AIIMS 2013 Chemistry Question Paper with Answer and Solution

(D) The extension $l$ in a wire is given by the formula $l = \frac{FL}{AY}$,where $F$ is the applied force,$L$ is the length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.
Since the material is the same,$Y$ is constant. Given that $F$ and $L$ are also constant,we have $l \propto \frac{1}{A}$.
Since $A = \pi r^2$ (where $r$ is the radius),$l \propto \frac{1}{r^2}$.
Given that the diameter of the second wire is $2$ times the diameter of the first wire,the radius $r_2 = 2r_1$.
Therefore,the ratio of extensions is $\frac{l_1}{l_2} = \left( \frac{r_2}{r_1} \right)^2 = (2)^2 = 4$.
Thus,the ratio $l_1 : l_2 = 4 : 1$.

(A) For the Lyman series,the transition occurs to the $n_1 = 1$ orbit from $n_2 = 2, 3, 4, \dots$ levels.
Using the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$,where $R \approx 1.097 \times 10^7 \text{ m}^{-1}$.
For the longest wavelength (minimum energy transition),$n_1 = 1$ and $n_2 = 2$:
$\frac{1}{\lambda_{\max}} = R \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = R \left[ 1 - \frac{1}{4} \right] = \frac{3R}{4}$.
$\lambda_{\max} = \frac{4}{3R} \approx \frac{4}{3 \times 1.097 \times 10^7} \approx 1216 \ \mathring{A}$ (approx $1213 \ \mathring{A}$).
For the shortest wavelength (maximum energy transition),$n_1 = 1$ and $n_2 = \infty$:
$\frac{1}{\lambda_{\min}} = R \left[ \frac{1}{1^2} - \frac{1}{\infty^2} \right] = R$.
$\lambda_{\min} = \frac{1}{R} \approx \frac{1}{1.097 \times 10^7} \approx 912 \ \mathring{A}$ (approx $910 \ \mathring{A}$).
Thus,the shortest wavelength is $910 \ \mathring{A}$ and the longest is $1213 \ \mathring{A}$.

(C) The ratio of radioactive isotope ${C^{14}}$ to stable isotope ${C^{12}}$ decreases over time due to radioactive decay.
Given that the current ratio is $\frac{1}{4}$ of the initial ratio,we use the radioactive decay formula: $N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}$.
Here,$\frac{N(t)}{N_0} = \frac{1}{4}$ and $T_{1/2} = 5700 \, years$.
Substituting the values: $\frac{1}{4} = \left( \frac{1}{2} \right)^{\frac{t}{5700}}$.
Since $\frac{1}{4} = \left( \frac{1}{2} \right)^2$,we equate the exponents: $2 = \frac{t}{5700}$.
Therefore,$t = 2 \times 5700 = 11400 \, years$.

(B) The family $Fabaceae$ (formerly $Papilionoideae$) is characterized by the following features:
$1$. Root system: Tap root with root nodules.
$2$. Leaves: Alternate,pinnately compound or simple,leaf base pulvinate,stipulate.
$3$. Inflorescence: Racemose.
$4$. Flower: Zygomorphic,bisexual,papilionaceous corolla (vexillary aestivation).
$5$. Androecium: $10$ stamens,diadelphous $(9+1)$,dithecous,introrse,basifixed.
$6$. Gynoecium: Monocarpellary,ovary superior,unilocular with many ovules,style single.
Option $B$ is incorrect because $Fabaceae$ flowers are zygomorphic (not actinomorphic),show vexillary aestivation (not twisted),and are polypetalous (not gamopetalous).

(D) The acceleration due to gravity $g$ at the surface of a planet is given by $g = \frac{GM}{R^2}$.
Since the mass $M$ of a planet can be expressed in terms of its density $\rho$ and radius $R$ as $M = \rho \times V = \rho \times \frac{4}{3} \pi R^3$,we substitute this into the formula for $g$:
$g = \frac{G (\rho \cdot \frac{4}{3} \pi R^3)}{R^2} = \frac{4}{3} \pi G \rho R$.
This shows that $g \propto \rho R$.
Therefore,the ratio of the accelerations due to gravity for the two planets is $\frac{g_1}{g_2} = \frac{\rho_1 R_1}{\rho_2 R_2}$.
Thus,$g_1 : g_2 = R_1 \rho_1 : R_2 \rho_2$.

(B) The total translational kinetic energy $(K)$ of an ideal gas is given by the formula $K = \frac{3}{2} n R T$.
From the ideal gas equation,we know that $PV = n R T$.
Substituting $PV$ for $n R T$ in the kinetic energy formula,we get $K = \frac{3}{2} PV = 1.5 PV$.
Thus,the Assertion is correct.
The Reason states that molecules collide and their velocities change,which is a fundamental postulate of the Kinetic Theory of Gases.
However,the collision of molecules is not the reason why the kinetic energy is equal to $1.5 PV$; that relationship is derived from the definition of temperature and the ideal gas law.
Therefore,both statements are correct,but the Reason is not the correct explanation of the Assertion.

(B) When a charged particle moves in a magnetic field with velocity $\vec{v}$ perpendicular to the magnetic field $\vec{B}$,the magnetic force acting on it is given by $\vec{F} = q(\vec{v} \times \vec{B})$.
Since the force $\vec{F}$ is always perpendicular to the velocity $\vec{v}$,it does no work on the particle $(W = \vec{F} \cdot \vec{d} = 0)$.
According to the work-energy theorem,the change in kinetic energy is zero,which implies that the speed of the particle remains constant.
However,because the direction of the force changes as the particle moves,both the velocity vector and the acceleration vector change continuously.

