AIIMS 2005 Chemistry Question Paper with Answer and Solution

(D) The $NOR$ gate is considered a universal gate.
This is because any other basic logic gate,such as $AND$,$OR$,or $NOT$,can be constructed using only $NOR$ gates.

(C) For an object to become a black hole,its escape velocity must be equal to or greater than the speed of light $(c)$.
The escape velocity $v_e$ from the surface of a spherical mass $m$ of radius $r$ is given by the formula:
$v_e = \sqrt{\frac{2Gm}{r}}$
For the object to be a black hole,the condition is $v_e \ge c$.
Substituting the expression for $v_e$,we get:
$\sqrt{\frac{2Gm}{r}} \ge c$
Therefore,the correct option is $C$.

(C) The most probable radius for an electron in a hydrogen-like species is given by the formula $r_{mp} = \frac{a_0}{Z}$, where $a_0$ is the Bohr radius $(52.9 \ pm)$ and $Z$ is the atomic number.
For the helium ion $(He^{+})$, the atomic number $Z = 2$.
Substituting the values: $r_{mp} = \frac{52.9 \ pm}{2} = 26.45 \ pm$.
Rounding to the nearest value, we get $26.5 \ pm$.

(D) To determine the isoelectronic pair,we calculate the total number of valence electrons for each species.
For $ClO_2^-$: $7 (Cl) + 2 \times 6 (O) + 1 (\text{charge}) = 20$ valence electrons.
For $ClF_2^+$: $7 (Cl) + 2 \times 7 (F) - 1 (\text{charge}) = 20$ valence electrons.
Since both $ClO_2^-$ and $ClF_2^+$ have $20$ valence electrons,they form an isoelectronic pair.
Therefore,the correct option is $D$.

(A) For the reaction,$CaCO_{3_{(s)}} \rightleftharpoons CaO_{(s)} + CO_{2_{(g)}}$,the equilibrium constant $K_p$ is given by $K_p = pCO_2 / p^{\circ}$.
According to the van't Hoff equation,$\ln K_p = -\frac{\Delta H_r^{\circ}}{RT} + C$.
Converting to base $10$ logarithm,we get $\log_{10} K_p = -\frac{\Delta H_r^{\circ}}{2.303 RT} + \text{constant}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log_{10} (pCO_2 / p^{\circ})$ and $x = 1 / T$,the slope $m = -\frac{\Delta H_r^{\circ}}{2.303 R}$.
Thus,$\Delta H_r^{\circ}$ can be determined from the slope of the plot of $\log_{10} (pCO_2 / p^{\circ})$ versus $1 / T$.

(D) The given reaction is $2NOCl_{(g)} \rightleftharpoons 2NO_{(g)} + Cl_{2(g)}$.
The change in the number of moles of gaseous products and reactants is $\Delta n = (2 + 1) - 2 = 1$.
The relationship between $K_P$ and $K_C$ is given by $K_P = K_C(RT)^{\Delta n}$.
Given $K_C = 3 \times 10^{-6}$,$R = 0.0821 \ L\ atm\ K^{-1}\ mol^{-1}$,and $T = 427 + 273 = 700 \ K$.
Substituting the values: $K_P = 3 \times 10^{-6} \times (0.0821 \times 700)^1$.
$K_P = 3 \times 10^{-6} \times 57.47 = 172.41 \times 10^{-6} = 1.7241 \times 10^{-4}$.
Rounding to the nearest value,$K_P \approx 1.75 \times 10^{-4}$.

(C) The reaction between a weak acid $(CH_3COOH)$ and a weak base $(NH_4OH)$ forms a salt $(CH_3COONH_4)$.
The $pH$ of a salt of a weak acid and a weak base is given by the formula:
$pH = \frac{1}{2} [pK_w + pK_a - pK_b]$
Given:
$pK_w = 14$
$pK_a = 5.0$
$pK_b = 5.0$
Substituting the values:
$pH = \frac{1}{2} [14 + 5.0 - 5.0]$
$pH = \frac{1}{2} [14] = 7$

(A) bomb calorimeter is a rigid,sealed container,which means the volume of the system remains constant throughout the reaction.
Since the volume change $\Delta V = 0$,the work done $w = -P_{ext} \Delta V = 0$.
The reaction between zinc $(Zn)$ and sulphuric acid $(H_2SO_4)$ is an exothermic process,which releases heat to the surroundings.
According to the first law of thermodynamics,$\Delta U = q + w$. Since $w = 0$,$\Delta U = q$.
Because the reaction is exothermic,$q < 0$,therefore $\Delta U < 0$.
Thus,the correct values are $\Delta U < 0$ and $w = 0$.

(B) The standard enthalpy of formation,$\Delta H^o_f$,is defined as the enthalpy change when $1 \ mol$ of a compound is formed from its constituent elements in their most stable standard states at $298 \ K$ and $1 \ bar$ pressure.
For methanol $(CH_3OH_{(l)})$,the constituent elements are carbon (in its most stable form,graphite),hydrogen $(H_{2(g)})$,and oxygen $(O_{2(g)})$.
The balanced thermochemical equation is:
$C_{(graphite)} + 1/2 O_{2(g)} + 2H_{2(g)} \to CH_3OH_{(l)}$
Therefore,option $B$ correctly represents the standard enthalpy of formation of methanol.

