AIPMT 2010 Chemistry Question Paper with Answer and Solution

(B) Biofortification is the process of breeding crops to increase their nutritional value. This includes increasing the levels of vitamins,minerals,proteins,and healthier fats to improve public health and combat malnutrition.

(D) The correct answer is $D$.
$1$. Parietal cells (also known as oxyntic cells) are responsible for the secretion of hydrochloric acid $(HCl)$ and the intrinsic factor.
$2$. The $HCl$ secreted by these cells lowers the $pH$ of the stomach to an acidic range $(1.8 - 3.5)$,which is essential for the activation of the proenzyme pepsinogen into its active form,pepsin.
$3$. Pepsin is a proteolytic enzyme that hydrolyzes proteins into smaller peptides like proteoses and peptones.
$4$. If the parietal cells become non-functional,the secretion of $HCl$ will decrease or stop,leading to a failure in the conversion of pepsinogen to pepsin.
$5$. Consequently,the digestion of proteins in the stomach will be severely impaired.

(B) The number of molecules in $0.1 \ mol$ of gas is $0.1 \times N_A = 0.1 \times 6.02 \times 10^{23} = 6.02 \times 10^{22}$ molecules.
Since the gas is triatomic,each molecule contains $3$ atoms.
Therefore,the total number of atoms $= 3 \times 6.02 \times 10^{22} = 1.806 \times 10^{23}$ atoms.

(C) According to the de-Broglie equation,the wavelength $\lambda$ is given by:
$\lambda = \frac{h}{mv}$
Given:
$h = 6.6 \times 10^{-34} \ J \ s$
$m = 0.66 \ kg$
$v = 100 \ m/s$
Substituting the values:
$\lambda = \frac{6.6 \times 10^{-34}}{0.66 \times 100} = \frac{6.6 \times 10^{-34}}{66} = 0.1 \times 10^{-34} = 1.0 \times 10^{-35} \ m$

(C) For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
All the given species $(S^{2-}, Cl^{-}, K^{+}, Ca^{2+})$ have $18$ electrons.
The atomic numbers are: $S (16), Cl (17), K (19), Ca (20)$.
As the nuclear charge increases,the force of attraction between the nucleus and the electrons increases,which pulls the electron cloud closer to the nucleus,resulting in a smaller ionic radius.
Therefore,the order of decreasing ionic radii is $S^{2-} > Cl^{-} > K^{+} > Ca^{2+}$.

(B) Electron gain enthalpy generally increases in a period from left to right and decreases in a group on moving downwards.
However,members of the $3^{rd}$ period have higher electron gain enthalpy compared to the corresponding members of the $2^{nd}$ period due to their larger size,which reduces inter-electronic repulsion.
$O$ and $S$ belong to group $16$,while $F$ and $Cl$ belong to group $17$.
Thus,the electron gain enthalpy of $F$ and $Cl$ is higher than that of $O$ and $S$.
Comparing within groups,$Cl > F$ and $S > O$ because the smaller size of $F$ and $O$ atoms leads to greater inter-electronic repulsion for the incoming electron.
Therefore,the correct order of increasing electron gain enthalpy is $O < S < F < Cl$.

(B) Atomic radius decreases across a period from left to right due to an increase in effective nuclear charge,while it increases down a group due to the addition of new shells.
The electronic configurations are:
$Mg (Z=12): [Ne] 3s^2$
$P (Z=15): [Ne] 3s^2 3p^3$
$Cl (Z=17): [Ne] 3s^2 3p^5$
$Ca (Z=20): [Ar] 4s^2$
$Mg, P,$ and $Cl$ belong to the $3^{rd}$ period. Within the same period,the atomic radius decreases as $Z$ increases: $Mg > P > Cl$.
$Ca$ belongs to the $4^{th}$ period,so it has a larger atomic radius than the $3^{rd}$ period elements.
Therefore,the increasing order of atomic radii is $Cl < P < Mg < Ca$.

(B) The ratio of cation to anion size is given by $\frac{r_{+}}{r_{-}}$.
To maximize this ratio,we need the largest cation $(r_{+})$ and the smallest anion $(r_{-})$.
Comparing the cations: $Li^{+} < Na^{+} < Cs^{+}$. Thus,$Cs^{+}$ is the largest cation.
Comparing the anions: $F^{-} < I^{-}$. Thus,$F^{-}$ is the smallest anion.
Therefore,the ratio is highest for $CsF$.

(B) For $sp^{2}$ hybridisation,the steric number (sum of $\sigma$-bonds and lone pairs) must be $3$.
$(I)$ $NO_{2}^{-}$: Central $N$ atom has $2$ $\sigma$-bonds and $1$ lone pair,so steric number $= 3$,which corresponds to $sp^{2}$ hybridisation.
$(II)$ $NH_{3}$: Central $N$ atom has $3$ $\sigma$-bonds and $1$ lone pair,so steric number $= 4$,which corresponds to $sp^{3}$ hybridisation.
$(III)$ $BF_{3}$: Central $B$ atom has $3$ $\sigma$-bonds and $0$ lone pairs,so steric number $= 3$,which corresponds to $sp^{2}$ hybridisation.
$(IV)$ $NH_{2}^{-}$: Central $N$ atom has $2$ $\sigma$-bonds and $2$ lone pairs,so steric number $= 4$,which corresponds to $sp^{3}$ hybridisation.
$(V)$ $H_{2}O$: Central $O$ atom has $2$ $\sigma$-bonds and $2$ lone pairs,so steric number $= 4$,which corresponds to $sp^{3}$ hybridisation.
Thus,the pair $BF_{3}$ and $NO_{2}^{-}$ both exhibit $sp^{2}$ hybridisation.

(B) The species $Be_2$ does not exist under normal conditions.
The electronic configuration of $Be_2$ ($Z=4$,total $8$ electrons) is: $\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2$.
The bond order is calculated as: $\text{Bond order} = \frac{N_b - N_a}{2} = \frac{4 - 4}{2} = 0$.
Since the bond order is $0$,the molecule is unstable and does not exist.

(C) The hybridization of the central atom can be determined using the formula $H = \frac{1}{2} [V + X - C + A]$,where $V$ is the number of valence electrons,$X$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
$1. SF_4: H = \frac{1}{2} [6 + 4] = 5 \rightarrow sp^3d$
$2. I_3^-: H = \frac{1}{2} [7 + 2 + 1] = 5 \rightarrow sp^3d$
$3. PCl_5: H = \frac{1}{2} [5 + 5] = 5 \rightarrow sp^3d$
$4. SbCl_5^{2-}: H = \frac{1}{2} [5 + 5 + 2] = 6 \rightarrow sp^3d^2$
Thus,$SbCl_5^{2-}$ has a different hybridization $(sp^3d^2)$ compared to the others $(sp^3d)$.

(B) The number of hybrid orbitals $(H)$ is calculated using the formula: $H = \frac{1}{2} [V + M - C + A]$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the total positive charge,and $A$ is the negative charge.
For $CH_4$: $H = \frac{1}{2} [4 + 4 - 0 + 0] = 4$,which corresponds to $sp^3$ hybridization.
For $SF_4$: $H = \frac{1}{2} [6 + 4 - 0 + 0] = 5$,which corresponds to $sp^3d$ hybridization.
For $BF_4^-$: $H = \frac{1}{2} [3 + 4 - 0 + 1] = 4$,which corresponds to $sp^3$ hybridization.
For $NH_4^+$: $H = \frac{1}{2} [5 + 4 - 1 + 0] = 4$,which corresponds to $sp^3$ hybridization.
Thus,the central atom in $SF_4$ does not have $sp^3$ hybridization.

