KCET 2021 Chemistry Question Paper with Answer and Solution

(B) Let the original separation between the plates be $d$ and the area of the plates be $A$. The original capacitance is $C = \frac{\epsilon_0 A}{d}$.
When a dielectric slab of thickness $t = 4 \times 10^{-3} \ m$ is introduced,the new capacitance $C'$ becomes $C' = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$.
To restore the capacity to its original value,the separation $d$ is increased by $\Delta d = 3.5 \times 10^{-3} \ m$. The new separation is $d' = d + \Delta d$.
The new capacitance is $C'' = \frac{\epsilon_0 A}{d' - t + \frac{t}{K}} = \frac{\epsilon_0 A}{d + \Delta d - t + \frac{t}{K}}$.
Since $C'' = C$,we have $d = d + \Delta d - t + \frac{t}{K}$.
This simplifies to $\Delta d = t(1 - \frac{1}{K})$.
Substituting the values: $3.5 \times 10^{-3} = 4 \times 10^{-3} (1 - \frac{1}{K})$.
$0.875 = 1 - \frac{1}{K} \implies \frac{1}{K} = 1 - 0.875 = 0.125$.
$K = \frac{1}{0.125} = 8$.

(C) Given $y = (\cos x^2)^2$.
Applying the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = 2(\cos x^2) \cdot \frac{d}{dx}(\cos x^2)$
$\frac{dy}{dx} = 2(\cos x^2) \cdot (-\sin x^2) \cdot \frac{d}{dx}(x^2)$
$\frac{dy}{dx} = 2(\cos x^2) \cdot (-\sin x^2) \cdot (2x)$
$\frac{dy}{dx} = -4x \sin x^2 \cos x^2$
Using the trigonometric identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we have:
$\frac{dy}{dx} = -2x(2 \sin x^2 \cos x^2) = -2x \sin 2x^2$.

(D) All the statements except $(d)$ are correct.
$\sigma$-bonds are formed by head-on overlap of orbitals,making them stronger than $\pi$-bonds formed by lateral overlap.
Hydrogen bonding is a strong dipole-dipole interaction,which is stronger than weak dispersion forces (London forces).
Ionic bonding is electrostatic in nature and is non-directional.
The statement in $(d)$ is incorrect because $\pi$-electrons,which are loosely held in the $\pi$-molecular orbital,are referred to as mobile electrons,not $\sigma$-electrons.

(C) For the given reaction,$A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}$,the change in the number of moles of gaseous species is $\Delta n_g = (1 + 1) - (1 + 1) = 0$.
Since $\Delta n_g = 0$,the pressure change does not affect the equilibrium position or the equilibrium constant.
The equilibrium constant $K_c$ is a function of temperature only.
Therefore,increasing the pressure will not disturb the equilibrium constant.

(B) The third ionisation enthalpy is highest in $alkaline \ earth \ metals$ because they achieve a stable noble gas configuration after the removal of two electrons.
Further removal of an electron from the completely filled orbital requires a very high ionisation enthalpy.

(A) Buckminster fullerene,also known as $C_{60}$,consists of $60$ carbon atoms arranged in a soccer ball-like structure.
It contains $20$ six-membered rings and $12$ five-membered rings.
Therefore,the number of six-membered and five-membered rings is $20$ and $12$ respectively.

(B) To determine the $IUPAC$ name,we identify the substituents on the benzene ring: an ethyl group $(-C_2H_5)$,a fluoro group $(-F)$,and a nitro group $(-NO_2)$.
According to $IUPAC$ rules,we number the ring to give the lowest possible locants to the substituents.
If we start numbering from the fluoro group as $1$,the nitro group is at $2$,and the ethyl group is at $4$.
The set of locants is $(1, 2, 4)$.
Alphabetical order of substituents: ethyl,fluoro,nitro.
Therefore,the name is $4-$ethyl$-1-$fluoro$-2-$nitrobenzene.

(D) The decreasing order of acidic behaviour is ethyne $>$ benzene $>$ $n$-hexane.
The acidity of a hydrocarbon depends on the $s$-character of the hybridised carbon atom to which the hydrogen is attached.
Higher $s$-character leads to higher electronegativity of the carbon atom,which makes the $C-H$ bond more polar and the resulting carbanion more stable.
The hybridisation of carbon atoms in these compounds is as follows:
Ethyne $(HC \equiv CH)$: $sp$ hybridised ($50\% \ s$-character).
Benzene $(C_6H_6)$: $sp^2$ hybridised ($\approx 33.3\% \ s$-character).
$n$-Hexane $(CH_3(CH_2)_4CH_3)$: $sp^3$ hybridised ($25\% \ s$-character).
Therefore,the order of acidity is ethyne $>$ benzene $>$ $n$-hexane.

