AIIMS 2001 Chemistry Question Paper with Answer and Solution

(D) According to Hooke's law,within the elastic limit,stress is directly proportional to strain.
Mathematically,$\text{Stress} \propto \text{Strain}$,which implies $\frac{\text{Stress}}{\text{Strain}} = \text{Constant}$.
This constant is known as the modulus of elasticity (e.g.,Young's modulus $Y$).
Since this ratio depends only on the nature of the material and not on the magnitude of the stress applied (as long as it is within the elastic limit),the ratio remains constant.

(B) In a purely inductive circuit (containing only $L$),the current lags behind the voltage by a phase difference of $\pi / 2$ or $90^{\circ}$.
In a purely capacitive circuit (containing only $C$),the current leads the voltage by a phase angle of $\pi / 2$ or $90^{\circ}$.
In a purely resistive circuit (containing only $R$),the current is in phase with the applied voltage.
Therefore,the correct option is $B$.

(C) The ratio of slit widths is equal to the ratio of intensities of the light waves emerging from them,i.e.,$\frac{I_1}{I_2} = \frac{w_1}{w_2} = 1:9$.
Since intensity $I \propto a^2$,where $a$ is the amplitude,we have $\frac{a_1}{a_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{1}{9}} = \frac{1}{3}$.
Let $a_1 = a$ and $a_2 = 3a$.
The intensity at maxima is $I_{\max} = (a_1 + a_2)^2 = (a + 3a)^2 = (4a)^2 = 16a^2$.
The intensity at minima is $I_{\min} = (a_1 - a_2)^2 = (a - 3a)^2 = (-2a)^2 = 4a^2$.
The ratio of intensity at minima to that at maxima is $\frac{I_{\min}}{I_{\max}} = \frac{4a^2}{16a^2} = \frac{1}{4}$.

(B) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$.
Given: Mass $(m) = 10^{-6} \ kg$,Velocity $(v) = 10 \ ms^{-1}$,and Planck's constant $(h) = 6.63 \times 10^{-34} \ J \cdot s$.
Substituting the values:
$\lambda = \frac{6.63 \times 10^{-34}}{10^{-6} \times 10} = \frac{6.63 \times 10^{-34}}{10^{-5}} = 6.63 \times 10^{-29} \ m$.
Therefore,the correct option is $(B)$.

(B) The total number of electrons in the given configuration is $2 + 2 + 1 + 1 + 1 = 7$.
An element with $7$ electrons is Nitrogen $(N)$.
The electronic configuration of Nitrogen $(Z = 7)$ is $1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1$.

(A) For a set of quantum numbers to be valid,the following rules must be satisfied:
$1$. $n$ is a positive integer $(1, 2, 3, \dots)$.
$2$. $l$ can have values from $0$ to $n-1$.
$3$. $m$ can have values from $-l$ to $+l$ (including $0$).
$4$. $s$ can be $+1/2$ or $-1/2$.
Checking the options:
$(A)$ $n=3, l=2, m=2, s=+1/2$: Here $l < n$ $(2 < 3)$ and $|m| \le l$ $(2 \le 2)$. This set is valid.
$(B)$ $n=3, l=4$: Invalid because $l$ must be less than $n$.
$(C)$ $n=4, l=0, m=2$: Invalid because $|m|$ must be $\le l$ $(2 \not\le 0)$.
$(D)$ $n=4, l=4$: Invalid because $l$ must be less than $n$.

(A) For a valid set of quantum numbers,the following rules must be satisfied:
$1$. $n$ (principal quantum number) must be a positive integer $(1, 2, 3, ...)$.
$2$. $l$ (azimuthal quantum number) can have values from $0$ to $n-1$.
$3$. $m$ (magnetic quantum number) can have values from $-l$ to $+l$ including $0$.
In option $A$,$n = 1$. The allowed value for $l$ is only $0$ $(n-1 = 1-1 = 0)$. Since $l = 2$ is given,this set is invalid because $l$ cannot be greater than or equal to $n$.