(A) For $NO_{3}^{-}$,the number of hybrid orbitals $H = \frac{1}{2} [5 + 0 - 0 + 1] = 3$. Thus,the central atom $N$ is $sp^{2}$ hybridized and the species has a trigonal planar geometry.
For $H_{3}O^{+}$,the number of hybrid orbitals $H = \frac{1}{2} [6 + 3 - 1 + 0] = 4$. Thus,the central atom $O$ is $sp^{3}$ hybridized and the species has a pyramidal geometry due to the presence of one lone pair.
Since they have different hybridization states ($sp^{2}$ vs $sp^{3}$) and different geometries (trigonal planar vs pyramidal),they are dissimilar in both hybridization and structure.

(C) Volume,$V = 0.03 \ m^3$
Temperature,$T = 129 + 273 = 402 \ K$
Mass of methane,$w = 6.0 \ g$
Molar mass of methane,$M = 12.01 + 4 \times 1.01 = 16.05 \ g \ mol^{-1}$
Using the ideal gas equation,$pV = nRT$,where $n = \frac{w}{M}$
$p = \frac{wRT}{MV} = \frac{6.0 \times 8.314 \times 402}{16.05 \times 0.03}$
$p = \frac{20065.392}{0.4815} \approx 41672.67 \ Pa$
Rounding to the nearest provided option,the value is $41648 \ Pa$.

(A) The reaction is $\frac{1}{2} X_2 + \frac{3}{2} Y_2 \rightleftharpoons XY_3$.
First,calculate the entropy change of the reaction $\Delta S^{\circ}_{rxn}$:
$\Delta S^{\circ}_{rxn} = S^{\circ}(XY_3) - [\frac{1}{2} S^{\circ}(X_2) + \frac{3}{2} S^{\circ}(Y_2)]$
$\Delta S^{\circ}_{rxn} = 50 - [(\frac{1}{2} \times 60) + (\frac{3}{2} \times 40)] \ J \ K^{-1} \ mol^{-1}$
$\Delta S^{\circ}_{rxn} = 50 - [30 + 60] = 50 - 90 = -40 \ J \ K^{-1} \ mol^{-1}$.
At equilibrium,$\Delta G = 0$,which implies $\Delta H = T \Delta S$.
Given $\Delta H = -30 \ kJ = -30000 \ J$.
$-30000 = T \times (-40)$.
$T = \frac{-30000}{-40} = 750 \ K$.

(C) $A. K_p > Q$: Since the reaction quotient $Q$ is less than the equilibrium constant $K_p$,the reaction proceeds in the forward direction,making it spontaneous. $(A-(iv))$
$B. \Delta G^\circ < RT \ln Q$: Using the relation $\Delta G = \Delta G^\circ + RT \ln Q$,if $\Delta G^\circ < -RT \ln Q$ (or $\Delta G^\circ < RT \ln Q$ in specific contexts),it implies $\Delta G > 0$,indicating a non-spontaneous process. $(B-(i))$
$C. K_p = Q$: The system is at equilibrium. $(C-(ii))$
$D. T > \frac{\Delta H}{\Delta S}$: For an endothermic process $(\Delta H > 0)$,if $T > \frac{\Delta H}{\Delta S}$,then $T \Delta S > \Delta H$,resulting in $\Delta G = \Delta H - T \Delta S < 0$,which means the process is spontaneous and endothermic. $(D-(iii))$

(D) The work done during the expansion of a gas is given by the formula $W = -P_{ext} \Delta V$.
Since the gas expands into a vacuum,the external pressure $P_{ext}$ is $0$.
Therefore,the work done $W = -0 \times \Delta V = 0 \ J$.

(C) According to the Gibbs equation:
$\Delta G = \Delta H - T \Delta S$
When $\Delta G = 0$,the system is at equilibrium,so:
$\Delta H = T \Delta S$
Given:
$\Delta H = 40.63 \ kJ \ mol^{-1} = 40.63 \times 10^{3} \ J \ mol^{-1}$
$\Delta S = 108.8 \ J \ K^{-1} \ mol^{-1}$
Substituting the values:
$T = \frac{\Delta H}{\Delta S}$
$T = \frac{40.63 \times 10^{3} \ J \ mol^{-1}}{108.8 \ J \ K^{-1} \ mol^{-1}}$
$T \approx 373.43 \ K$
Thus,the temperature is approximately $373.4 \ K$.

(D) Given reactions:
$(I) \ Fe_2O_{3(s)} + 3CO_{(g)} \rightarrow 2Fe_{(s)} + 3CO_{2(g)}; \Delta H_1 = -26.8 \ kJ$
$(II) \ FeO_{(s)} + CO_{(g)} \rightarrow Fe_{(s)} + CO_{2(g)}; \Delta H_2 = -16.5 \ kJ$
We need to find $\Delta H$ for:
$(III) \ Fe_2O_{3(s)} + CO_{(g)} \rightarrow 2FeO_{(s)} + CO_{2(g)}$
To obtain reaction $(III)$,we perform the operation: $(I) - 2 \times (II)$
$\Delta H = \Delta H_1 - 2(\Delta H_2)$
$\Delta H = -26.8 - 2(-16.5)$
$\Delta H = -26.8 + 33.0 = +6.2 \ kJ$

(C) Given,$pH$ of $Ba(OH)_2 = 12$.
Since $pH + pOH = 14$,we have $pOH = 14 - 12 = 2$.
Therefore,$[OH^-] = 10^{-pOH} = 10^{-2} \, M$.
The dissociation of $Ba(OH)_2$ is: $Ba(OH)_2 \rightleftharpoons Ba^{2+} + 2OH^-$.
From the stoichiometry,$[OH^-] = 2S$,where $S$ is the solubility of $Ba(OH)_2$.
So,$S = \frac{[OH^-]}{2} = \frac{10^{-2}}{2} = 0.5 \times 10^{-2} \, M$.
The solubility product constant $K_{sp}$ is given by: $K_{sp} = [Ba^{2+}][OH^-]^2 = (S)(2S)^2 = 4S^3$.
Substituting the value of $S$: $K_{sp} = 4 \times (0.5 \times 10^{-2})^3 = 4 \times (0.125 \times 10^{-6}) = 0.5 \times 10^{-6} = 5.0 \times 10^{-7} \, M^3$.

(D) The solution contains a weak acid $(CH_3COOH)$ and its conjugate salt $(CH_3COONa)$,which forms an acidic buffer.
For an acidic buffer,the concentration of $[H^{+}]$ is given by the Henderson-Hasselbalch equation:
$[H^{+}] = K_a \times \frac{[acid]}{[salt]}$
Given:
$K_a = 1.8 \times 10^{-5}$
$[acid] = [CH_3COOH] = 0.10\, M$
$[salt] = [CH_3COONa] = 0.20\, M$
Substituting the values:
$[H^{+}] = 1.8 \times 10^{-5} \times \frac{0.10}{0.20}$
$[H^{+}] = 1.8 \times 10^{-5} \times 0.5$
$[H^{+}] = 0.9 \times 10^{-5} = 9.0 \times 10^{-6}\, mol/L$

(D) Key Idea: The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n_g}$,where $\Delta n_g$ is the change in the number of moles of gaseous species.
$K_p$ and $K_c$ are not equal when $\Delta n_g \neq 0$.
$(a)$ $\Delta n_g = (1+1) - 2 = 0$,so $K_p = K_c$.
$(b)$ $\Delta n_g = (1+1) - (1+1) = 0$,so $K_p = K_c$.
$(c)$ $\Delta n_g = 2 - (1+1) = 0$,so $K_p = K_c$.
$(d)$ $\Delta n_g = 2 - 1 = 1$,so $K_p = K_c(RT)^1$,which means $K_p \neq K_c$.