(D) According to Snell's law at each interface for parallel surfaces:
$\mu_1 \sin i_1 = \mu_2 \sin i_2 = \mu_3 \sin i_3 = \mu_4 \sin i_4$
Since the emergent ray $CD$ is parallel to the incident ray $AB$,the angle of incidence $i_1$ must be equal to the angle of emergence $i_4$.
Therefore,$\mu_1 \sin i_1 = \mu_4 \sin i_4$.
Since $i_1 = i_4$,we have $\mu_1 = \mu_4$.

(B) For $100\%$ modulation,the modulation index $m_a = 1$.
The total power radiated in an $AM$ wave is given by $P_t = P_c(1 + \frac{m_a^2}{2})$,where $P_c$ is the carrier power.
The useful power is the power contained in the sidebands,given by $P_{sb} = P_c(\frac{m_a^2}{2})$.
The ratio of useful power to total power is $\frac{P_{sb}}{P_t} = \frac{m_a^2/2}{1 + m_a^2/2} = \frac{m_a^2}{2 + m_a^2}$.
Substituting $m_a = 1$:
$\text{Ratio} = \frac{1^2}{2 + 1^2} = \frac{1}{3}$.
Thus,the useful part of the total power radiated is $1/3$ of the total power.

(C) From the balanced chemical equation,$2 \ \text{moles}$ of $KMnO_4$ react with $5 \ \text{moles}$ of $H_2C_2O_4$.
Using the concept of milliequivalents (Meq):
$\text{Meq of } KMnO_4 = \text{Molarity} \times n\text{-factor} \times \text{Volume (mL)}$
For $KMnO_4$,the $n$-factor is $5$ $(Mn^{+7} \to Mn^{+2})$.
$\text{Meq of } KMnO_4 = 0.1 \times 5 \times 20 = 10 \ \text{Meq}$.
For $H_2C_2O_4$,the $n$-factor is $2$ $(C_2^{+3} \to 2C^{+4} + 2e^-)$.
We need $10 \ \text{Meq}$ of $H_2C_2O_4$.
$\text{Meq} = \text{Molarity} \times 2 \times \text{Volume (mL)} = 10$.
Checking option $(C)$: $50 \ mL$ of $0.1 \ M \ H_2C_2O_4$:
$\text{Meq} = 0.1 \times 2 \times 50 = 10 \ \text{Meq}$.
Thus,$20 \ mL$ of $0.1 \ M \ KMnO_4$ is equivalent to $50 \ mL$ of $0.1 \ M \ H_2C_2O_4$.

(A) The relationship between volume strength and normality $(N)$ of $H_2O_2$ is given by the formula: $N = \frac{\text{Volume strength}}{5.6}$.
Given,volume strength = $10 \, V$.
Therefore,$N = \frac{10}{5.6} \approx 1.78 \, N$.

(C) The normality of a solution is given by the formula: $\text{Normality} = \text{Molarity} \times \text{n-factor}$.
For $H_3PO_3$ (phosphorous acid),the structure contains two $P-OH$ bonds,making it a dibasic acid. Thus,its n-factor is $2$.
Normality $= 0.3 \ M \times 2 = 0.6 \ N$. Therefore,the Assertion is correct.
The equivalent weight of an acid is defined as $\frac{\text{Molecular weight}}{\text{Basicity}}$.
Since the basicity of $H_3PO_3$ is $2$,its equivalent weight is $\frac{\text{Molecular weight}}{2}$.
The Reason states the denominator is $3$,which is incorrect. Therefore,the Reason is incorrect.

(B) For a set of quantum numbers to be allowed,they must satisfy the following conditions:
$1$. $n$ must be a positive integer $(1, 2, 3, ...)$.
$2$. $l$ must be an integer ranging from $0$ to $n-1$.
$3$. $m$ must be an integer ranging from $-l$ to $+l$.
$4$. $m_s$ must be either $+\frac{1}{2}$ or $-\frac{1}{2}$.
Evaluating the options:
Option $A$: $m_s = 0$ is not allowed,as $m_s$ can only be $\pm\frac{1}{2}$.
Option $B$: $n=2, l=0, m=0, m_s = -\frac{1}{2}$. All values satisfy the conditions ($l < n$ and $|m| \le l$). This is allowed.
Option $C$: $l = -3$ is not allowed,as $l$ must be $\ge 0$.
Option $D$: $m = 1$ is not allowed when $l = 0$,as $|m| \le l$ must hold.

(C) Isoelectronic species are those that have the same number of electrons.
For $CO$: $6 + 8 = 14$ electrons.
For $NO^{+}$: $7 + 8 - 1 = 14$ electrons.
For $CN^{-}$: $6 + 7 + 1 = 14$ electrons.
For $C_2^{2-}$: $(6 \times 2) + 2 = 14$ electrons.
Since all these species contain $14$ electrons,they are isoelectronic.

(B) The assertion states that the angular momentum of an electron in an orbit is given by $mvr = \frac{nh}{2\pi}$,which is a fundamental postulate of the Bohr model of the atom. This statement is correct.
The reason states that the principal quantum number,$n$,can have any integral value $(n = 1, 2, 3, \dots)$. This is also a correct statement regarding the definition of the principal quantum number.
However,the fact that $n$ can take any integral value is not the reason why the angular momentum is quantized as $\frac{nh}{2\pi}$. The quantization of angular momentum is a separate postulate derived from the wave nature of the electron. Therefore,the reason is not the correct explanation of the assertion.

(B) The correct order for ionization energy is $B < Be < C < O < N$.
On moving from left to right in a period,with an increase in the atomic number,the ionization enthalpy generally increases.
However,there are two main exceptions due to electronic configuration:
$(1)$ Ionization energy of $B < $ ionization energy of $Be$. This is because $Be$ $(1s^2 2s^2)$ has a fully filled $2s$ subshell,which is more stable than the $2p^1$ electron in $B$ $(1s^2 2s^2 2p^1)$.
$(2)$ Ionization energy of $O < $ ionization energy of $N$. Nitrogen $(1s^2 2s^2 2p^3)$ has a half-filled $2p$ subshell,which is extra stable. Therefore,more energy is required to remove an electron from $N$ compared to $O$ $(1s^2 2s^2 2p^4)$.

