AIPMT 2011 Chemistry Question Paper with Answer and Solution

(A) Nitrifying bacteria,such as $Nitrosomonas$ and $Nitrobacter$,play a crucial role in the nitrogen cycle.
They oxidize ammonia $(NH_3)$ into nitrites $(NO_2^-)$ and subsequently into nitrates $(NO_3^-)$.
This process is known as nitrification,which makes nitrogen available in a form that plants can readily absorb from the soil.

(C) The filiform apparatus is a specialized cellular thickening of the plasma membrane in the synergid cells of the embryo sac.
It consists of finger-like projections that extend into the cytoplasm of the synergids.
Its primary function is to guide the entry of the pollen tube into the synergid during the process of fertilization.
Therefore,the filiform apparatus is a characteristic feature of the synergids.

(C) Wind pollination,also known as anemophily,is a common mode of pollination in grasses.
Grasses possess lightweight,non-sticky pollen grains that can be easily transported by wind currents.
Additionally,their flowers are often small,inconspicuous,and lack nectar or scent,which are adaptations to attract insects.
In contrast,legumes,lilies,and orchids typically rely on biotic agents like insects for pollination.

(D) An amphoteric oxide is one that can react with both acids and bases.
Among the given options,$CaO$ is basic,$CO_2$ is acidic,and $SiO_2$ is acidic.
$SnO_2$ is amphoteric in nature as it reacts with both strong acids and strong bases.

(C) As the size of the halogen atom increases,the ionic character decreases and the covalent character increases due to polarization (Fajans' rule),which leads to a decrease in the melting point.
The correct order of melting points is:
$CaF_2 > CaCl_2 > CaBr_2 > CaI_2$.
Therefore,$CaI_2$ has the lowest melting point.

(C) Let $\phi$ be the elevation angle of the projectile at its highest point as seen from the point of projection $O$,and $\theta$ be the angle of projection with the horizontal.
From the figure,$\tan \phi = \frac{H}{R/2} = \frac{2H}{R}$.
In projectile motion,the maximum height $H$ is given by $H = \frac{u^2 \sin^2 \theta}{2g}$ and the horizontal range $R$ is given by $R = \frac{u^2 \sin 2\theta}{g}$.
Substituting these values into the expression for $\tan \phi$:
$\tan \phi = \frac{2 \left( \frac{u^2 \sin^2 \theta}{2g} \right)}{\left( \frac{u^2 \sin 2\theta}{g} \right)} = \frac{\sin^2 \theta}{\sin 2\theta} = \frac{\sin^2 \theta}{2 \sin \theta \cos \theta} = \frac{1}{2} \tan \theta$.
Given $\theta = 45^\circ$,we have $\tan \phi = \frac{1}{2} \tan 45^\circ = \frac{1}{2} \times 1 = \frac{1}{2}$.
Therefore,$\phi = \tan^{-1}\left(\frac{1}{2}\right)$.

(A) According to Gauss's Law,the net electric flux $\phi$ through any closed surface is given by $\phi = \frac{Q_{\text{enclosed}}}{\varepsilon_0}$.
Here,$Q_{\text{enclosed}} = Q$. Therefore,$\phi = \frac{Q}{\varepsilon_0}$.
Since the electric flux depends only on the charge enclosed by the surface and not on the shape or size of the Gaussian surface,changing the radius from $R$ to $2R$ does not affect the net electric flux.
Thus,the outward electric flux will remain the same.

(b) : Chilli is the member of Solanaceae, in which flowers are bisexual , actinomorphic $(\bigoplus )$ calyx -$5$ and gamosepalous, corolla -$5$ and gamopetalous ; androecium -$5$, free, epipetalous basifixed, inferior; gynoecium -bicarpellary, syncarpous and ovary superior.
So, floral formula of chilli is :

(A) The correct answer is $A$.
At the tissue site,the partial pressure of $CO_2$ is high due to catabolic processes.
$CO_2$ diffuses into the blood ($RBCs$ and plasma) and reacts with water to form carbonic acid $(H_2CO_3)$,which then dissociates into bicarbonate ions $(HCO_3^-)$ and hydrogen ions $(H^+)$.
Approximately $70\%$ of $CO_2$ is transported in the form of bicarbonates.
At the alveolar site,where the partial pressure of $CO_2$ is low,the reaction proceeds in the opposite direction,leading to the formation of $CO_2$ and $H_2O$,which is then exhaled.

(D) : Apomixis is a form of asexual reproduction that mimics sexual reproduction in plants,but it does not involve the fusion of gametes (fertilization).
In apomixis,the embryo develops directly from a diploid cell within the ovule without meiosis or syngamy.
Consequently,the offspring produced are genetically identical to the parent plant.
Similarly,vegetative reproduction is a type of asexual reproduction where new plants arise from vegetative parts of the parent plant,resulting in clones that are genetically identical to the parent.

(C) The number of molecules is directly proportional to the number of moles $(n = \text{mass} / \text{molar mass})$.
$A) \ 44 \ g \ CO_2 = 44 / 44 = 1 \ \text{mole}$
$B) \ 48 \ g \ O_3 = 48 / 48 = 1 \ \text{mole}$
$C) \ 8 \ g \ H_2 = 8 / 2 = 4 \ \text{moles}$
$D) \ 64 \ g \ SO_2 = 64 / 64 = 1 \ \text{mole}$
Since $8 \ g \ H_2$ has the highest number of moles $(4 \ \text{moles})$,it contains the maximum number of molecules.

(B) The total number of atomic orbitals in a given energy level (shell) with principal quantum number $n$ is given by the formula $n^{2}$.
For the fourth energy level,$n = 4$.
Therefore,the number of atomic orbitals $= 4^{2} = 16$.

(B) The energy of a radiation is given by the formula $E = \frac{hc}{\lambda}$,which implies $E \propto \frac{1}{\lambda}$.
Given $E_1 = 25 \ eV$ and $E_2 = 50 \ eV$.
Therefore,$\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1}$.
Substituting the values: $\frac{25}{50} = \frac{\lambda_2}{\lambda_1}$.
$\frac{1}{2} = \frac{\lambda_2}{\lambda_1}$,which simplifies to $\lambda_1 = 2 \lambda_2$.