(D) For a basic buffer solution,the $pOH$ is given by the Henderson-Hasselbalch equation:
$pOH = pK_b + \log \frac{[salt]}{[base]}$
Given that $[salt] = [B^{-}]$ and $[base] = [HB]$,and their concentrations are equal,we have $\frac{[B^{-}]}{[HB]} = 1$.
$pK_b = -\log(K_b) = -\log(10^{-10}) = 10$.
Substituting these values into the equation:
$pOH = 10 + \log(1) = 10 + 0 = 10$.
Since $pH + pOH = 14$ at $25^{\circ}C$:
$pH = 14 - pOH = 14 - 10 = 4$.

(B) The reaction is $2A_{(g)} + B_{(g)} \rightleftharpoons 3C_{(g)} + D_{(g)}$.
Initial concentrations: $[A] = 1.00 \ M$,$[B] = 1.00 \ M$,$[C] = 0 \ M$,$[D] = 0 \ M$.
At equilibrium,$[D] = 0.25 \ M$. Since the stoichiometry of $D$ is $1$,the change in concentration for $D$ is $+0.25 \ M$.
Using stoichiometry:
$[A]_{eq} = 1.00 - 2(0.25) = 0.50 \ M$
$[B]_{eq} = 1.00 - 0.25 = 0.75 \ M$
$[C]_{eq} = 3(0.25) = 0.75 \ M$
$[D]_{eq} = 0.25 \ M$
The equilibrium constant expression is $K_c = \frac{[C]^3[D]}{[A]^2[B]}$.
Substituting the values: $K_c = \frac{(0.75)^3(0.25)}{(0.50)^2(0.75)}$.

(A) Heavy water $(D_2O)$ is used as a moderator in nuclear reactors to slow down fast-moving neutrons.
Heavy water has a higher boiling point $(374.42 \ K)$ compared to ordinary water $(373 \ K)$,which indicates that it is more associated due to stronger intermolecular forces.
The dielectric constant of ordinary water is higher than that of heavy water,making ordinary water a better solvent than heavy water.
Therefore,statements $(i)$ and $(ii)$ are correct.

(B) The solubility of alkaline earth metal sulphates depends on the balance between hydration enthalpy and lattice enthalpy.
Hydration enthalpy decreases as the size of the cation increases,while lattice enthalpy remains relatively constant for these sulphates due to the large size of the sulphate anion.
The ionic size order is $Be^{2+} < Mg^{2+} < Ca^{2+} < Sr^{2+} < Ba^{2+}$.
Since $Be^{2+}$ is the smallest ion,it has the highest hydration enthalpy.
$BeSO_4$ is the only sulphate among the options where the hydration enthalpy is sufficiently high to overcome the lattice enthalpy,making it highly soluble in water.

(A) As we move down the group in alkaline earth metals,the lattice energy decreases more rapidly than the hydration energy for hydroxides,leading to an increase in the solubility of their hydroxides in water.
Conversely,the solubility of sulphates decreases down the group.
Ionization energy and electronegativity also decrease down the group due to the increase in atomic size.

(B) In a peroxide,the oxidation number of $O$ is $-1$,and they contain a peroxide linkage $(-O-O-)$. They react with dilute acids to produce $H_2O_2$.
For $KO_2$: Let the oxidation state of $O$ be $x$. Then $1 + 2x = 0$,so $x = -\frac{1}{2}$. This is a superoxide.
For $BaO_2$: Let the oxidation state of $O$ be $x$. Then $2 + 2x = 0$,so $x = -1$. This is a peroxide.
For $MnO_2$ and $NO_2$: These are dioxides where the oxidation state of $O$ is $-2$. Therefore,they are not peroxides.
Thus,$BaO_2$ is the correct answer.

(A) The reactions are as follows:
$A \xrightarrow{\Delta} \text{colourless gas} (CO_2) + \text{residue} (CaO)$
$\text{Residue} (CaO) + H_2O \rightarrow Ca(OH)_2 (B)$
$Ca(OH)_2 + \text{excess } CO_2 \rightarrow Ca(HCO_3)_2 (C)$
$Ca(HCO_3)_2 \xrightarrow{\Delta} CaCO_3 (A) + H_2O + CO_2$
Thus,the compound $A$ is $CaCO_3$.

(C) Electron-deficient molecules act as Lewis acids.
Among the given molecules,only diborane $(B_2H_6)$ is electron-deficient,meaning it does not have a complete octet.
Thus,it acts as a Lewis acid.
$NH_3$ and $H_2O$ are electron-rich molecules and behave as Lewis bases.

(B) The Lewis acid strength of boron trihalides depends on the extent of $p\pi - p\pi$ back bonding from the halogen atom to the boron atom.
In $BF_3$,the $2p - 2p$ back bonding is most effective due to the small size of the $F$ atom,which significantly reduces the electron deficiency of boron.
As the size of the halogen increases from $F$ to $Cl$ to $Br$,the effectiveness of $p\pi - p\pi$ back bonding decreases $(F > Cl > Br)$.
Consequently,the electron deficiency on the boron atom increases in the order $BF_3 < BCl_3 < BBr_3$.
Therefore,the Lewis acid strength follows the sequence $BBr_3 > BCl_3 > BF_3$.

(C) Electrophiles are electron-deficient species that act as Lewis acids by accepting an electron pair.
$Cl^{\oplus}$,$BH_3$,and $NO_2^{\oplus}$ are electron-deficient and act as electrophiles.
In $H_3O^{\oplus}$,the oxygen atom has a complete octet and a lone pair of electrons,which it can donate.
Therefore,$H_3O^{\oplus}$ is not electron-deficient and does not behave as an electrophile.

(B) The given compound is $CH_3-CH=CH-C\equiv CH$.
According to $IUPAC$ rules,the numbering of the carbon chain starts from the end that gives the lowest possible locants to the multiple bonds.
If there is a choice between a double bond and a triple bond,the double bond is given priority for the lower number.
Here,the double bond starts at $C-3$ and the triple bond starts at $C-1$.
Thus,the correct name is pent$-3-$en$-1-$yne.

(C) The structure is $CH \equiv C-CH=CH_2$.
According to $IUPAC$ rules for naming hydrocarbons containing both double and triple bonds,if both are at the same position from either end,the double bond is given priority in numbering.
Numbering from right to left: $CH_2(1)=CH(2)-C(3) \equiv CH(4)$.
The double bond is at position $1$ and the triple bond is at position $3$.
Therefore,the name is but$-1-$ene$-3-$yne.

(B) The stability of a conformation depends on the minimization of steric repulsion between the groups attached to adjacent carbon atoms.
In the $anti$ conformation of $n-$butane,the two bulky $-CH_{3}$ groups are at a dihedral angle of $180^{\circ}$,which is the maximum possible distance between them. This minimizes the steric repulsion and torsional strain,making it the most stable conformation.
Therefore,the $anti$ conformation is the most stable.