(A) compound is considered polar if there is a significant difference in electronegativity between the bonded atoms,resulting in a permanent dipole moment.
In the given options,all are polar,but $HF$ exhibits the highest degree of polarity due to the largest electronegativity difference between $H$ $(2.1)$ and $F$ $(4.0)$.
Therefore,$HF$ is the most polar compound among the choices.

(A) $IF_7$ exhibits $sp^3d^3$ hybridization with a pentagonal bipyramidal geometry.
In this structure,the equatorial bond angles are $72^o$,which is the smallest among the given molecules.
$CH_4$ has a bond angle of $109.5^o$,$BeF_2$ has $180^o$,and $BF_3$ has $120^o$.

(D) The given reaction is $2SO_3 \rightleftharpoons 2SO_2 + O_2$ with $K_p = 1.3 \times 10^{-3} \ atm$.
For this reaction,$\Delta n = (2+1) - 2 = 1$.
We need to find $K_c$ for the reverse reaction $2SO_2 + O_2 \rightleftharpoons 2SO_3$.
Let $K_p'$ be the equilibrium constant for the reverse reaction,so $K_p' = \frac{1}{K_p} = \frac{1}{1.3 \times 10^{-3}} \ atm^{-1} \approx 769.23 \ atm^{-1}$.
For the reverse reaction,$\Delta n = 2 - (2+1) = -1$.
Using the relation $K_p = K_c(RT)^{\Delta n}$,we get $K_c = K_p(RT)^{-\Delta n} = K_p(RT)^1$.
$K_c = 769.23 \times (0.0821 \times 700) = 769.23 \times 57.47 \approx 44208$.
Wait,re-evaluating the standard approach: The question asks for $K_c$ of $2SO_2 + O_2 \rightleftharpoons 2SO_3$. Given $K_p$ for $2SO_3 \rightleftharpoons 2SO_2 + O_2$ is $1.3 \times 10^{-3}$.
$K_c = K_p(RT)^{-\Delta n}$. For $2SO_2 + O_2 \rightleftharpoons 2SO_3$,$\Delta n = 2 - 3 = -1$.
$K_p$ for this reaction is $1 / (1.3 \times 10^{-3}) = 769.23$.
$K_c = 769.23 / (0.0821 \times 700)^{-1} = 769.23 \times 57.47 = 44208$.
Given the options,the calculation $K_c = K_p / (RT)^{\Delta n}$ with $\Delta n = -1$ leads to $1.3 \times 10^{-3} \times (0.0821 \times 700) = 7.46 \times 10^{-2}$. Thus,option $D$ is correct.

(D) strong base is a substance that completely dissociates into its constituent ions in an aqueous solution.
$NaOH$ (sodium hydroxide) is a strong base because it undergoes complete ionization in water to provide $OH^{-}$ ions.
Therefore,both the ability to ionize easily and the production of $OH^{-}$ ions are characteristic of its strength as a base.

(A) The strength of a conjugate base is inversely proportional to the strength of its corresponding acid.
The order of acidic strength of the oxyacids of chlorine is $HClO < HClO_2 < HClO_3 < HClO_4$.
Since $HClO$ is the weakest acid among the given options,its conjugate base,$ClO^-$,is the strongest Bronsted base.

(C) Work is not a state function because its value depends on the path followed during a process.
Internal energy,enthalpy,and entropy are state functions because their values depend only on the state of the system and not on the path taken to reach that state.

(A) The internal energy $(U)$ of a substance is a state function that depends on the temperature of the system.
For most substances,as the temperature increases,the kinetic energy of the particles increases,which leads to an increase in the total internal energy of the substance.
Therefore,the internal energy increases with an increase in temperature.

(A) For an adiabatic process,the work done $(W)$ is given by the change in internal energy: $W = -\Delta U = -nC_v(T_2 - T_1) = nC_v(T_1 - T_2)$.
Given: $n = 1 \, mol$,$C_v = 20 \, J \, K^{-1} \, mol^{-1}$,$T_1 = 27 + 273 = 300 \, K$,and $W = 3 \, kJ = 3000 \, J$.
Substituting the values: $3000 = 1 \times 20 \times (300 - T_2)$.
$3000 = 6000 - 20T_2$.
$20T_2 = 3000$.
$T_2 = 150 \, K$.