(A) The given reaction is $IO_3^- + aI^{-} + bH^{+} \to cH_2O + dI_2$.
Step $1$: Identify oxidation and reduction half-reactions.
Oxidation: $I^{-} \to I_2$
Reduction: $IO_3^- \to I_2$
Step $2$: Balance the reduction half-reaction: $2IO_3^- + 12H^{+} + 10e^{-} \to I_2 + 6H_2O$.
Step $3$: Balance the oxidation half-reaction: $2I^{-} \to I_2 + 2e^{-}$.
Step $4$: Multiply the oxidation half-reaction by $5$ to balance electrons: $10I^{-} \to 5I_2 + 10e^{-}$.
Step $5$: Add the two half-reactions: $2IO_3^- + 10I^{-} + 12H^{+} \to 6I_2 + 6H_2O$.
Step $6$: Simplify by dividing by $2$: $IO_3^- + 5I^{-} + 6H^{+} \to 3H_2O + 3I_2$.
Comparing this with the given equation $IO_3^- + aI^{-} + bH^{+} \to cH_2O + dI_2$,we get $a = 5, b = 6, c = 3, d = 3$.

(D) The correct answer is $(D)$.
Both $Be(OH)_2$ and $Zn(OH)_2$ are amphoteric in nature,meaning they can react with both acids and bases.

(B) In diborane $(B_2H_6)$,there are two types of hydrogen atoms: terminal and bridging.
The terminal $H-B-H$ bond angle is approximately $122^{\circ}$.
The bridging $H-B-H$ bond angle is approximately $97^{\circ}$.
Therefore,the two angles are nearly $97^{\circ}$ and $122^{\circ}$.

(B) The hydrolysis of magnesium carbide $(Mg_2C_3)$ yields propyne $(CH_3C \equiv CH)$.
The chemical reaction is as follows:
$Mg_2C_3 + 4H_2O \to CH_3C \equiv CH + 2Mg(OH)_2$
Therefore,the correct option is $(B)$.

(D) The stability of substituted cyclohexanes depends on steric hindrance and intramolecular interactions.
In $1,2-$cyclohexanediol,the $OH$ groups are closer,leading to higher steric repulsion.
In $1,3-$cyclohexanediol,the $OH$ groups are further apart,reducing steric repulsion.
Between $cis$ and $trans$ isomers,the $trans-1,3-$cyclohexanediol allows both $OH$ groups to occupy equatorial positions in the chair conformation,which significantly minimizes steric strain compared to other isomers.
Therefore,$trans-1,3-$cyclohexanediol is the most stable compound.

(C) Rutherford first used a zinc sulphide $(ZnS)$ screen as a phosphor for the detection of $\alpha$-particles because it produces a flash of light (scintillation) when struck by an $\alpha$-particle.

(D) The reaction involves the bromination of isobutane $(CH_3-CH(CH_3)-CH_3)$ using $CH_3OBr$ as a brominating agent,which proceeds via a free radical mechanism.
The tertiary hydrogen atom is the most reactive,leading to the formation of tert-butyl bromide $(CH_3-C(CH_3)_2-Br)$ as the intermediate.
Subsequently,the tert-butyl bromide undergoes $S_N1$ solvolysis in the presence of the solvent methanol $(CH_3OH)$ to form the major product,$2$-methoxy-$2$-methylpropane $(CH_3-C(OCH_3)(CH_3)-CH_3)$.

(D) Plastids are organelles found primarily in plant cells and algae. They are responsible for photosynthesis (chloroplasts),storage of starch (amyloplasts),and pigment synthesis. However,many cells,such as animal cells,fungal cells,and certain specialized plant cells,do not contain plastids. These cells are still capable of performing all essential metabolic functions and undergoing mitosis for cell division,as these processes rely on other organelles like the nucleus,mitochondria,and ribosomes.

(B) The graph shows that as the substrate concentration increases,the velocity of the enzymatic reaction initially increases,reaches a maximum,and then decreases. This decline in reaction velocity at very high substrate concentrations is typically indicative of the presence of an enzyme inhibitor in the reaction mixture,which interferes with the enzyme's activity. Therefore,option $B$ is the correct interpretation of the provided graph.

(C) Option $C$ is correct. The anticoagulant hirudin is a protein that prevents blood clotting. The gene encoding hirudin was isolated from the medicinal leech $(Hirudo \text{ } medicinalis)$ and expressed in transgenic $Brassica \text{ } napus$ (rapeseed) seeds to produce the protein on a large scale.
Option $A$ is incorrect because $Bt$ stands for the bacterium $Bacillus \text{ } thuringiensis$, not biotechnology.
Option $B$ is incorrect because somatic hybridization involves the fusion of isolated protoplasts, not complete plant cells (which have cell walls).
Option $D$ is incorrect because the Flavr Savr tomato was engineered to delay ripening by inhibiting the production of ethylene, not enhancing it.

(A) To determine the configuration,assign priorities to the groups attached to each chiral center using Cahn-Ingold-Prelog $(CIP)$ rules.
For the top chiral center $(C1)$: The priorities are $-OH$ $(1)$,$-CH(OH)CH_3$ $(2)$,$-CH_3$ $(3)$,and $-H$ $(4)$. Since the lowest priority group $(-H)$ is on a horizontal bond,the configuration is reversed. The sequence $1$ $\rightarrow 2$ $\rightarrow 3$ is clockwise $(R)$,so it becomes $S$.
For the bottom chiral center $(C2)$: The priorities are $-OH$ $(1)$,$-CH(OH)CH_3$ $(2)$,$-CH_3$ $(3)$,and $-H$ $(4)$. Since the lowest priority group $(-H)$ is on a vertical bond,the configuration is maintained. The sequence $1$ $\rightarrow 2$ $\rightarrow 3$ is counter-clockwise $(S)$,so it remains $S$.
Thus,the configuration is $1S, 2S$.