(B) Lower hydrocarbons exist in the gaseous state,while higher ones exist in the liquid or solid state.
On cracking or pyrolysis,a hydrocarbon with a higher molecular mass breaks down into a mixture of hydrocarbons having lower molecular masses.
Therefore,by the process of cracking,a liquid hydrocarbon can be converted into a mixture of gaseous hydrocarbons.

(C) The gauche conformer of ethylene glycol is the most stable conformer. This is due to the formation of intramolecular hydrogen bonding between the two hydroxyl $(-OH)$ groups,which stabilizes this specific conformation despite the steric repulsion between the groups.

(D) The oxidation state of $P$ $(x)$ is calculated as follows:
For $H_{4}P_{2}O_{5}: 4(+1) + 2x + 5(-2) = 0$ $\Rightarrow 2x - 6 = 0$ $\Rightarrow x = +3$
For $H_{4}P_{2}O_{6}: 4(+1) + 2x + 6(-2) = 0$ $\Rightarrow 2x - 8 = 0$ $\Rightarrow x = +4$
For $H_{4}P_{2}O_{7}: 4(+1) + 2x + 7(-2) = 0$ $\Rightarrow 2x - 10 = 0$ $\Rightarrow x = +5$
Thus,the oxidation states are $+3, +4, +5$.

(D) To determine the bond angle,we analyze the hybridization and the number of lone pairs on the central atom:
$1$. $ClO_2^-$: The central $Cl$ atom has $2$ lone pairs and $2$ bond pairs. The bond angle is approximately $111^\circ$.
$2$. $Cl_2O$: The central $O$ atom has $2$ lone pairs and $2$ bond pairs. Due to the lower electronegativity of $Cl$ compared to $O$,the bond pairs are further from the central atom,resulting in a smaller bond angle of approximately $110.9^\circ$.
$3$. $ClO_2$: The central $Cl$ atom has $1$ lone pair and $2$ bond pairs. With fewer lone pairs,there is less repulsion,leading to a larger bond angle of approximately $117.5^\circ$.
Thus,the correct order of increasing bond angles is $ClO_2^- < Cl_2O < ClO_2$.

(A) In the structure of $P_4O_{10}$,there are four phosphorus atoms at the corners of a tetrahedron. Each phosphorus atom is bonded to one terminal oxygen atom via a double bond $(P=O)$.
Additionally,there are six oxygen atoms that act as bridges between the phosphorus atoms,forming $P-O-P$ linkages.
Therefore,the total number of bridging oxygen atoms in $P_4O_{10}$ is $6$.

(B) Ions that possess unpaired electrons exhibit colour in aqueous solutions due to $d-d$ or $f-f$ transitions.
$Ti^{3+} (Z = 22)$ has an electronic configuration of $[Ar] 3d^{1}$,which contains $1$ unpaired electron. Therefore,it exhibits colour.
$Sc^{3+} (Z = 21)$ has a configuration of $[Ar] 3d^{0}$,meaning it has no unpaired electrons.
$La^{3+} (Z = 57)$ has a configuration of $[Xe] 4f^{0}$,and $Lu^{3+} (Z = 71)$ has a configuration of $[Xe] 4f^{14}$. Both lack unpaired electrons in the $f$-subshell,so no $f-f$ transitions are possible.
Thus,only $Ti^{3+}$ is coloured.

(D) Fructose is a ketose. In the presence of a base,it undergoes enolisation to form an enediol intermediate,which is then converted into an aldose (glucose and mannose) via tautomerization. These aldoses contain the $-CHO$ group,which is responsible for reducing Tollen's reagent to give a silver mirror test.

(A) In human pedigree analysis,standard symbols are used to represent different individuals and their relationships:
$1$. $A$ square represents a male.
$2$. $A$ circle represents a female.
$3$. $A$ horizontal line connecting a square and a circle represents mating.
$4$. $A$ double horizontal line connecting a square and a circle represents mating between relatives (consanguineous mating).
$5$. Shading or filling a symbol indicates that the individual is affected by the trait or disease being studied.
$6$. Therefore,the symbol showing a square and a circle connected by a double horizontal line correctly represents 'mating between relatives'.

(C) The law of dominance states that in a heterozygous organism,only one allele (the dominant one) expresses its phenotype,while the other (the recessive one) remains masked.
Option $A$ describes the concept of factors (genes).
Option $B$ describes the core principle of dominance.
Option $D$ describes the pairing of factors.
Option $C$ describes the law of segregation (or purity of gametes),which states that alleles do not blend and separate during gamete formation,allowing both traits to reappear in the $F_2$ generation. This phenomenon is explained by the law of segregation,not the law of dominance.

(A) Nitrogen fixation is the process of converting atmospheric nitrogen $(N_2)$ into ammonia $(NH_3)$.
This process is catalyzed by the enzyme nitrogenase.
Nitrogenase is a complex metalloprotein that requires both $Mo$ (Molybdenum) and $Fe$ (Iron) as essential cofactors for its catalytic activity.
Therefore,Molybdenum is an essential element for nitrogen fixation in plants,particularly in legumes.

(D) Goblet cells are specialized epithelial cells found in the mucosal lining of the digestive and respiratory tracts.
Their primary function is to secrete mucus,which is a glycoprotein-rich substance.
In the digestive tract,this mucus acts as a lubricant,protecting the intestinal wall from mechanical abrasion and facilitating the smooth movement of food (chyme) through the intestine.
If goblet cells become non-functional,the lack of mucus will lead to increased friction,causing difficulty in the passage of food and potentially damaging the intestinal lining.

(D) The tricuspid valve is located between the right atrium and the right ventricle of the human heart.
Chordae tendineae are fibrous cords that attach the valve leaflets to the papillary muscles,preventing the valves from prolapsing into the atria during ventricular systole.
If the chordae tendineae of the tricuspid valve are damaged or non-functional,the valve will not close properly during ventricular contraction.
This leads to the regurgitation of blood from the right ventricle back into the right atrium.
Since the right ventricle pumps blood into the pulmonary artery,a failure of the tricuspid valve results in a reduced volume of blood being pumped into the pulmonary artery.
Therefore,the correct answer is $D$.

(A) The pancreas contains the Islets of Langerhans, which consist of different types of cells:
$1$. Alpha ($\alpha$) cells secrete Glucagon.
$2$. Beta ($\beta$) cells secrete Insulin.
$3$. Delta ($\delta$) cells secrete Somatostatin.
In option $A$, Glucagon is stated to be secreted by Beta cells, which is incorrect as Beta cells secrete Insulin. Therefore, the pair $A$ is incorrectly matched.

(B) Membrane-bound organelles (such as mitochondria,chloroplasts,endoplasmic reticulum,and nucleus) are characteristic features of eukaryotic cells.
$Saccharomyces$ (yeast) is a fungus,$Chlamydomonas$ is an alga,and $Plasmodium$ is a protozoan; all of these are eukaryotes.
$Streptococcus$ is a bacterium,which belongs to the kingdom $Monera$.
Bacteria are prokaryotic organisms that lack a well-defined nucleus and membrane-bound organelles.
Therefore,the correct answer is $Streptococcus$.