(C) process is defined as reversible in thermodynamics if the system and surroundings are always in equilibrium with each other throughout the entire process. This implies that the driving force is infinitesimally small,allowing the process to be reversed at any stage.

(A) spectator ion is an ion that exists in the same form on both the reactant and product sides of a chemical equation.
In the given reaction: $Zn(s) + 2H^{+}(aq) + 2Cl^{-}(aq) \to Zn^{2+}(aq) + 2Cl^{-}(aq) + H_2(g)$.
The $Cl^{-}$ ion appears on both sides of the equation without undergoing any change in its oxidation state or chemical form.
Therefore,$Cl^{-}$ is the spectator ion.

(A) The size of an ion depends on the effective nuclear charge and the number of electrons.
$1$. $A$ cation $(O_2^+)$ is formed by the loss of an electron,which increases the effective nuclear charge per electron,resulting in a smaller size compared to the parent molecule $(O_2)$.
$2$. Anions ($O_2^-$ and $O_2^{2-}$) are formed by the gain of electrons,which increases inter-electronic repulsion and decreases the effective nuclear charge,resulting in a larger size compared to the parent molecule.
$3$. Therefore,the order of size is $O_2^{2-} > O_2^- > O_2 > O_2^+$.
$4$. Thus,$O_2^+$ is the smallest ion.

(D) The correct answer is $D$.
Calcium is an alkaline earth metal,which is highly reactive and cannot be obtained by carbon reduction.
It is extracted by the electrolysis of molten $CaCl_2$ (fused salt) because electrolysis of aqueous $CaCl_2$ would result in the evolution of hydrogen gas at the cathode instead of calcium metal.
At the cathode: $Ca^{+2} + 2e^{-} \to Ca$
At the anode: $2Cl^{-} \to Cl_2 + 2e^{-}$

(A) The solubility of alkaline earth metal hydroxides increases as we move down the group from $Be$ to $Ba$.
$Be(OH)_2$ is amphoteric and has very low solubility in water compared to the other hydroxides in the group.
Therefore,$Be(OH)_2$ is considered insoluble in water.

(A) The $C-C$ bond length in benzene is $1.39 \ \mathring{A}$.
This value is intermediate between the single bond length of $C-C$ $(1.54 \ \mathring{A})$ and the double bond length of $C=C$ $(1.34 \ \mathring{A})$.
This is due to the delocalization of $\pi$-electrons in the benzene ring,known as resonance.

(D) The stability of carbocations follows the order: $3^\circ > 2^\circ > 1^\circ > CH_3^+$.
This is due to the inductive effect ($+I$ effect) and hyperconjugation from the alkyl groups.
$CH_3-C^+(CH_3)-CH_3$ is a tertiary $(3^\circ)$ carbocation,which is the most stable among the given options.

(C) Given: $98 \ \%$ $H_2SO_4$ by weight means $98 \ g$ of $H_2SO_4$ is present in $100 \ g$ of the solution.
Step $1$: Calculate the volume of the solution.
$Volume = \frac{Mass}{Density} = \frac{100 \ g}{1.84 \ g/cc} \approx 54.35 \ cc = 0.05435 \ L$.
Step $2$: Calculate the moles of solute $(H_2SO_4)$.
Molar mass of $H_2SO_4 = 2(1) + 32 + 4(16) = 98 \ g/mol$.
Moles of $H_2SO_4 = \frac{98 \ g}{98 \ g/mol} = 1 \ mol$.
Step $3$: Calculate Molarity $(M)$.
$M = \frac{\text{Moles of solute}}{\text{Volume of solution in } L} = \frac{1 \ mol}{0.05435 \ L} \approx 18.4 \ M$.

(A) . Heavy water $(D_2O)$ is used as a moderator in a nuclear reactor.
It slows down the speed of fast-moving neutrons to thermal energy levels.
It also acts as a coolant in some reactor designs.

(C) The nitration of benzene involves the reaction of benzene with a nitrating mixture $(conc. \ HNO_3 + conc. \ H_2SO_4)$.
The rate-determining step of this electrophilic aromatic substitution reaction is the formation of the electrophile,the nitronium ion $(NO_2^+)$,from the nitrating mixture.