(B) Let the density of the candle be $\rho_C$ and the density of the liquid be $\rho_L$. Let the total length of the candle be $2L$. Initially,the candle floats with length $L$ submerged.
By the principle of flotation,the weight of the candle equals the weight of the displaced liquid:
$\rho_C \cdot A \cdot (2L) = \rho_L \cdot A \cdot L$
$\Rightarrow \frac{\rho_C}{\rho_L} = \frac{1}{2} \dots (i)$
When the candle burns by a length $\Delta h = 2 \ cm$,let its new total length be $2L - 2$ and the new submerged length be $L'$.
By the principle of flotation: $\rho_C \cdot A \cdot (2L - 2) = \rho_L \cdot A \cdot L'$
Using $(i)$,we get: $\frac{1}{2} \rho_L \cdot A \cdot (2L - 2) = \rho_L \cdot A \cdot L'$
$L - 1 = L'$
The change in submerged length is $\Delta L' = L - L' = 1 \ cm$.
The top of the candle is at a height $h_{top} = L' + (2L - 2 - L') = 2L - 2$ above the bottom of the candle. However,relative to the liquid surface,the top is at height $H = 2L - 2 - L' = L - 1$.
Since $L$ is constant,the change in height of the top relative to the liquid surface is $\Delta H = 1 \ cm$.
Thus,the top of the candle falls at a rate of $1 \ cm/hour$.

(B) The phenomenon of brilliant colours seen on a thin oil film on water is due to the interference of light.
When white light falls on a thin film of oil,the light waves reflect from both the top and bottom surfaces of the film.
These reflected waves undergo superposition,leading to constructive or destructive interference depending on the path difference and the wavelength of the light.
Since the thickness of the oil film varies,different wavelengths (colours) undergo constructive interference at different points,resulting in the observation of brilliant colours.

(B) The reaction of $CH_3CH_2CH(F)CH_3$ with a strong base like $CH_3O^-$ in $CH_3OH$ proceeds via an $E2$ elimination mechanism.
According to Saytzeff's rule,the more substituted alkene is the major product.
Here,the base abstracts a proton from the $\beta$-carbon.
Removing a proton from $C_3$ leads to $CH_3CH=CHCH_3$ (but$-2-$ene),which is more substituted and thus more stable.
Removing a proton from $C_1$ leads to $CH_3CH_2CH=CH_2$ (but$-1-$ene),which is less substituted.
Therefore,$CH_3CH=CHCH_3$ is the major product.

(A) For the solid sphere to climb the inclined surface to a height $h$,its total initial kinetic energy must be greater than or equal to the potential energy gained at height $h$.
The total kinetic energy $(K.E.)_{total}$ is the sum of translational kinetic energy $(K.E.)_{T}$ and rotational kinetic energy $(K.E.)_{R}$:
$(K.E.)_{total} = (K.E.)_{T} + (K.E.)_{R} \ge P.E.$
$\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \ge mgh$
For a solid sphere,the moment of inertia $I = \frac{2}{5}mr^2$ and for pure rolling,$\omega = \frac{v}{r}$.
Substituting these values:
$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 \ge mgh$
$\frac{1}{2}mv^2 + \frac{1}{5}mv^2 \ge mgh$
$\frac{7}{10}mv^2 \ge mgh$
$v^2 \ge \frac{10gh}{7}$
$v \ge \sqrt{\frac{10gh}{7}}$

(A) Let the side of the square be $L$. The distance of each charge $Q$ from the centre of the square is $r = \frac{L}{\sqrt{2}}$.
When a negative charge $-q$ is placed at a distance $z$ from the centre along the $Z$-axis, the distance of the charge from each corner is $d = \sqrt{r^2 + z^2} = \sqrt{\frac{L^2}{2} + z^2}$.
The force exerted by each charge $Q$ on $-q$ is $F = \frac{kQq}{d^2}$.
The component of this force along the $Z$-axis is $F_z = -F \cos \theta$, where $\cos \theta = \frac{z}{d}$.
Thus, the net force on the charge $-q$ is $F_{net} = 4 \times \left( -\frac{kQq}{d^2} \times \frac{z}{d} \right) = -\frac{4kQqz}{(\frac{L^2}{2} + z^2)^{3/2}}$.
Since $z << L$, we can approximate $d \approx \frac{L}{\sqrt{2}}$.
Then $F_{net} \approx -\frac{4kQqz}{(L^2/2)^{3/2}} = -\left( \frac{4kQq}{L^3 / 2\sqrt{2}} \right) z = -kz$, where $k$ is a positive constant.
Since $F_{net} \propto -z$, the force is a restoring force, and the negative charge will perform simple harmonic motion (oscillate) along the $Z$-axis about the centre of the square.

(B) Let the length of the candle be $L$ and its density be $\rho_c$. Let the density of the liquid be $\rho_l$.
For the candle to float,the weight of the candle must be equal to the buoyant force:
$A \cdot L \cdot \rho_c \cdot g = A \cdot h \cdot \rho_l \cdot g$,where $h$ is the submerged length and $A$ is the cross-sectional area of the candle.
This gives $h = L \cdot (\rho_c / \rho_l)$.
Since the candle is floating,the ratio $\rho_c / \rho_l$ is constant. Let $k = \rho_c / \rho_l$. Then $h = k \cdot L$.
The height of the top of the candle above the liquid surface is $H = L - h = L - k \cdot L = L(1 - k)$.
When the candle burns,its length $L$ decreases at a rate $v = dL/dt = 2 \, cm/hour$.
The rate of change of the height of the top of the candle is $dH/dt = (dL/dt) \cdot (1 - k)$.
Assuming the candle is made of wax with density approximately half that of water (or generally,for a candle to float,$k < 1$),if we assume the candle floats with half its length submerged $(k = 0.5)$,then $dH/dt = 2 \cdot (1 - 0.5) = 1 \, cm/hour$.
Thus,the top of the candle falls at the rate of $1 \, cm/hour$.

(A) When the tube rotates with angular velocity $\omega$,the water in the horizontal arms experiences a centrifugal force directed outwards from the axis of rotation.
This force causes the water to be pushed towards the outer ends of the tube.
As a result,the water level in both vertical arms $A$ and $B$ rises.
The pressure gradient established in the rotating fluid provides the necessary centripetal force for the water to maintain circular motion.
Since both arms are at a distance from the axis of rotation,the water level in both sections $A$ and $B$ will rise.

