AIIMS 2009 Chemistry Question Paper with Answer and Solution

(C) The element $(X)$ forms $XCl_3$,$X_2O_5$,and $Ca_3X_2$.
Nitrogen $(N)$ forms $NCl_3$,$N_2O_5$,and $Ca_3N_2$.
Nitrogen does not form $NCl_5$ because it lacks $d$-orbitals in its valence shell,which prevents it from expanding its covalency beyond $4$.
Therefore,the element $(X)$ is $N$.

(C) Reaction $(I): \ SeO_3^{2-} + BrO_3^{-} + H^{+} \to SeO_4^{2-} + Br_2 + H_2O$ (n-factor for $SeO_3^{2-} = 2$,for $BrO_3^{-} = 5$)
Reaction $(II): \ BrO_3^{-} + AsO_2^{-} + H_2O \to Br^{-} + AsO_4^{3-} + H^{+}$ (n-factor for $BrO_3^{-} = 6$,for $AsO_2^{-} = 2$)
Step $1$: Calculate milliequivalents of $AsO_2^{-}$ used in back titration:
$meq \ of \ AsO_2^{-} = 12.5 \ mL \times (\frac{1}{25} \ M) \times 2 = 1 \ meq$
Step $2$: Calculate millimoles of excess $BrO_3^{-}$:
$meq \ of \ BrO_3^{-} = meq \ of \ AsO_2^{-} = 1 \ meq$
$mmol \ of \ BrO_3^{-} = \frac{1}{6} \ mmol$
Step $3$: Calculate millimoles of $BrO_3^{-}$ consumed in reaction $(I)$:
Total $mmol \ of \ BrO_3^{-} = 70 \ mL \times \frac{1}{60} \ M = \frac{7}{6} \ mmol$
$mmol \ of \ BrO_3^{-} \ consumed = \frac{7}{6} - \frac{1}{6} = 1 \ mmol$
Step $4$: Calculate millimoles of $SeO_3^{2-}$:
$meq \ of \ SeO_3^{2-} = meq \ of \ BrO_3^{-} \ consumed$
$mmol \ of \ SeO_3^{2-} \times 2 = 1 \ mmol \times 5$
$mmol \ of \ SeO_3^{2-} = 2.5 \ mmol = 2.5 \times 10^{-3} \ mol$

(A) The assertion is correct because $Be$ can expand its coordination number up to $4$ due to the presence of only $s$ and $p$ orbitals in its valence shell $(n=2)$.
$Al$ can expand its coordination number to $6$ because it has vacant $3d$ orbitals in its valence shell $(n=3)$.
Since $Be$ lacks vacant $d-$ orbitals,it cannot form $BeF_6^{3-}$,which would require a coordination number of $6$.
Thus,the reason correctly explains the assertion.

(D) When a solid melts,it undergoes a phase transition from a highly ordered state to a less ordered state.
Since entropy $(S)$ is a measure of the randomness or disorder of a system,the transition from solid to liquid results in an increase in the randomness of the particles.
Therefore,$S$ increases during the melting process.

(D) For an isothermal reversible expansion,the work done is given by the formula: $W = -2.303 \times nRT \times \log(\frac{V_2}{V_1})$.
Given values: $n = 6 \ mol$,$T = 27 \ ^oC = 300 \ K$,$V_1 = 1 \ L$,$V_2 = 10 \ L$,and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Substituting the values: $W = -2.303 \times 6 \times 8.314 \times 300 \times \log(\frac{10}{1})$.
$W = -2.303 \times 6 \times 8.314 \times 300 \times 1$.
$W = -34465.3 \ J = -34.465 \ kJ$.
The magnitude of maximum work done is $34.465 \ kJ$.

(B) The required reaction is $C_6H_{10} + H_2 \to C_6H_{12}$,$\Delta H_1 = ? \dots (1)$
Given combustion reactions:
$H_2 + \frac{1}{2}O_2 \to H_2O$,$\Delta H_2 = -241 \, kJ/mol \dots (2)$
$C_6H_{10} + \frac{17}{2}O_2 \to 6CO_2 + 5H_2O$,$\Delta H_3 = -3800 \, kJ/mol \dots (3)$
$C_6H_{12} + 9O_2 \to 6CO_2 + 6H_2O$,$\Delta H_4 = -3920 \, kJ/mol \dots (4)$
Applying Hess's Law,the required reaction $(1)$ is obtained by $(2) + (3) - (4)$:
$\Delta H_1 = (\Delta H_2 + \Delta H_3) - \Delta H_4$
$\Delta H_1 = (-241 - 3800) - (-3920)$
$\Delta H_1 = -4041 + 3920 = -121 \, kJ/mol$.