(D) The total number of individuals of a species per unit area or volume in a given habitat is known as absolute density. It represents the actual count of the population size within a specific area.

(B) Auxins,such as $NAA$ (Naphthalene Acetic Acid),are known to inhibit the process of leaf abscission (the shedding of leaves). $GA_3$ (Gibberellic acid) generally promotes growth,while ethylene is a plant hormone that promotes senescence and abscission. Zeatin is a type of cytokinin that promotes cell division but is not primarily associated with the inhibition of abscission compared to auxins. Therefore,$NAA$ is the correct answer.

(C) Work is not a state function.
It is a path function because its value depends on the path taken during the process.

(D) According to Heisenberg's uncertainty principle,the product of uncertainty in position $(\Delta x)$ and uncertainty in momentum $(\Delta p)$ is given by: $\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$.
Since the uncertainty in position $(\Delta x)$ is the same for both the electron and the helium atom $(1.0 \, nm)$,the uncertainty in momentum $(\Delta p)$ must also be the same for both particles to satisfy the inequality.
Given that the uncertainty in the momentum of the electron is $50 \times 10^{-26} \, kg \, m \, s^{-1}$,the minimum uncertainty in the measurement of the momentum of the helium atom will also be $50 \times 10^{-26} \, kg \, m \, s^{-1}$.

(D) Baking soda is $NaHCO_3$.
Baking powder is a mixture of baking soda $(NaHCO_3)$ and a mild edible acid,such as potassium hydrogen tartrate $(KHC_4H_4O_6)$,also known as cream of tartar.
When mixed with water,the acid reacts with the baking soda to release $CO_2$ gas,which helps the dough rise.
Therefore,$KHC_4H_4O_6$ transforms baking soda into baking powder.

(A) The distance traveled by a body in the $n^{th}$ second is given by $S_n = u + \frac{a}{2}(2n - 1)$.
For body $A$,it starts from rest $(u=0)$ and we consider the $5^{th}$ second of its motion. Thus,$S_{A} = 0 + \frac{a_1}{2}(2 \times 5 - 1) = \frac{9a_1}{2}$.
Body $B$ starts $2 \ s$ after $A$. Therefore,the $5^{th}$ second after the start of $A$ corresponds to the $(5-2) = 3^{rd}$ second of the motion of body $B$.
For body $B$,$S_{B} = 0 + \frac{a_2}{2}(2 \times 3 - 1) = \frac{5a_2}{2}$.
Given that the distances are equal,$S_A = S_B$.
$\frac{9a_1}{2} = \frac{5a_2}{2}$.
$9a_1 = 5a_2 \Rightarrow \frac{a_1}{a_2} = \frac{5}{9}$.

(D) Work done is defined as the dot product of the force vector $\vec F$ and the displacement vector $\vec d$.
Given,$\vec F = (3\hat i + 4\hat j)\, N$ and $\vec d = (3\hat i + 4\hat j)\, m$.
The formula for work done is $W = \vec F \cdot \vec d$.
Substituting the values,we get $W = (3\hat i + 4\hat j) \cdot (3\hat i + 4\hat j)$.
Using the dot product property $\hat i \cdot \hat i = 1$,$\hat j \cdot \hat j = 1$,and $\hat i \cdot \hat j = 0$,we calculate:
$W = (3 \times 3) + (4 \times 4) = 9 + 16 = 25\, J$.

(B) For an adiabatic process,the relationship between pressure $P$ and temperature $T$ is given by $P^{1-\gamma} T^{\gamma} = \text{constant}$,which can be rewritten as $P \propto T^{\gamma / (\gamma - 1)}$.
For a monoatomic gas,the adiabatic index $\gamma = 5/3$.
Substituting the value of $\gamma$ into the expression for $C$:
$C = \frac{\gamma}{\gamma - 1} = \frac{5/3}{5/3 - 1} = \frac{5/3}{2/3} = 5/2$.
Therefore,the value of $C$ is $5/2$.