(B) According to Moseley's law,the frequency $(f)$ of the characteristic $X-$rays emitted from a target is given by the relation:
$f = a(Z - b)^2$
where $a$ and $b$ are constants depending on the specific series of $X-$rays (e.g.,$K_{\alpha}, L_{\alpha}$).
For high atomic numbers $(Z)$,the term $b$ becomes negligible compared to $Z$.
Therefore,the relationship simplifies to:
$f \propto Z^2$
Thus,the frequency of characteristic $X-$rays is proportional to the square of the atomic number $(Z)$.

(D) The first ionization energy of $N$ $(1s^2, 2s^2, 2p^3)$ is higher than that of $O$ $(1s^2, 2s^2, 2p^4)$ because $N$ has a stable,exactly half-filled $2p$ subshell.
Across a period,the effective nuclear charge increases,not decreases,due to the addition of protons in the same shell while shielding remains relatively constant.
Therefore,the Assertion is incorrect and the Reason is also incorrect.

(C) In $SeCl_4$,the central atom $Se$ has $6$ valence electrons. It forms $4$ bonds with $Cl$ atoms and has $1$ lone pair of electrons.
Thus,the total number of electron pairs is $4 + 1 = 5$,which corresponds to $sp^3d$ hybridization.
Due to the presence of one lone pair,the geometry is a see-saw shape,not tetrahedral.
Therefore,the Assertion is correct,but the Reason is incorrect because $Se$ has only one lone pair,not two.

(D) The total number of electrons in a $B_2$ molecule is $10$.
The molecular orbital configuration is $\sigma(1s)^2 \sigma^*(1s)^2 \sigma(2s)^2 \sigma^*(2s)^2 \pi 2p_x^1 \pi 2p_y^1$.
Due to the presence of two unpaired electrons in the $\pi 2p$ orbitals,the $B_2$ molecule is paramagnetic.
The highest occupied molecular orbital $(HOMO)$ is of $\pi$ type,not $\sigma$ type.
Therefore,both the Assertion and the Reason are incorrect.

(A) For hydrogen gas $(H_2)$,the compressibility factor $Z$ is defined as $Z = \frac{PV}{nRT}$.
For $H_2$,$Z$ is always greater than $1$ at all pressures,and it increases linearly with pressure.
This indicates that the gas is more difficult to compress than an ideal gas,which is due to the dominance of repulsive forces between the molecules even at low pressures.
Therefore,both the Assertion and the Reason are correct,and the Reason correctly explains the Assertion.

(D) The presence of $NH_4Cl$ (a strong electrolyte) suppresses the dissociation of $NH_4OH$ (a weak base) due to the common ion effect ($NH_4^+$ ions).
This significantly reduces the concentration of $OH^-$ ions in the solution.
For the precipitation of $Ba(OH)_2$ to occur,the ionic product $[Ba^{2+}][OH^-]^2$ must exceed the solubility product constant $(K_{sp})$ of $Ba(OH)_2$.
Since the concentration of $OH^-$ is kept very low,the ionic product does not exceed $K_{sp}$,and no precipitation occurs.
Furthermore,$Ba(OH)_2$ is actually a strong base and is soluble in water.
Therefore,the Assertion is incorrect and the Reason is incorrect.

(C) $SiF_6^{2-}$ is known because $F$ has a small size,which allows six $F$ atoms to fit around the $Si$ atom without significant steric hindrance.
In $SiCl_6^{2-}$,the $Cl$ atoms are much larger than $F$ atoms,leading to significant inter-electronic repulsions and steric hindrance,which makes the ion unstable.
The reason provided is incorrect because the stability of $SiF_6^{2-}$ is primarily due to the small size of $F$ minimizing steric hindrance,not due to lone pair interaction with $d-$orbitals.

(D) The rate of nitration of benzene and hexadeuterobenzene is the same because the rate-determining step (formation of the sigma complex or carbocation) is the same in both cases.
This step does not involve the cleavage of the $C-H$ or $C-D$ bond,which occurs in the subsequent fast step.
Furthermore,the $C-D$ bond is actually stronger than the $C-H$ bond due to lower zero-point energy,making the Reason statement also incorrect.
Therefore,both the Assertion and the Reason are incorrect.

(A) The cyclopentadienyl anion contains $6\pi$ electrons ($4n+2$ where $n=1$),which satisfies $H$ückel's rule for aromaticity. It is planar,cyclic,and fully conjugated,making it highly stable due to aromaticity.
The allyl anion $(CH_2=CH-CH_2^-)$ is stabilized by resonance but is not aromatic.
Since the cyclopentadienyl anion is aromatic,it is significantly more stable than the non-aromatic allyl anion. Thus,both the Assertion and the Reason are correct,and the Reason correctly explains the Assertion.

(A) Galvanised iron is iron coated with a thin layer of zinc.
Since zinc has a more negative standard electrode potential $(E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V)$ compared to iron $(E^{\circ}_{Fe^{2+}/Fe} = -0.44 \ V)$,zinc acts as a sacrificial anode.
It undergoes oxidation in preference to iron,thereby protecting the iron from rusting.
Thus,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(B) Gibberellins promote seed germination in cereals by inducing the synthesis of hydrolyzing enzymes,such as $\alpha$-amylase and proteases. These enzymes break down stored starch and proteins into simpler,soluble forms that the developing embryo can use for growth.