(C) For the weak acid $HA$,the degree of ionization $\alpha = 0.1$ at concentration $C = 0.01 \ M$.
The dissociation constant $K_a$ is given by $K_a = \frac{C \alpha^2}{1 - \alpha} = \frac{0.01 \times (0.1)^2}{1 - 0.1} = \frac{0.0001}{0.9} = 1.11 \times 10^{-4}$.
Calculate $pK_a$: $pK_a = -\log(1.11 \times 10^{-4}) = 4 - \log(1.11) \approx 4 - 0.0458 = 3.9542$.
Using the Henderson-Hasselbalch equation for the buffer solution containing $HA$ $(0.1 \ M)$ and $NaA$ $(0.05 \ M)$:
$pH = pK_a + \log \left( \frac{[Salt]}{[Acid]} \right) = 3.9542 + \log \left( \frac{0.05}{0.1} \right)$.
$pH = 3.9542 + \log(0.5) = 3.9542 - 0.3010 = 3.6532 \approx 3.653$.

(A) The solubility of $PbF_2$ in water is $S = 10^{-3}\, M$.
The solubility product constant $K_{sp}$ is given by $K_{sp} = [Pb^{2+}][F^-]^2 = (S)(2S)^2 = 4S^3$.
$K_{sp} = 4 \times (10^{-3})^3 = 4 \times 10^{-9}$.
In $0.05\, M\, NaF$ solution,$NaF$ dissociates completely to provide $[F^-] = 0.05\, M$.
Let the solubility of $PbF_2$ in this solution be $S'\, M$. Then $[Pb^{2+}] = S'$ and $[F^-] = (2S' + 0.05)\, M$.
$K_{sp} = S'(2S' + 0.05)^2 = 4 \times 10^{-9}$.
Since $S'$ is very small,we assume $2S' \ll 0.05$,so $(2S' + 0.05) \approx 0.05$.
$S' \times (0.05)^2 = 4 \times 10^{-9}$.
$S' \times 2.5 \times 10^{-3} = 4 \times 10^{-9}$.
$S' = \frac{4 \times 10^{-9}}{2.5 \times 10^{-3}} = 1.6 \times 10^{-6}\, M$.

(A) The reaction is $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$.
Initially,let the total moles be $1$. Thus,$n(N_2) = 0.79$ and $n(O_2) = 0.21$.
Let the degree of dissociation be $\alpha$. At equilibrium:
$n(N_2) = 0.79 - \alpha/2$,$n(O_2) = 0.21 - \alpha/2$,$n(NO) = \alpha$.
However,using the standard approximation where $\alpha$ is small,we assume the total moles remain constant at $1$.
$K_p = \frac{P_{NO}^2}{P_{N_2} \cdot P_{O_2}} = \frac{n_{NO}^2}{n_{N_2} \cdot n_{O_2}} = \frac{\alpha^2}{0.79 \times 0.21} = 1.1 \times 10^{-3}$.
$\alpha^2 = 1.1 \times 10^{-3} \times 0.79 \times 0.21 = 0.18249 \times 10^{-3} = 1.8249 \times 10^{-4}$.
$\alpha = \sqrt{1.8249 \times 10^{-4}} = 0.0135$.
Volume percent of $NO = \alpha \times 100 = 1.35 \% \approx 1.33 \%$ (given the approximation in the provided solution steps).

(A) In the titration of a weak acid $(HA)$ with a strong base $(NaOH)$,the reaction is $HA + OH^- \rightarrow A^- + H_2O$.
At the half equivalence point,half of the acid has been neutralized,meaning $[HA] = [A^-]$.
According to the Henderson-Hasselbalch equation,$pH = pK_a + \log(\frac{[salt]}{[acid]})$.
Since $[salt] = [acid]$ at this point,$\log(1) = 0$,so $pH = pK_a$.
This mixture acts as an acidic buffer,and the buffer capacity is indeed maximum when the concentration of the acid equals the concentration of its conjugate base (salt).

(B) The common table salt $(NaCl)$ is purified by passing $HCl$ gas through a saturated solution of salt.
This process is based on the common ion effect,where the increase in the concentration of $Cl^-$ ions forces the equilibrium to shift towards the precipitation of $NaCl$.
The reaction is: $NaCl(aq) + HCl(g) \to NaCl(s) + H^+(aq) + Cl^-(aq)$.