(D) The Assertion is incorrect because molecules in a gas possess a distribution of speeds (Maxwell-Boltzmann distribution) at a given temperature,not the same speed.
The Reason is also incorrect because,for a pure gas,all molecules are identical in size and shape.
Therefore,both Assertion and Reason are incorrect.

(C) Magnesium $(Mg)$ continues to burn in nitric oxide $(NO)$ because the heat evolved during the combustion of $Mg$ is sufficient to decompose $NO$ into $N_2$ and $O_2$.
The reaction is: $2Mg + 2NO \rightarrow 2MgO + N_2$.
Since the heat evolved decomposes $NO$,the Reason provided is incorrect.

(A) The chlorination of $CH_4$ is a free radical substitution reaction.
In the dark,there is no energy source (like $h\nu$ or high temperature) to initiate the homolytic cleavage of the $Cl-Cl$ bond to produce $Cl$ free radicals.
Therefore,the reaction does not occur in the dark.
Sunlight provides the necessary energy to initiate the formation of $Cl$ free radicals,which then propagate the reaction.
Thus,both the Assertion and Reason are correct,and the Reason is the correct explanation of the Assertion.

(D) $H_2O_2$ is prepared by the action of $H_2SO_4$ on hydrated $BaO_2$ $(BaO_2 \cdot 8H_2O)$.
Anhydrous $BaO_2$ is not used because it forms a protective layer of $BaSO_4$ on its surface,which prevents further reaction.
Therefore,the Assertion is incorrect because anhydrous $BaO_2$ is not used.
The Reason is also incorrect because hydrated $BaO_2$ is readily available and is the preferred reagent.
The reaction is: $BaO_2 \cdot 8H_2O + H_2SO_4 \to BaSO_4 + H_2O_2 + 8H_2O$.

(A) solution of $FeCl_3$ undergoes hydrolysis in water to form $Fe(OH)_3$,which is a brown precipitate.
The chemical reaction is: $FeCl_3 + 3H_2O \to Fe(OH)_3 + 3HCl$.
Since the formation of the brown precipitate is directly caused by the hydrolysis of $FeCl_3$,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(A) When both nitrogen and sulphur are present in an organic compound,sodium thiocyanate $(NaSCN)$ is formed during the preparation of Lassaigne's extract.
$Na + C + N + S \xrightarrow{\Delta} NaSCN$
On adding $FeCl_3$ to this extract,the $Fe^{3+}$ ions react with $SCN^-$ ions to form ferric thiocyanate,which gives a blood-red coloration.
$Fe^{3+} + 3SCN^- \longrightarrow [Fe(SCN)_3]$ (blood-red color)
Thus,both the Assertion and the Reason are correct,and the Reason explains the Assertion.

(C) In a sinusoidal wave,the time taken to travel from the maximum displacement (amplitude) to the zero displacement (equilibrium position) corresponds to one-quarter of the time period $(T)$.
Given,$\frac{T}{4} = 0.170 \,s$.
Therefore,the total time period $T = 4 \times 0.170 \,s = 0.680 \,s$.
The frequency $(f)$ is the reciprocal of the time period:
$f = \frac{1}{T} = \frac{1}{0.680} \approx 1.47 \,Hz$.

(B) State functions are thermodynamic properties that depend only on the initial and final states of the system,not on the path taken to reach that state.
$Internal \ Energy$,$Enthalpy$,and $Entropy$ are state functions.
$Work$ and $Heat$ are path functions,meaning their values depend on the process path taken.

(D) The first ionisation energies of $Ti$,$V$,$Cr$,and $Mn$ are $656 \ kJ/mol$,$650 \ kJ/mol$,$652 \ kJ/mol$,and $717 \ kJ/mol$ respectively.
Ionisation energy generally increases from left to right in a period,although there are irregularities due to electronic configurations.
Among the given elements,$Mn$ $(3d^5 4s^2)$ has a stable half-filled $d$-subshell configuration,which contributes to its higher ionisation energy.
Therefore,$Mn$ has the maximum first ionisation potential.

(B) The chemical formula of Mohr's salt is $FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$.
It is a double salt containing $6$ water molecules of crystallization.

