AIIMS 2014 Chemistry Question Paper with Answer and Solution

(D) The relation between Young's modulus $(Y)$,rigidity modulus $(\eta)$,and Poisson's ratio $(\sigma)$ is given by:
$Y = 2\eta(1 + \sigma)$
Dividing both sides by $2\eta$:
$\frac{Y}{2\eta} = 1 + \sigma$
$\frac{Y}{2\eta} - 1 = \sigma$
$\sigma = \frac{Y - 2\eta}{2\eta}$
$\sigma = \frac{0.5Y - \eta}{\eta}$
Thus,option $(d)$ is correct.

(B) Resonance occurs when an external periodic force is applied to a system at a frequency equal to its natural frequency.
Since the system is driven by an external periodic force,it is a specific case of forced vibration.
Therefore,resonance is an example of forced vibration.
The correct option is $B$.

(A) Magnetic lines of force represent the direction of the magnetic field at any point. If two magnetic lines of force were to intersect at a point,there would be two different directions for the magnetic field at that same point. Since a point in space can only have a single net magnetic field vector at any given time,it is physically impossible for magnetic lines of force to intersect.

(C) According to Le Chatelier's principle,increasing the temperature favors the endothermic direction of a reaction.
$A$. $CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)}$ is exothermic $(\Delta H < 0)$.
$B$. $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$ is exothermic $(\Delta H < 0)$.
$C$. $H_2O_{(g)} \rightleftharpoons H_{2(g)} + \frac{1}{2}O_{2(g)}$ is endothermic $(\Delta H > 0)$.
$D$. $4HCl_{(g)} + O_{2(g)} \rightleftharpoons 2H_2O_{(g)} + 2Cl_{2(g)}$ is exothermic $(\Delta H < 0)$.
Therefore,increasing the temperature will shift the equilibrium to the right for reaction $C$.

(A) Chloramphenicol is obtained from $Streptomyces \text{ } venezuelae$ $(1947)$.
Erythromycin is obtained from $Streptomyces \text{ } erythraeus$ $(1953)$.
Both these antibiotics are produced by species of the genus $Streptomyces$, which are filamentous bacteria known for producing a wide variety of secondary metabolites, including many clinically important antibiotics.

(A) Electromagnetic waves are waves that are created as a result of vibrations between an electric field and a magnetic field.
$\gamma -$ rays and $X-$ rays are part of the electromagnetic spectrum.
$\beta -$ rays consist of high-energy,high-speed electrons or positrons emitted by certain types of radioactive nuclei.
Since $\beta -$ rays are streams of charged particles (matter),they are not electromagnetic waves.
Therefore,the correct option is $A$.

(C) Electromagnetic waves are waves that are created as a result of vibrations between an electric field and a magnetic field.
Cosmic rays,$\gamma$-rays,and $X$-rays are all part of the electromagnetic spectrum.
$\beta$-rays consist of high-energy,high-speed electrons or positrons emitted by certain types of radioactive nuclei.
Since $\beta$-rays are streams of charged particles (matter),they are not electromagnetic waves.
Therefore,the correct option is $C$.

(A) Carbon dioxide $(CO_2)$ is transported in the blood in three main forms:
$1$. As bicarbonate ions $(HCO_3^-)$: This is the primary method,accounting for approximately $70\%$ of total $CO_2$ transport.
$2$. As carbamino-haemoglobin: About $20-25\%$ of $CO_2$ binds to haemoglobin to form carbamino-haemoglobin.
$3$. In dissolved state: About $7\%$ of $CO_2$ is carried in a dissolved state through the blood plasma.
Therefore,the correct answer is as bicarbonate ions.

(B) Meiosis $I$ is divided into four stages: Prophase $I,$ Metaphase $I,$ Anaphase $I,$ and Telophase $I.$
Prophase $I$ is further subdivided into five stages: Leptotene,Zygotene,Pachytene,Diplotene,and Diakinesis.
During the Zygotene stage of Prophase $I,$ homologous chromosomes begin to pair together. This process of association is called synapsis.
Therefore,the pairing of chromosomes starts at the Zygotene stage.

(C) Electromagnetic waves are waves that are created as a result of vibrations between an electric field and a magnetic field. $\gamma$-rays and $X$-rays are part of the electromagnetic spectrum. Cosmic rays consist of high-energy charged particles (mostly protons and atomic nuclei) and are not electromagnetic waves. $\beta$-rays consist of high-energy electrons or positrons,which are also charged particles and not electromagnetic waves. Among the given options,$\beta$-rays are specifically identified as streams of charged particles,making them not electromagnetic in nature. Note: Cosmic rays are also not electromagnetic waves,but in standard physics curricula,$\beta$-rays are the classic example of particle radiation contrasted with electromagnetic radiation.

(A) The reaction sequence is as follows:
$1$. $CH_3CH_2COOH \xrightarrow{PCl_3} CH_3CH_2COCl$ (Propionyl chloride,$I$)
$2$. $CH_3CH_2COCl \xrightarrow[AlCl_3]{C_6H_6} C_6H_5COCH_2CH_3$ (Propiophenone,$II$)
$3$. $C_6H_5COCH_2CH_3 \xrightarrow[base, heat]{NH_2NH_2} C_6H_5CH_2CH_2CH_3$ (Propylbenzene,$III$)
This is a Wolff-Kishner reduction where the ketone group is reduced to a methylene group. The final product is propylbenzene.