(B) The Assertion is correct because when we touch ice,heat flows from our body (at a higher temperature) to the ice (at a lower temperature),causing a sensation of coldness.
The Reason is also correct because ice is indeed the solid state of $H_2O$.
However,the Reason does not explain why we feel cold; the sensation of cold is due to the heat transfer process,not merely the state of matter.

(A) The $s-$block elements are highly electropositive in nature,which makes them very reactive.
Due to their high reactivity,they readily react with other elements and do not occur in a free state in nature.
They are typically found in nature in the form of compounds like halides,carbonates,and sulphates.
Therefore,the Reason correctly explains the Assertion.

(A) The electronic configuration of $Be$ is $1s^2 2s^2$. It has only $s$ and $p$ orbitals in its valence shell $(n=2)$.
Since there are no vacant $d-$orbitals available in the valence shell of $Be$,it cannot expand its coordination number beyond $4$.
Therefore,$Be$ can form $BeF_4^{2-}$ but cannot form $BeF_6^{3-}$.
$Al$ has $3d$ orbitals available,allowing it to form $AlF_6^{3-}$.
Thus,both the assertion and the reason are correct,and the reason correctly explains why $BeF_6^{3-}$ is not formed.

(B) Graphite is thermodynamically the most stable allotrope of carbon. That is why $\Delta_fH^o$ (graphite) is taken as $0 \ kJ \ mol^{-1}$.
$\Delta_fH^o$ (diamond) $= +1.90 \ kJ \ mol^{-1}$
$\Delta_fH^o$ (fullerene) $= +38.1 \ kJ \ mol^{-1}$
Since the standard enthalpy of formation of graphite is the lowest (zero),it is the most stable form.

(A) $(x)$ is a conjugated diene,which is more stable due to resonance.
$(w)$ is an isolated diene,which is less stable than a conjugated diene.
$(z)$ is a cumulated diene (allene),which is highly unstable due to steric strain and $sp$ hybridization of the central carbon.
$(y)$ is cyclobutadiene,which is an antiaromatic system and is highly unstable.
Therefore,the stability order is $(x) > (w) > (z) > (y)$.

(A) Pyrolysis (or cracking) is the process of breaking down higher alkanes into lower alkanes and alkenes by heating them in the absence of air.
For example: $C_6H_{14} \xrightarrow[\Delta]{\text{Pyrolysis}} C_2H_4 + C_4H_{10}$
This is the most effective industrial method for obtaining hydrocarbons with a lower carbon number from higher ones.

(C) The reaction of $trans-2-Butene$ with $Br_2$ proceeds via the formation of a cyclic bromonium ion intermediate.
The bromide ion then attacks the intermediate from the side opposite to the bromonium bridge,resulting in $anti-addition$.
$trans-2-Butene$ undergoes $anti-addition$ of $Br_2$ to yield $meso-2,3-dibromobutane$.
Therefore,the Assertion is correct,but the Reason is incorrect because the reaction involves $anti-addition$,not $syn-addition$.

(A) The reaction of $1$-butene with $HBr$ in the presence of peroxide follows the anti-Markovnikov addition rule.
$CH_3-CH_2-CH=CH_2 + HBr \xrightarrow{\text{Peroxide}} CH_3-CH_2-CH_2-CH_2Br$ ($1$-bromobutane).
This reaction proceeds via a free radical mechanism,which explains the anti-Markovnikov regioselectivity. Therefore,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(C) The addition of $Br_2$ to an alkene is an $\text{anti-addition}$ process,not $\text{syn-addition}$.
When $\text{trans-}2\text{-butene}$ undergoes $\text{anti-addition}$ of $Br_2$,it forms $\text{meso-}2,3\text{-dibromobutane}$.
Since the Assertion is correct but the Reason is incorrect,the correct option is $C$.

(C) The $S_8$ molecule adopts a puckered ring structure (crown shape).
In this structure,the $S-S-S$ bond angle is approximately $107^o$ (often cited in the range of $102^o-108^o$).
Therefore,the assertion is correct.
However,$S_8$ does not have a $V$-shape; it has a crown-like puckered ring structure.
Thus,the reason is incorrect.

(B) The assertion is true because $H_2SO_4$ molecules are associated through a large number of intermolecular hydrogen bonds,which increases its viscosity compared to water.
The reason is also true because concentrated $H_2SO_4$ is highly hygroscopic and has a strong affinity for water.
However,the high viscosity is due to hydrogen bonding,not the affinity for water. Thus,the reason is not the correct explanation of the assertion.

