetic polymer produced from the monomer acrylonitrile $(CH_2=CH-CN)$.
Its structure is represented as $(-CH_2-CH(CN)-)_n$.
It is known for being hard,horny,and having a high melting point.

(D) Glycolysis is a metabolic pathway that occurs in the cytosol of the cell.
In this process,one molecule of glucose $(C_6H_{12}O_6)$ is broken down into two molecules of pyruvate $(CH_3COCOOH)$ through a series of enzymatic reactions.
It is an anaerobic process that yields a net gain of $2$ $ATP$ and $2$ $NADH$ molecules.

(D) Phospholipids are esters of glycerol with two carboxylic acid residues and one phosphate group.
Hence,phospholipids may be regarded as derivatives of glycerol in which two hydroxyl groups are esterified with fatty acids,while the third is esterified with phosphoric acid.

(B) According to Chargaff's rule,the $DNA$ of any species from any organism should have a $1:1$ stoichiometric ratio of purine and pyrimidine bases.
Specifically,the amount of adenine $(A)$ is equal to that of thymine $(T)$,and the amount of guanine $(G)$ is equal to that of cytosine $(C)$.
Therefore,the correct option is $(B)$.

(D) Vitamin $B_{12}$,also known as cyanocobalamin,contains a central metal ion.
The metal present in the coordination complex of vitamin $B_{12}$ is Cobalt in the $+3$ oxidation state,denoted as $Co(III)$.
Therefore,the correct option is $(D)$.

(C) The combination of $Liquid \ H_2$ and $Liquid \ O_2$ is widely used as a high-performance rocket propellant.
In this system,$Liquid \ H_2$ acts as the fuel and $Liquid \ O_2$ acts as the oxidizer.
The reaction is highly exothermic,providing the necessary thrust for the rocket.

(C) The ionic radii of lanthanoids decrease with an increase in atomic number due to lanthanide contraction.
$La^{3+}$ $(Z=57)$ has the largest size,while $Lu^{3+}$ $(Z=71)$ has the smallest size among the lanthanoids.
$Y^{3+}$ $(Z=39)$ has an ionic radius comparable to the heavier lanthanoids (like $Ho^{3+}$ or $Er^{3+}$),but it is smaller than $La^{3+}$ and $Eu^{3+}$.
Comparing the given ions: $La^{3+} (1.03 \ \mathring{A}) > Eu^{3+} (0.95 \ \mathring{A}) > Y^{3+} (0.90 \ \mathring{A}) > Lu^{3+} (0.86 \ \mathring{A})$.
Therefore,the correct order is $Lu^{3+} < Y^{3+} < Eu^{3+} < La^{3+}$.

(D) The reaction of acetaldehyde $(CH_3CHO)$ with $HCN$ is a nucleophilic addition reaction.
Since the carbonyl carbon in $CH_3CHO$ is $sp^2$ hybridized and planar,the cyanide ion $(CN^-)$ can attack from either side of the plane with equal probability.
This leads to the formation of a racemic mixture of the cyanohydrin intermediate.
Subsequent hydrolysis of the cyanohydrin to the carboxylic acid ($CH_3CH(OH)COOH$,lactic acid) preserves this racemic nature.
Therefore,the final product is a $50\% \ D$ and $50\% \ L$ mixture,which is a racemic mixture.

(D) The two polynucleotide chains of $DNA$ molecules are twisted around a common axis but run in opposite directions to form a right-handed helix.
These two chains are held together by specific hydrogen bonds between the complementary nitrogenous bases ($A$ pairs with $T$,and $G$ pairs with $C$).

(A) Grignard reagents $(RMgX)$ are strong bases and highly reactive nucleophiles.
They react rapidly with acidic hydrogen atoms,such as those present in the hydroxyl $(-OH)$ group of hydroxyketones,to form an alkoxide and an alkane $(R-H)$.
Therefore,to perform a Grignard reaction on a hydroxyketone,the hydroxyl group must first be protected (e.g.,by silylation) to prevent this side reaction.