(B) In metal carbonyls,the $CO$ bond strength is inversely proportional to the extent of $\pi$-back bonding from the metal to the $CO$ ligand.
As the positive charge on the central metal atom increases,its ability to donate electron density into the $\pi^*$ anti-bonding orbitals of the $CO$ ligand decreases.
The oxidation states of the metals are $V$ in $[V(CO)_6]^-$ is $-1$,$Cr$ in $[Cr(CO)_6]$ is $0$,and $Mn$ in $[Mn(CO)_6]^+$ is $+1$.
Therefore,the extent of back-bonding decreases in the order $[V(CO)_6]^- > [Cr(CO)_6] > [Mn(CO)_6]^+$.
Consequently,the $CO$ bond strength increases in the order $[V(CO)_6]^- < [Cr(CO)_6] < [Mn(CO)_6]^+$.

(B) The $CO$ bond strength in metal carbonyls is inversely proportional to the extent of $M \rightarrow CO$ $\pi$-back bonding.
Greater negative charge on the metal center increases the electron density on the metal,which enhances the $\pi$-back donation to the $CO$ antibonding $\pi^*$ orbitals,thereby weakening the $CO$ bond.
For the isoelectronic series: $[V(CO)_6]^-$ (charge $-1$),$[Cr(CO)_6]$ (charge $0$),and $[Mn(CO)_6]^+$ (charge $+1$).
The extent of back donation follows the order: $[V(CO)_6]^- > [Cr(CO)_6] > [Mn(CO)_6]^+$.
Therefore,the $CO$ bond strength follows the reverse order: $[V(CO)_6]^- < [Cr(CO)_6] < [Mn(CO)_6]^+$.
Thus,the correct option is $B$.

(C) The number of molecules is calculated as: $\text{Number of molecules} = \text{Moles} \times N_A$.
For $A$: $\text{Moles of } CO_2 = \frac{44 \ g}{44 \ g/mol} = 1 \ mol$. Number of molecules = $1 \times N_A$.
For $B$: $\text{Moles of } O_3 = \frac{48 \ g}{48 \ g/mol} = 1 \ mol$. Number of molecules = $1 \times N_A$.
For $C$: $\text{Moles of } H_2 = \frac{8 \ g}{2 \ g/mol} = 4 \ mol$. Number of molecules = $4 \times N_A$.
For $D$: $\text{Moles of } SO_2 = \frac{64 \ g}{64 \ g/mol} = 1 \ mol$. Number of molecules = $1 \times N_A$.
Comparing the values,$8 \ g \ H_2$ contains the maximum number of molecules $(4 \times N_A)$.

(A) According to the $(n+l)$ rule,the higher the value of $(n+l)$,the higher is the energy of the orbital.
If the $(n+l)$ values are the same,the orbital with the lower value of $n$ has lower energy.
Calculating $(n+l)$ values:
$(i) n=4, l=1 \implies n+l = 5$
$(ii) n=4, l=0 \implies n+l = 4$
$(iii) n=3, l=2 \implies n+l = 5$
$(iv) n=3, l=1 \implies n+l = 4$
Comparing the values:
For $(iv)$ and $(ii)$,both have $(n+l) = 4$. Since $n=3 < n=4$,$(iv) < (ii)$.
For $(i)$ and $(iii)$,both have $(n+l) = 5$. Since $n=3 < n=4$,$(iii) < (i)$.
Combining these,the order of increasing energy is $(iv) < (ii) < (iii) < (i)$.

(C) The series limit corresponds to the transition from $n_2 = \infty$ to $n_1$.
The Rydberg formula for the wavenumber is $\bar{v} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the series limit,$n_2 = \infty$,so $\bar{v} = \frac{R}{n_1^2}$.
Given $\bar{v} = 12186.3 \ cm^{-1}$ and $R = 109677.76 \ cm^{-1}$,we have:
$12186.3 = \frac{109677.76}{n_1^2}$
$n_1^2 = \frac{109677.76}{12186.3} \approx 9$
$n_1 = 3$.
Since $n_1 = 3$,the series is the Paschen series.

(C) The Assertion is correct because the spin quantum number $(m_s)$ describes the intrinsic angular momentum of an electron,which can only take two values: $+\frac{1}{2}$ and $-\frac{1}{2}$.
The Reason is incorrect because the $+$ and $-$ signs in spin quantum numbers do not refer to the sign of the wave function; rather,they represent the two opposite orientations of the electron's spin angular momentum.

(C) The atomic number of the last noble gas in the $7^{th}$ period is $118$ (Oganesson).
Since the element has an atomic number of $117$,it is located exactly one position before the noble gas in the $7^{th}$ period.
Elements in the group immediately preceding the noble gases are known as the halogens (Group $17$).
Therefore,the element with atomic number $117$ (Tennessine) belongs to the halogen family.

(C) $LiCl$ is a covalent compound due to the high polarizing power of the small $Li^{+}$ cation,which causes polarization of the large $Cl^{-}$ anion (Fajans' rule).
The electronegativity difference between $Li$ $(1.0)$ and $Cl$ $(3.0)$ is $2.0$,which is relatively large,not small.
Therefore,the Assertion is correct,but the Reason is incorrect.