(A) According to the $(n + l)$ rule,electrons fill orbitals in order of increasing $(n + l)$ values.
For $n = 6$:
$ns$ corresponds to $6s$ $(n+l = 6+0 = 6)$
$(n-2)f$ corresponds to $4f$ $(n+l = 4+3 = 7)$
$(n-1)d$ corresponds to $5d$ $(n+l = 5+2 = 7)$
$np$ corresponds to $6p$ $(n+l = 6+1 = 7)$
Comparing the $(n+l)$ values and the $n$ values for orbitals with the same $(n+l)$,the filling order is $6s$ $\rightarrow 4f$ $\rightarrow 5d$ $\rightarrow 6p$.

(C) The energy of a photon emitted during an electronic transition is given by the formula: $E = \Delta E = 13.6 \ eV \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
As the principal quantum number $n$ increases,the energy gap between consecutive shells decreases.
The energy difference between shells $n$ and $n+1$ is given by $\Delta E = 13.6 \left[ \frac{1}{n^2} - \frac{1}{(n+1)^2} \right]$.
Comparing the transitions:
$(A)$ $n=6 \to n=1$: Large energy gap.
$(B)$ $n=5 \to n=4$: $\Delta E \propto (1/16 - 1/25) = 0.0225$.
$(C)$ $n=6 \to n=5$: $\Delta E \propto (1/25 - 1/36) = 0.0122$.
$(D)$ $n=5 \to n=3$: $\Delta E \propto (1/9 - 1/25) = 0.0711$.
The transition $n=6 \to n=5$ involves the shells with the highest principal quantum numbers,resulting in the smallest energy difference and thus the least energetic photon.

(A) The ionization energy $(IE_{1})$ is the energy required to remove an electron from a neutral atom: $Na(g) \longrightarrow Na^{ }(g) e^{-}(g) ; IE_{1} = 5.1 \ eV$.
The electron gain enthalpy $(\Delta H_{eg})$ is the energy change when an electron is added to a gaseous ion to form a neutral atom,which is the reverse process of ionization: $Na^{ }(g) e^{-}(g) \longrightarrow Na(g)$.
According to the law of conservation of energy,the enthalpy change for the reverse process is the negative of the ionization energy: $\Delta H_{eg} = -IE_{1} = -5.1 \ eV$.

(A) The bond order is inversely proportional to the bond length.
Bond order of $O_2^+ = \frac{10-5}{2} = 2.5$
Bond order of $O_2 = \frac{10-6}{2} = 2.0$
Bond order of $O_2^- = \frac{10-7}{2} = 1.5$
Bond order of $O_2^{2-} = \frac{10-8}{2} = 1.0$
Since $O_2^+$ has the highest bond order $(2.5)$,it has the minimum bond length.

(A) To determine the hybridization,we calculate the steric number $(SN)$ using the formula: $SN = \frac{1}{2} (V + M - C + A)$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
$1. NO_2^-: SN = \frac{1}{2} (5 + 0 - 0 + 1) = 3 \rightarrow sp^2$ hybridization.
$2. NO_3^-: SN = \frac{1}{2} (5 + 0 - 0 + 1) = 3 \rightarrow sp^2$ hybridization.
$3. NH_2^-: SN = \frac{1}{2} (5 + 2 - 0 + 1) = 4 \rightarrow sp^3$ hybridization.
$4. NH_4^+: SN = \frac{1}{2} (5 + 4 - 1 + 0) = 4 \rightarrow sp^3$ hybridization.
$5. SCN^-: SN = \frac{1}{2} (4 + 0 - 0 + 1) = 2 \rightarrow sp$ hybridization.
Comparing the results,$NO_2^-$ and $NO_3^-$ both exhibit $sp^2$ hybridization.

(A) The bond lengths for the given bonds are as follows:
$C - H \approx 0.109 \ nm$
$C=C \approx 0.134 \ nm$
$C - O \approx 0.143 \ nm$
$C - C \approx 0.154 \ nm$
Comparing these values,the order of increasing bond length is $C - H < C=C < C - O < C - C$.

(D) Formal charges help in the selection of the lowest energy structure from a number of possible Lewis structures for a given species.
Generally,the lowest energy structure is the one with the smallest formal charges on the atoms.
Formal charge on an atom $=$ (total number of valence electrons) $-$ (non-bonding electrons) $-$ $\frac{1}{2} \times$ (bonding electrons).
For the Lewis structure of $SO_3$ with three double bonds:
Formal charge on $S$ atom $= 6 - 0 - \frac{1}{2} \times 12 = 0$.
Formal charge on each of the three $O$ atoms $= 6 - 4 - \frac{1}{2} \times 4 = 0$.
Since all atoms have a formal charge of $0$,this structure is the most preferred and has the lowest energy.

(C) The electronic configuration of $O_2^+$ ($13$ electrons) is: $KK \sigma 2s^2 \sigma^* 2s^2 \sigma 2pz^2 (\pi 2px^2 = \pi 2py^2) (\pi^* 2px^1)$. It has one unpaired electron,so it is paramagnetic.
The electronic configuration of $O_2$ ($16$ electrons) is: $KK \sigma 2s^2 \sigma^* 2s^2 \sigma 2pz^2 (\pi 2px^2 = \pi 2py^2) (\pi^* 2px^1 = \pi^* 2py^1)$. It has two unpaired electrons,so it is paramagnetic.
Since both $O_2^+$ and $O_2$ contain unpaired electrons in their $\pi^*$ antibonding molecular orbitals,both are paramagnetic.

(D) The formula for the average velocity $(V_{av})$ of a gaseous molecule is given by $V_{av} = \sqrt{\frac{8RT}{\pi M}}$.
From this expression,it is clear that $V_{av} \propto \sqrt{T}$.
When the temperature is doubled,the new temperature $T' = 2T$.
Therefore,the ratio of the new average velocity to the initial average velocity is $\frac{V_{av}'}{V_{av}} = \sqrt{\frac{T'}{T}} = \sqrt{\frac{2T}{T}} = \sqrt{2} \approx 1.414$.
Thus,the average velocity increases by a factor of approximately $1.4$.

(B) According to Graham's Law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$ for gases at the same volume and pressure: $\frac{r_A}{r_B} = \sqrt{\frac{M_B}{M_A}}$.
Since $r = \frac{V}{t}$ and $V_A = V_B$,the equation becomes $\frac{t_B}{t_A} = \sqrt{\frac{M_B}{M_A}}$.
Given $t_A = 20 \ s$,$t_B = 10 \ s$,and $M_A = 49 \ u$,we substitute these values:
$\frac{10}{20} = \sqrt{\frac{M_B}{49}}$.
$\frac{1}{2} = \frac{\sqrt{M_B}}{7}$.
$\sqrt{M_B} = \frac{7}{2} = 3.5$.
$M_B = (3.5)^2 = 12.25 \ u$.