(D) The electronic configuration of $N_2$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. Bond order = $(10-4)/2 = 3$.
In $N_2^+$,one electron is removed from the bonding molecular orbital $(\sigma 2p_z)$,so the configuration becomes $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^1$. Bond order = $(9-4)/2 = 2.5$. Since the bond order decreases,the $N-N$ bond weakens and $N_2^+$ is paramagnetic.
For $O_2$,the configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. Bond order = $(10-6)/2 = 2$.
In $O_2^+$,one electron is removed from the antibonding molecular orbital $(\pi^*)$,so the configuration becomes $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Bond order = $(10-5)/2 = 2.5$. The bond order increases,and paramagnetism decreases as the number of unpaired electrons reduces from $2$ to $1$.
Therefore,the statement that $N_2^+$ becomes diamagnetic is wrong.

(C) The bond angle of $H_2S$ $(92^\circ)$ is smaller than that of $H_2O$ $(104^\circ 31')$.
According to the $VSEPR$ theory,as the electronegativity of the central atom decreases,the bond pairs move further away from the central atom,reducing the repulsion between them and consequently decreasing the bond angle.
In $H_2O$,the central atom is $O$ (more electronegative),and in $H_2S$,it is $S$ (less electronegative).
Therefore,the Reason statement is incorrect because the bond angle actually increases as the electronegativity of the central atom increases (due to bond pairs being closer to the central atom,increasing repulsion).
Thus,the Assertion is correct,but the Reason is incorrect.

(C) The bond angle of $H_2S$ $(92^o)$ is smaller than that of $H_2O$ $(104.5^o)$.
As the electronegativity of the central atom decreases,the bond pairs move further away from the central atom,reducing the repulsion between them and leading to a smaller bond angle.
Since oxygen is more electronegative than sulfur,the bond pairs in $H_2O$ are closer to the central atom,resulting in greater repulsion and a larger bond angle.
Therefore,the Assertion is correct,but the Reason is incorrect because the bond angle actually increases as the electronegativity of the central atom increases.

(B) According to Graham's law of diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.
First,calculate the molar masses $(M)$ of the given gases:
$M(SO_2) = 32 + 2 \times 16 = 64 \ g/mol$
$M(CO_2) = 12 + 2 \times 16 = 44 \ g/mol$
$M(PCl_3) = 31 + 3 \times 35.5 = 137.5 \ g/mol$
$M(SO_3) = 32 + 3 \times 16 = 80 \ g/mol$
Comparing the molar masses: $M(PCl_3) > M(SO_3) > M(SO_2) > M(CO_2)$.
Since the rate of diffusion is inversely proportional to the square root of the molar mass,the order of the rate of diffusion is: $CO_2 > SO_2 > SO_3 > PCl_3$.

(B) The standard enthalpy change for the reaction is calculated using the formula: $\Delta H^o = \sum \Delta H_f^o(\text{products}) - \sum \Delta H_f^o(\text{reactants})$.
For the reaction $CO_{2(g)} + H_{2(g)} \to CO_{(g)} + H_2O_{(g)}$,the enthalpy change is:
$\Delta H^o = [\Delta H_f^o(CO_{(g)}) + \Delta H_f^o(H_2O_{(g)})] - [\Delta H_f^o(CO_{2(g)}) + \Delta H_f^o(H_{2(g)})]$.
Given $\Delta H_f^o(H_{2(g)}) = 0 \ kJ \ mol^{-1}$ (standard state of an element).
$\Delta H^o = [-110.5 + (-241.8)] - [-393.5 + 0]$.
$\Delta H^o = -352.3 - (-393.5) = 41.2 \ kJ$.

(C) For an isothermal process,the temperature remains constant. Since internal energy $(\Delta U)$ of an ideal gas is a function of temperature only,$\Delta U = 0$.
According to the first law of thermodynamics,$\Delta U = Q + W$.
Substituting $\Delta U = 0$,we get $0 = Q + W$,which implies $Q = -W$.
This means the heat absorbed by the system is equal to the work done by the system.
However,for an isothermal process,enthalpy change $\Delta H = \Delta U + \Delta(PV)$. For an ideal gas,$\Delta H = \Delta U + \Delta(nRT) = 0 + nR\Delta T = 0$. While $\Delta H$ is zero for an ideal gas,the reason provided is not the fundamental explanation for $Q = -W$,which is based on the first law and $\Delta U = 0$. In general,$\Delta H$ is not necessarily zero for all isothermal processes (e.g.,real gases). Thus,the Assertion is correct,but the Reason is incorrect.

(D) Let the given reactions be:
$(1) XeF_{6(g)} + H_2O_{(g)} \rightleftharpoons XeOF_{4(g)} + 2HF_{(g)} ; K_1$
$(2) XeO_{4(g)} + XeF_{6(g)} \rightleftharpoons XeOF_{4(g)} + XeO_3F_{2(g)} ; K_2$
To obtain the target reaction $XeO_{4(g)} + 2HF_{(g)} \rightleftharpoons XeO_3F_{2(g)} + H_2O_{(g)}$,we perform the operation: $(2) - (1)$.
This is equivalent to adding reaction $(2)$ and the reverse of reaction $(1)$.
The equilibrium constant for the reverse of reaction $(1)$ is $\frac{1}{K_1}$.
Therefore,the equilibrium constant $K$ for the target reaction is $K = K_2 \times \frac{1}{K_1} = \frac{K_2}{K_1}$.