(B) The peroxide effect (Kharasch effect) is observed with the addition of $HBr$ to unsymmetrical alkenes because the $H-Br$ bond can be homolytically cleaved by free radicals generated from the peroxide.
However,in the case of $HI$,the $H-I$ bond is very weak.
When iodine free radicals are generated,they are highly unstable and rapidly combine with each other to form stable $I_2$ molecules instead of attacking the alkene double bond.
The reaction is:
$I^{\bullet} + I^{\bullet} \rightarrow I_2$
(Unstable) (Stable)
Therefore,the anti-Markovnikov addition is not observed for $HI$.

(C) Permanent hardness is caused by the presence of soluble salts of magnesium and calcium in the form of chlorides and sulphates in water.
Clark's method involves the addition of calculated amounts of lime $(Ca(OH)_2)$ to water to remove temporary hardness caused by calcium bicarbonate $(Ca(HCO_3)_2)$ and magnesium bicarbonate $(Mg(HCO_3)_2)$.
It is ineffective against permanent hardness,which requires methods like washing soda treatment,Calgon's method,or ion exchange method.

(D) The strength of an acid is directly proportional to its $K_{a}$ value,while the strength of its conjugate base is inversely proportional to the $K_{a}$ value of the parent acid.
Therefore,the acid with the lowest $K_{a}$ value will produce the strongest conjugate base.
Comparing the given $K_{a}$ values: $1.3 \times 10^{-2} > 4 \times 10^{-4} > 1.8 \times 10^{-5} > 4 \times 10^{-10}$.
Since $HCN$ has the lowest $K_{a}$ value $(4 \times 10^{-10})$,it produces the strongest conjugate base.

(B) The boiling point depends on the extent of intermolecular hydrogen bonding.
$H_{2}O$ has the highest boiling point because each molecule can form four hydrogen bonds due to the presence of two hydrogen atoms and two lone pairs on the oxygen atom.
$HF$ has a high boiling point due to strong hydrogen bonding,but it forms fewer hydrogen bonds per molecule compared to $H_{2}O$.
$NH_{3}$ has the lowest boiling point among these three because nitrogen is less electronegative than oxygen and fluorine,resulting in weaker hydrogen bonding.
Therefore,the correct order is $H_{2}O > HF > NH_{3}$.

(D) Oxidation is defined as the addition of oxygen,the addition of an electronegative element,or the removal of hydrogen.
Removal of an electronegative element is actually a characteristic of reduction.
Therefore,option $(D)$ is not true for oxidation.

(C) The solubility of ionic compounds,such as the hydroxides of alkaline earth metals,depends on two factors: $(i)$ lattice energy and $(ii)$ hydration energy.
As we move down the group,the lattice energy decreases more significantly compared to the hydration energy.
Consequently,the solubility of alkaline earth metal hydroxides increases with an increase in their atomic number.
In contrast,other properties such as ionisation enthalpy,electronegativity,and the solubility of their sulphates in water decrease as the atomic number increases.

(D) Moles of carbon $= \frac{2.4 \ g}{12 \ g/mol} = 0.2 \ mol$
Moles of $H = \frac{1.2 \times 10^{23}}{6.022 \times 10^{23}} \approx 0.2 \ mol$
Moles of oxygen atoms $= 0.2 \ mol$
The molar ratio of $C:H:O$ is $0.2:0.2:0.2$,which simplifies to $1:1:1$.
Therefore,the empirical formula is $CHO$.

(B) According to the ideal gas law,$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$.
Given: $T_{2} = 2 T_{1}$ and $p_{2} = \frac{p_{1}}{2}$.
Substituting these values into the equation:
$\frac{p_{1} V_{1}}{T_{1}} = \frac{(p_{1} / 2) \times V_{2}}{2 T_{1}}$.
Canceling $p_{1}$ and $T_{1}$ from both sides:
$V_{1} = \frac{V_{2}}{4}$.
Therefore,$V_{2} = 4 V_{1}$.
The volume becomes $4$ times the original volume.

(B) For the $3p$ orbital,the principal quantum number $n = 3$ and the azimuthal quantum number $l = 1$.
The number of angular nodes is given by $l = 1$.
The number of radial nodes is given by the formula $(n - l - 1)$.
Substituting the values: $3 - 1 - 1 = 1$.
Therefore,the number of angular and radial nodes in the $3p$ orbital are $1$ and $1$ respectively.