(C) In humans,sex determination is based on the type of sperm that fertilizes the egg.
Human males are heterogametic,producing two types of gametes: $50\%$ with an $X$ chromosome and $50\%$ with a $Y$ chromosome.
If a sperm carrying an $X$ chromosome fertilizes the egg,the zygote develops into a female $(XX)$.
If a sperm carrying a $Y$ chromosome fertilizes the egg,the zygote develops into a male $(XY)$.
Therefore,the assertion is correct.
However,sex in humans is determined by the presence or absence of the $Y$ chromosome (specifically the $SRY$ gene),not by a polygenic trait involving cumulative effects of genes on $X$ and $Y$ chromosomes.
Thus,the reason is incorrect.

(A) In recombinant $DNA$ technology,human genes are frequently inserted into host organisms like bacteria (e.g.,$E. coli$) or yeast because these organisms have a very short generation time and multiply rapidly.
This rapid multiplication allows for the production of a large population of cells,which in turn facilitates the mass production of the desired gene product (protein).
Therefore,the Assertion is correct because these hosts are ideal for cloning and expression,and the Reason correctly explains that their rapid multiplication is the primary reason for their selection.

(C) $1$. In $Bt$-cotton, $Bt$ stands for the bacterium $Bacillus \text{ } thuringiensis$, not just biotechnology. Thus, option $A$ is incorrect.
$2$. Somatic hybridization involves the fusion of isolated protoplasts, not complete plant cells (which have cell walls). Thus, option $B$ is incorrect.
$3$. Hirudin is an anticoagulant protein originally found in leeches. It is produced commercially using transgenic $Brassica \text{ } napus$ (rapeseed) seeds. Thus, option $C$ is correct.
$4$. $Flavr \text{ } Savr$ tomato was engineered to delay ripening by inhibiting the production of ethylene (polygalacturonase enzyme inhibition), not enhancing it. Thus, option $D$ is incorrect.

(C) $Bt$ in '$Bt$ cotton' stands for $Bacillus$ $thuringiensis$,a soil bacterium from which the $Bt$ gene (encoding $Bt$ toxin) is obtained.
Somatic hybridization involves the fusion of protoplasts (i.e.,cells minus cell walls) of two different plant varieties.
$Flavr$ $savr$ is a transgenic tomato variety where the production of ethylene is delayed (not enhanced) to increase shelf life.
The anticoagulant hirudin is indeed produced from the seeds of transgenic $Brassica$ $napus$ at a commercial scale.

(A) The structure of phosphorus trioxide $(P_4O_6)$ consists of four phosphorus atoms at the corners of a tetrahedron,with six oxygen atoms bridging the edges between them. Thus,it has $6$ $P-O-P$ bridges.
In phosphorus pentoxide $(P_4O_{10})$,each phosphorus atom is further bonded to an additional terminal oxygen atom via a coordinate bond. The core structure remains the same as $P_4O_6$,meaning it also contains $6$ $P-O-P$ bridges.
Therefore,the number of $P-O-P$ bridges in both $P_4O_{10}$ and $P_4O_6$ is $6$ each.

(D) To determine the number of lone pairs on the central $Xe$ atom in each molecule,we use the formula: $\text{Lone pairs} = \frac{1}{2} \times (V - M - C + A)$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
$1.$ For $XeO_3$:
$Xe$ has $8$ valence electrons. Oxygen is divalent,so $M = 0$.
$\text{Lone pairs} = \frac{1}{2} \times (8 - 0) = 4$ electrons,which corresponds to $1$ lone pair.
$2.$ For $XeOF_4$:
$Xe$ has $8$ valence electrons. $F$ is monovalent $(M = 4)$,$O$ is divalent $(M = 0)$.
$\text{Lone pairs} = \frac{1}{2} \times (8 - 4) = 2$ electrons,which corresponds to $1$ lone pair.
$3.$ For $XeF_6$:
$Xe$ has $8$ valence electrons. $F$ is monovalent $(M = 6)$.
$\text{Lone pairs} = \frac{1}{2} \times (8 - 6) = 2$ electrons,which corresponds to $1$ lone pair.
Since all three molecules ($XeO_3$,$XeOF_4$,and $XeF_6$) have $1$ lone pair on the $Xe$ atom,the correct option is $(D)$.

(A) The correct answer is $A$.
Nucleophilicity is the ability of a species to donate an electron pair to an electrophile.
Factors affecting nucleophilicity include charge,electronegativity,and solvent effects.
Comparing the given species:
$1$. $C_2H_5SH$ (Ethanethiol): Sulfur is less electronegative than oxygen and nitrogen,making its lone pair more available for donation.
$2$. $CH_3COO^-$ (Acetate ion): The negative charge is delocalized over two oxygen atoms via resonance,which decreases its nucleophilicity.
$3$. $CH_3NH_2$ (Methylamine): Nitrogen is more electronegative than sulfur,making its lone pair less available.
$4$. $NCCH_2^-$ (Cyanomethyl anion): The negative charge is stabilized by the strong electron-withdrawing $-I$ and $-M$ effect of the cyano $(-CN)$ group,significantly reducing its nucleophilicity.
Therefore,$C_2H_5SH$ acts as the strongest nucleophile among the given options due to the lower electronegativity of sulfur compared to oxygen and nitrogen,and the lack of resonance stabilization or strong electron-withdrawing groups.

(B) $3$-Phenylpropene $(C_6H_5CH_2CH=CH_2)$ reacts with $HBr$ via electrophilic addition.
According to Markownikoff's rule,the electrophile $(H^+)$ adds to the terminal carbon to form a more stable secondary benzylic carbocation $(C_6H_5CH^+CH_2CH_3)$.
The nucleophile $(Br^-)$ then attacks this carbocation to form the major product,$1$-bromo-$1$-phenylpropane $(C_6H_5CH(Br)CH_2CH_3)$.

(D) The basicity of amines is primarily due to the availability of a lone pair of electrons on the nitrogen atom for donation to a proton or Lewis acid.
In triethylamine,the lone pair on the nitrogen is localized and readily available for donation.
In pyridine,the nitrogen atom is $sp^2$ hybridized,and its lone pair is part of the aromatic $\pi$-electron system (delocalized) to satisfy $H$ückel's rule.
Because the lone pair is delocalized,it is significantly less available for protonation,making pyridine less basic than triethylamine.