(B) According to Graham's Law of Diffusion,the rate of diffusion $r \propto \frac{1}{\sqrt{M}}$,where $M$ is the molar mass of the gas.
The molar mass of $NH_3$ is $17 \ g/mol$ and the molar mass of $HCl$ is $36.5 \ g/mol$.
Since $M_{HCl} > M_{NH_3}$,the rate of diffusion of $NH_3$ is faster than that of $HCl$.
Therefore,the $NH_3$ gas travels a greater distance in the same amount of time,and the white ring of $NH_4Cl$ is formed closer to the $HCl$ bottle.

(D) The critical temperature $(T_c)$ of a gas depends on the magnitude of intermolecular forces of attraction.
$CO_2$ has the highest critical temperature among the given options,which is $304.2 \ K$,due to stronger van der Waals forces compared to $H_2$,$He$,and $N_2$.

(A) The van der Waal's constant $a$ represents the magnitude of intermolecular attractive forces in a gas.
Higher values of $a$ indicate stronger attractive forces between gas molecules,which makes it easier for the gas to be liquefied.
In the van der Waal's equation,the pressure correction term is $\frac{an^2}{V^2}$,where $a$ accounts for these attractive forces.
Therefore,both the assertion and the reason are correct,and the reason is the correct explanation of the assertion.

(A) The reaction is: $2H_2O_{2(l)} \to 2H_2O_{(l)} + O_{2(g)}$
$\Delta H = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})$
$\Delta H = [2 \times \Delta H_f(H_2O_{(l)}) + \Delta H_f(O_{2(g)})] - [2 \times \Delta H_f(H_2O_{2(l)})]$
Given: $\Delta H_f(H_2O_{(l)}) = -286 \ kJ/mol$,$\Delta H_f(H_2O_{2(l)}) = -188 \ kJ/mol$,and $\Delta H_f(O_{2(g)}) = 0 \ kJ/mol$ (standard state).
$\Delta H = [2 \times (-286) + 0] - [2 \times (-188)]$
$\Delta H = [-572] - [-376]$
$\Delta H = -572 + 376 = -196 \ kJ/mol$

(C) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by the equation: $\Delta H = \Delta E + \Delta n_g RT$.
For the reaction $2NH_{3(g)} \to N_{2(g)} + 3H_{2(g)}$,the change in the number of gaseous moles is $\Delta n_g = (1 + 3) - 2 = 2$.
Since $\Delta n_g = 2$ (which is positive),$\Delta H = \Delta E + 2RT$,implying $\Delta H > \Delta E$. Thus,the Assertion is correct.
The Reason states that enthalpy change is always greater than internal energy change,which is false because if $\Delta n_g$ is negative or zero,$\Delta H$ can be less than or equal to $\Delta E$.

(C) For a salt of the type $M(OH)_2$,the solubility product is given by $K_{sp} = [M^{2+}][OH^-]^2 = (S)(2S)^2 = 4S^3$.
Given $K_{sp} = 3.2 \times 10^{-11}$,we have $4S^3 = 3.2 \times 10^{-11}$,which implies $S^3 = 0.8 \times 10^{-11} = 8 \times 10^{-12}$.
Thus,$S = 2 \times 10^{-4} \ M$.
The concentration of hydroxide ions is $[OH^-] = 2S = 2 \times (2 \times 10^{-4}) = 4 \times 10^{-4} \ M$.
$pOH = -\log[OH^-] = -\log(4 \times 10^{-4}) = 4 - \log 4 = 4 - 0.60 = 3.40$.
Since $pH + pOH = 14$,we have $pH = 14 - 3.40 = 10.60$.

(C) According to Le Chatelier's principle,an increase in temperature favors the endothermic direction of a reaction.
Reaction $(a)$ is the water-gas shift reaction,which is exothermic $(\Delta H < 0)$.
Reaction $(b)$ is the contact process step,which is exothermic $(\Delta H < 0)$.
Reaction $(c)$ represents the thermal decomposition of water,which is highly endothermic $(\Delta H > 0)$.
Reaction $(d)$ is the Deacon process,which is exothermic $(\Delta H < 0)$.
Therefore,only reaction $(c)$ will shift to the right side upon increasing the temperature.

(A) To determine the number of electrons transferred,we calculate the change in the oxidation number $(O.N.)$ of the central metal atom.
For option $A$: In $MnO_4^-$,the $O.N.$ of $Mn$ is $+7$. In $Mn^{2+}$,the $O.N.$ is $+2$. The change is $|7 - 2| = 5$ electrons.
For option $B$: In $CrO_4^{2-}$,the $O.N.$ of $Cr$ is $+6$. In $Cr^{3+}$,the $O.N.$ is $+3$. The change is $3$ electrons.
For option $C$: In $MnO_4^{2-}$,the $O.N.$ of $Mn$ is $+6$. In $MnO_2$,the $O.N.$ of $Mn$ is $+4$. The change is $2$ electrons.
For option $D$: In $Cr_2O_7^{2-}$,the $O.N.$ of $Cr$ is $+6$. In $Cr^{3+}$,the $O.N.$ is $+3$. Since there are two $Cr$ atoms,the total change is $2 \times (6 - 3) = 6$ electrons.
Thus,the reaction involving the transfer of $5$ electrons is $MnO_4^- \to Mn^{2+}$.

(D) The Assertion is incorrect because the $HOF$ bond angle $(101^\circ)$ is smaller than the $HOCl$ bond angle $(103^\circ)$.
This occurs because fluorine is more electronegative than oxygen,pulling the bonding electron pair away from the oxygen atom,which reduces the repulsion between the bonding pairs.
In $HClO$,oxygen is more electronegative than chlorine,so the bonding electron pair is closer to the oxygen atom,leading to greater repulsion and a larger bond angle.
The Reason is also incorrect because while oxygen is more electronegative than most halogens,it is less electronegative than fluorine $(O = 3.44, F = 3.98)$.