(B) The balanced chemical equation for the reaction is: $3Fe_{(s)} + 4H_2O_{(g)} \leftrightarrow Fe_3O_{4(s)} + 4H_{2(g)}$.
In the expression for the equilibrium constant $(K_p)$,only the partial pressures of gaseous species are included.
Solid species like $Fe_{(s)}$ and $Fe_3O_{4(s)}$ are considered to have unit activity and are omitted from the expression.
Therefore,the equilibrium constant expression is $K_p = \frac{(P_{H_2})^4}{(P_{H_2O})^4}$.

(A) The precipitation of metal sulfides depends on the solubility product $(K_{sp})$ and the concentration of sulfide ions $(S^{2-})$.
In an acidic medium,the dissociation of $H_2S$ is suppressed due to the common ion effect of $H^+$ ions,resulting in a very low concentration of $S^{2-}$ ions.
Since the $K_{sp}$ of $As_2S_3$ is extremely low,even this low concentration of $S^{2-}$ is sufficient to exceed the ionic product of $As_2S_3$,causing it to precipitate.
Conversely,the $K_{sp}$ of $ZnS$ is relatively higher,so the low concentration of $S^{2-}$ in the acidic medium is insufficient to exceed the $K_{sp}$ of $ZnS$,preventing its precipitation.

(C) redox reaction involves a change in oxidation number $(O.N.)$ of the reacting species.
In the reaction $N_2 + O_2 \to 2NO$,the $O.N.$ of $N$ changes from $0$ to $+2$ (oxidation) and the $O.N.$ of $O$ changes from $0$ to $-2$ (reduction).
Therefore,this is a redox reaction.

(C) The chemical $A$ is calcium hydroxide,$Ca(OH)_2$,also known as slaked lime.
$1$. It is used for water softening to remove temporary hardness by reacting with calcium bicarbonate: $Ca(HCO_3)_2 + Ca(OH)_2 \to 2CaCO_3 \downarrow + 2H_2O$.
$2$. It reacts with sodium carbonate $(Na_2CO_3)$ to produce caustic soda $(NaOH)$: $Ca(OH)_2 + Na_2CO_3 \to 2NaOH + CaCO_3$.
$3$. When $CO_2$ is passed through lime water $(Ca(OH)_2)$,it turns milky due to the formation of insoluble calcium carbonate: $Ca(OH)_2 + CO_2 \to CaCO_3 \downarrow + H_2O$.

(A) The diagonal relationship between $Be$ and $Al$ arises due to their similar ionic potential (charge/size ratio) and similar electronegativity values.
The ionization energy of $Be$ $(9.32 \ eV)$ is very close to that of $Al$ ($5.98 \ eV$ for first,but the overall properties are governed by charge density).
Actually,the similarity in their charge-to-size ratio (ionic potential) is the primary reason for their diagonal relationship,which is directly related to their ionization energies and electronegativities.
Therefore,both the assertion and the reason are correct,and the reason explains the assertion.

(D) The stability of a carbanion is increased by electron-withdrawing groups $(EWG)$ and decreased by electron-donating groups $(EDG)$ attached to the carbon bearing the negative charge.
$1.$ Option $A$ is an alkyl carbanion with a phenyl group further away,providing minimal stabilization.
$2.$ Option $B$ is a benzyl carbanion,stabilized by resonance with the phenyl ring.
$3.$ Option $C$ has a $-OCH_3$ group,which is an electron-donating group (via $+M$ effect),thus destabilizing the carbanion.
$4.$ Option $D$ has a $-NO_2$ group,which is a strong electron-withdrawing group (via $-M$ and $-I$ effects). It effectively disperses the negative charge through resonance,making it the most stable carbanion among the given options.

(B) The structure $HOH_2CCH(OH)CHO$ is glyceraldehyde.
It contains one primary $(1^o)$ alcoholic group $(-CH_2OH)$ and one secondary $(2^o)$ alcoholic group $(-CH(OH)-)$.
It does not contain a tertiary $(3^o)$ alcoholic group.
Therefore,the statement that it contains a tertiary alcoholic group is incorrect.

(D) $1$. Identify the longest carbon chain. The longest chain in the given structure contains $8$ carbon atoms,so the parent alkane is octane.
$2$. Number the chain from the end that gives the lowest locants to the substituents.
$3$. Numbering from left to right gives substituents at positions $3$ and $4$. Numbering from right to left gives substituents at positions $5$ and $6$.
$4$. Following the lowest locant rule,we number from left to right.
$5$. At position $3$,there is a methyl group $(-CH_3)$,and at position $4$,there is an ethyl group $(-CH_2CH_3)$.
$6$. According to $IUPAC$ rules,substituents are listed in alphabetical order. Therefore,ethyl comes before methyl.
$7$. The correct name is $4-$ethyl$-3-$methyloctane.

(B) The dipole moment depends on the molecular symmetry and the polarity of bonds.
$1$. $CH_3-C \equiv C-CH_3$ (but$-2-$yne) is a linear and symmetrical molecule. Due to its high symmetry,the dipole moments of the $C-CH_3$ bonds cancel each other out,resulting in a net dipole moment of zero.
$2$. $CH_3-CH_2-C \equiv CH$ (but$-1-$yne) and $CH_2=CH-C \equiv CH$ (but$-1-$en$-3-$yne) are asymmetric,leading to a non-zero net dipole moment.
$3$. cis-but$-2-$ene has a non-zero dipole moment due to its polar nature and lack of center of symmetry.
Therefore,$CH_3-C \equiv C-CH_3$ has the lowest dipole moment.

(A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.
The reaction proceeds as follows:
$M + HNO_3 \to MNO_3 + [H]$ (Nascent hydrogen)
$2HNO_3 + 2[H] \to 2NO_2 + 2H_2O$
Overall,the nascent hydrogen produced by the metal-acid reaction reduces the nitric acid to nitrogen dioxide $(NO_2)$.