(B) The reaction follows the Markownikoff rule,where the electrophilic part $(NO^+)$ adds to the carbon atom with more hydrogen atoms,and the nucleophilic part $(Br^-)$ adds to the carbon atom with fewer hydrogen atoms.
In the reactant $(CH_3)_2C = CHCH_3$,the carbon at position $2$ has no hydrogen atoms,while the carbon at position $3$ has one hydrogen atom.
Therefore,the $Br$ atom attaches to the carbon at position $2$,and the $NO$ group attaches to the carbon at position $3$.
The product formed is $(CH_3)_2C(Br) - CH(NO)CH_3$.

(D) $1$. Step $1$: Free radical bromination of methylcyclopentane with $Br_2/hv$ occurs preferentially at the tertiary carbon to form $1-$bromo$-1-$methylcyclopentane $(X)$ as the major product.
$2$. Step $2$: Dehydrohalogenation of $(X)$ with alcoholic $KOH/\Delta$ follows the $E2$ mechanism to form the more substituted alkene,$1$-methylcyclopentene $(Y)$,as the major product.
$3$. Step $3$: Anti-Markovnikov addition of $HBr$ to $1-$methylcyclopentene $(Y)$ in the presence of peroxide yields $1-$bromo$-2-$methylcyclopentane $(Z)$ as the major product.

(D) $R'R$ and $R'S$ are diastereomers and have different physical properties like water solubility,$B.P.$,$M.P.$,etc.
Since they are diastereomers,they have different physical properties,meaning they do not have the same solubility in water.
Therefore,the statement that $R'R$ and $R'S$ have the same solubility in water is incorrect.

(A) In the condition of obstructive jaundice,the entry of bile into the small intestine is prevented due to an obstruction in the bile duct.
Bile salts are essential for the digestion of fats through emulsification and for their absorption by forming water-soluble droplets called micelles.
From these micelles,fatty acids,glycerides,sterols,and fat-soluble vitamins are absorbed into the intestinal cells.
Therefore,in the absence of bile,fats remain unabsorbed and are consequently eliminated from the body in the faeces.

(B) Graphite is the thermodynamically most stable allotrope of carbon at standard temperature and pressure.

(D) Comparative anatomy is the study of similarities and differences in the structures of different organisms. Homologous structures,which are organs or skeletal elements of animals and organisms that,by virtue of their similarity,suggest their connection to a common ancestor,are a primary focus of comparative anatomy. Comparative cytology involves the study of cellular structures,while biochemistry focuses on molecular comparisons like $DNA$ and proteins. Geology deals with the study of the Earth's physical structure and substance.

(C) When the temperature is raised,the molecules of the dissolved gas present in a liquid gain kinetic energy.
This higher kinetic energy allows the gas molecules to overcome the intermolecular forces and escape from the solution.
Therefore,the solubility of gases in liquids decreases as the temperature increases.
Thus,the Assertion is correct,but the Reason is incorrect.

(A) Borax $(Na_2B_4O_7 \cdot 10H_2O)$ when heated,loses water and swells up,then melts to form a transparent glassy bead consisting of sodium metaborate $(NaBO_2)$ and boric anhydride $(B_2O_3)$.
$Na_2B_4O_7 \cdot 10H_2O$ $\xrightarrow{\Delta} Na_2B_4O_7$ $\xrightarrow{\Delta} 2NaBO_2 + B_2O_3$
When this glassy bead is heated with coloured transition metal salts,it forms coloured metaborates,which are characteristic of the metal cation. Thus,the borax bead test is used to identify coloured cations.
Since the formation of the transparent bead and its subsequent reaction with cations to form coloured beads explains why the test works,the Reason is the correct explanation of the Assertion.

(D) In graphite,the carbon atoms are arranged in regular hexagons in flat parallel layers. This structure corresponds to a $Hexagonal$ crystal system.

(A) The stability of a crystal depends upon the strength of the interparticle attractive force.
The melting point of a solid is a measure of the energy required to overcome these attractive forces.
Therefore,a higher melting point indicates stronger interparticle forces,which corresponds to higher crystal stability.
Thus,the stability of a crystal is reflected in the magnitude of its melting point,and the Reason correctly explains the Assertion.

(B) An aqueous solution of $HCl$ shows a negative deviation from Raoult's law.
In this solution,the intermolecular forces of attraction between the solute $(HCl)$ and solvent $(H_2O)$ molecules are stronger than the forces between the pure components.
This stronger attraction reduces the escaping tendency of the molecules into the vapour phase.
Consequently,the observed partial vapour pressure is lower than the value predicted by Raoult's law,which is the characteristic definition of a negative deviation.