(A) The mixture of $Na_2O_2$ and $HCl$ is commercially known as oxone.
It is primarily used for the bleaching of delicate fibres.

(C) $LiCl$ is a covalent compound due to the high polarizing power of the small $Li^+$ cation,which causes polarization of the electron cloud of the large $Cl^-$ anion (Fajans' rule).
The electronegativity difference between $Li$ $(1.0)$ and $Cl$ $(3.0)$ is $2.0$,which is significant,not small.
Therefore,the Assertion is correct,but the Reason is incorrect.

(C) The Assertion is correct because $K$,$Rb$,and $Cs$ are large alkali metals that can stabilize the superoxide ion $(O_2^-)$ in their superoxides ($KO_2$,$RbO_2$,$CsO_2$).
The Reason is incorrect because the ionic radii of alkali metals increase down the group as the number of shells increases. Therefore,the correct trend is $K^{+} < Rb^{+} < Cs^{+}$.

(A) Silica $(SiO_2)$ on heating with carbon $(C)$ at an elevated temperature produces carborundum,which is silicon carbide $(SiC)$.
$SiO_2 + 3C \xrightarrow{\Delta} SiC + 2CO$
Carborundum is an extremely hard substance used as an abrasive.

(A) The thermal decomposition of ammonium nitrate $(NH_4NO_3)$ is given by the following reaction:
$NH_4NO_3(s) \xrightarrow{\Delta} N_2O(g) + 2H_2O(l)$
Here,$(A)$ is $NH_4NO_3$,$(B)$ is $N_2O$ (a colourless neutral gas),and $(C)$ is $H_2O$ (a liquid at room temperature which is neutral to litmus).
Thus,the correct option is $(A)$.

(C) The Assertion is correct because $Pb^{4+}$ is a strong oxidizing agent as it tends to reduce to $Pb^{2+}$,which is more stable due to the inert pair effect.
The Reason is incorrect because the inert pair effect causes the lower oxidation state (e.g.,$+2$) to be more stable for heavier elements in group $14$,not the higher oxidation state $(+4)$.

(C) The assertion is correct because,due to the 'inert pair effect',the stability of the $+4$ oxidation state decreases down the group $14$ $(C > Si > Ge > Sn > Pb)$.
Consequently,$Pb^{4+}$ is highly unstable and acts as a strong oxidizing agent to reduce itself to the more stable $+2$ state.
The reason is incorrect because the higher oxidation states (like $+4$) become less stable,not more stable,for the heavier members of the group due to the 'inert pair effect'.

(C) In $(I)$ and $(III)$,the positive charge is delocalized through resonance,and the carbocation character is stabilized by the presence of an adjacent alkyl group.
Specifically,$(I)$ is a secondary $(2^\circ)$ allylic carbocation,and $(III)$ is a primary $(1^\circ)$ allylic carbocation.
However,in $(I)$,the positive charge is on a secondary carbon,while in $(III)$,it is on a primary carbon.
In $(II)$,the positive charge is on a primary carbon and is allylic.
Comparing the resonance structures and the inductive effects,$(I)$ and $(III)$ show greater stability due to the nature of the allylic system and the electron-donating effects of the methyl group.
Thus,the correct order is $(I) \approx (III) > (II)$.

(B) Diphenylmethane consists of two phenyl rings attached to a central methylene group $(-CH_2-)$.
Due to the symmetry of the molecule,the two phenyl rings are equivalent.
Within each phenyl ring,there are three distinct types of hydrogen atoms: ortho,meta,and para positions relative to the $-CH_2-$ group.
Additionally,there are the hydrogen atoms on the central methylene group.
Thus,there are $4$ distinct positions for substitution:
$1$. Ortho position on the ring.
$2$. Meta position on the ring.
$3$. Para position on the ring.
$4$. The central methylene carbon.
Therefore,there are $4$ possible structural isomers.

(A) The compound $CHCl=CHCHOHCOOH$ exhibits:
$1$. Geometric isomerism: Due to the presence of the $C=C$ double bond,it can exist as $cis$ and $trans$ isomers.
$2$. Optical isomerism: Due to the presence of a chiral carbon atom $(CHOH)$,it can exist as $d$ and $l$ enantiomers.
$3$. Position isomerism: The $Cl$ atom,$OH$ group,or the double bond can occupy different positions in the carbon chain.
$4$. Functional isomerism: Different functional groups (e.g.,carboxylic acid vs. ester) can be formed for the same molecular formula $C_4H_5O_3Cl$.
Therefore,it exhibits geometric,optical,position,and functional isomerism.

(C) The Assertion is true because $NH_4Cl$ is added to $NH_4OH$ to provide a common ion $(NH_4^+)$,which suppresses the dissociation of $NH_4OH$ due to the common ion effect.
This reduction in the concentration of $OH^-$ ions ensures that only the hydroxides of the third group cations (like $Fe^{3+}$,$Al^{3+}$,$Cr^{3+}$) are precipitated,preventing the precipitation of hydroxides of subsequent groups.
The Reason is false because the purpose is not to convert ions into chlorides,but to control the $OH^-$ concentration for selective precipitation.