(C) Given,$(BE)_{A_{2}} : (BE)_{B_{2}} : (BE)_{AB} = 2 : 1 : 2$,where $BE$ is bond enthalpy.
Let $(BE)_{A_{2}} = 2x$,$(BE)_{B_{2}} = x$,and $(BE)_{AB} = 2x$.
The reaction for the formation of $AB$ is: $\frac{1}{2} A_{2}(g) + \frac{1}{2} B_{2}(g) \rightarrow AB(g)$.
The enthalpy of formation is given by: $\Delta H^{\circ}_{f} = \Sigma (BE)_{\text{reactants}} - \Sigma (BE)_{\text{products}}$.
Substituting the values: $-100 = [\frac{1}{2}(2x) + \frac{1}{2}(x)] - 2x$.
$-100 = [x + 0.5x] - 2x$.
$-100 = 1.5x - 2x$.
$-100 = -0.5x$.
$x = \frac{100}{0.5} = 200 \ kJ \ mol^{-1}$.
Thus,the bond enthalpy of $B_{2}$ is $200 \ kJ \ mol^{-1}$.

(A) $LiAlH_{4} / \text{ether}$ is a strong reducing agent. It is capable of reducing carboxylic acids to primary alcohols.
The reaction is: $CH_{3}COOH \xrightarrow{LiAlH_{4} / \text{ether}} CH_{3}CH_{2}OH$.

(B) $1$. The compound $A$ $(C_{7}H_{8}O)$ is insoluble in $NaHCO_{3}$ (meaning it is not a carboxylic acid) but soluble in $NaOH$ and gives a characteristic color with neutral $FeCl_{3}$ solution,which indicates the presence of a phenolic $-OH$ group.
$2$. The reaction with bromine water to form a tribromo derivative $(C_{7}H_{5}OBr_{3})$ is a characteristic reaction of phenols where the ortho and para positions are substituted by bromine atoms.
$3$. Among the given options,$m$-cresol ($3$-methylphenol) is a phenol that reacts with bromine water to form $2,4,6$-tribromo-$3$-methylphenol $(C_{7}H_{5}OBr_{3})$.
$4$. Therefore,compound $A$ is $m$-cresol.

(A) The given reactions are standard methods for the preparation of acetaldehyde $(CH_3CHO)$:
$1$. $A$ is ethyne $(HC \equiv CH)$. Hydration of ethyne in the presence of $HgSO_4 / H_2SO_4$ gives vinyl alcohol $(CH_2=CH-OH)$,which tautomerises to form acetaldehyde $(CH_3CHO)$.
$2$. $B$ is ethanol $(CH_3CH_2OH)$. $PCC$ (Pyridinium chlorochromate) is a mild oxidising agent that oxidises primary alcohols to aldehydes. Thus,$CH_3CH_2OH \xrightarrow{PCC} CH_3CHO$.
$3$. $C$ is ethane nitrile $(CH_3CN)$. The Stephen reduction of nitriles using $SnCl_2 / HCl$ followed by acid hydrolysis $(H_3O^+)$ yields acetaldehyde $(CH_3CHO)$.
Therefore,$A$ is ethyne,$B$ is ethanol,and $C$ is ethane nitrile.

(C) The given reaction involves the oxidation of a methyl ketone group $(-COCH_3)$ to a carboxylic acid group $(-COOH)$ while keeping the carbon-carbon double bond intact.
This transformation is characteristic of the haloform reaction.
When a methyl ketone is treated with $I_2$ and $NaOH$,it undergoes the iodoform reaction to form a carboxylate salt,which upon subsequent acidification yields the corresponding carboxylic acid.
The reaction is:
$CH_3-CH=CH-CH_2-COCH_3 \xrightarrow[(ii) \text{ Acidification}]{(i) I_2/NaOH} CH_3-CH=CH-CH_2-COOH + CHI_3$

(A) The given reaction proceeds as follows:
$1$. The cumene hydroperoxide process yields phenol $(A)$ and acetone $(B)$.
$2$. Phenol $(A)$ gives a white precipitate with bromine water due to the formation of $2,4,6-$tribromophenol.
$3$. Acetone $(B)$ undergoes aldol condensation in the presence of barium hydroxide $(Ba(OH)_2)$ to form $4-$hydroxy$-4-$methylpentan$-2-$one $(C)$.
$4$. Upon strong heating,$C$ undergoes dehydration to form $4-$methylpent$-3-$en$-2-$one $(D)$,which is mesityl oxide.
Therefore,the product $D$ is $4-$methylpent$-3-$en$-2-$one.