(C) For the initial solution: $n_A = 3, n_B = 2$. Total moles = $5$.
$P_1 = p_A^o \left( \frac{3}{5} \right) + p_B^o \left( \frac{2}{5} \right) = 600 \ torr$.
$3p_A^o + 2p_B^o = 3000$ (Equation $1$).
After adding $1.5 \ mol$ of $A$ and $0.5 \ mol$ of $C$: $n_A = 4.5, n_B = 2, n_C = 0.5$. Total moles = $7$.
Since $C$ is non-volatile,$p_C^o = 0$.
$P_2 = 600 + 30 = 630 \ torr$.
$P_2 = p_A^o \left( \frac{4.5}{7} \right) + p_B^o \left( \frac{2}{7} \right) = 630$.
$4.5p_A^o + 2p_B^o = 4410$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(4.5 - 3)p_A^o = 4410 - 3000 \implies 1.5p_A^o = 1410 \implies p_A^o = 940 \ torr$.
Substituting $p_A^o$ in Equation $1$: $3(940) + 2p_B^o = 3000 \implies 2820 + 2p_B^o = 3000 \implies 2p_B^o = 180 \implies p_B^o = 90 \ torr$.

(B) According to Kohlrausch's law of independent migration of ions,the equivalent conductance at infinite dilution is the sum of the ionic conductances of the cation and the anion.
For $NaF$,$\Lambda^{\infty}_{NaF} = \lambda^{\infty}_{Na^+} + \lambda^{\infty}_{F^-} = 90.1 \, \Omega^{-1} \, cm^2$.
For $KF$,$\Lambda^{\infty}_{KF} = \lambda^{\infty}_{K^+} + \lambda^{\infty}_{F^-}$.
Since the ionic mobility of $K^+$ is higher than $Na^+$,the equivalent conductance of $KF$ will be higher than $NaF$.
However,in many simplified textbook contexts,if the question implies a comparison where the values are expected to be similar or if it is a trick question regarding the nature of the electrolyte,we must note that $KF$ has a higher value than $90.1$.
Given the options provided,$111.2$ is the only value greater than $90.1$ that fits the expected trend for the substitution of $Na^+$ with $K^+$.

(C) Using the Arrhenius equation: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303 \ R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$
Given $T_1 = 727 + 273 = 1000 \ K$,$T_2 = 1571 + 273 = 1844 \ K$,$T_3 = 1150 + 273 = 1423 \ K$.
For the first interval: $\log \left( \frac{1.667 \times 10^{-4}}{1.667 \times 10^{-6}} \right) = \frac{E_a}{2.303 \ R} \left( \frac{1844 - 1000}{1844 \times 1000} \right)$
$2 = \frac{E_a}{2.303 \ R} \left( \frac{844}{1844000} \right) \dots (1)$
For the second interval: $\log \left( \frac{k_3}{1.667 \times 10^{-6}} \right) = \frac{E_a}{2.303 \ R} \left( \frac{1423 - 1000}{1423 \times 1000} \right) \dots (2)$
Dividing $(2)$ by $(1)$:
$\frac{\log \left( \frac{k_3}{1.667 \times 10^{-6}} \right)}{2} = \frac{423}{1423 \times 1000} \times \frac{1844000}{844} \approx 0.6495$
$\log \left( \frac{k_3}{1.667 \times 10^{-6}} \right) = 1.299$
$\frac{k_3}{1.667 \times 10^{-6}} = 19.9$
$k_3 = 19.9 \times 1.667 \times 10^{-6} = 3.318 \times 10^{-5} \ s^{-1}$

(A) The rate law is determined experimentally and depends on the slowest step (rate-determining step) of the reaction mechanism.
In contrast,the equilibrium constant expression is derived from the stoichiometry of the overall balanced chemical equation.
Therefore,the exponents in the rate law do not necessarily correspond to the stoichiometric coefficients of the reactants in the overall balanced equation.
Since the reaction rate is governed by the reaction mechanism,the Reason correctly explains the Assertion.

(C) In chemisorption,adsorption involves the formation of chemical bonds between the adsorbate and the adsorbent,which requires an initial activation energy.
Initially,as temperature increases,more molecules gain sufficient energy to overcome this activation barrier,leading to an increase in adsorption.
However,once the chemical bonds are formed,the process is typically exothermic. According to Le Chatelier's principle,an increase in temperature at higher values favors the reverse (desorption) process,causing adsorption to decrease.
Therefore,the Assertion is correct,but the Reason is incorrect because heat does not provide 'more and more' activation energy indefinitely; rather,it facilitates the initial bond formation,and excessive heat eventually destabilizes the formed bonds.