(A) Hypophosphorus acid is $H_3PO_2$.
In its structure,the central phosphorus atom is bonded to one oxygen atom by a double bond $(P=O)$,one hydroxyl group $(-OH)$,and two hydrogen atoms directly attached to the phosphorus atom $(P-H)$.
This structure is represented by option $A$.

(B) Molality is defined as the number of moles of solute per $1 \ kg$ of solvent.
Since mass does not change with temperature,molality is independent of temperature.
In contrast,molarity,formality,and normality involve volume,which changes with temperature.

(C) The isotope $_{88}Ra^{226}$ is radioactive.
For this nucleus,the number of protons $(p)$ is $88$ and the number of neutrons $(n)$ is $226 - 88 = 138$.
The $\frac{n}{p}$ ratio is $\frac{138}{88} \approx 1.568$.
Since the $\frac{n}{p}$ ratio is significantly greater than $1.5$,the nucleus is unstable and undergoes radioactive decay.

(A) The binding energy per nucleon is a measure of the stability of a nucleus.
Elements with mass numbers in the range of $40$ to $100$ are the most stable.
Among the given options,$Fe$ (Iron-$56$) has a mass number of $56$,which falls in the region of maximum stability.
Therefore,$Fe$ has the highest binding energy per nucleon.

(B) For a first order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 45 \, \text{min}$,so $k = \frac{0.693}{45} \, \text{min}^{-1}$.
For $99.9 \%$ completion,the remaining concentration is $a - 0.999a = 0.001a$.
The time taken is $t = \frac{2.303}{k} \log \left( \frac{a}{0.001a} \right) = \frac{2.303}{k} \log(10^3) = \frac{2.303 \times 3}{k}$.
Substituting $k = \frac{0.693}{45}$,we get $t = \frac{2.303 \times 3 \times 45}{0.693} \approx 448.5 \, \text{min}$.
Converting to hours: $t = \frac{448.5}{60} \approx 7.475 \, \text{hr} \approx 7.5 \, \text{hr}$.

(A) For a first order reaction,the rate constant $k$ is given by the formula:
$k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$
Given: $[A]_0 = 2.00 \, M$,$[A]_t = 0.15 \, M$,and $t = 200 \, \min$.
Substituting the values:
$k = \frac{2.303}{200} \log \left( \frac{2.00}{0.15} \right)$
$k = \frac{2.303}{200} \log(13.333)$
$k = \frac{2.303}{200} \times 1.1249$
$k \approx 1.29 \times 10^{-2} \, \min^{-1}$
Therefore,the correct option is $A$.

(B) The given reactions are oxidation half-reactions. The standard reduction potentials $(E^o_{red})$ are the negative of the standard oxidation potentials $(E^o_{ox})$.
$E^o_{red}(Zn^{2+}/Zn) = -0.76 \ V$
$E^o_{red}(Fe^{2+}/Fe) = -0.41 \ V$
In the cell reaction $Fe^{2+} + Zn \to Zn^{2+} + Fe$,$Zn$ is oxidized (anode) and $Fe^{2+}$ is reduced (cathode).
$E^o_{cell} = E^o_{cathode} - E^o_{anode}$
$E^o_{cell} = E^o_{red}(Fe^{2+}/Fe) - E^o_{red}(Zn^{2+}/Zn)$
$E^o_{cell} = -0.41 \ V - (-0.76 \ V)$
$E^o_{cell} = -0.41 \ V + 0.76 \ V = +0.35 \ V$.

(A) The movement of charged colloidal particles under the influence of an applied electric field is known as electrophoresis. When an electric potential is applied across two platinum electrodes dipping in a colloidal solution,the colloidal particles move towards one or the other electrode depending on the charge they carry.

(C) Generally,$d$-block elements are called transition elements because they contain an inner partially filled $d$-subshell.
Thus,their general electronic configuration is represented as $(n - 1)d^{1 - 10}ns^{1 - 2}$.

(B) The complex ion is given as $[Ag(CN)_2]^-$.
The superscript outside the square brackets represents the net charge on the coordination entity.
Therefore,the charge on the $[Ag(CN)_2]^-$ complex is $-1$.