(C) When non-terminal alkynes are heated with $NaNH_2$ in an inert solvent,the triple bond undergoes isomerization to form the more stable terminal alkyne.
The reaction proceeds as follows:
$CH_3-C\equiv C-CH_3 \xrightarrow[heat]{NaNH_2} CH_3-CH_2-C\equiv CH$
Thus,the product $P$ is $CH_3-CH_2-C\equiv CH$.

(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 Tropylium cation $(C_7H_7^+)$ is a seven-membered ring with $6\pi$ electrons ($n=1$ in $H$ückel's rule $(4n+2)\pi$).
It is planar and fully conjugated,making it aromatic.
Therefore,the Assertion is correct.
However,the Reason is incorrect because aromaticity is determined by multiple factors,including cyclic structure,planarity,complete conjugation,and the presence of $(4n+2)\pi$ electrons,not just planarity alone.

(B) During $Zygotene$,a substage of $Prophase-I$ of meiosis $I$,chromosomes start pairing together,a process known as $synapsis$.
Such paired chromosomes are referred to as homologous chromosomes.
$A$ complex structure,known as the $synaptonemal$ $complex$,is formed by a pair of synapsed homologous chromosomes,which is called a $bivalent$ or a $tetrad$.

(A) $CO_2$ is transported from the respiratory tissues to the lungs by the blood in $3$ ways:
$(i)$ In dissolved state or as a physical solution: $A$ very small amount is physically dissolved in plasma ($7\%$ i.e.,$0.3\, ml$ of $CO_2$ per $100\, ml$ of blood).
$(ii)$ Bicarbonate ions: About $70\%$ (i.e.,$2.5\, ml$ per $100\, ml$ of blood) of $CO_2$ diffuses into plasma and then into $RBCs$,where it combines with $H_2O$ in the presence of the enzyme carbonic anhydrase to form carbonic acid,which then dissociates into hydrogen ions and bicarbonate ions.
$(iii)$ Carbaminohaemoglobin: $23\%$ (i.e.,$1\, ml$ of $CO_2$ per $100\, ml$ of blood) combines with haemoglobin to form an unstable compound called carbaminohaemoglobin.

(B) The rate at which consumers synthesize new organic matter (biomass) from the food they consume is known as secondary productivity.
Since a rabbit is a primary consumer (herbivore) in a grassland ecosystem,the organic matter it forms is categorized as secondary productivity.

(D) Photochemical smog is formed in warm and dry climates due to the action of sunlight on nitrogen oxides and hydrocarbons.
$NO_2$ and $O_3$ are strong oxidizing agents.
They react with unburnt hydrocarbons in the atmosphere to produce secondary pollutants such as formaldehyde,acrolein,and $PAN$ (peroxyacetyl nitrate).
Therefore,$O_3$ and $PAN$ are considered the secondary pollutants or secondary precursors formed during the process.

(B) The figure represents adipose tissue. Adipose tissue is a type of loose connective tissue located mainly beneath the skin. The cells of this tissue are specialized to store fats. The excess of nutrients which are not used immediately are converted into fats and are stored in this tissue. It also acts as a heat insulator.

(A) Biofertilizers are organisms that enrich the nutrient quality of the soil.
$Azolla$ is a water fern that has a symbiotic association with the nitrogen-fixing cyanobacterium $Anabaena$ $azollae$.
$BGA$ (Blue-Green Algae) such as $Nostoc$ and $Anabaena$ are also well-known biofertilizers that fix atmospheric nitrogen.
Therefore,the pair $Azolla$ and $BGA$ represents biofertilizers.

(D) The electronic configuration of $Cu$ $(Z=29)$ is $[Ar] 3d^{10} 4s^1$.
For $Cu^+$ $(Cu^+)$,the configuration is $[Ar] 3d^{10}$,which has no unpaired electrons.
For $Cu^{2+}$ $(Cu^{2+})$,the configuration is $[Ar] 3d^9$,which has one unpaired electron.
Therefore,the Assertion is incorrect because $Cu^+$ has no unpaired electrons,while $Cu^{2+}$ has one.
Regarding the Reason,$Cu^+$ is indeed colourless due to the absence of unpaired electrons ($d^{10}$ configuration),and $Cu^{2+}$ is blue due to $d-d$ transitions facilitated by the presence of an unpaired electron.
Since the Assertion is incorrect but the Reason is correct,the correct option is $D$.

(B) Carbon cannot be used to produce magnesium by chemical reduction of $MgO$ because magnesium reacts with carbon to form carbides at high temperatures.
$2 Mg + 3 C \xrightarrow{2000^{\circ} C} Mg_2 C_3$

(C) Friedel-Crafts alkylation of benzene with $n$-butyl chloride $(CH_3CH_2CH_2CH_2Cl)$ in the presence of $AlCl_3$ leads to rearrangement of the carbocation. The primary carbocation rearranges to a more stable secondary carbocation,resulting in $2-$phenylbutane as the major product.
Similarly,$but-1-ene + HF$ and $butan-2-ol + H_2SO_4$ both generate the $sec-butyl$ carbocation,which yields $2-$phenylbutane.
However,the reaction of $Butanoyl chloride + AlCl_3$ is a Friedel-Crafts acylation,which produces $1-$phenylbutan$-1-$one. Subsequent Clemmensen reduction $(Zn(Hg)/HCl)$ of this ketone yields $n$-butylbenzene ($1-$phenylbutane),not $2-$phenylbutane.