(B) $1$. Reaction of $CH_3CHO$ with $CH_3MgBr$ followed by hydrolysis gives $A$,which is $CH_3-CH(OH)-CH_3$ (Propan$-2-$ol).
$2$. Dehydration of $A$ with $Conc. H_2SO_4$ at high temperature gives $B$,which is $CH_3-CH=CH_2$ (Prop$-1-$ene).
$3$. Hydroboration-oxidation of $B$ $(CH_3-CH=CH_2)$ with $(i) B_2H_6$ and $(ii) H_2O_2/OH^-$ follows anti-Markovnikov addition of water to give $C$,which is $CH_3-CH_2-CH_2OH$ (Propan$-1-$ol).
$4$. Comparing $A$ $(CH_3-CH(OH)-CH_3)$ and $C$ $(CH_3-CH_2-CH_2OH)$,the position of the $-OH$ group is different (carbon-$2$ vs carbon-$1$). Therefore,they are position isomers.

(C) Compound $(Y)$ gives the iodoform test,which means it contains an acetyl group $(CH_{3}-C=O)$.
$PCC$ (Pyridinium chlorochromate) is an oxidizing agent that oxidizes primary alcohols to aldehydes and secondary alcohols to ketones.
Since compound $Y$ is formed from $X$ using $PCC$ and $Y$ gives the iodoform test,$Y$ must be acetaldehyde $(CH_{3}CHO)$,as it is the only aldehyde that gives the iodoform test.
Therefore,the starting material $X$ must be ethanol $(CH_{3}CH_{2}OH)$.
The reactions are as follows:
$CH_{3}CH_{2}OH$ $\xrightarrow{PCC} CH_{3}CHO$ $\xrightarrow{I_{2}/OH^{-}} CHI_{3} + HCOONa$

(A) $RNA$ and $DNA$ are chiral molecules because they contain sugar units ($D$-ribose in $RNA$ and $D$-2-deoxyribose in $DNA$).
These sugar molecules possess multiple chiral carbon atoms,which impart chirality to the entire $RNA$ and $DNA$ structure.

(C) In $DNA$,the base pairing rules are based on hydrogen bonding between specific nitrogenous bases: $A$ (Adenine) pairs with $T$ (Thymine),and $G$ (Guanine) pairs with $C$ (Cytosine).
Given the sequence $G-A-T-G-C$,the complementary sequence is formed by replacing each base with its pair:
$G \rightarrow C$
$A \rightarrow T$
$T \rightarrow A$
$G \rightarrow C$
$C \rightarrow G$
Thus,the complementary chain is $CTACG...$.

(C) Higher order reactions $(>3)$ are rare because the probability of simultaneous collision of more than three reacting species is extremely low,making the frequency of such effective collisions negligible.

(A) For the reaction,$A + 2B \rightarrow$ Products.
Since the half-life $(t_{1/2})$ remains constant when the concentration of $B$ is increased,the reaction is of $1^{st}$ order with respect to $B$.
Since the rate remains constant when the concentration of $A$ is doubled,the reaction is of $0^{th}$ order with respect to $A$.
The rate law expression is: $\text{Rate} = k[A]^0[B]^1$.
The overall order of the reaction is $0 + 1 = 1$.
The unit of the rate constant $(k)$ for a $1^{st}$ order reaction is $s^{-1}$.

(C) The rate law is given by $r = k[A][B]^{2}$.
When the volume of the vessel is reduced to half $(V_{2} = V_{1}/2)$,the concentration of the gaseous reactants doubles because concentration is inversely proportional to volume $(C = n/V)$.
Thus,the new concentrations are $[A]' = 2[A]$ and $[B]' = 2[B]$.
The new rate $r'$ is calculated as:
$r' = k[A]'([B]')^{2} = k(2[A])(2[B])^{2}$.
$r' = k \times 2[A] \times 4[B]^{2} = 8 \times k[A][B]^{2}$.
Since $r = k[A][B]^{2}$,we have $r' = 8r$.
Therefore,the reaction rate increases by $8$ times.