(A) The reaction shown is hydroboration-oxidation of $1-$pentene.
$(i)$ The first step involves the addition of $BH_3$ across the double bond,which follows anti-Markovnikov addition.
(ii) The second step involves oxidation with $H_2O_2$ in the presence of $OH^-$,which replaces the boron atom with a hydroxyl group $(-OH)$ with retention of configuration.
Therefore,$CH_3CH_2CH_2CH=CH_2$ reacts to form $CH_3CH_2CH_2CH_2CH_2OH$,which is $1-$pentanol.

(B) The reaction of ethanol with $PCl_5$ is: $C_2H_5OH + PCl_5 \rightarrow C_2H_5Cl (A) + POCl_3 + HCl$.
Here,$A$ is ethyl chloride $(C_2H_5Cl)$.
The reaction of $A$ with silver nitrite $(AgNO_2)$ is: $C_2H_5Cl + AgNO_2 \rightarrow C_2H_5NO_2 (B) + AgCl$.
Here,$B$ is nitroethane $(C_2H_5NO_2)$.
Therefore,$A$ and $B$ are $C_2H_5Cl$ and $C_2H_5NO_2$ respectively.

(D) Basidiomycetes,commonly known as club fungi,include mushrooms,bracket fungi,puffballs,and bird's nest fungi.
$A$. Puffballs are Basidiomycetes,but $Claviceps$ belongs to Ascomycetes.
$B$. $Peziza$ belongs to Ascomycetes and $Alternaria$ belongs to Deuteromycetes.
$C$. $Morchella$ belongs to Ascomycetes,while mushrooms are Basidiomycetes.
$D$. Both bird's nest fungi and puffballs are members of the class Basidiomycetes.

(C) The $P$-wave represents the electrical excitation (or depolarization) of the atria,which leads to the contraction of both the atria.
The $QRS$ complex represents the depolarization of the ventricles,which initiates the ventricular contraction. The contraction starts shortly after $Q$ and marks the beginning of the systole.
The $T$-wave represents the return of the ventricles from an excited to a normal state (repolarization). The end of the $T$-wave marks the end of systole.
Therefore,none of the options $A, B,$ or $D$ are correctly interpreted. However,in many contexts,the $QRS$ complex is used to determine the heart rate by counting the number of $QRS$ complexes that occur in a given time period,which corresponds to one complete pulse.

(D) The correct statement is that nearly $99$ percent of the glomerular filtrate is reabsorbed by the renal tubules.
- The ascending limb of the Loop of Henle is permeable to electrolytes but impermeable to water.
- The descending limb of the Loop of Henle is permeable to water but impermeable to electrolytes.
- The distal convoluted tubule $(DCT)$ is capable of reabsorbing $HCO_3^-$ and water,and it also performs conditional reabsorption of $Na^+$ and water.
- The glomerular filtration rate $(GFR)$ is approximately $180$ liters per day,while the volume of urine released is only about $1.5$ liters,which confirms that about $99$ percent of the filtrate is reabsorbed.

(C) An acidic buffer is formed by a mixture of a weak acid and its salt with a strong base.
In option $C$,we have $100 \ mL$ of $0.1 \ M \ HCl$ $(10 \ mmol)$ and $200 \ mL$ of $0.1 \ M \ CH_3COONa$ $(20 \ mmol)$.
The reaction is: $HCl + CH_3COONa \rightarrow CH_3COOH + NaCl$
Initial moles: $10 \ mmol \ HCl$ and $20 \ mmol \ CH_3COONa$.
After reaction: $0 \ mmol \ HCl$,$10 \ mmol \ CH_3COONa$ remaining,and $10 \ mmol \ CH_3COOH$ formed.
Since the final mixture contains a weak acid $(CH_3COOH)$ and its conjugate base ($CH_3COO^-$ from $CH_3COONa$),it acts as an acidic buffer.

(A) The taxonomic hierarchy follows the order: $Species < Genus < Family < Order < Class < Phylum/Division < Kingdom$.
As we move from species to kingdom,the number of common characteristics decreases,meaning the groups become more general.
Conversely,as we move from kingdom to species,the number of common characteristics increases,meaning the groups become less general.
Since $Species$ is the lowest taxonomic category,it possesses the most specific characters and is therefore less general than $Genus$.

(A) The taxonomic hierarchy follows the order: $Kingdom > Phylum/Division > Class > Order > Family > Genus > Species$.
As we move down the hierarchy from higher categories to lower categories, the number of common characteristics shared by the organisms decreases in generality and increases in specificity.
Since $Species$ is a lower taxonomic category than $Genus$, it possesses more specific characters and is less general in its definition compared to $Genus$.

(D) Plasmodesmata are narrow channels that act as intercellular cytoplasmic bridges to facilitate communication and transport of materials between plant cells.
Adhering,tight,and gap junctions are specialized junctions that provide both structural and functional links between the individual cells of animal tissues.
Therefore,plasmodesmata are exclusively found in plant cells and not in animal tissues.

(C) Step $1$: Reaction of ethanol with $PCl_5$:
$C_2H_5OH + PCl_5 \rightarrow C_2H_5Cl (A) + POCl_3 + HCl$
Thus,$A$ is $C_2H_5Cl$ (ethyl chloride).
Step $2$: Reaction of $A$ with silver nitrite $(AgNO_2)$:
$C_2H_5Cl + AgNO_2 \rightarrow C_2H_5NO_2 (B) + AgCl$
Thus,$B$ is $C_2H_5NO_2$ (nitroethane).
Therefore,$A$ and $B$ are $C_2H_5Cl$ and $C_2H_5NO_2$ respectively.