(B) Step $1$: Calculate the molarity of $Na_2CO_3$ solution.
Molarity $(M) = \frac{\text{mass of solute}}{\text{molar mass} \times \text{volume of solution in } L}$
$M = \frac{25.3 \ g}{106 \ g \ mol^{-1} \times 0.250 \ L} = 0.955 \ M$
Step $2$: Write the dissociation equation.
$Na_2CO_3(aq) \rightarrow 2Na^{+}(aq) + CO_3^{2-}(aq)$
Step $3$: Determine the concentration of ions.
Since $1 \ mol$ of $Na_2CO_3$ produces $2 \ mol$ of $Na^{+}$ and $1 \ mol$ of $CO_3^{2-}$,
$[Na^{+}] = 2 \times 0.955 \ M = 1.910 \ M$
$[CO_3^{2-}] = 1 \times 0.955 \ M = 0.955 \ M$
Thus,the concentrations are $1.910 \ M$ and $0.955 \ M$ respectively.

(C) In an endothermic reaction,the energy of the reactants is less than that of the products.
From the potential energy diagram for an endothermic reaction,the relationship between the activation energy of the forward reaction $(E_a)$,the activation energy of the backward reaction $(E_a')$,and the enthalpy of reaction $(\Delta H)$ is given by:
$E_a = E_a' + \Delta H$
Since the activation energy of the backward reaction $(E_a')$ must always be a positive value $(E_a' > 0)$,it follows that:
$E_a > \Delta H$
Therefore,the minimum value of $E_a$ is greater than $\Delta H$.

(C) $1$. When toluene reacts with $Cl_2$ in the presence of a Lewis acid catalyst like $FeCl_3$,electrophilic aromatic substitution occurs at the ring (nuclear halogenation),resulting in $o-$ and $p-$chlorotoluene as $X$.
$2$. When toluene reacts with $Cl_2$ in the presence of light $(hv)$,free radical substitution occurs at the side chain (side chain halogenation),resulting in benzyl chloride,benzal chloride,and finally trichloromethyl benzene (benzotrichloride) as $Y$.

(C) $1$. Oxidation: Ethylbenzene reacts with alkaline $KMnO_4$ to form benzoic acid $(B)$. The alkyl group is oxidized to a $-COOH$ group.
$2$. Bromination: Benzoic acid is a meta-directing group. Reaction with $Br_2/FeCl_3$ yields $3-$bromobenzoic acid $(C)$.
$3$. Esterification: $3-$bromobenzoic acid reacts with ethanol $(C_2H_5OH)$ in the presence of an acid catalyst $(H^+)$ to form ethyl $3-$bromobenzoate $(D)$.

(A) For a body-centred cubic $(bcc)$ lattice, the distance $(d)$ between two oppositely charged ions (located at the corner and the body center) is given by the formula:
$d = \frac{\sqrt{3}}{2} a$
Given $a = 387 \ pm$,
$d = \frac{1.732 \times 387}{2} \ pm$
$d = 0.866 \times 387 \ pm$
$d \approx 335.14 \ pm$
Therefore, the closest value is $335 \ pm$.

(D) The vapour pressure of a solution is inversely proportional to the concentration of the solute particles.
Adding water to an aqueous solution of $KI$ dilutes the solution,which decreases the concentration of solute particles.
According to Raoult's law,a decrease in solute concentration leads to an increase in the vapour pressure of the solvent.
In contrast,adding $NaCl$,$Na_2SO_4$,or more $KI$ increases the total number of solute particles,which further lowers the vapour pressure due to the colligative property of vapour pressure lowering.

(A) Given: $w_2 = 68.5 \, g$,$M_2 = 342 \, g \, mol^{-1}$,$w_1 = 1000 \, g = 1 \, kg$,$K_f = 1.86 \, K \, kg \, mol^{-1}$.
$\Delta T_f = K_f \times m = K_f \times \frac{w_2}{M_2 \times w_1(kg)}$
$\Delta T_f = 1.86 \times \frac{68.5}{342 \times 1} = 0.3725 \, K$ or $^oC$.
Freezing point of solution $T_f = T_f^o - \Delta T_f = 0 \, ^oC - 0.3725 \, ^oC = -0.3725 \, ^oC \approx -0.37 \, ^oC$.

(A) The standard Gibbs energy change is given by the formula: $\Delta G^{\circ} = -nFE^{\circ}_{cell}$
The cell reaction is: $2 Ag^{+} + Cu \rightarrow Cu^{2+} + 2 Ag$
Here,the number of electrons transferred,$n = 2$.
Given,$E^{\circ}_{cell} = + 0.46 \ V$ and $F = 96500 \ C \ mol^{-1}$.
Substituting the values: $\Delta G^{\circ} = -2 \times 96500 \times 0.46 \ J \ mol^{-1}$
$\Delta G^{\circ} = -88780 \ J \ mol^{-1}$
Converting to $kJ \ mol^{-1}$: $\Delta G^{\circ} = -88.78 \ kJ \ mol^{-1}$
Rounding to the nearest value,we get $\Delta G^{\circ} \approx -89.0 \ kJ \ mol^{-1}$.

(A) For a strong electrolyte,the number of ions is already fixed because it is completely dissociated at all concentrations.
Equivalent conductance $(\lambda_{eq})$ is defined as $\lambda_{eq} = \kappa \times V$,where $\kappa$ is the specific conductance and $V$ is the volume containing $1 \ g$-equivalent of the electrolyte.
Upon dilution,the inter-ionic attractions decrease,which leads to an increase in the ionic mobility of the ions.
Since the number of ions remains constant for a strong electrolyte,the increase in equivalent conductance is primarily due to the increase in ionic mobility.

(A) The equivalent conductance at infinite dilution $(\Lambda_{eq}^{\infty})$ of an electrolyte is the sum of the equivalent conductances of its constituent ions.
By definition,the equivalent conductance of any electrolyte at infinite dilution is the sum of the equivalent conductances of its cation and anion.
For $Al_2(SO_4)_3$,the equivalent conductance at infinite dilution is given by:
$\Lambda_{eq}^{\infty} = \Lambda_{Al^{3+}}^o + \Lambda_{SO_4^{2-}}^o$
This is because the equivalent conductance of an ion is defined as the molar conductance divided by its charge (valence factor),which inherently accounts for the stoichiometry of the salt.

(D) The $EMF$ of a cell is defined as:
$EMF_{cell} = E_{cathode} (reduction) - E_{anode} (reduction)$
Since $E_{anode} (reduction) = -E_{anode} (oxidation)$,we can write:
$EMF_{cell} = E_{cathode} (reduction) + E_{anode} (oxidation)$
Also,since $E_{cathode} (reduction) = -E_{cathode} (oxidation)$,we can write:
$EMF_{cell} = E_{anode} (oxidation) - E_{cathode} (oxidation)$
Comparing these with the given relations:
$(ii)$ $EMF$ of cell = (Oxidation potential of anode) $+$ (Reduction potential of cathode) is correct.
$(iv)$ $EMF$ of cell = (Oxidation potential of anode) $-$ (Oxidation potential of cathode) is correct.
Therefore,the correct option is $D$.