(A) For a first order reaction,the integrated rate equation is $t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}$.
Given that $99 \%$ of the reaction is complete,the remaining concentration $[R]$ is $100 \% - 99 \% = 1 \%$ of the initial concentration $[R]_0$.
So,$[R] = 0.01 [R]_0 = \frac{[R]_0}{100}$.
Substituting these values into the equation:
$t = \frac{2.303}{k} \log \frac{[R]_0}{[R]_0 / 100} = \frac{2.303}{k} \log(100)$.
Since $\log(100) = 2$,we get $t = \frac{2.303 \times 2}{k} = \frac{4.606}{k}$.

(C) homoleptic complex is a coordination compound in which all the ligands attached to the central metal atom or ion are identical.
In $K_3[Al(C_2O_4)_3]$,the central metal $Al^{3+}$ is bonded to three identical oxalate $(C_2O_4)^{2-}$ ligands.
In $[CoCl_2(en)_2]^{+}$,the central metal $Co^{3+}$ is bonded to two different types of ligands: chloride $(Cl^-)$ and ethylenediamine $(en)$. Thus,it is a heteroleptic complex.
In $K_2[Zn(OH)_4]$,the central metal $Zn^{2+}$ is bonded to four identical hydroxide $(OH)^-$ ligands.
Therefore,$K_3[Al(C_2O_4)_3]$ and $K_2[Zn(OH)_4]$ are homoleptic complexes.

(A) In the complex $[Co(NH_3)_5(CO_3)]Cl$,the ligands are ammine $(NH_3)$ and carbonato $(CO_3^{2-})$.
According to $IUPAC$ nomenclature rules,ligands are named in alphabetical order. Thus,'ammine' precedes 'carbonato'.
The oxidation state of cobalt $(Co)$ is calculated as follows: $x + 5(0) + 1(-2) = +1 \Rightarrow x = +3$.
Since the complex ion $[Co(NH_3)_5(CO_3)]^+$ is cationic,the metal name remains 'cobalt'.
Combining these,the $IUPAC$ name is pentaamminecarbonatocobalt $(III)$ chloride.

(B) In an octahedral complex,the five degenerate $d$-orbitals split into two sets due to the approach of ligands along the axes.
The two orbitals $d_{x^2-y^2}$ and $d_{z^2}$ (collectively called $e_{g}$ orbitals) point directly towards the ligands and experience greater repulsion,thus their energy increases by $+(3/5) \Delta_{o}$.
The three orbitals $d_{xy}$,$d_{yz}$,and $d_{zx}$ (collectively called $t_{2g}$ orbitals) point between the axes and experience less repulsion,thus their energy decreases by $-(2/5) \Delta_{o}$ relative to the barycenter.

(A) The wavelength of light absorbed $(\lambda)$ is inversely proportional to the crystal field splitting energy $(\Delta_o)$.
According to the spectrochemical series, the strength of the ligands is $Cl^{-} < NH_{3} < CN^{-}$.
Since $\Delta_o$ is directly proportional to the ligand field strength, the order of $\Delta_o$ is $[CoCl(NH_{3})_{5}]^{2+} < [Co(NH_{3})_{6}]^{3+} < [Co(CN)_{6}]^{3-}$.
Therefore, the order of wavelength of light absorbed $(\lambda \propto 1/\Delta_o)$ is $[CoCl(NH_{3})_{5}]^{2+} > [Co(NH_{3})_{6}]^{3+} > [Co(CN)_{6}]^{3-}$.

(C) The property stated against option $(C)$ is incorrect.
The electronic configurations of the given ions are:
$Cr^{2+} = [Ar] 3d^4$ ($4$ unpaired electrons)
$Mn^{2+} = [Ar] 3d^5$ ($5$ unpaired electrons)
$Fe^{2+} = [Ar] 3d^6$ ($4$ unpaired electrons)
Paramagnetic behaviour is directly proportional to the number of unpaired electrons.
Since $Mn^{2+}$ has $5$ unpaired electrons,it shows the maximum paramagnetic behaviour among the three.
Therefore,the correct order of paramagnetic behaviour is $Cr^{2+} = Fe^{2+} < Mn^{2+}$.

(C) The elements from $Sc$ $(Z=21)$ to $Cu$ $(Z=29)$ are transition elements belonging to the $3d$ series.
For all these elements,the $3d$ subshell is partially filled in their ground state electronic configuration.
For example:
$Sc: [Ar] 3d^1 4s^2$
$Ti: [Ar] 3d^2 4s^2$
...
$Cu: [Ar] 3d^{10} 4s^1$
Since the $3d$ orbital is not completely filled for all elements from $Sc$ to $Cu$ in their ground state (except $Cu$,but the question asks for a general property of the series elements),option $(c)$ is the most appropriate description of the $3d$ series characteristics.