(A) According to Dalton's Law of Partial Pressures,the partial pressure of a gas is proportional to its mole fraction in the mixture.
Given that the moles of $CO$ and $N_2$ are equal,let $n_{CO} = n_{N_2} = n$.
The mole fraction of $N_2$ is given by $x_{N_2} = \frac{n_{N_2}}{n_{CO} + n_{N_2}} = \frac{n}{n + n} = \frac{n}{2n} = 0.5$.
The partial pressure of $N_2$ is calculated as $p_{N_2} = x_{N_2} \times P_{total}$.
Given $P_{total} = 1 \ atm$,we have $p_{N_2} = 0.5 \times 1 \ atm = 0.5 \ atm$.

(A) Using the combined gas law,$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$,we can express the ratio of volumes as $\frac{V_2}{V_1} = \frac{P_1 T_2}{P_2 T_1}$.
Given initial conditions: $T_1 = 15 + 273 = 288 \ K$ and $P_1 = 1.5 \ bar$.
Given final conditions: $T_2 = 25 + 273 = 298 \ K$ and $P_2 = 1.0 \ bar$.
Substituting the values: $\frac{V_2}{V_1} = \frac{1.5 \times 298}{1.0 \times 288} = \frac{447}{288} \approx 1.552$.
Rounding to one decimal place,the volume increases by a factor of $1.6$.

(D) The process is the phase transition of liquid water to steam at a constant temperature.
Given: $\Delta H_{vap} = 30 \ kJ \ mol^{-1} = 30000 \ J \ mol^{-1}$.
Temperature $T = 27 \ ^oC = 27 + 273 = 300 \ K$.
The entropy change is given by the formula: $\Delta S_{vap} = \frac{\Delta H_{vap}}{T}$.
Substituting the values: $\Delta S_{vap} = \frac{30000 \ J \ mol^{-1}}{300 \ K} = 100 \ J \ mol^{-1} \ K^{-1}$.

(A) The given reaction is $4H_{(g)} \rightarrow 2H_{2_{(g)}}$ with $\Delta H = -869.6 \ kJ$.
This reaction represents the formation of $2$ moles of $H_2$ molecules from $4$ moles of $H$ atoms.
The dissociation energy of a bond is defined as the energy required to break $1$ mole of bonds in the gaseous state.
First,reverse the reaction to represent the dissociation of $2$ moles of $H_2$ molecules:
$2H_{2_{(g)}} \rightarrow 4H_{(g)}$; $\Delta H = +869.6 \ kJ$.
Now,for $1$ mole of $H_2$ dissociation:
$H_{2_{(g)}} \rightarrow 2H_{(g)}$; $\Delta H = \frac{869.6}{2} = 434.8 \ kJ$.
Thus,the dissociation energy of the $H-H$ bond is $434.8 \ kJ$.

(C) For the free expansion of an ideal gas,the process is adiabatic,so $q = 0$.
Since the expansion occurs against zero external pressure $(P_{ext} = 0)$,the work done is $W = -P_{ext} \Delta V = 0$.
According to the first law of thermodynamics,$\Delta U = q + W$. Since $q = 0$ and $W = 0$,$\Delta U = 0$.
For an ideal gas,internal energy $U$ is a function of temperature only,so $\Delta U = 0$ implies $\Delta T = 0$.

(B) To obtain the target reaction $B + D \rightarrow E + 2C$,we manipulate the given equations:
$1. \ 2 \times (I): A \rightarrow 2B, \Delta H = 2 \times 150 = 300 \ kJ/mol$
$2. \ (II): 3B \rightarrow 2C + D, \Delta H = -125 \ kJ/mol$
$3. \ -(III): 2D \rightarrow E + A, \Delta H = -350 \ kJ/mol$
Adding these equations:
$(A + 3B + 2D) \rightarrow (2B + 2C + D + E + A)$
Simplifying by canceling common terms ($A$ and $2B$ and $D$ from both sides):
$B + D \rightarrow E + 2C$
$\Delta H = 300 - 125 - 350 = -175 \ kJ/mol$

(D) The given reaction is $X_{2(g)} + 4Y_{2(g)} \rightleftharpoons 2XY_{4(g)}$.
Given that $\Delta H < 0$,the reaction is exothermic.
Calculate the change in the number of moles of gaseous species: $\Delta n_g = n_{products} - n_{reactants} = 2 - (1 + 4) = 2 - 5 = -3$.
Since $\Delta n_g < 0$,the forward reaction is favoured at high pressure according to Le Chatelier's principle.
Since the reaction is exothermic $(\Delta H < 0)$,the forward reaction is favoured at low temperature.
Therefore,the formation of $XY_{4(g)}$ is favoured at high pressure and low temperature.

(B) For a basic buffer,the $pOH$ is calculated using the Henderson-Hasselbalch equation:
$pOH = pK_b + \log \frac{[Salt]}{[Base]}$
Given: $[Base] = [NH_3] = 0.30 \ M$,$[Salt] = [NH_4^+] = 0.20 \ M$,and $K_b = 1.8 \times 10^{-5}$.
$pK_b = -\log(1.8 \times 10^{-5}) = 5 - \log(1.8) = 5 - 0.255 = 4.745$.
$pOH = 4.745 + \log \frac{0.20}{0.30} = 4.745 + \log(0.667) = 4.745 - 0.176 = 4.569$.
Since $pH + pOH = 14$,
$pH = 14 - 4.569 = 9.431 \approx 9.43$.

(C) Lewis base is a substance that can donate a lone pair of electrons.
$H_2O$,$NH_3$,and $OH^{-}$ all possess lone pairs of electrons that can be donated.
$BF_3$ has an incomplete octet around the central Boron atom,making it an electron-deficient species.
Therefore,$BF_3$ acts as a Lewis acid rather than a Lewis base.

(C) Given reactions:
$1) N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)} ; K_1$
$2) 2NO_{(g)} + O_{2(g)} \rightleftharpoons 2NO_{2(g)} ; K_2$
Adding equations $(1)$ and $(2)$ gives:
$N_{2(g)} + 2O_{2(g)} \rightleftharpoons 2NO_{2(g)}$
The equilibrium constant for this combined reaction is $K_{eq} = K_1 \times K_2$.
Now,to obtain the target reaction $NO_{2(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + O_{2(g)}$,we reverse the combined reaction and multiply the coefficients by $\frac{1}{2}$.
Reversing the reaction gives $K' = \frac{1}{K_1 K_2}$.
Multiplying coefficients by $\frac{1}{2}$ gives $K = (K')^{1/2} = \left[ \frac{1}{K_1 K_2} \right]^{1/2}$.