(A) Due to the low ignition temperature of white phosphorus,it undergoes oxidation in the presence of air,which slowly raises its temperature,and after a few moments,it catches fire spontaneously. Due to this reason,it is stored under water.
Ignition temperature of red phosphorus is high.
Black phosphorus is crystalline in nature.
Phosphorus forms a number of hydrides,such as $PH_3$ and $P_2H_4$.

(C) $N$ forms $NCl_3$,$N_2O_5$ and $Ca_3N_2$.
Nitrogen,due to the absence of empty $d$-orbitals,cannot extend its covalency beyond $3$ and hence does not form $NCl_5$.
Due to its small size and high electronegativity,it can accept $3$ electrons to form the $N^{3-}$ ion in $Ca_3N_2$.

(B) The reaction of $SO_3^{2-}$ with $Al$ in the presence of $NaOH$ is a reduction reaction where $SO_3^{2-}$ is reduced to $S^{2-}$.
The chemical equation is: $SO_3^{2-} + 2Al + 2OH^- + 3H_2O \to S^{2-} + 2[Al(OH)_4]^-$.
In this reaction,$H_2S$ is not evolved because the sulfide ion $(S^{2-})$ formed is immediately trapped in the alkaline medium as $Na_2S$ (or $S^{2-}$ ions),and no acidic conditions are present to release $H_2S$ gas.
In contrast,reactions with dilute acids like $HCl$ or $H_2SO_4$ typically provide the $H^+$ ions necessary to convert $S^{2-}$ into $H_2S$ gas.

(B) The hydrides $MH_3$ have a pyramidal shape with a lone pair of electrons on the central atom. The $H-M-H$ bond angle is less than the tetrahedral angle of $109^o 28'$ due to lone pair-bond pair repulsion.
As the electronegativity of the central atom $M$ decreases from $N$ to $Sb$,the bond pair of electrons shifts further away from the central atom.
Consequently,the bond angle decreases and approaches $90^o$.
This indicates that the bonding involves almost pure $p-$ orbitals of the central atom,with very little $s-$ character involved in the hybridization.

(D) The acidity of oxides generally increases with an increase in the oxidation state of the central atom and with an increase in the electronegativity of the central atom.
$Ag_2O$ is basic,$V_2O_5$ is amphoteric,$CO$ is neutral,and $N_2O_5$ is strongly acidic.
In $N_2O_5$,the oxidation state of $N$ is $+5$ and $N$ has high electronegativity,which makes the $N-O$ bond more polar and the oxide more acidic compared to the others.

(B) Iodised salt is common table salt $(NaCl)$ that has been fortified with a small amount of iodine-containing salts.
These salts are typically $KI$ (potassium iodide),$KIO_3$ (potassium iodate),$NaI$ (sodium iodide),or $NaIO_3$ (sodium iodate).
Among the given options,$NaIO_3$ is the correct chemical form present in iodised salt.

(B) When chlorine water $(Cl_2)$ is added to a solution of $KBr$,the following reaction occurs: $2KBr + Cl_2 \to 2KCl + Br_2$.
Initially,the solution turns brown due to the formation of $Br_2$.
However,in the presence of excess chlorine water,$Br_2$ reacts further with $Cl_2$ to form $BrCl$ (bromine monochloride),which is orange-red in color: $Br_2 + Cl_2 \to 2BrCl$.

(A) In $KMnO_4$,Manganese $(Mn)$ is in the $+7$ oxidation state with a $d^0$ electronic configuration.
Since there are no $d$ electrons,$d-d$ transitions are not possible.
The intense purple color arises from Ligand-to-Metal Charge Transfer $(LMCT)$,where an electron is promoted from the $O^{2-}$ ligand to the empty $d$-orbitals of the $Mn^{7+}$ center.
Therefore,the transition is $L \to M$ charge transfer.

(B) $Fe$ and $Pt$ do not form amalgams with mercury.
$\text{Silver amalgam} (Ag-Hg) \xrightarrow[\Delta]{Fe \text{ vessel}} Ag + Hg \uparrow$
If the vessel were made of other metals like $Cu$,$Ni$,or $Zn$,they would react with the liberated mercury to form their own amalgams,contaminating the silver.

(C) $K_3[Cu(CN)_4]$: $Cu$ is in $+1$ oxidation state $(3d^{10})$,so it has no unpaired electrons,making it diamagnetic and colourless.
$(NH_4)_2[TiCl_6]$: $Ti$ is in $+4$ oxidation state $(3d^0)$,so it has no unpaired electrons,making it diamagnetic and colourless.
$VOSO_4$: $V$ is in $+4$ oxidation state $(3d^1)$,so it has one unpaired electron,making it paramagnetic and coloured.
$K_2Cr_2O_7$: $Cr$ is in $+6$ oxidation state $(3d^0)$,so it has no unpaired electrons. It is diamagnetic,but appears coloured due to ligand-to-metal charge transfer $(LMCT)$.