(B) The rate of reaction is given by the expression: $-\frac{d[N_{2}O_{5}]}{dt} = \frac{1}{2} \frac{d[NO_{2}]}{dt} = 2 \frac{d[O_{2}]}{dt}$.
Given,$-\frac{d[N_{2}O_{5}]}{dt} = 6.25 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.
For $NO_{2}$ formation: $\frac{d[NO_{2}]}{dt} = 2 \times (-\frac{d[N_{2}O_{5}]}{dt}) = 2 \times 6.25 \times 10^{-3} = 1.25 \times 10^{-2} \ mol \ L^{-1} \ s^{-1}$.
For $O_{2}$ formation: $\frac{d[O_{2}]}{dt} = \frac{1}{2} \times (-\frac{d[N_{2}O_{5}]}{dt}) = \frac{6.25 \times 10^{-3}}{2} = 3.125 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.

(D) Let the order of reaction with respect to $A$ be $x$ and with respect to $B$ be $y$.
The rate law is given by: $\text{Rate} = k[A]^x[B]^y$
Using the data from the table:
$I. \ 6.0 \times 10^{-3} = k(0.1)^x(0.1)^y$
$II. \ 7.2 \times 10^{-2} = k(0.3)^x(0.2)^y$
$III. \ 2.88 \times 10^{-1} = k(0.3)^x(0.4)^y$
$IV. \ 2.40 \times 10^{-2} = k(0.4)^x(0.1)^y$
Dividing Eq. $(I)$ by Eq. $(IV)$:
$\frac{6.0 \times 10^{-3}}{2.40 \times 10^{-2}} = \left(\frac{0.1}{0.4}\right)^x \left(\frac{0.1}{0.1}\right)^y$
$0.25 = (0.25)^x \implies x = 1$
Dividing Eq. $(II)$ by Eq. $(III)$:
$\frac{7.2 \times 10^{-2}}{2.88 \times 10^{-1}} = \left(\frac{0.3}{0.3}\right)^x \left(\frac{0.2}{0.4}\right)^y$
$0.25 = (0.5)^y \implies (0.5)^2 = (0.5)^y \implies y = 2$
Therefore,the rate law is: $\text{Rate} = k[A]^1[B]^2 = k[A][B]^2$.

(A) The rate constant $k$ is a characteristic property of a reaction at a given temperature.
It is independent of the concentrations of the reactants.
According to the Arrhenius equation,$k = Ae^{-E_a/RT}$,the rate constant $k$ increases with an increase in temperature.

(D) The electronic configuration of neutral atoms is:
$Mn (25) = [Ar] 3d^5 4s^2$
$Fe (26) = [Ar] 3d^6 4s^2$
$Co (27) = [Ar] 3d^7 4s^2$
$Ni (28) = [Ar] 3d^8 4s^2$
For the ions:
$Ni^{3+} = [Ar] 3d^7$
$Mn^{3+} = [Ar] 3d^4$
$Fe^{3+} = [Ar] 3d^5$
$Co^{3+} = [Ar] 3d^6$
Thus,$Co^{3+}$ has the $[Ar] 3d^6$ configuration.

(C) In general,the atomic and ionic radii increase on moving down a group.
However,the elements of the second transition series $(Zr, Nb, Mo, \dots)$ have almost the same radii as the elements of the third transition series $(Hf, Ta, W, \dots)$.
This phenomenon is due to lanthanoid contraction,which arises from the imperfect shielding of one $4f$ electron by another,leading to a smaller increase in size than expected when moving from the $4d$ to the $5d$ series.
Therefore,$Zr^{4+}$ and $Hf^{4+}$ have nearly identical ionic radii.

(D) The correct matches are as follows:
$A$. Sulphuric acid is manufactured by the $iv$. Contact process.
$B$. Steel is manufactured by the $ii$. Bessemer's process.
$C$. Sodium hydroxide was historically manufactured by the $iii$. Leblanc process.
$D$. Ammonia is manufactured by the $i$. Haber's process.
Therefore,the correct matching is $A-iv, B-ii, C-iii, D-i$.

(D) The most common oxidation state exhibited by lanthanoids is $+3$.

(A) transition metal complex absorbs visible light only if it has unpaired electrons,which allows for $d-d$ transitions.
$1$. In $[Ni(CN)_4]^{2-}$,$Ni$ is in the $+2$ oxidation state $(3d^8)$. Since $CN^-$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbitals,resulting in no unpaired electrons ($d^8$ configuration: $t_{2g}^6 e_g^2$ is not applicable here as it is square planar,but all electrons are paired).
$2$. $[Cr(NH_3)_6]^{3+}$ has $Cr^{3+}$ $(3d^3)$,which has $3$ unpaired electrons.
$3$. $[Fe(H_2O)_6]^{2+}$ has $Fe^{2+}$ $(3d^6)$,which has $4$ unpaired electrons.
$4$. $[Ni(H_2O)_6]^{2+}$ has $Ni^{2+}$ $(3d^8)$,which has $2$ unpaired electrons.
Since $[Ni(CN)_4]^{2-}$ has no unpaired electrons,it is not expected to absorb visible light.

(D) In an octahedral complex,the $d$-orbitals split into $t_{2g}$ (lower energy) and $e_g$ (higher energy) sets.
For a high spin $d^4$ configuration,the electrons occupy the orbitals according to Hund's rule because the crystal field splitting energy $\Delta_o$ is less than the pairing energy $P$.
The electronic configuration is $t_{2g}^3 e_g^1$.
The Crystal Field Stabilization Energy $(CFSE)$ is calculated as:
$CFSE = (n_{t_{2g}} \times -0.4 \Delta_o) + (n_{e_g} \times 0.6 \Delta_o)$
$CFSE = (3 \times -0.4 \Delta_o) + (1 \times 0.6 \Delta_o)$
$CFSE = -1.2 \Delta_o + 0.6 \Delta_o = -0.6 \Delta_o$.

(D) Complexes of $[MA_4B_2]$ type exhibit geometrical isomerism.
The complex $[Co(NH_3)_4Cl_2]^+$ is a $[MA_4B_2]$ type complex and thus,fulfills the conditions that are necessary to exhibit geometrical isomerism.
Hence,it has two geometrical isomers (cis and trans) of different colours.
For linkage isomerism,the presence of an ambidentate ligand is necessary.
For coordination isomerism,both the cation and anion of the complex must be complex ions.
For ionisation isomerism,an anion different from the ligands must be present outside the coordination sphere.
All these conditions are not satisfied by this complex. Hence,it does not exhibit other given isomerisms.

(C) The complex $[Ni(NH_3)_2Cl_2]$ adopts a tetrahedral geometry because $Ni^{2+}$ is a $d^8$ ion and $NH_3$ and $Cl^-$ are weak field ligands in this context,leading to $sp^3$ hybridization.
Tetrahedral complexes of the type $[M(A)_2(B)_2]$ do not exhibit geometric isomerism because all positions in a tetrahedron are equivalent relative to each other.
In contrast,$[Pt(NH_3)_2Cl_2]$ is a square planar complex which exhibits cis-trans isomerism.
$[Ni(NH_3)_4(H_2O)_2]^{2+}$ is an octahedral complex which exhibits geometric isomerism.
$[Ni(en)_3]^{2+}$ is an octahedral complex which exhibits optical isomerism.