(C) For $AgCl$ precipitation: $K_{sp} = [Ag^{+}][Cl^{-}]$.
Given $[Cl^{-}] = 0.10 \, M = 10^{-1} \, M$.
$[Ag^{+}] = \frac{K_{sp}}{[Cl^{-}]} = \frac{1.8 \times 10^{-10}}{0.10} = 1.8 \times 10^{-9} \, M$.
For $PbCl_2$ precipitation: $K_{sp} = [Pb^{2+}][Cl^{-}]^2$.
$[Pb^{2+}] = \frac{K_{sp}}{[Cl^{-}]^2} = \frac{1.7 \times 10^{-5}}{(0.10)^2} = \frac{1.7 \times 10^{-5}}{0.01} = 1.7 \times 10^{-3} \, M$.

(C) The melting point of ionic compounds is directly related to their lattice energy.
As the size of the anion increases $(F^- < Cl^- < Br^- < I^-)$,the lattice energy decreases due to weaker electrostatic attraction between the cation and the larger,more polarizable anion.
Therefore,the melting point decreases in the order: $CaF_2 > CaCl_2 > CaBr_2 > CaI_2$.
Thus,$CaI_2$ has the lowest melting point.

(A) . Plaster of Paris is $CaSO_4 \cdot \frac{1}{2} H_2O$ $(i)$.
$B$. Epsomite is $MgSO_4 \cdot 7H_2O$ $(ii)$.
$C$. Kieserite is $MgSO_4 \cdot H_2O$ $(iii)$.
$D$. Gypsum is $CaSO_4 \cdot 2H_2O$ $(iv)$.
Therefore,the correct matching is $A-i, B-ii, C-iii, D-iv$.

(C) In a pyrosilicate structure,two $[SiO_4]^{4-}$ tetrahedral units are joined by sharing one oxygen atom at one corner.
This results in the formula $[Si_2O_7]^{6-}$.
Therefore,the correct answer is pyrosilicate.

(C) Aluminium reacts with excess $NaOH$ to form sodium tetrahydroxoaluminate$(III)$,which is a soluble complex,not $Al(OH)_3$.
The reaction is: $2 Al(s) + 2 NaOH(aq) + 6 H_2O(l) \longrightarrow 2 Na[Al(OH)_4](aq) + 3 H_2(g)$.
Therefore,the statement in option $C$ is incorrect.

(A) $SnO_2$ is an amphoteric oxide because it reacts with both acids and bases.
Reaction with acid:
$SnO_2 + 4 HCl \longrightarrow SnCl_4 + 2 H_2O$
Reaction with base:
$SnO_2 + 2 NaOH \longrightarrow Na_2SnO_3 + H_2O$

(B) The geometry of a molecule depends on the hybridization of its carbon atoms.
$sp^{3}$ hybridized carbon atoms have a tetrahedral geometry with a bond angle of $109^{\circ} 28^{\prime}$.
$sp^{2}$ hybridized carbon atoms have a trigonal planar geometry with a bond angle of $120^{\circ}$.
$sp$ hybridized carbon atoms have a linear geometry with a bond angle of $180^{\circ}$.
In $CH_3 - C \equiv C - CH_3$ (but$-2-$yne),the two central carbon atoms are $sp$ hybridized,making the central part of the molecule linear.

(A) $1$. Identify the longest carbon chain containing the double bond. The longest chain has $6$ carbons,making it a hexene derivative.
$2$. Number the chain starting from the end closest to the double bond. The double bond starts at $C-1$.
$3$. The substituents are an ethyl group at $C-3$ and a propyl group at $C-4$.
$4$. Combining these,the correct $IUPAC$ name is $4-$ethyl$-3-$propylhex$-1-$ene.

(B) The pressure of dry nitrogen gas is $p_1 = p - p_{aq} = 715 \ mm - 15 \ mm = 700 \ mm$.
Using the ideal gas equation $\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$ to find the volume of $N_2$ at $STP$ $(p_2 = 760 \ mm, T_2 = 273 \ K)$:
$V_2 = \frac{p_1 V_1 T_2}{T_1 p_2} = \frac{700 \times 55 \times 273}{300 \times 760} = 48.098 \ mL$.
The percentage of nitrogen is calculated as:
$\% \ N = \frac{28}{22400} \times \frac{V_2 \ (mL)}{W \ (g)} \times 100$
$\% \ N = \frac{28}{22400} \times \frac{48.098}{0.35} \times 100 = 16.45 \ \%$.

(A) If $Na_2S$ and $NaCN$ are present in the extract,they will interfere with the test for halogens by forming precipitates with $AgNO_3$.
Boiling the extract with conc. $HNO_3$ decomposes these compounds into volatile $H_2S$ and $HCN$ gases:
$NaCN + HNO_3 \rightarrow NaNO_3 + HCN \uparrow$
$Na_2S + 2HNO_3 \rightarrow 2NaNO_3 + H_2S \uparrow$
These gases escape from the solution,ensuring they do not interfere with the halogen test.

$(A)$ The longest carbon chain contains $5$ carbon atoms with a $C=C$ bond starting at the second carbon atom, so it is a $2$-pentene derivative.
The substituents are a chlorine atom $(Cl)$ at position $2$ and an iodine atom $(I)$ at position $3$.
Looking at the geometry across the $C=C$ bond, the higher priority groups (based on atomic number, $Cl$ vs $CH_3$ on one side, and $I$ vs $CH_2CH_3$ on the other) are on opposite sides of the double bond, which corresponds to the $E$ configuration, often referred to as $trans$ in simple cases where similar groups are on opposite sides.
Specifically, the $Cl$ and the $CH_2CH_3$ group are on opposite sides of the double bond, and the $CH_3$ and $I$ are on opposite sides. Thus, the configuration is $trans-2-chloro-3-iodo-2-pentene$.

(B) The concentration of $DO$ (Dissolved Oxygen) in water is crucial for aquatic life. The growth of fish is inhibited if the concentration of $DO$ falls below $6 \ ppm$. Therefore,the statement that a concentration below $6 \ ppm$ is good for the growth of fish is false.