(B) In a zinc blende $(ZnS)$ structure,the anions $(I^{-})$ form a face-centered cubic $(fcc)$ lattice.
Each unit cell contains $4$ anions $(I^{-})$.
The number of tetrahedral voids in an $fcc$ lattice is $2 \times$ (number of atoms) = $2 \times 4 = 8$.
Since the formula of the compound is $AgI$,the ratio of $Ag^{+}$ to $I^{-}$ is $1:1$.
Therefore,there are $4$ $Ag^{+}$ ions in the unit cell.
The fraction of tetrahedral voids occupied by $Ag^{+}$ ions is $\frac{4}{8} = 0.5$,which is $50\%$.

(C) Ferromagnetic and ferrimagnetic substances possess ordered magnetic domains at room temperature.
On heating,the thermal energy increases,which leads to the randomisation of the magnetic moments (spins).
This randomisation results in the loss of the ordered magnetic structure,causing the substance to become paramagnetic.
The reason provided is incorrect because the electrons do not change their intrinsic spin; rather,the alignment of the spins becomes random due to thermal agitation.

(A) The depression in freezing point $\Delta T_f$ is directly proportional to the van't Hoff factor $i$ for solutions of the same molality: $\Delta T_f = i \times K_f \times m$.
Since the molality $m$ is the same for all,the solution with the highest value of $i$ will have the greatest depression in freezing point and thus the lowest freezing point.
$1$. $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$ $(i = 5)$
$2$. $C_6H_{12}O_6$ (glucose) is a non-electrolyte $(i = 1)$
$3$. $KCl \rightarrow K^{+} + Cl^{-}$ $(i = 2)$
$4$. $C_{12}H_{22}O_{11}$ (sucrose) is a non-electrolyte $(i = 1)$
Since $Al_2(SO_4)_3$ has the highest van't Hoff factor $(i = 5)$,it will show the maximum depression in freezing point,resulting in the lowest freezing point.

(A) Two solutions are isotonic if they have the same osmotic pressure,which implies they have the same molar concentration at the same temperature.
Molar concentration of urea $= \frac{10 \ g \ L^{-1}}{60 \ g \ mol^{-1}} = \frac{1}{6} \ mol \ L^{-1}$.
$A$ $5 \%$ solution of a non-volatile solute means $5 \ g$ of solute in $100 \ mL$ of solution,which is $50 \ g$ of solute in $1 \ L$ $(dm^3)$ of solution.
Let the molecular mass of the non-volatile solute be $M$.
Molar concentration of the non-volatile solute $= \frac{50 \ g \ L^{-1}}{M \ g \ mol^{-1}} = \frac{50}{M} \ mol \ L^{-1}$.
Since the solutions are isotonic,$\frac{1}{6} = \frac{50}{M}$.
Therefore,$M = 50 \times 6 = 300 \ g \ mol^{-1}$.

(D) Specific conductance (conductivity,$\kappa$) is defined as the conductance of $1 \ cm^3$ of the solution.
On increasing dilution,the number of ions per unit volume decreases,which leads to a decrease in specific conductance.
Therefore,the Assertion is incorrect.
Regarding the Reason,while the degree of ionisation of a weak electrolyte increases with dilution,the number of ions per unit volume decreases,not increases.
Thus,both the Assertion and the Reason are incorrect.

(A) During the electrolysis of $CuSO_{4(aq)}$ with copper electrodes,the copper anode undergoes oxidation: $Cu_{(s)} \to Cu^{2+}_{(aq)} + 2e^-$.
At the cathode,the copper ions from the solution undergo reduction: $Cu^{2+}_{(aq)} + 2e^- \to Cu_{(s)}$.
Thus,copper dissolves at the anode and deposits at the cathode.
Since oxidation occurs at the anode and reduction at the cathode,the Reason correctly explains the Assertion.

(B) For a first order reaction,the rate is given by $Rate_1 = k_1 [A]_0$,where $k_1 = \frac{0.693}{t_{1/2}}$. Thus,$Rate_1 = \frac{0.693 [A]_0}{t_{1/2}}$.
For a zero order reaction,the rate is given by $Rate_0 = k_0$,where $k_0 = \frac{[A]_0}{2 t_{1/2}}$.
Given that the half-lives $(t_{1/2})$ are the same for both reactions,the ratio of the initial rates is:
$\frac{Rate_1}{Rate_0} = \frac{\frac{0.693 [A]_0}{t_{1/2}}}{\frac{[A]_0}{2 t_{1/2}}} = 0.693 \times 2 = 1.386$.

(A) Since the sol particles migrate towards the cathode,they are positively charged.
According to the Hardy-Schulze law,the coagulating power of an ion is directly proportional to its valency.
For a positively charged sol,the effective coagulating ions are the anions: $Cl^-$,$SO_4^{2-}$,and $PO_4^{3-}$.
The coagulating power order is $PO_4^{3-} > SO_4^{2-} > Cl^-$.
Since the coagulating value is inversely proportional to the coagulating power,the order of coagulating values is $NaCl > Na_2SO_4 > Na_3PO_4$.

(C) The ability of halogens to act as oxidizing agents decreases down the group $(F_2 > Cl_2 > Br_2 > I_2)$.
$Br_2$ can only oxidize the iodide ion $(I^-)$ to iodine $(I_2)$ because $I^-$ is a stronger reducing agent than $Br^-$,$Cl^-$,and $F^-$.
$Br_2$ cannot oxidize $Cl^-$ or $F^-$ because $Br_2$ is a weaker oxidizing agent than $Cl_2$ and $F_2$.
Therefore,the reaction is: $2NaI + Br_2 \to 2NaBr + I_2$.