(B) The $ABCABC$ packing sequence corresponds to the face-centered cubic $(FCC)$ or cubic close-packed $(CCP)$ structure.
In any close-packed structure,if the number of atoms per unit cell is $Z$,the number of octahedral voids is $Z$ and the number of tetrahedral voids is $2Z$.
Therefore,the number of tetrahedral voids is $2Z$.

(C) The nuclear reaction is: $_{92}U^{238} \xrightarrow{-8\alpha, -6\beta} _{Z}X^{A}$.
Change in mass number $(A)$: $238 - (8 \times 4) = 238 - 32 = 206$.
Change in atomic number $(Z)$: $92 - (8 \times 2) + (6 \times 1) = 92 - 16 + 6 = 82$.
The product nucleus is $_{82}X^{206}$.
Number of protons $(p)$ $= 82$.
Number of neutrons $(n)$ $= A - Z = 206 - 82 = 124$.
Neutron/proton ratio $= n/p = 124/82 = 62/41$.

(B) The rate law for the reaction is given by: $\text{Rate} = k[A]^n$,where $n$ is the order of reaction.
For the first condition: $2.4 = k(2.2)^n$ ... $(i)$
For the second condition,$[A] = 2.2/2 = 1.1 \, mM$ and $\text{Rate} = 0.6 \, mM \, s^{-1}$:
$0.6 = k(1.1)^n$ ... $(ii)$
Dividing equation $(i)$ by equation $(ii)$:
$\frac{2.4}{0.6} = \frac{k(2.2)^n}{k(1.1)^n}$
$4 = (2)^n$
$2^2 = 2^n$
Therefore,$n = 2$.
The order of reaction with respect to $A$ is $2$.

(C) In an endothermic reaction,the potential energy of the products $(P)$ is higher than that of the reactants $(R)$.
Activation energy $(E_a)$ is the energy difference between the transition state (peak of the curve) and the reactants $(R)$.
Diagram $C$ shows that the energy of the products is higher than the reactants (endothermic) and the energy barrier (peak) is significantly higher than the reactant energy,indicating high activation energy.

(C) The given cell reaction is $2AgCl_{(s)} + H_{2(g)} \to 2HCl_{(aq)} + 2Ag_{(s)}$.
In this reaction,$H_2$ is oxidized to $H^+$ at the anode,and $AgCl$ is reduced to $Ag$ at the cathode.
The anode is a standard hydrogen electrode represented as $Pt_{(s)} | H_{2(g)}, 1 \ bar | H^+_{(aq)}$.
The cathode involves the $Ag/AgCl$ electrode,represented as $Cl^-_{(aq)} | AgCl_{(s)} | Ag_{(s)}$.
Combining these,the cell notation is $Pt_{(s)} | H_{2(g)}, 1 \ bar | 1 \ M \ HCl_{(aq)} | AgCl_{(s)} | Ag_{(s)}$.

(C) Benzene is a non-polar molecule. According to the principle of 'like dissolves like',non-polar substances are better dispersed by other non-polar substances.
$m$-methyl nonylbenzene is a non-polar molecule with a long hydrocarbon chain and a substituted benzene ring,making it highly lipophilic and suitable for dispersing non-polar benzene.
In contrast,options $(A)$,$(B)$,and $(D)$ contain polar functional groups (carboxylate ions or chlorine atoms),which make them amphiphilic or polar,thus less suitable for dispersing pure non-polar benzene compared to the purely non-polar $m$-methyl nonylbenzene.

(B) The chemical formulas for the given ores are:
$1$. Pyrolusite: $MnO_2$ (Oxide ore)
$2$. Malachite: $CuCO_3 \cdot Cu(OH)_2$ (Carbonate ore)
$3$. Diaspore: $Al_2O_3 \cdot H_2O$ (Oxide/Hydroxide ore)
$4$. Cassiterite: $SnO_2$ (Oxide ore)
Therefore,Malachite is a carbonate ore.

(A) In the case of $Tl$ (Thallium),the $Tl^{+}$ ion is more stable than the $Tl^{3+}$ ion due to the inert pair effect,where the $6s^{2}$ electrons are reluctant to participate in bonding.
Consequently,$Tl^{3+}$ acts as a strong oxidizing agent and gets reduced to $Tl^{+}$:
$Tl^{3+} + 2e^{-} \to Tl^{+}$
In contrast,for $Cu$,$Cr$,and $V$,the higher oxidation states ($Cu^{2+}$,$Cr^{3+}$,$VO^{2+}$) are generally more stable in aqueous solution due to higher hydration energy or electronic configuration stability.

(A) The reaction is: $K_2MnF_6 + 2SbF_5 \rightarrow 2KSbF_6 + MnF_3 + \frac{1}{2}F_2$.
In this reaction,the stronger Lewis acid $SbF_5$ displaces the weaker one,$MnF_4$,from its salt.
$MnF_4$ is unstable and readily decomposes to give $MnF_3$ and fluorine gas $(F_2)$.

(C) The aqueous solution of $CoCl_2$ contains the octahedral complex $[Co(H_2O)_6]^{2+}$,which is pink in color.
Upon adding excess concentrated $HCl$,the chloride ions $(Cl^-)$ act as ligands and displace the water molecules.
This leads to the formation of the tetrahedral complex $[CoCl_4]^{2-}$,which is blue in color.
The reaction is: $[Co(H_2O)_6]^{2+} + 4Cl^- \rightarrow [CoCl_4]^{2-} + 6H_2O$.