(B) The molar mass of oxalic acid dihydrate $(H_2C_2O_4 \cdot 2H_2O)$ is $126 \ g/mol$.
Since it is a dibasic acid,its equivalent mass is $\frac{126}{2} = 63 \ g/eq$.
Normality of the oxalic acid solution = $\frac{\text{mass}}{\text{equivalent mass} \times \text{volume in L}} = \frac{6.3}{63 \times 0.250} = 0.4 \ N$.
Using the titration formula $N_1 V_1 = N_2 V_2$ for $10 \ mL$ of the solution:
$0.4 \ N \times 10 \ mL = 0.1 \ N \times V_2$.
$V_2 = \frac{4}{0.1} = 40 \ mL$.

(D) The Bronsted-Lowry definition defines acids as proton $(H^+)$ donors and bases as proton acceptors.
In options $A$,$B$,and $C$,there is a clear transfer of a proton $(H^+)$ from the acid to the base.
In the reaction $[Cu(H_2O)_4]^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+} + 4H_2O$,there is no proton transfer.
Instead,$NH_3$ acts as a Lewis base by donating a lone pair of electrons to the $Cu^{2+}$ ion (Lewis acid) to form a coordinate covalent bond.
Therefore,this reaction is an acid-base reaction according to the Lewis definition but not the Bronsted-Lowry definition.

(B) For a $bcc$ structure,the number of atoms per unit cell,$z = 2$.
Given: Atomic mass $M = 100 \ g/mol$,edge length $a = 400 \ pm = 400 \times 10^{-10} \ cm$,and Avogadro's number $N_A = 6.023 \times 10^{23} \ mol^{-1}$.
The formula for density is $\rho = \frac{z \times M}{N_A \times a^3}$.
Substituting the values: $\rho = \frac{2 \times 100}{6.023 \times 10^{23} \times (400 \times 10^{-10})^3}$.
$\rho = \frac{200}{6.023 \times 10^{23} \times 64 \times 10^{-24}} = \frac{200}{38.5472} \approx 5.188 \ g/cm^3$.

(A) According to Raoult's law for an ideal solution,the total vapour pressure $P_{total}$ is given by:
$P_{total} = P_A^o x_A + P_B^o x_B$
Given: $P_A^o = 120 \, mm \, Hg$,$P_B^o = 180 \, mm \, Hg$,$n_A = 2 \, moles$,$n_B = 3 \, moles$.
Total moles $= 2 + 3 = 5 \, moles$.
Mole fraction of $A$ $(x_A)$ $= \frac{2}{5} = 0.4$.
Mole fraction of $B$ $(x_B)$ $= \frac{3}{5} = 0.6$.
$P_{total} = (120 \times 0.4) + (180 \times 0.6) = 48 + 108 = 156 \, mm \, Hg$.

(D) The depression in freezing point is given by the formula $\Delta T_f = i \times K_f \times m$.
For equimolal solutions,$\Delta T_f$ is directly proportional to the van't Hoff factor $(i)$.
$C_6H_{12}O_6$ (glucose) is a non-electrolyte,so its $i = 1$.
$C_6H_5NH_3^+Cl^-$ dissociates into $2$ ions $(i = 2)$.
$Ca(NO_3)_2$ dissociates into $3$ ions $(i = 3)$.
$La(NO_3)_3$ dissociates into $4$ ions $(i = 4)$.
Since glucose has the lowest $i$ value,it will have the minimum depression in freezing point $(\Delta T_f)$.
Therefore,the freezing point $(T_f = T_f^0 - \Delta T_f)$ will be the highest for $C_6H_{12}O_6$.

(D) According to the thermodynamic relationship between Raoult's law and Henry's law,if one component of a binary solution obeys Raoult's law $(P_i = x_i P_i^o)$ over the entire range of composition,the other component must also obey Raoult's law.
However,in dilute solutions,the solvent obeys Raoult's law $(P_1 = x_1 P_1^o)$ while the solute obeys Henry's law $(P_2 = K_H x_2)$.
The Assertion is incorrect because if one component obeys Raoult's law,the other component often obeys Henry's law in the dilute range.
The Reason is correct because Raoult's law is indeed a special case of Henry's law where the Henry's constant $(K_H)$ becomes equal to the pure component vapor pressure $(P_i^o)$.

(A) reducing agent is a substance that undergoes oxidation,and its strength is determined by its standard oxidation potential. The standard oxidation potential is the negative of the standard reduction potential $(E^\circ_{ox} = -E^\circ_{red})$.
$1$. For $Zn_{(s)} \rightarrow Zn^{2+} + 2e^-$,$E^\circ_{ox} = -(-0.762 \ V) = +0.762 \ V$
$2$. For $Cr_{(s)} \rightarrow Cr^{3+} + 3e^-$,$E^\circ_{ox} = -(-0.740 \ V) = +0.740 \ V$
$3$. For $H_{2(g)} \rightarrow 2H^+ + 2e^-$,$E^\circ_{ox} = -(0.00 \ V) = 0.00 \ V$
$4$. For $Fe^{2+} \rightarrow Fe^{3+} + e^-$,$E^\circ_{ox} = -(0.770 \ V) = -0.770 \ V$
Since $Zn_{(s)}$ has the highest standard oxidation potential $(+0.762 \ V)$,it is the strongest reducing agent among the given options.

(A) For a first-order reaction,the integrated rate equation is given by:
$t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$
Given:
$k = 6 \, min^{-1}$
$[A]_0 = 0.5 \, mol \, L^{-1}$
$[A]_t = 0.05 \, mol \, L^{-1}$
Substituting the values:
$t = \frac{2.303}{6} \log \frac{0.5}{0.05}$
$t = \frac{2.303}{6} \log(10)$
Since $\log(10) = 1$,
$t = \frac{2.303}{6} \approx 0.384 \, min$.