(D) In the given complex,there are two $en$ ligands and one $C_2O_4$ ligand. Both are bidentate ligands.
Coordination number = $(2 \times 2) + (1 \times 2) = 4 + 2 = 6$.
Let the oxidation state of $E$ be $x$.
The complex is $[E(en)_2(C_2O_4)]NO_2$. Since $NO_2$ is a counter ion with a charge of $-1$,the complex cation $[E(en)_2(C_2O_4)]^+$ has a charge of $+1$.
$x + 2(0) + 1(-2) = +1$
$x - 2 = +1$
$x = +3$.
Therefore,the coordination number is $6$ and the oxidation state is $+3$. Thus,option $(d)$ is correct.

(B) The $CO$ bond strength is inversely proportional to the extent of back-bonding from the metal to the $CO$ ligand.
Greater negative charge on the metal center increases the electron density,which enhances the back-donation into the $\pi^*$ antibonding orbitals of $CO$.
This weakens the $C-O$ bond.
The order of negative charge is $[V(CO)_6]^- > [Cr(CO)_6] > [Mn(CO)_6]^+$.
Therefore,the extent of back-bonding follows the order: $[V(CO)_6]^- > [Cr(CO)_6] > [Mn(CO)_6]^+$.
Consequently,the $CO$ bond strength increases in the order: $[V(CO)_6]^- < [Cr(CO)_6] < [Mn(CO)_6]^+$.
Thus,the correct option is $B$.

(D) To determine if a complex is paramagnetic,we check for the presence of unpaired electrons in the central metal ion.
$1$. In $[Fe(CN)_6]^{4-}$,$Fe$ is in $+2$ oxidation state $(3d^6)$. $CN^-$ is a strong field ligand,causing pairing of electrons,resulting in $0$ unpaired electrons (diamagnetic).
$2$. In $[Ni(CO)_4]$,$Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing,resulting in $0$ unpaired electrons (diamagnetic).
$3$. In $[Ni(CN)_4]^{2-}$,$Ni$ is in $+2$ oxidation state $(3d^8)$. $CN^-$ is a strong field ligand,causing pairing,resulting in $0$ unpaired electrons (diamagnetic).
$4$. In $[CoF_6]^{3-}$,$Co$ is in $+3$ oxidation state $(3d^6)$. $F^-$ is a weak field ligand,so no pairing occurs. The configuration is $t_{2g}^4 e_g^2$,which has $4$ unpaired electrons,making it paramagnetic.

(A) The $EDTA^{4-}$ ion is a hexadentate ligand,meaning it has $6$ donor atoms (two nitrogen atoms and four oxygen atoms).
These $6$ donor atoms coordinate simultaneously to the central metal ion,forming a stable octahedral complex.
Therefore,the Reason correctly explains why the Assertion is true.

(B) $1$. $CH_3CH_2CH_2Br$ reacts with aqueous $NaOH$ (nucleophilic substitution) to form propan-$1$-ol $(X)$: $CH_3CH_2CH_2Br + NaOH(aq) \rightarrow CH_3CH_2CH_2OH + NaBr$.
$2$. Propan-$1$-ol $(X)$ undergoes dehydration with $Al_2O_3$ at high temperature to form propene $(Y)$: $CH_3CH_2CH_2OH \xrightarrow{Al_2O_3, \Delta} CH_3CH=CH_2 + H_2O$.
$3$. Propene $(Y)$ reacts with $Cl_2/H_2O$ (hypochlorous acid,$HOCl$) to form chlorohydrin $(Z)$: $CH_3CH=CH_2 + Cl_2 + H_2O \rightarrow CH_3CH(OH)CH_2Cl + HCl$. The hydroxyl group attaches to the more substituted carbon (Markovnikov addition).

(A) $1$. The molecular formula $C_4H_{10}O$ corresponds to an alcohol.
$2$. Lucas reagent $(conc. HCl + ZnCl_2)$ reacts with tertiary alcohols instantaneously at room temperature to form alkyl chlorides.
$3$. Compound $B$ on heating with alcoholic $KOH$ (dehydrohalogenation) gives isobutene $(CH_3-C(CH_3)=CH_2)$.
$4$. The structure of isobutene indicates that the precursor alkyl chloride $(B)$ must be $2-$chloro$-2-$methylpropane.
$5$. Therefore,compound $A$ must be $2-$methylpropan$-2-$ol ($tert-$butyl alcohol).
$6$. The reaction is: $(CH_3)_3COH + HCl \rightarrow (CH_3)_3CCl + H_2O$.
$7$. Thus,$A$ is $2-$methylpropan$-2-$ol and $B$ is $2-$chloro$-2-$methylpropane.

(D) The dehydration of alcohols to form ethers using a trace of sulphuric acid follows an $S_N2$ mechanism.
Primary alcohols are the least sterically hindered and therefore undergo $S_N2$ substitution most readily to form ethers.
Secondary and tertiary alcohols are more prone to elimination reactions (forming alkenes) due to steric hindrance and the stability of the carbocation intermediate.
Among the given options,$1-$Pentanol is a primary alcohol,while the others are secondary or tertiary alcohols.
Therefore,$1-$Pentanol gives the best yield of dialkyl ether.