(B) In $[Ni(CN)_4]^{2-}$,the oxidation state of $Ni$ is $+2$ ($3d^8$ configuration).
Since $CN^-$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbitals.
This results in one vacant $3d$ orbital,one $4s$ orbital,and two $4p$ orbitals,which undergo $dsp^2$ hybridisation to form a square planar geometry.

(D) The reaction of ethyl alcohol with thionyl chloride in the presence of pyridine is known as the Darzens process.
The reaction is: $C_2H_5OH + SOCl_2 \xrightarrow{\text{Pyridine}} C_2H_5Cl + SO_2 + HCl$.
Pyridine is used to neutralize the $HCl$ produced in the reaction,which drives the reaction to completion.

(D) The correct answer is $(D)$.
Raw rubber is plastic in nature,meaning it deforms easily.
It becomes soft at high temperatures and brittle at low temperatures.
It has low durability and a high water-absorption capacity,making it unsuitable for many industrial applications without vulcanization.

(C) Yeast cells derive their energy from glucose primarily through the process of fermentation.
In the absence of oxygen,yeast performs alcoholic fermentation,converting glucose into ethanol and $CO_2$,while generating $ATP$.
Even in the presence of oxygen,yeast often prefers fermentation (the Crabtree effect).
Therefore,the most appropriate answer among the given choices is fermentation.

(B) In many proteins,the linear polypeptide chains are cross-linked to provide structural stability.
The most common type of cross-link is the disulfide bond $(-S-S-)$.
This bond is formed by the oxidation of the thiol $(-SH)$ groups present in a pair of cysteine residues.
Therefore,cysteine is the amino acid responsible for cross-linking peptide chains.

(A) The protein that destroys the antigen when it enters a body cell is $Antibodies$.
$Antibodies$ are protective proteins produced by the immune system in response to the presence of a foreign substance,known as an antigen.

(B) Given: Current $(I) = 5 \ A$,Time $(t) = 965 \ s$,Molar mass of $Cu = 63.5 \ g/mol$,Valency of $Cu = 2$.
According to Faraday's first law of electrolysis,the mass $(w)$ deposited is given by:
$w = \frac{Z \times I \times t}{96500}$,where $Z = \frac{\text{Equivalent weight}}{96500} = \frac{\text{Molar mass}}{n \times 96500}$.
$w = \frac{63.5 \times 5 \times 965}{2 \times 96500}$.
$w = \frac{63.5 \times 4825}{193000} = \frac{306387.5}{193000} = 1.5875 \ g$.

(B) The formula for charge $(Q)$ is given by $Q = I \times t$.
Given current $(I)$ = $1.8 \, A$.
Time $(t)$ = $1.36 \, \text{minutes} = 1.36 \times 60 \, \text{seconds} = 81.6 \, \text{seconds}$.
Therefore,$Q = 1.8 \, A \times 81.6 \, \text{s} = 146.88 \, C$.
Rounding to the nearest whole number,we get $147 \, C$.

(B) The conversion of an alkyl halide $(R-X)$ to an alcohol $(R-OH)$ is a nucleophilic substitution reaction.
When an alkyl halide is treated with an aqueous alkali like $Aq. KOH$ or $NaOH$,the halide ion $(X^-)$ is replaced by the hydroxyl group $(OH^-)$.
Reaction: $R-X + KOH (aq) \rightarrow R-OH + KX$.
This is a nucleophilic substitution reaction ($S_N1$ or $S_N2$ depending on the substrate).
Therefore,the correct option is $B$.

(A) Fluorine is the most electronegative element and has a small atomic size.
It has the electronic configuration $1s^2 2s^2 2p^5$.
Due to the absence of vacant $d-$orbitals in the valence shell,it cannot expand its octet or show positive oxidation states.
Therefore,it exhibits only a $-1$ oxidation state.
The reason provided ($2s^2 2p^5$ configuration) is the fundamental cause for its inability to show other oxidation states,making it the correct explanation for the assertion.

(D) Iron is a moderately reactive metal and is not found in a free state in nature; it is typically found in the form of oxides,carbonates,or sulfides.
Therefore,the Assertion is incorrect,and the Reason is correct.