(A) The reaction between acidified $K_2Cr_2O_7$ and $Na_2SO_3$ is a redox reaction where $Cr_2O_7^{2-}$ acts as an oxidizing agent and $SO_3^{2-}$ acts as a reducing agent.
The balanced chemical equation is:
$K_2Cr_2O_7 + 3Na_2SO_3 + 4H_2SO_4 \rightarrow K_2SO_4 + 3Na_2SO_4 + Cr_2(SO_4)_3 + 4H_2O$
In this reaction,the orange-colored dichromate ion $(Cr_2O_7^{2-})$ is reduced to the green-colored chromium$(III)$ sulfate $(Cr_2(SO_4)_3)$,which causes the solution to turn green.

(D) Guard cells are specialized epidermal cells that surround the stomatal pore.
Unlike typical epidermal cells,which are generally transparent and lack photosynthetic machinery,guard cells contain chloroplasts.
These chloroplasts are essential for the production of $ATP$ via photosynthesis,which provides the energy required for the active transport of ions (like $K^+$) to regulate stomatal opening and closing.
While both guard cells and other epidermal cells contain mitochondria,endoplasmic reticulum,and a cytoskeleton,the presence of chloroplasts is the distinguishing feature that allows guard cells to perform their unique physiological function.

(D) Companion cells are specialized parenchyma cells associated with sieve tube elements in the phloem.
They maintain a pressure gradient in the sieve tubes by actively loading sucrose into them.
This process is known as phloem loading,which requires energy in the form of $ATP$ because it occurs against the concentration gradient (active transport).
Therefore,the primary function is the active loading of sucrose into sieve elements.

(D) Guttation is the process of exudation of liquid droplets from the margins of leaves through specialized pores called hydathodes.
It occurs due to the development of high positive hydrostatic pressure in the xylem of roots,known as root pressure.
This phenomenon is typically observed during the night or early morning when transpiration is low and soil moisture is high.

(B) Calcium hypochlorite,with the formula $Ca(OCl)_2$,is the primary active ingredient in bleaching powder.
It is responsible for the bleaching action and disinfection properties of the compound.
The chemical reaction for the preparation of bleaching powder is:
$3Ca(OH)_2 + 2Cl_2 \xrightarrow{\text{below } 35 \ ^\circ C} Ca(OCl)_2 \cdot CaCl_2 \cdot Ca(OH)_2 \cdot 2H_2O$.

(C) In this reaction,the bromine atom $(Br)$ in the alkyl halide is replaced by the amino group $(-NH_2)$ from ammonia $(NH_3)$.
Since a nucleophile $(NH_3)$ replaces a leaving group $(Br^{-})$,it is a nucleophilic substitution reaction.
Option $(a)$ is an electrophilic addition reaction.
Options $(b)$ and $(d)$ are nucleophilic addition reactions.

(A) Nucleophilic aromatic substitution reactions are facilitated by the presence of electron-withdrawing groups on the benzene ring.
These groups stabilize the intermediate carbanion (Meisenheimer complex) formed during the reaction.
The $-NO_2$ group is a strong electron-withdrawing group due to its $-I$ and $-M$ effects.
In contrast,$-CH_3$ and $-OCH_3$ groups are electron-donating groups,which destabilize the intermediate carbanion and make the reaction more difficult.
Therefore,$p$-nitrochlorobenzene undergoes nucleophilic substitution most easily.

(C) For a compound with $NaCl$ structure, the limiting radius ratio is $0.414$.
The radius ratio formula is given by $\frac{r^+}{r^-} = 0.414$.
Given that the radius of the cation $r^+ = 100 \ pm$, we substitute this into the equation:
$\frac{100}{r^-} = 0.414$
$r^- = \frac{100}{0.414} \approx 241.54 \ pm$.
Rounding to the nearest option, the radius of the anion is $241.5 \ pm$.

(C) $1.00 \, m$ solution means $1 \, mol$ of solute is present in $1000 \, g$ of water.
The number of moles of water is $n_{H_{2}O} = \frac{1000 \, g}{18 \, g/mol} = 55.5 \, mol$.
The mole fraction of the solute is given by $X_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{H_{2}O}}$.
Substituting the values,$X_{\text{solute}} = \frac{1}{1 + 55.5} = \frac{1}{56.5} \approx 0.0177$.

(B) The formula for freezing point depression is $\Delta T_f = i \times K_f \times m.$
Here,$\Delta T_f = 3.82 \, ^oC,$ $K_f = 1.86 \, ^oC \, kg \, mol^{-1},$ $W_{solute} = 5.00 \, g,$ $W_{solvent} = 45.0 \, g,$ and the molar mass of $Na_2SO_4$ is $M = 142 \, g \, mol^{-1}.$
The molality $m$ is given by $\frac{W_{solute} \times 1000}{M \times W_{solvent}} = \frac{5.00 \times 1000}{142 \times 45.0} \approx 0.782 \, m.$
Substituting these values into the equation: $3.82 = i \times 1.86 \times 0.782.$
Solving for $i$: $i = \frac{3.82}{1.86 \times 0.782} \approx 2.63.$

(C) The van't Hoff factor $i$ is defined as the ratio of the observed colligative property to the calculated colligative property.
For a compound that undergoes dissociation,the number of particles in the solution increases,so $i > 1$.
For a compound that undergoes association,the number of particles in the solution decreases,so $i < 1$.
Therefore,for dissociation and association respectively,$i$ is greater than $1$ and less than $1$.

(D) For the dissociation of weak acid $HX$: $HX \rightleftharpoons H^+ + X^-$
Initial moles: $1 \quad 0 \quad 0$
At equilibrium: $(1-\alpha) \quad \alpha \quad \alpha$
Total moles at equilibrium $= 1 - \alpha + \alpha + \alpha = 1 + \alpha$
Van't Hoff factor $i = 1 + \alpha = 1 + 0.3 = 1.3$
Depression in freezing point $\Delta T_f = i \times K_f \times m$
$\Delta T_f = 1.3 \times 1.86 \times 0.1 = 0.2418 \ ^{\circ}C$
Freezing point of solution $T_f = T_f^{\circ} - \Delta T_f = 0 - 0.2418 \ ^{\circ}C = - 0.24 \ ^{\circ}C$

(D) The formula for osmotic pressure is $\pi V = nRT$,where $n = \frac{w}{M}$.
Given:
$\pi = 2.57 \times 10^{-3} \ bar$
$V = 200 \ mL = 0.2 \ L$
$w = 1.26 \ g$
$T = 300 \ K$
$R = 0.083 \ L \ bar \ mol^{-1} \ K^{-1}$
Substituting the values into the equation $\pi V = \frac{w}{M} RT$:
$2.57 \times 10^{-3} \times 0.2 = \frac{1.26}{M} \times 0.083 \times 300$
$M = \frac{1.26 \times 0.083 \times 300}{2.57 \times 10^{-3} \times 0.2}$
$M = \frac{31.374}{0.000514} \approx 61039 \ g \ mol^{-1}$.