(C) The movement of colloidal particles towards the anode indicates that the sol is negatively charged.
According to the Hardy-Schulze rule,the coagulating power of an electrolyte depends on the valency of the ion carrying a charge opposite to that of the colloidal particles.
Since the sol is negative,the coagulating power depends on the charge of the cation ($Na^+$,$Ba^{2+}$,$Al^{3+}$).
Greater the charge on the cation,higher is its coagulating power.
Thus,the order of coagulating power is $Al^{3+} > Ba^{2+} > Na^+$,which corresponds to $AlCl_3 > BaCl_2 > NaCl$.

(C) The Assertion is correct because the Freundlich adsorption isotherm is indeed given by the equation $\frac{x}{m} = k \cdot p^{1/n}$.
The Reason is incorrect. The Freundlich isotherm is typically represented by plotting $\log(\frac{x}{m})$ versus $\log(p)$,which yields a straight line with a slope of $\frac{1}{n}$ and an intercept of $\log(k)$.
If several substances have the same value of $\frac{1}{n}$,their plots will have the same slope,meaning the lines will be parallel to each other and will not meet at a single point.

(B) The correct matches are as follows:
$I$. Cyanide process is used for the extraction of $Au$ $(I-D)$.
$II$. Froth floatation process uses pine oil as a frother $(II-B)$.
$III$. Electrolytic reduction is used for the extraction of $Al$ $(III-C)$.
$IV$. Zone refining is used for obtaining ultrapure $Ge$ $(IV-A)$.
Thus,the correct sequence is $I-D, II-B, III-C, IV-A$.

(A) During the $Bessemerization$ process,the molten copper is poured into moulds and allowed to cool.
As the metal solidifies,the dissolved $SO_2$ gas escapes,creating bubbles or blisters on the surface of the solidified copper.
Hence,the copper obtained is called blister copper.
Both the assertion and the reason are correct,and the reason is the correct explanation of the assertion.

(A) The reaction of chlorine with cold and dilute sodium hydroxide is a disproportionation reaction.
The balanced chemical equation is:
$2NaOH + Cl_2 \to NaCl + NaOCl + H_2O$
In this reaction,chlorine is both oxidized and reduced.
The ionic products formed are chloride $(Cl^{-})$ and hypochlorite $(OCl^{-})$.

(D) The spin-only magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
Given $\mu = \sqrt{15} \ BM$.
Equating the two expressions: $\sqrt{15} = \sqrt{n(n+2)}$.
Squaring both sides: $15 = n(n+2)$.
$n^2 + 2n - 15 = 0$.
Solving the quadratic equation: $(n+5)(n-3) = 0$.
Since $n$ must be a positive integer,$n = 3$.
Therefore,the number of unpaired electrons is $3$.

(B) The magnetic moment values of actinides are lower than the theoretically predicted values because the $5f$ electrons are less effectively shielded by the outer shells. This leads to the quenching of the orbital contribution to the magnetic moment,making the observed values smaller than those calculated using the spin-only formula. While actinides are paramagnetic,the reason provided does not explain why the magnetic moments are lower than predicted.

(A) The name of the complex is triamminediaquachlorocobalt $(III)$ chloride.
$1$. The central metal ion is $Co^{3+}$.
$2$. The ligands are three ammine $(NH_3)$,two aqua $(H_2O)$,and one chloro $(Cl^-)$ groups.
$3$. The coordination sphere is $[Co(NH_3)_3(H_2O)_2Cl]$.
$4$. To balance the charge of $Co^{3+}$ $(+3)$ and $Cl^-$ $(-1)$,the total charge of the coordination sphere is $+2$. Thus,two chloride ions are required outside the sphere.
$5$. The formula is $[CoCl(NH_3)_3(H_2O)_2]Cl_2$.

(D) In $[FeF_6]^{3-}$,the oxidation state of $Fe$ is $+3$ ($3d^5$ configuration).
$F^{-}$ is a weak field ligand,which does not cause pairing of electrons.
Therefore,it forms a high spin complex with $5$ unpaired electrons.
Thus,the Assertion is incorrect,while the Reason is a correct statement regarding the definition of low spin complexes.

(A) $S_N2$ (Substitution Nucleophilic Bimolecular) reactions are highly favored by methyl halides and primary $(1^\circ)$ alkyl halides because they have the least steric hindrance,allowing the nucleophile to attack the carbon atom easily from the backside.
$CH_3Br$ is a methyl halide and undergoes $S_N2$ reaction most readily among the given options.
Option $(b)$ is a secondary halide,option $(c)$ is an elimination reaction,and option $(d)$ is a tertiary halide which typically follows the $S_N1$ mechanism.

(D) The Assertion is incorrect because alkylbenzene $CAN$ be prepared by Friedel-Crafts alkylation,although it has limitations such as polyalkylation and rearrangement.
The Reason is also incorrect because alkyl halides are generally more reactive or comparable in reactivity to acyl halides in the context of Friedel-Crafts reactions,but the primary issue with alkylation is the formation of polyalkylated products due to the activating nature of the alkyl group added to the benzene ring.
Therefore,both the Assertion and the Reason are incorrect.

(D) The Assertion is false because aryl halides do not undergo nucleophilic substitution reactions under ordinary conditions. This is due to the resonance effect,which gives the $C-Cl$ bond partial double bond character,making it shorter and stronger,and thus difficult to replace by nucleophiles.
The Reason is also false because $S_{N}2$ reactions proceed with inversion of configuration,not retention.

(B) Vinyl alcohol is unstable and tautomerizes to acetaldehyde. Therefore,it cannot be polymerized directly. Polyvinyl alcohol is prepared by the alkaline hydrolysis (or alcoholysis) of polyvinyl acetate. The reaction is: $-(CH_2-CH(OCOCH_3))_n- + nH_2O \xrightarrow{OH^-} -(CH_2-CH(OH))_n- + nCH_3COOH$.