(A) The energy of absorption $(E)$ is inversely proportional to the wavelength of absorption $(\lambda)$,i.e.,$E = \frac{hc}{\lambda}$.
According to the spectrochemical series,the field strength of the ligands follows the order: $H_2O < NH_3 < NO_2^-$.
Stronger field ligands cause a larger crystal field splitting energy $(\Delta_o)$,which corresponds to higher energy absorption and thus a shorter wavelength of absorption.
Therefore,the order of field strength is $[Ni(H_2O)_6]^{2+} < [Ni(NH_3)_6]^{2+} < [Ni(NO_2)_6]^{4-}$.
Consequently,the order of wavelength of absorption is the inverse: $[Ni(NO_2)_6]^{4-} < [Ni(NH_3)_6]^{2+} < [Ni(H_2O)_6]^{2+}$.

(B) Optical isomerism is shown by complexes that lack a plane of symmetry and a center of inversion (i.e.,they are chiral).
$1$. $[Co(en)_3]Cl_3$: This complex contains three bidentate ethylenediamine $(en)$ ligands. It exists as a pair of non-superimposable mirror images (enantiomers) and thus shows optical isomerism.
$2$. $cis-[Co(en)_2Cl_2]Cl$: In the $cis$ isomer,the two $Cl$ ligands are adjacent. This configuration lacks a plane of symmetry,making it chiral and optically active.
Therefore,both complexes in option $B$ exhibit optical isomerism.

(B) The reaction of $CH_3CH_2CH(F)CH_3$ with a strong base like $CH_3O^-$ in $CH_3OH$ proceeds via an $E2$ elimination mechanism.
Two possible alkenes can be formed: $CH_3CH=CHCH_3$ (but-$2$-ene) and $CH_3CH_2CH=CH_2$ (but-$1$-ene).
According to Saytzeff's rule,the more substituted alkene is the major product because it is more stable.
$CH_3CH=CHCH_3$ has two alkyl substituents on the double-bonded carbons,while $CH_3CH_2CH=CH_2$ has only one.
Therefore,$CH_3CH=CHCH_3$ is the major product.

(C) Step $I$: Ethyl acetate $(CH_3CO_2C_2H_5)$ undergoes Claisen condensation in the presence of sodium ethoxide $(NaOC_2H_5)$ to form ethyl acetoacetate $(CH_3COCH_2COOC_2H_5)$,which is compound $A$.
Step $II$: Ethyl acetoacetate,when heated in the presence of an acid,undergoes decarboxylation and dehydration/rearrangement to form the cyclic compound shown in the provided image (a derivative of cyclobutanone or similar cyclic structure as per the reaction mechanism provided in the image).

(C) The conversion of an amide $(C_6H_5CONHCH_3)$ to an amine $(C_6H_5CH_2NHCH_3)$ involves the reduction of the carbonyl group $(C=O)$ to a methylene group $(CH_2)$.
$LiAlH_4$ is a strong reducing agent that reduces amides to amines,but it typically reduces the carbonyl group to a methylene group $(C_6H_5CONHCH_3 \xrightarrow{LiAlH_4} C_6H_5CH_2NHCH_3)$.
However,the Clemmensen reduction $(Zn-Hg/HCl)$ is specifically used to reduce aldehydes and ketones to alkanes,but it is not the standard reagent for reducing amides to amines.
In the context of standard organic chemistry reagents,$LiAlH_4$ is the correct reagent for the reduction of amides to amines.

(C) Methyl isocyanate $(CH_3NCO)$ is industrially prepared by the reaction of methylamine $(CH_3NH_2)$ with phosgene $(COCl_2)$.
The chemical reaction is:
$CH_3NH_2 + COCl_2 \rightarrow CH_3NHCOCl + HCl$
$CH_3NHCOCl \rightarrow CH_3NCO + HCl$
Thus,the chemicals used are $(i)$ $Methylamine$ and $(ii)$ $Phosgene$.

(D) In the Hofmann rearrangement (Hofmann bromamide degradation),the reaction proceeds through the formation of several intermediates:
$1.$ $N$-bromoamide $(R-CONHBr)$
$2.$ Acyl nitrene $(R-CO-\ddot{N})$
$3.$ Alkyl isocyanate $(R-N=C=O)$
Alkyl isocyanide $(R-NC)$ is not formed as an intermediate in this reaction.

(A) . $\alpha$-Keratin is a fibrous protein. Fibrous proteins are generally insoluble in water due to their long,thread-like structure and extensive hydrogen bonding.

(A) To determine the magnetic property,we look at the electronic configuration of the central metal ion and the nature of the ligand:
$1$. In $[Ni(CN)_4]^{2-}$,$Ni$ is in the $+2$ oxidation state $(3d^8)$. $CN^-$ is a strong field ligand,causing pairing of electrons in the $3d$ orbitals,resulting in no unpaired electrons. Thus,it is diamagnetic.
$2$. In $[NiCl_4]^{2-}$,$Ni^{2+}$ $(3d^8)$ is bonded to $Cl^-$ (a weak field ligand),which does not cause pairing,resulting in two unpaired electrons. Thus,it is paramagnetic.
$3$. In $[CoCl_4]^{2-}$,$Co^{2+}$ $(3d^7)$ is bonded to $Cl^-$ (a weak field ligand),resulting in three unpaired electrons. Thus,it is paramagnetic.
$4$. In $[CoF_6]^{2-}$,$Co^{4+}$ $(3d^5)$ is bonded to $F^-$ (a weak field ligand),resulting in five unpaired electrons. Thus,it is paramagnetic.
Therefore,the correct option is $A$.

(B) The correct statement is that the third base of the codon is less specific,which is known as the $Wobble$ $Hypothesis$.
$m-RNA$ does not directly recognize amino acids; instead,$t-RNA$ acts as an adaptor molecule.
Most amino acids are coded by more than one codon (degeneracy of genetic code).
Every $t-RNA$ molecule has only one specific amino acid attachment site at its $3'$-end.