(C) The Kolbe reaction involves the reaction of sodium phenoxide with $CO_2$ under pressure followed by acidification to form salicylic acid. This reaction is specific to phenols because the phenoxide ion is resonance-stabilized,making the ring electron-rich and susceptible to electrophilic attack by $CO_2$.
Ethanol does not undergo this reaction because it does not form a stable phenoxide-like intermediate. Thus,the Assertion is true.
Regarding the Reason,the phenoxide ion $(C_6H_5O^-)$ is significantly less basic than the ethoxide ion $(C_2H_5O^-)$. This is because the negative charge on the phenoxide ion is delocalized into the benzene ring via resonance,whereas the ethoxide ion has no such stabilization and the alkyl group $(C_2H_5-)$ is electron-donating,which increases the electron density on the oxygen atom.
Therefore,the Reason is incorrect.

(A) $CH_3COOH \xrightarrow[\Delta]{NH_3} CH_3CONH_2 (A)$ (Ethanamide)
$CH_3CONH_2 \xrightarrow{Br_2/NaOH} CH_3NH_2 (B)$ (Methanamine) via Hofmann bromamide degradation.
$CH_3NH_2 \xrightarrow{NaNO_2/dil. HCl} CH_3OH (C)$ (Methanol) via diazotization followed by hydrolysis.

(A) The reaction sequence is as follows:
$1$. $CH_3-CH_2-COOH + PCl_3 \rightarrow CH_3-CH_2-COCl$ (Compound $I$,Propanoyl chloride).
$2$. $CH_3-CH_2-COCl + C_6H_6 \xrightarrow{AlCl_3} C_6H_5-CO-CH_2-CH_3$ (Compound $II$,Propiophenone,which is also known as ethyl phenyl ketone).
$3$. $C_6H_5-CO-CH_2-CH_3 \xrightarrow{NH_2-NH_2/\text{base/heat}} C_6H_5-CH_2-CH_2-CH_3$ (Compound $III$,Propylbenzene).
This is a Friedel-Crafts acylation followed by a Wolff-Kishner reduction.

(B) The reaction between benzaldehyde $(C_6H_5CHO)$ and acetaldehyde $(CH_3CHO)$ in the presence of dilute alkali is a Claisen-Schmidt condensation.
Acetaldehyde acts as the nucleophile by forming an enolate ion,which attacks the carbonyl carbon of benzaldehyde.
The initial aldol product formed is $C_6H_5-CH(OH)-CH_2-CHO$ ($3$-hydroxy$-3-$phenylpropanal).
This product undergoes spontaneous dehydration in the presence of base and heat to form the more stable conjugated product,cinnamaldehyde $(C_6H_5-CH=CH-CHO)$.

(D) The Assertion is incorrect because benzaldehyde is less reactive than ethanol towards nucleophilic attack. This is due to the steric hindrance of the bulky phenyl group and the resonance stabilization of the carbonyl carbon by the phenyl ring.
The Reason is also incorrect because the $+R$ effect of the phenyl group is electron-donating,which increases the electron density on the carbonyl carbon,thereby reducing its electrophilicity.

(A) The reaction of acetamide $(CH_3CONH_2)$ with $Br_2$ in the presence of a base like $CH_3ONa$ is a variation of the Hofmann bromamide degradation.
In this specific medium,the intermediate formed is methyl isocyanate $(CH_3-N=C=O)$.
This methyl isocyanate then undergoes nucleophilic attack by the methanol $(CH_3OH)$ present in the medium to form methyl $N$-methylcarbamate $(CH_3NHCOOCH_3)$.
Therefore,both the Assertion and the Reason are correct,and the Reason correctly explains the mechanism of the reaction.

(B) Osazone formation involves the reaction of phenylhydrazine with the first two carbon atoms of a reducing sugar.
The necessary structural unit is an $\alpha$-hydroxy carbonyl group,which is represented by the structure $-CHOH-CO-$ or $-CHOH-CHOH-$.
Among the given options,the structural unit $-CHOH-CHOH-$ (or similar $\alpha$-hydroxy carbonyl precursors) is essential for the reaction to proceed.
Option $(b)$ represents the $\alpha$-hydroxy carbonyl moiety found in sugars.

(A) Glucose and fructose differ only in the configuration at $C_1$ and $C_2$ atoms.
During the reaction with excess phenylhydrazine,the $C_1$ and $C_2$ atoms are involved in the formation of the osazone structure.
Since the stereochemistry at $C_3, C_4, C_5,$ and $C_6$ remains identical in both glucose and fructose,they yield the same osazone product.
Therefore,the Reason correctly explains the Assertion.

(A) Nylon $-66$ is a polyamide formed by the condensation polymerization of adipic acid $(HOOC(CH_2)_4COOH)$ and hexamethylenediamine $(H_2N(CH_2)_6NH_2)$.
Therefore,the correct set of compounds is $HOOC(CH_2)_4COOH + H_2N(CH_2)_6NH_2$.

(D) Chloramphenicol is a broad-spectrum antibiotic that functions by inhibiting bacterial protein synthesis. It is effective against a wide range of Gram-positive bacteria,Gram-negative bacteria,and anaerobes.

(A) Antioxidants are chemical substances that retard the action of oxygen on food,thereby preventing oxidation and helping in preservation. $BHA$ (Butylated hydroxy anisole) is a well-known antioxidant added to butter to prevent rancidity caused by oxidation. Therefore,the addition of $BHA$ increases the storage life of butter,and the reason correctly explains the assertion.