(A) Ethers have two lone pairs of electrons on the oxygen atom,which makes them act as Lewis bases.
In the presence of mineral acids (like $HCl$),the oxygen atom donates a lone pair to the $H^+$ ion,forming an oxonium salt.
The reaction is: $R-O-R + HCl \rightarrow [R_2O^+-H]Cl^-$.
Since the basic behavior is directly caused by the lone pairs on the oxygen atom,the Reason correctly explains the Assertion.

(A) The reaction of benzaldehyde $(C_6H_5CHO)$ with a Grignard reagent $(CH_3MgBr)$ follows a nucleophilic addition mechanism.
First,the nucleophilic $CH_3^-$ group attacks the carbonyl carbon of benzaldehyde to form an intermediate alkoxide: $C_6H_5CHO + CH_3MgBr \rightarrow C_6H_5CH(OMgBr)CH_3$.
Second,acid hydrolysis of this intermediate yields the final product: $C_6H_5CH(OMgBr)CH_3 + H_2O/H^+ \rightarrow C_6H_5CH(OH)CH_3 + Mg(OH)Br$.
The product $C_6H_5CH(OH)CH_3$ is $1$-phenylethanol,which is a secondary $(2^o)$ alcohol.

(B) The reaction of carboxylic acids having an $\alpha$-hydrogen with halogen ($Cl_2$ or $Br_2$) in the presence of a small amount of red phosphorus to give $\alpha$-halo carboxylic acids is known as the Hell-Volhard-Zelinsky $(HVZ)$ reaction.

(D) Both Assertion and Reason are incorrect.
Acetoacetic ester $(CH_3COCH_2COOC_2H_5)$ does not give a positive iodoform test under standard conditions because the active methylene group $(CH_2)$ between the two carbonyl groups is highly acidic and reactive,leading to cleavage rather than the formation of iodoform.
The Reason is also incorrect because the structure of acetoacetic ester clearly contains the $CH_3CO-$ (acetyl) group.

(A) The reduction of nitroalkanes $(R-NO_2)$ yields primary amines $(R-NH_2)$.
For example,$CH_3CH_2NO_2 + 6[H] \xrightarrow{Sn/HCl} CH_3CH_2NH_2 + 2H_2O$.
Option $B$ is an alkyl nitrite,which on reduction gives alcohol and ammonia.
Option $C$ is azobenzene,which on reduction gives aniline (a primary amine),but nitroalkanes are the standard textbook example for primary amine preparation via reduction.
Option $D$ is an isocyanide,which on reduction gives a secondary amine $(R-NH-CH_3)$.

(A) One molecule of glucose reacts with $3$ molecules of phenylhydrazine to form glucosazone.
The reaction proceeds as follows:
$1.$ The first molecule of phenylhydrazine reacts with the aldehyde group at $C-1$ to form phenylhydrazone.
$2.$ The second molecule of phenylhydrazine oxidizes the secondary alcohol group at $C-2$ to a carbonyl group $(C=O)$.
$3.$ The third molecule of phenylhydrazine reacts with the newly formed carbonyl group at $C-2$ to yield the final osazone.

(A) At the isoelectric point $(pH = pI)$,an amino acid exists as a zwitterion,which carries both a positive and a negative charge,resulting in a net charge of zero. Because the molecule is electrically neutral,it does not migrate towards either the cathode or the anode when an electric field is applied. Therefore,the Reason correctly explains the Assertion.

(A) The reaction of $D$-glucose with dilute alkali is known as the Lobry de Bruyn-van Ekenstein transformation.
This process involves the formation of an enediol intermediate.
During the formation of the enediol,the $C_2$ carbon atom changes its hybridization from $sp^3$ to $sp^2$.
This intermediate allows for the interconversion between glucose,mannose,and fructose.
Therefore,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(C) The first magic bullet was fired at syphilis on this day in $1909$. Although specific diseases responded better to some drugs than to others,before the early $1900s$ development of $Salvarsan$,an arsenic-based drug to treat syphilis,drugs were not developed to target a specific disease.

(A) When $KCN$ is added to a mixture of $Cu^{2+}$ and $Cd^{2+}$,$Cu^{2+}$ is reduced to $Cu^{+}$ by $CN^-$ ions,forming the stable complex $K_3[Cu(CN)_4]$.
$Cd^{2+}$ also forms a complex,$K_2[Cd(CN)_4]$,but it is less stable than the copper complex.
When $H_2S$ gas is passed through this solution,the $Cd^{2+}$ complex dissociates to provide enough $Cd^{2+}$ ions to react with $S^{2-}$ and form a yellow precipitate of $CdS$.
However,the $Cu^{+}$ complex is highly stable and does not dissociate to provide sufficient $Cu^{+}$ ions to form $Cu_2S$.
Thus,$Cu^{2+}$ and $Cd^{2+}$ are separated. Both the assertion and the reason are correct,and the reason explains why $Cu^{2+}$ does not precipitate.

(C) The reaction of ethanol with $PCl_5$ is as follows:
$C_2H_5OH + PCl_5 \rightarrow C_2H_5Cl (A) + POCl_3 + HCl$
Thus,$A$ is $C_2H_5Cl$ (ethyl chloride).
Next,the reaction of ethyl chloride $(A)$ with silver nitrite $(AgNO_2)$ is:
$C_2H_5Cl + AgNO_2 \rightarrow C_2H_5NO_2 (B) + AgCl$
Thus,$B$ is $C_2H_5NO_2$ (nitroethane).
Therefore,$A$ and $B$ are $C_2H_5Cl$ and $C_2H_5NO_2$ respectively.