(A) The relationship between standard Gibbs free energy change $(\Delta G^{\circ})$ and standard cell potential $(E^{\circ}_{cell})$ is given by $\Delta G^{\circ} = -nFE^{\circ}_{cell}$.
If $E^{\circ}_{cell} < 0$,then $\Delta G^{\circ} > 0$,which indicates that the reaction is non-spontaneous.
Furthermore,the relationship between $E^{\circ}_{cell}$ and the equilibrium constant $(K_{eq})$ is given by $E^{\circ}_{cell} = \frac{RT}{nF} \ln K_{eq}$.
If $E^{\circ}_{cell} < 0$,then $\ln K_{eq} < 0$,which implies $K_{eq} < 1$.

(C) The standard reduction potentials are given as:
$E_{X}^{0} = -1.2 \, V$
$E_{Y}^{0} = +0.5 \, V$
$E_{Z}^{0} = -3.0 \, V$
Reducing power is inversely proportional to the standard reduction potential.
Therefore,the metal with the lowest (most negative) reduction potential has the highest reducing power.
Comparing the values: $-3.0 \, V < -1.2 \, V < +0.5 \, V$.
Thus,the order of reducing power is $Z > X > Y$.

(B) For the reaction $Cu^{2+} + e^- \rightarrow Cu^{+}$,$E^0_1 = 0.15 \ V$. The Gibbs free energy change is $\Delta G^0_1 = -n_1 F E^0_1 = -1 \times F \times 0.15 = -0.15 F$.
For the reaction $Cu^{+} + e^- \rightarrow Cu$,$E^0_2 = 0.50 \ V$. The Gibbs free energy change is $\Delta G^0_2 = -n_2 F E^0_2 = -1 \times F \times 0.50 = -0.50 F$.
For the overall reaction $Cu^{2+} + 2e^- \rightarrow Cu$,the total Gibbs free energy change is $\Delta G^0 = \Delta G^0_1 + \Delta G^0_2 = -0.15 F - 0.50 F = -0.65 F$.
Using the relation $\Delta G^0 = -n F E^0_{Cu^{2+}/Cu}$,where $n = 2$:
$-2 F E^0_{Cu^{2+}/Cu} = -0.65 F$.
Therefore,$E^0_{Cu^{2+}/Cu} = \frac{0.65}{2} = 0.325 \ V$.

(B) The standard reduction potential for the cathode $(Sn^{4+}/Sn^{2+})$ is $E_{cathode}^{0} = +0.15 \ V$.
The standard reduction potential for the anode $(Cr^{3+}/Cr)$ is $E_{anode}^{0} = -0.74 \ V$.
The standard cell potential is calculated using the formula: $E_{cell}^{0} = E_{cathode}^{0} - E_{anode}^{0}$.
Substituting the values: $E_{cell}^{0} = 0.15 \ V - (-0.74 \ V) = 0.15 + 0.74 = +0.89 \ V$.

(C) The standard reduction potential $(E^{\circ})$ for $Fe^{3+}/Fe^{2+}$ is $+0.77 \ V$ and for $I_2/2I^{-}$ is $+0.536 \ V$.
Since the reduction potential of $Fe^{3+}/Fe^{2+}$ is higher than that of $I_2/2I^{-}$,$Fe^{3+}$ has a greater tendency to get reduced to $Fe^{2+}$.
Conversely,$I^{-}$ has a greater tendency to get oxidised to $I_2$.
Therefore,the spontaneous reaction is $2Fe^{3+} + 2I^{-} \rightarrow 2Fe^{2+} + I_2$.
Thus,$I^{-}$ will be oxidised to $I_2$.

(D) The order of a reaction is defined as the sum of the powers of the concentration terms in the rate law expression.
It is an experimental quantity and is not necessarily related to the stoichiometric coefficients of the balanced chemical equation.
The order of a reaction can be zero,a whole number,or even a fraction.
Therefore,the statement that the order of reaction is always a whole number is incorrect.

(B) The rate of reaction is defined as:
$-\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$
Substituting the given rate expressions:
$\frac{1}{2} k[N_2O_5] = \frac{1}{4} k'[N_2O_5] = k''[N_2O_5]$
Dividing by $[N_2O_5]$:
$\frac{k}{2} = \frac{k'}{4} = k''$
From $\frac{k}{2} = \frac{k'}{4}$,we get $k' = 2k$.
From $\frac{k}{2} = k''$,we get $k'' = \frac{k}{2}$.

(A) The rate law expression for a zero order reaction is given by: $\text{Rate} = k[A]^0$.
Since the rate of reaction is expressed in units of concentration per unit time $(mol\, L^{-1}\, s^{-1})$,we have:
$mol\, L^{-1}\, s^{-1} = k \times (mol\, L^{-1})^0$.
Since any value raised to the power of $0$ is $1$,the unit of $k$ is equal to the unit of the rate of reaction.
Therefore,the unit of the rate constant for a zero order reaction is $mol\, L^{-1}\, s^{-1}$.

(C) Enzyme-catalysed reactions follow first-order kinetics.
The concentration of the substance decreases from $1.28 \; mg \; L^{-1}$ to $0.04 \; mg \; L^{-1}$.
We can calculate the number of half-lives $(n)$ as follows:
$1.28$ $\xrightarrow{t_{1/2}} 0.64$ $\xrightarrow{t_{1/2}} 0.32$ $\xrightarrow{t_{1/2}} 0.16$ $\xrightarrow{t_{1/2}} 0.08$ $\xrightarrow{t_{1/2}} 0.04$
This sequence shows that the concentration reduces by half $5$ times,so $n = 5$.
The total time required is $t = n \times t_{1/2}$.
$t = 5 \times 138 \; s = 690 \; s$.

(D) The relation $x/m = p \times T$ is incorrect for the adsorption process.
Adsorption is typically described by the Freundlich or Langmuir isotherms,where the extent of adsorption $(x/m)$ is a function of pressure $(p)$ at constant temperature $(T)$,or a function of temperature $(T)$ at constant pressure $(p)$.
The relationship $p = f(T)$ at constant $x/m$ represents isosteres.
The expression $x/m = p \times T$ does not represent any standard adsorption isotherm or thermodynamic relation.