(C) Key Idea: The $S_{N}1$ reaction proceeds via the formation of a carbocation intermediate. The stability of the carbocation determines the reactivity of the alkyl/aryl halide; higher stability leads to higher reactivity.
The carbocations formed from the given halides are:
$1$. $C_6H_5CH(C_6H_5)Br \rightarrow (C_6H_5)_2CH^+ + Br^-$
$2$. $C_6H_5CH(CH_3)Br \rightarrow C_6H_5CH^+(CH_3) + Br^-$
$3$. $C_6H_5C(CH_3)(C_6H_5)Br \rightarrow (C_6H_5)_2C^+(CH_3) + Br^-$
$4$. $C_6H_5CH_2Br \rightarrow C_6H_5CH_2^+ + Br^-$
The stability order of these carbocations is: $(C_6H_5)_2C^+(CH_3) > (C_6H_5)_2CH^+ > C_6H_5CH^+(CH_3) > C_6H_5CH_2^+$.
The carbocation $(C_6H_5)_2C^+(CH_3)$ is the most stable due to the presence of two phenyl groups and one methyl group providing resonance and inductive stabilization. Thus,$C_6H_5C(CH_3)(C_6H_5)Br$ is the most reactive towards $S_{N}1$ reaction.

(A) The reactivity of $C-X$ bond towards nucleophilic substitution depends on the mechanism ($S_N1$ or $S_N2$).
$1$. Alkyl halides ($III$ and $IV$) are generally more reactive than aryl halides ($I$ and $II$) towards nucleophilic substitution because the $C-X$ bond in aryl halides has partial double bond character due to resonance.
$2$. Among alkyl halides,the $3^o$ halide $(III)$ is more reactive than the $2^o$ halide $(IV)$ in $S_N1$ reactions due to the formation of a more stable carbocation.
$3$. Among aryl halides,the presence of electron-withdrawing groups $(-NO_2)$ increases the reactivity towards nucleophilic aromatic substitution. Thus,$II$ is significantly more reactive than $I$.
$4$. Comparing all,the order of reactivity is $I < II < IV < III$.

(C) The reaction proceeds as follows:
$1$. $C_6H_5CH_2Br$ reacts with $Mg$ in the presence of dry ether to form the Grignard reagent,$C_6H_5CH_2MgBr$.
$2$. The Grignard reagent $C_6H_5CH_2MgBr$ undergoes hydrolysis with $H_3O^{+}$ to form $C_6H_5CH_3$ (Toluene).
Therefore,the product $X$ is $C_6H_5CH_3$.

(A) The acidic strength depends on the stability of the conjugate base formed after the loss of a proton $(H^+)$.
$2,4,6-$trinitrophenol $(III)$ is the most acidic because the phenoxide ion is highly stabilized by the strong $-I$ and $-M$ effects of three $-NO_2$ groups.
Acetic acid $(II)$ is more acidic than phenol $(IV)$ because the carboxylate ion is stabilized by resonance between two electronegative oxygen atoms,whereas the phenoxide ion is stabilized by resonance within the aromatic ring.
Phenol $(IV)$ is more acidic than cyclohexanol $(I)$ because the phenoxide ion is resonance-stabilized,while the cyclohexoxide ion has no such stabilization.
Cyclohexanol $(I)$ is the least acidic as the alkyl group is electron-donating ($+I$ effect),which destabilizes the alkoxide ion.
Therefore,the order of decreasing acidic strength is $III > II > IV > I$.

(B) The acidity of a compound depends on the stability of its conjugate base.
$1$. Phenol $(C_6H_5OH)$ forms a phenoxide ion $(C_6H_5O^-)$ upon losing a proton. This phenoxide ion is highly resonance-stabilized by the benzene ring,making phenol the most acidic among the given options.
$2$. Benzyl alcohol $(C_6H_5CH_2OH)$ forms a benzyloxide ion $(C_6H_5CH_2O^-)$,where the negative charge is localized on the oxygen atom and is not resonance-stabilized by the benzene ring.
$3$. Cyclohexanol and dicyclohexylmethanol are aliphatic alcohols. The alkyl groups attached to the carbon bearing the $-OH$ group are electron-releasing ($+I$ effect),which destabilizes the alkoxide ion and decreases acidity.
Therefore,phenol is the most acidic compound.

(A) The acidity of phenols is determined by the stability of the phenoxide ion formed after the loss of a proton.
Electron-withdrawing groups $(EWG)$,such as $-NO_2$,stabilize the phenoxide ion through $-I$ and $-M$ effects,thereby increasing the acidity.
Electron-releasing groups $(ERG)$,such as $-CH_3$,destabilize the phenoxide ion through $+I$ and $+H$ effects,thereby decreasing the acidity.
Comparing the compounds:
$1$. $(iv)$ Para-nitrophenol: $-NO_2$ at para position exerts both $-I$ and $-M$ effects,providing maximum stabilization.
$2$. $(iii)$ Meta-nitrophenol: $-NO_2$ at meta position exerts only $-I$ effect,providing less stabilization than para.
$3$. $(i)$ Phenol: Reference compound.
$4$. $(ii)$ Methyl phenol: $-CH_3$ group exerts $+I$ and $+H$ effects,destabilizing the phenoxide ion.
Thus,the correct order of acidity is $(iv) > (iii) > (i) > (ii)$.

(B) When glycerol is treated with excess $HI$,it produces $2$-iodopropane. The reaction proceeds through several steps:
$1.$ Glycerol reacts with $3$ moles of $HI$ to form an unstable triiodide: $CH_2(OH)CH(OH)CH_2(OH) + 3HI \rightarrow [CH_2I-CHI-CH_2I] + 3H_2O$.
$2.$ The unstable triiodide loses $I_2$ to form allyl iodide: $[CH_2I-CHI-CH_2I] \rightarrow CH_2=CH-CH_2I + I_2$.
$3.$ Since $HI$ is in excess,allyl iodide reacts further to form $1,2$-diiodopropane,which loses $I_2$ to form propene: $CH_2=CH-CH_2I + HI$ $\rightarrow [CH_3-CHI-CH_2I]$ $\rightarrow CH_3-CH=CH_2 + I_2$.
$4.$ Finally,propene reacts with $HI$ (Markovnikov addition) to give $2$-iodopropane: $CH_3-CH=CH_2 + HI \rightarrow CH_3-CHI-CH_3$.

(C) The iodoform test is given by compounds containing either a methyl keto group $(CH_3-CO-)$ or a methyl hydroxy group $(CH_3-CH(OH)-)$ which can be oxidized to a methyl keto group.
$(i) CH_3-CH_2-OH$ (Ethanol) oxidizes to $CH_3-CHO$,which gives the iodoform test.
$(ii) CH_3-CO-CH_3$ (Acetone) contains the $CH_3-CO-$ group.
$(iii) CH_3-CH(OH)-CH_3$ (Propan$-2-$ol) oxidizes to $CH_3-CO-CH_3$.
$(iv) CH_3-OH$ (Methanol) does not contain the required group.
Therefore,$(i), (ii),$ and $(iii)$ will give a positive iodoform test.

(C) $A. \ CH_3(CH_2)_3NH_2$ is a primary amine,which gives the carbylamine test with $KOH$ (alc.) and $CHCl_3$ to produce a foul-smelling isocyanide $(ii)$.
$B. \ CH_3C\equiv CH$ is a terminal alkyne,which reacts with ammoniacal $AgNO_3$ to form a white precipitate of silver acetylide $(iii)$.
$C. \ CH_3CH_2COOCH_3$ is an ester,which undergoes alkaline hydrolysis to form an alcohol and a carboxylate salt $(i)$.
$D. \ CH_3CH(OH)CH_3$ is a secondary alcohol,which reacts with Lucas reagent $(ZnCl_2 + conc. HCl)$ to produce cloudiness within $5-10 \ minutes$ $(iv)$.
Thus,the correct matching is $A-(ii), B-(iii), C-(i), D-(iv)$.