(D) Globular proteins are proteins where the polypeptide chain is folded into a spherical shape.
They are found in various biological systems:
$1$. In blood,albumins and globulins are globular proteins.
$2$. In milk,caseinogen is a globular protein.
$3$. In egg white,albumin is a globular protein.
Therefore,all of the given options contain globular proteins.

(B) $Ozone$ $(O_3)$ is a powerful oxidising agent because it is thermodynamically unstable and readily decomposes to give nascent oxygen $(O_3 \to O_2 + [O])$.
The magnetic property of $O_3$ is diamagnetic,whereas $O_2$ is paramagnetic due to the presence of two unpaired electrons in its antibonding molecular orbitals.
While both statements are factually correct,the magnetic property of $O_3$ is not the reason for its high oxidising power. Therefore,the Reason is not the correct explanation of the Assertion.

(A) Acetic acid $(CH_3COOH)$ undergoes dimerization in nonpolar solvents like benzene due to hydrogen bonding,resulting in an observed molecular weight that is double the theoretical value.
In polar solvents like water,acetic acid undergoes dissociation into ions ($CH_3COO^-$ and $H^+$),resulting in an observed molecular weight that is lower than the theoretical value.
Since the extent of association or dissociation depends on the polarity of the solvent,the molecular weight determined by the depression in freezing point method is different in benzene and water.
Thus,both the Assertion and Reason are correct,and the Reason correctly explains why the molecular weight differs due to the solvent's polarity.

(C) The extraction of iron from $Fe_2O_3$ is carried out in a blast furnace by heating with coke $(C)$.
The coke reacts with oxygen to form $CO$,which acts as the reducing agent:
$Fe_2O_3 3CO \to 2Fe 3CO_2$
Thus,the assertion is correct.
The reaction $Fe_2O_{3(s)} \to 2Fe_{(s)} 3/2 O_{2(g)}$ is not spontaneous because it involves the breaking of strong metal-oxygen bonds,which is highly endothermic $(\Delta H > 0)$ and results in an increase in entropy $(\Delta S > 0)$,but at standard temperatures,the Gibbs free energy change $(\Delta G = \Delta H - T\Delta S)$ remains positive.
Therefore,the reason is incorrect.

(A) The standard reduction potential of $Zn^{2+}/Zn$ is $-0.76 \ V$,while that of $Fe^{2+}/Fe$ is $-0.44 \ V$.
Since zinc has a more negative electrode potential than iron,it acts as a sacrificial anode.
Zinc gets oxidized in preference to iron,thereby protecting the iron from rusting.
Thus,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(C) The reaction between $H_2S$ and $SO_2$ in the presence of $Fe_2O_3$ catalyst is given by: $2H_2S + SO_2 \to 2H_2O + 3S \downarrow$.
In this reaction,$H_2S$ acts as a reducing agent and is oxidized to elemental sulphur $(S)$,while $SO_2$ acts as an oxidizing agent and is reduced.
Therefore,the Assertion is correct,but the Reason is incorrect because $SO_2$ acts as an oxidizing agent in this reaction,not a reducing agent.

(C) In potassium ferrocyanide,$K_4[Fe(CN)_6]$,$Fe$ is in the $+2$ oxidation state ($Fe^{2+}$: $3d^6$). Since $CN^-$ is a strong field ligand,it causes pairing of electrons,resulting in $t_{2g}^6 e_g^0$ configuration with $n=0$ unpaired electrons,making it diamagnetic.
In potassium ferricyanide,$K_3[Fe(CN)_6]$,$Fe$ is in the $+3$ oxidation state ($Fe^{3+}$: $3d^5$). $CN^-$ causes pairing,resulting in $t_{2g}^5 e_g^0$ configuration with $n=1$ unpaired electron,making it paramagnetic.
The Assertion is correct.
The Reason states that crystal field splitting in ferrocyanide is greater than in ferricyanide. This is incorrect because crystal field splitting energy $(\Delta_o)$ increases with the oxidation state of the metal ion. Since $Fe^{3+}$ has a higher oxidation state than $Fe^{2+}$,the crystal field splitting in the ferricyanide ion is greater than in the ferrocyanide ion. Thus,the Reason is incorrect.

(D) Friedel-Crafts reactions (alkylation and acylation) do not occur with nitrobenzene because the strong electron-withdrawing $-NO_2$ group deactivates the benzene ring towards electrophilic attack.
Furthermore,the lone pair on the nitrogen atom of the $-NO_2$ group coordinates with the Lewis acid catalyst (e.g.,$AlCl_3$),which further deactivates the ring.
Therefore,the Assertion is incorrect.
Since nitrobenzene is highly deactivated,it does not easily undergo electrophilic substitution reactions,making the Reason also incorrect.

(A) Alkyl isocyanides $(R-NC)$ undergo hydrolysis in the presence of acid to form alkyl formamides $(R-NH-CHO)$.
In the mechanism,the terminal carbon atom of the isocyanide group possesses a lone pair and a partial negative charge,allowing it to act as a nucleophile and get protonated by $H^+$.
After protonation,the carbon atom becomes electron-deficient (electrophilic) and is subsequently attacked by water $(H_2O)$.
The reaction sequence is: $R-N \equiv C$ $\xrightarrow{H^+} R-N^+ \equiv CH$ $\xrightarrow{H_2O} R-N=C(H)(OH) \rightleftharpoons R-NH-CHO$.

(B) The borax bead test is used to identify transition metal ions by the formation of characteristic coloured metaborates. $Al^{3+}$ ions do not form coloured metaborates,which is why the test is not suitable for $Al(III)$.
$Al_2O_3$ is indeed insoluble in water,but this property is unrelated to the failure of the borax bead test for aluminium. Therefore,both statements are correct,but the reason does not explain the assertion.