(D) The magnetic moment (spin only) is given by $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons. Zero magnetic moment implies $n = 0$.
$1$. $[Ni(NH_3)_6]Cl_2$: $Ni^{2+}$ is $3d^8$. $NH_3$ is a weak field ligand,resulting in $sp^3d^2$ hybridization with $2$ unpaired electrons.
$2$. $Na_3[FeF_6]$: $Fe^{3+}$ is $3d^5$. $F^-$ is a weak field ligand,resulting in $sp^3d^2$ hybridization with $5$ unpaired electrons.
$3$. $[Cr(H_2O)_6]SO_4$: $Cr^{2+}$ is $3d^4$. $H_2O$ is a weak field ligand,resulting in $sp^3d^2$ hybridization with $4$ unpaired electrons.
$4$. $K_4[Fe(CN)_6]$: $Fe^{2+}$ is $3d^6$. $CN^-$ is a strong field ligand,causing pairing of electrons. It undergoes $d^2sp^3$ hybridization with $0$ unpaired electrons.
Thus,$K_4[Fe(CN)_6]$ has a zero magnetic moment.

(B) For coordination complexes with a coordination number of $4$,the geometry can be either tetrahedral or square planar.
Square planar complexes typically exhibit low-spin configurations because the crystal field splitting energy is large enough to cause pairing.
Tetrahedral complexes,due to smaller crystal field splitting,almost always exhibit high-spin configurations.
Therefore,measuring the magnetic moment to determine the number of unpaired electrons helps distinguish between these two geometries.

(A) For an octahedral complex of the type $[M(AB)(CD)ef]$,where $AB$ and $CD$ are unsymmetrical bidentate ligands and $e, f$ are monodentate ligands,the total number of geometrical isomers is $20$.
Each geometrical isomer is chiral because the complex lacks a plane of symmetry and a center of inversion.
Since each geometrical isomer exists as a pair of enantiomers (a $d$ and $l$ form),the total number of enantiomeric pairs is equal to the total number of geometrical isomers.
Therefore,there are $20$ pairs of enantiomers possible.

(B) In the qualitative analysis of nitrate,a brown ring is formed due to the formation of $[Fe(H_2O)_5(NO)]^{2+}$. The chemical reaction is: $FeSO_4 + NO + 5H_2O \to [Fe(H_2O)_5(NO)]SO_4$.
In this complex,the central metal ion is $Fe^{2+}$. The ligands are five $H_2O$ molecules and one $NO$ molecule.
The coordination number is the total number of sigma bonds formed by the ligand atoms with the central metal ion,which is $5 + 1 = 6$.
Both the Assertion and the Reason are correct,but the Reason does not explain why the complex is brown-coloured (which is due to charge transfer). Therefore,the Reason is not the correct explanation of the Assertion.

(D) The Assertion is incorrect because the oxidation of phenol with $KMnO_4$ does not yield meso-tartaric acid; it typically leads to the formation of $p$-benzoquinone or other oxidation products depending on conditions.
The Reason is correct because pure phenol is a colourless crystalline solid that turns pink or red upon exposure to air and light due to the formation of phenoquinone (a type of quinone) via atmospheric oxidation.
Since the Assertion is incorrect and the Reason is correct,the correct option is $D$.

(C) The formation of diethyl ether from ethanol $(C_2H_5OH)$ in the presence of concentrated $H_2SO_4$ at $413 \ K$ $(140^\circ C)$ is an example of an intermolecular dehydration reaction.
$2 \ C_2H_5OH \xrightarrow{H_2SO_4, 413 \ K} C_2H_5-O-C_2H_5 + H_2O$

(C) In the Cannizzaro reaction,the rate-determining step involves the transfer of a hydride ion $(H^-)$ from the gem-diol dianion intermediate to another molecule of the aldehyde.
The ability of the intermediate to act as a hydride donor depends on the electron density at the carbon atom attached to the hydrogen. An electron-donating group (like $-OCH_3$) increases the electron density on the carbon,making the $C-H$ bond more polarized and facilitating the release of the hydride ion.
Conversely,an electron-withdrawing group (like $-NO_2$) decreases the electron density,making it a poorer hydride donor.
Comparing the options:
- Option $A$ is a mono-anion,which is not the primary hydride donor.
- Option $B$ is the unsubstituted benzaldehyde gem-diol dianion.
- Option $C$ has a methoxy group $(-OCH_3)$,which is an electron-donating group ($+M$ effect),making it the best hydride donor.
- Option $D$ has a nitro group $(-NO_2)$,which is an electron-withdrawing group ($-M$ effect),making it the poorest hydride donor.
Therefore,the intermediate with the electron-donating group is the best hydride donor.