(B) Pig iron is the most impure form of iron,obtained directly from the blast furnace.
It contains about $3-4.5 \%$ of carbon as the major impurity.
Other impurities like $S, P, Si,$ and $Mn$ are present in much smaller amounts compared to carbon.

(B) $Zr$ and $Ti$ are purified by the van Arkel method.
$\underset{\text{Impure}}{Zr + 2I_2}$ $\xrightarrow{600\,^{\circ}C} ZrI_4$ $\xrightarrow{1800\,^{\circ}C} \underset{\text{Pure}}{Zr + 2I_2}$
This method is useful for removing all the oxygen and nitrogen present as impurities in certain metals like $Zr$ and $Ti$.

(C) In the blast furnace,the limestone $(CaCO_3)$ decomposes to form calcium oxide $(CaO)$,which acts as a flux.
This flux reacts with the silica $(SiO_2)$ impurity (gangue) present in the ore to form a fusible slag called calcium silicate $(CaSiO_3)$.
The reaction is: $CaO_{(s)} + SiO_{2(s)} \rightarrow CaSiO_{3(l)}$.

(A) The stability of the $+2$ oxidation state in transition metals is influenced by the electronic configuration and exchange energy.
$1$. $Mn^{2+}$ has a $d^5$ configuration,which is half-filled and highly stable due to maximum exchange energy.
$2$. $Fe^{2+}$ has a $d^6$ configuration. While it has one electron paired,it still possesses significant exchange energy,making it more stable than $Cr^{2+}$ $(d^4)$.
$3$. $Cr^{2+}$ has a $d^4$ configuration,which is less stable than $d^5$ and $d^6$ due to lower exchange energy.
$4$. $Co^{2+}$ has a $d^7$ configuration,which is the least stable among these due to increased inter-electronic repulsion.
Thus,the order of stability is $Mn^{2+} > Fe^{2+} > Cr^{2+} > Co^{2+}$.

(B) The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$. In $[Ni(CN)_4]^{2-}$,$Ni$ is in the $+2$ oxidation state,so its configuration is $Ni^{2+} = 3d^8$.
Since $CN^-$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbitals.
This results in $d^8$ configuration with all electrons paired,making the complex diamagnetic.
In contrast,$[NiCl_4]^{2-}$,$[CuCl_4]^{2-}$,and $[CoF_6]^{3-}$ contain unpaired electrons and are paramagnetic.

(C) The complexes $[Co(NH_3)_6][Cr(CN)_6]$ and $[Cr(NH_3)_6][Co(CN)_6]$ are examples of coordination isomerism.
This type of isomerism occurs in complexes where both the cation and the anion are complex ions.
It arises due to the interchange of ligands between the cationic and anionic coordination spheres.

(A) The complex $[Pt(Py)(NH_3)BrCl]$ is a square planar complex of the type $[M(abcd)]$,where $M = Pt$,$a = Py$,$b = NH_3$,$c = Br$,and $d = Cl$.
For a square planar complex of the type $[M(abcd)]$,there are $3$ possible geometrical isomers.
These isomers are formed by fixing one ligand (e.g.,$Py$) and varying the positions of the other three ligands $(NH_3, Br, Cl)$ relative to it.
Thus,the total number of geometrical isomers is $3$.

(C) The paramagnetic behaviour depends on the number of unpaired electrons $(n)$.
$1$. For $[Cr(H_2O)_6]^{2+}$,$Cr^{2+}$ is $d^4$. In a weak field ligand like $H_2O$,$n = 4$.
$2$. For $[Mn(H_2O)_6]^{2+}$,$Mn^{2+}$ is $d^5$. In a weak field ligand like $H_2O$,$n = 5$.
$3$. For $[Fe(H_2O)_6]^{2+}$,$Fe^{2+}$ is $d^6$. In a weak field ligand like $H_2O$,$n = 4$.
$4$. For $[Co(H_2O)_6]^{2+}$,$Co^{2+}$ is $d^7$. In a weak field ligand like $H_2O$,$n = 3$.
Since $[Co(H_2O)_6]^{2+}$ has the minimum number of unpaired electrons $(n = 3)$,it exhibits the minimum paramagnetic behaviour.
The magnetic moment is given by $\mu = \sqrt{n(n+2)} \text{ BM}$.

(A) The strength of the $C-O$ bond in metal carbonyls depends on the extent of back-bonding from the metal $d$-orbitals to the $\pi^*$-antibonding orbitals of the $CO$ ligand.
As the negative charge on the metal complex increases,the metal atom becomes more electron-rich and donates more electron density into the $\pi^*$-orbitals of $CO$,which weakens the $C-O$ bond.
Conversely,as the positive charge on the central metal atom increases,the metal's ability to donate electron density decreases.
Therefore,the $C-O$ bond is strongest in the species with the highest positive charge or lowest negative charge,as there is minimal back-bonding.
Among the given options,$Mn(CO)_6^+$ has the highest positive charge,resulting in the least back-bonding and the strongest $C-O$ bond.

(B) $Ti$ $\rightarrow [Ar] 3d^{2} 4s^{2}, Ti^{3+}$ $\rightarrow [Ar] 3d^{1} 4s^{0}$ ($1$ unpaired electron)
$Cr$ $\rightarrow [Ar] 3d^{5} 4s^{1}, Cr^{3+}$ $\rightarrow [Ar] 3d^{3} 4s^{0}$ ($3$ unpaired electrons)
$Co$ $\rightarrow [Ar] 3d^{7} 4s^{2}, Co^{3+}$ $\rightarrow [Ar] 3d^{6} 4s^{0}$ ($0$ unpaired electrons because $NH_3$ is a strong field ligand causing pairing)
$Zn$ $\rightarrow [Ar] 3d^{10} 4s^{2}, Zn^{2+}$ $\rightarrow [Ar] 3d^{10}$ ($0$ unpaired electrons)
Since $[Cr(NH_3)_6]^{3+}$ has the highest number of unpaired electrons $(3)$,it exhibits the highest paramagnetic behaviour.

(C) Reaction $(i)$ involves a primary alkyl halide reacting with a weak nucleophile $(C_2H_5OH)$,which proceeds via the $S_N1$ mechanism.
Reaction $(ii)$ involves a primary alkyl halide reacting with a strong nucleophile $(C_2H_5O^-)$,which proceeds via the $S_N2$ mechanism.
Therefore,the mechanisms are $S_N1$ and $S_N2$ respectively.