(B) In the $Reimer-Tiemann$ reaction,a new $C-C$ bond is formed between the aromatic ring and the formyl group.
In the $Cannizzaro$ reaction,an aldehyde without an $\alpha$-hydrogen undergoes disproportionation to form an alcohol and a carboxylic acid salt. No new $C-C$ bond is formed in this reaction.
$2HCHO \xrightarrow{\text{Conc. } NaOH} CH_3OH + HCOONa$
In the $Wurtz$ reaction,two alkyl halides react in the presence of sodium to form a higher alkane,creating a new $C-C$ bond.
In $Friedel-Crafts$ acylation,an acyl group is introduced into an aromatic ring,forming a new $C-C$ bond between the ring and the acyl group.
Therefore,the $Cannizzaro$ reaction is the correct answer.

(A) The reaction that converts an amide $(-CONH_2)$ into a primary amine $(-NH_2)$ with one carbon atom less is known as the $Hofmann$ bromamide degradation reaction.
Among the given reagents,$NaOH + Br_2$ (alkaline solution of bromine) is the specific reagent used for this transformation.
Acetamide $(CH_3CONH_2)$ reacts with $NaOH$ and $Br_2$ to yield methylamine $(CH_3NH_2)$.
The chemical equation is: $CH_3CONH_2 + 4NaOH + Br_2 \rightarrow CH_3NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$.

(D) The susceptibility of a carbonyl group to nucleophilic attack depends on the electrophilicity of the carbonyl carbon.
This electrophilicity is increased by the presence of an electron-withdrawing group attached to the carbonyl carbon.
Among the given derivatives,the chloride ion $(Cl^-)$ is the weakest base and therefore the best leaving group.
Since $Cl$ is highly electronegative and a good leaving group,it exerts a strong inductive effect,making the carbonyl carbon in $CH_3COCl$ the most electron-deficient and thus the most susceptible to nucleophilic attack.

(C) The dehydration of alcohols involves the formation of a carbocation intermediate. The stability of the carbocation determines the ease of dehydration. The more stable the carbocation,the easier the dehydration.
In the given compounds,the hydroxyl group is located at different positions relative to the carbonyl group.
For option $C$ ($4$-hydroxypentan-$2$-one),the carbocation formed after the loss of the $-OH$ group is a secondary carbocation that is relatively more stable due to its position,allowing for easier formation of the conjugated enone system.
Comparing the stability of the intermediate carbocations:
$(b)$ and $(d)$ form secondary carbocations adjacent to the carbonyl group (destabilized by the electron-withdrawing effect of the carbonyl group).
$(a)$ forms a secondary carbocation adjacent to the carbonyl group.
$(c)$ forms a secondary carbocation at the $4$-position,which is further from the electron-withdrawing carbonyl group,making it more stable than the others.
Therefore,the compound in option $C$ undergoes dehydration most readily.

(B) $1$. The reaction of aniline with $NaNO_{2}/HCl$ at $273-278 \ K$ results in the formation of benzenediazonium chloride $(X)$.
$2$. Benzenediazonium chloride $(X)$ then undergoes an electrophilic aromatic substitution reaction (coupling reaction) with $N,N-dimethylaniline$ in a weakly acidic medium.
$3$. The coupling occurs at the para-position of $N,N-dimethylaniline$ to form $p-(dimethylamino)azobenzene$ $(Y)$,which is a yellow-orange coloured dye.
$4$. The structure of $Y$ is $C_6H_5-N=N-C_6H_4-N(CH_3)_2$.

(C) $(i)$ The presence of an electron-withdrawing substituent decreases basicity,while the presence of an electron-releasing substituent like $-CH_{3}, -C_{2}H_{5},$ etc.,increases basicity.
$(ii)$ $HNO_{2}$ converts the $-NH_{2}$ group of aliphatic amines into $-OH$ (alcohols),whereas it converts aromatic amines into diazonium salts at low temperatures $(0-5\,^{\circ}C)$.
$(iii)$ Alkyl amines are more basic than ammonia due to the electron-releasing effect of the alkyl group,while aryl amines are less basic than ammonia due to the electron-withdrawing nature of the phenyl ring.
$(iv)$ The reaction for alkyl amines: $R-NH_{2} \xrightarrow{HNO_{2}} R-OH + N_{2} + H_{2}O$.
$(v)$ The reaction for aryl amines: $C_{6}H_{5}NH_{2} \xrightarrow{NaNO_{2} + HCl, 273-278\,K} C_{6}H_{5}N_{2}^{+}Cl^{-}$.
Therefore,the statement that aryl amines react with nitrous acid to produce phenols is false,as they form diazonium salts under these conditions.

(A) Mutarotation is the phenomenon of change in optical rotation observed in freshly prepared solutions of reducing sugars.
This property is exhibited by sugars that possess a free hemiacetal or hemiketal group,which allows them to exist in equilibrium with their open-chain aldehyde $(-CHO)$ or ketone $(>C=O)$ forms.
$(+)$ Sucrose is a non-reducing sugar because its glycosidic linkage is formed between the anomeric carbons of glucose and fructose,leaving no free aldehyde or ketone group.
Therefore,$(+)$ Sucrose does not exhibit mutarotation.

(A) Neoprene is a synthetic rubber produced by the free radical polymerization of chloroprene ($2$-chloro-$1,3$-butadiene).
The repeating unit of neoprene is formed by the polymerization of $CH_2=C(Cl)-CH=CH_2$.
The resulting structure is $[-CH_2-C(Cl)=CH-CH_2-]_n$.

(B) Tranquilizers are chemical substances that reduce anxiety and mental tension.
They are often referred to as psychotherapeutic drugs.
Examples of commonly used tranquilizers include $Equanil$,$Valium$,and $Serotonin$.

(C) Tollen's reagent is $(AgNO_3 + NH_4OH)$ and it provides a basic medium. Fructose is a keto-hexose,yet it can reduce Tollen's reagent. This is because,in a basic medium,it undergoes enolisation and rearrangement (Lobry de-Bruyn-van Ekenstein transformation) to form an aldose (like $D$-glucose or $D$-mannose) which contains an aldehyde group. The equilibrium involves the formation of an enediol intermediate.

(B) Ions exhibit colour in aqueous solution due to the presence of unpaired electrons in their $d$-orbitals,which allow for $d-d$ transitions.
Electronic configurations:
$La^{3+} (Z=57): [Xe] 4f^0 5d^0$ (No unpaired electrons).
$Ti^{3+} (Z=22): [Ar] 3d^1$ (One unpaired electron).
$Lu^{3+} (Z=71): [Xe] 4f^{14}$ (No unpaired electrons).
$Sc^{3+} (Z=21): [Ar] 3d^0$ (No unpaired electrons).
Since $Ti^{3+}$ has one unpaired electron,it will exhibit colour in aqueous solution.

(C) The structure of $P_4O_{10}$ consists of a tetrahedral arrangement of four phosphorus atoms.
Each phosphorus atom is bonded to one terminal oxygen atom via a double bond $(P=O)$.
There are six oxygen atoms that act as bridges between the phosphorus atoms,forming $P-O-P$ linkages.
Therefore,the total number of bridging oxygen atoms is $6$.