(B) Acetone reacts with ethylene glycol in the presence of dry $HCl$ gas to form a cyclic ketal,specifically $2,2-dimethyl-1,3-dioxolane$.
The reaction involves the elimination of a water molecule between the carbonyl oxygen of the ketone and the hydroxyl hydrogens of the diol.
The reaction is: $CH_3-CO-CH_3 + HO-CH_2-CH_2-OH \xrightarrow{HCl} \text{Cyclic Ketal} + H_2O$.

(C) The Assertion is correct: $RCOCl$,$(RCO)_2O$,and $RCOOR'$ all react with excess Grignard reagents $(R''MgX)$ to form $3^o$ alcohols after hydrolysis.
The Reason is incorrect: While $RCOCl$ reacts with $R_2Cd$ to form ketones,$(RCO)_2O$ also reacts with $R_2Cd$ to form ketones. Therefore,the statement that $(RCO)_2O$ does not react at all is false.

(B) The Assertion is correct because protonation of the carbonyl oxygen atom increases the positive charge on the carbon atom,thereby enhancing its electrophilic character.
The Reason is also correct because the oxygen atom of the carbonyl group acts as a nucleophile (due to lone pairs) and accepts a proton $(H^+)$,which is an electrophile.
However,the Reason is not the correct explanation for the Assertion,as the increased electrophilicity is a consequence of the resonance effect and inductive effect induced by the positive charge on oxygen,not simply the fact that an electrophile was added.

(C) The $-COOH$ group is a strong electron-withdrawing group due to both $-I$ and $-M$ effects.
It deactivates the benzene ring towards electrophilic substitution.
The electron density is decreased at all positions,but the decrease is relatively less at the $m-$position compared to the $o-$ and $p-$positions.
Therefore,the incoming electrophile $(NO_2^+)$ attacks the $m-$position.
The Assertion is correct,but the Reason is incorrect because the carboxyl group decreases,not increases,the electron density.

(B) Diethyl oxalate is used for distinguishing $1^o, 2^o$ and $3^o$ amines because they react differently.
$1^o$ amines form crystalline substituted oxamides.
$2^o$ amines form liquid diethyl oxamic esters.
$3^o$ amines do not react with diethyl oxalate because they lack a replaceable hydrogen atom.
$RNH_2 + (COOC_2H_5)_2 \to (CONHR)_2 + 2 C_2H_5OH$ ($1^o$ amine,crystalline oxamide)
$R_2NH + (COOC_2H_5)_2 \to (CONR_2)(COOC_2H_5) + C_2H_5OH$ ($2^o$ amine,liquid oxamic ester)
$R_3N + (COOC_2H_5)_2 \to \text{No reaction}$ ($3^o$ amine)

(B) Fluorescein,also known as resorcinolphthalein,is a classic example of a phthalein dye.
It is synthesized by the condensation reaction of phthalic anhydride with resorcinol in the presence of a catalyst like anhydrous zinc chloride $(ZnCl_2)$.
Due to its structure containing the phthalein backbone,it is classified under phthalein dyes.

(C) According to the base pairing rules in $DNA$,$Adenine$ $(A)$ pairs with $Thymine$ $(T)$ and $Cytosine$ $(C)$ pairs with $Guanine$ $(G)$.
Given strand: $ATCGTATG$
Complementary strand: $TAGCATAC$

(C) The structural features of soaps and detergents are similar,except that the polar end in detergents is $-OSO_3^-Na^+$ or $-SO_3^-Na^+$,while in soaps,the polar end is $-COO^-Na^+$.
Detergents have an advantage over soaps because their polar ends (sulphate or sulphonate groups) retain their efficiency in hard water.
This is because the corresponding $Ca^{2+}$ and $Mg^{2+}$ salts of detergents are soluble in water,whereas the $Ca^{2+}$ and $Mg^{2+}$ salts of soaps are insoluble and form scum.
Therefore,the correct reason is that their $Ca^{2+}$ and $Mg^{2+}$ salts are water soluble.

(B) Procaine is a well-known local anaesthetic used in medical procedures to numb specific areas of the body.
Chloramphenicol and Penicillin $-G$ are antibiotics used to treat bacterial infections.
Diazepam is a sedative belonging to the benzodiazepine class,used to treat anxiety and muscle spasms.