(B) In reaction $(i)$,dehydration of $3\text{-methylbutan-}2\text{-ol}$ occurs via a carbocation intermediate.
The initially formed $2^\circ$ carbocation $CH_3-CH(CH_3)-C^+H-CH_3$ undergoes a $1,2\text{-hydride shift}$ to form a more stable $3^\circ$ carbocation $CH_3-C^+(CH_3)-CH_2-CH_3$.
Elimination of a proton from this $3^\circ$ carbocation according to Saytzeff's rule gives the major product $A$,which is $2\text{-methylbut-}2\text{-ene}$ $(CH_3-C(CH_3)=CH-CH_3)$.
In reaction $(ii)$,Markovnikov addition of $HBr$ to $A$ in the absence of peroxide yields the major product $C$,which is $2\text{-bromo-}2\text{-methylbutane}$ $(CH_3-C(Br)(CH_3)-CH_2-CH_3)$.

(B) The reducing agent used in Clemmensen reduction is $Zn-Hg$ (zinc amalgam) and $HCl$.
In this reaction,the carbonyl group $(>C=O)$ of an aldehyde or ketone is reduced to a methylene group $(>CH_2)$.
The reaction is represented as: $>C=O \xrightarrow{Zn-Hg / HCl} >CH_2$.

(A) The reaction sequence is as follows:
$1$. $m$-bromobenzoic acid reacts with $SOCl_2$ to form $m$-bromobenzoyl chloride $(B)$.
$2$. $m$-bromobenzoyl chloride reacts with $NH_3$ to form $m$-bromobenzamide $(C)$.
$3$. $m$-bromobenzamide undergoes the Hofmann bromamide degradation reaction with $Br_2$ and $NaOH$ to form $m$-bromoaniline $(D)$.
Thus,the product $D$ is $m$-bromoaniline.

(D) The reactivity of carbonyl compounds towards nucleophilic addition reactions depends on two main factors:
$1$. Electronic factors: Alkyl groups exhibit a $+I$-effect,which increases the electron density on the carbonyl carbon,thereby reducing its electrophilicity. Aryl groups exhibit a $+R$-effect,which also decreases the electrophilicity of the carbonyl carbon.
$2$. Steric factors: As the size and number of groups attached to the carbonyl carbon increase,steric hindrance increases,which hinders the approach of the nucleophile.
In the given compounds:
$(I)$ is acetaldehyde $(CH_3CHO)$,which has one alkyl group and one hydrogen atom.
$(II)$ is acetone $(CH_3COCH_3)$,which has two alkyl groups.
$(III)$ is benzophenone $(PhCOPh)$,which has two bulky phenyl groups.
Due to the combined effect of increased electron density and higher steric hindrance,the reactivity decreases in the order: $I > II > III$.

(D) . Benzaldehyde undergoes $ii$. Benzoin condensation in the presence of cyanide ions.
$B$. Phthalic anhydride reacts with phenol to form $i$. Phenolphthalein.
$C$. Phenyl benzoate undergoes $iv$. Fries rearrangement to form hydroxybenzophenones.
$D$. Methyl salicylate is known as $iii$. Oil of wintergreen.
Therefore,the correct matching is $A-ii, B-i, C-iv, D-iii$.

(D) The reaction sequence is as follows:
$CH_3CH_2COOH (A)$ $\xrightarrow{NH_3} CH_3CH_2COONH_4 (B)$ $\xrightarrow{\Delta} CH_3CH_2CONH_2 (C)$ $\xrightarrow{Br_2/KOH} CH_3CH_2NH_2$ (Ethylamine).
The last step is the Hoffmann Bromamide degradation reaction,where an amide is converted into an amine with one less carbon atom.
Since the product is ethylamine $(CH_3CH_2NH_2)$,which has $2$ carbon atoms,the amide $(C)$ must have $3$ carbon atoms.
Therefore,compound $A$ is propanoic acid $(CH_3CH_2COOH)$.

(A) The reduction of nitrobenzene with $Zn$ in the presence of $NH_4Cl$ is a controlled reduction process.
Nitrobenzene $(C_6H_5NO_2)$ reacts with $Zn/NH_4Cl$ to form $N$-phenylhydroxylamine $(C_6H_5NHOH)$.
Therefore,the correct product is phenylhydroxylamine.

(B) The basicity of an amine depends on the availability of the lone pair of electrons on the nitrogen atom for protonation.
In $C_6H_5-CH_2NH_2$ (benzylamine),the lone pair on the nitrogen atom is localised because the $-NH_2$ group is attached to an $sp^3$ hybridized carbon atom,which is not directly attached to the benzene ring.
In the other compounds ($p$-nitroaniline,acetanilide,and aniline),the lone pair on the nitrogen atom is delocalized over the benzene ring due to resonance,making it less available for donation.
Therefore,benzylamine is the most basic compound among the given options.

(C) $(+)$ Lactose is a reducing sugar because it contains a hemiacetal group,which allows it to exist in equilibrium with its open-chain form.
All reducing sugars exhibit mutarotation in aqueous solution.
Therefore,the statement that $(+)$ lactose does not exhibit mutarotation is false.

(A) The correct option is $A$.
$1.$ Vitamins are classified into two categories based on their solubility: water-soluble and fat-soluble vitamins.
$2.$ Fat-soluble vitamins include Vitamin $A$,$D$,$E$,and $K$,which are stored in the liver and fatty tissues.
$3.$ Water-soluble vitamins include Vitamin $B$ complex and Vitamin $C$,which are not stored in the body and must be replenished regularly.
$4.$ Therefore,Vitamin $B$ complex is water-soluble,not fat-soluble.

(C) Statement $1$ is correct: Denaturation of proteins involves the unfolding of the polypeptide chain,leading to the loss of secondary and tertiary structures,while the primary structure remains intact.
Statement $2$ is correct: Denaturation of $DNA$ involves the separation of the two strands of the double helix into single strands due to the breaking of hydrogen bonds.
Statement $3$ is incorrect: Denaturation does not affect the primary structure of proteins,as the peptide bonds remain intact.
Therefore,statements $1$ and $2$ are correct.

(A) $Terylene$ is a polyester because it contains ester linkages $(-COO-)$ in its backbone.
It is formed by the condensation polymerization of monomer units,terephthalic acid $(C_6H_4(COOH)_2)$ and ethylene glycol $(HO-CH_2-CH_2-OH)$.

(B) An antihistamine is a type of pharmaceutical drug that opposes the activity of histamine receptors in the body.
Diphenylhydramine is employed as an antihistamine.
Omeprazole is used in the treatment of peptic ulcer diseases.
Chloramphenicol is an antibiotic.
Norethindrone is an antifertility drug.
Hence,option $B$ is correct.