(D) Ions are coloured in aqueous solution if they have unpaired electrons in their $d$-orbitals (i.e.,$d^1$ to $d^9$ configuration).
$1$. $Sc^{3+}$ $(3d^0)$: Colourless (no unpaired electrons).
$2$. $Ti^{3+}$ $(3d^1)$: Coloured (one unpaired electron).
$3$. $Mn^{2+}$ $(3d^5)$: Coloured (five unpaired electrons).
$4$. $Ni^{2+}$ $(3d^8)$: Coloured (two unpaired electrons).
$5$. $Cu^{+}$ $(3d^{10})$: Colourless (no unpaired electrons).
$6$. $Ti^{4+}$ $(3d^0)$: Colourless (no unpaired electrons).
Comparing the options:
- Option $(A)$: $Sc^{3+}$ is colourless.
- Option $(B)$: $Ti^{4+}$ is colourless.
- Option $(C)$: $Cu^{+}$ is colourless.
- Option $(D)$: Both $Mn^{2+}$ and $Ti^{3+}$ have unpaired electrons and are coloured.
Therefore,the correct pair is $(Mn^{2+}, Ti^{3+})$.

(C) The standard reduction potential of $Zn^{2+} / Zn$ half-cell is $E^{\circ} = -0.76 \ V$.
The Nernst equation for the reduction half-cell reaction $Zn^{2+}_{(aq)} + 2e^{-} \longrightarrow Zn_{(s)}$ is:
$E_{red} = E^{\circ}_{red} - \frac{0.059}{n} \log \frac{1}{[Zn^{2+}]}$
$E_{red} = -0.76 + \frac{0.059}{2} \log [Zn^{2+}]$
Since the term $\frac{0.059}{2} \log [Zn^{2+}]$ is positive,a higher concentration of $Zn^{2+}$ leads to a higher (less negative) reduction potential.
Comparing the concentrations: $[Q] = 0.1 \ M$,$[R] = 0.01 \ M$,$[S] = 0.001 \ M$,$[P] = 0.0001 \ M$.
Therefore,the order of electrode potentials is $Q > R > S > P$.

(A) For the given reaction,the number of electrons transferred $n = 2$.
Given $E^{\circ}_{cell} = 0.22 \ V$ at $25^{\circ} C$.
At equilibrium,the relationship between $E^{\circ}_{cell}$ and equilibrium constant $K_{C}$ is given by:
$E^{\circ}_{cell} = \frac{0.0591}{n} \log K_{C}$
Substituting the values:
$0.22 = \frac{0.0591}{2} \log K_{C}$
$\log K_{C} = \frac{0.22 \times 2}{0.0591} \approx 7.445$
$K_{C} = \text{antilog}(7.445) \approx 2.786 \times 10^{7} \approx 2.8 \times 10^{7}$

(A) Given: Resistance $(R) = 1500 \ \Omega$
Conductivity $(\kappa) = 0.1466 \times 10^{-3} \ S \ cm^{-1}$
The formula for cell constant $(G^*)$ is:
$G^* = \kappa \times R$
Substituting the values:
$G^* = (0.1466 \times 10^{-3} \ S \ cm^{-1}) \times (1500 \ \Omega)$
$G^* = 0.2199 \ \approx 0.219 \ cm^{-1}$

(B) The hydrocarbon $A$ is $but-2-ene$ $(CH_{3}-CH=CH-CH_{3})$.
Reaction with $HCl$ follows Markovnikov's rule to give $B$,which is $2-chlorobutane$ $(CH_{3}-CHCl-CH_{2}-CH_{3})$.
Reaction of $B$ with $NH_{3}$ gives $C$,which is $butan-2-amine$ $(CH_{3}-CH(NH_{2})-CH_{2}-CH_{3})$.
Reaction of $C$ with $NaNO_{2}/HCl$ forms a diazonium salt,which upon hydrolysis with water yields $D$,which is $butan-2-ol$ $(CH_{3}-CH(OH)-CH_{2}-CH_{3})$.
$Butan-2-ol$ contains a chiral carbon atom and is therefore optically active.

(A) Benzyl chloride $(C_6H_5CH_2Cl)$ reacts with alcoholic $NH_3$ to form benzylamine ($C_6H_5CH_2NH_2$,$A$) via nucleophilic substitution.
Benzylamine $(A)$ then undergoes alkylation with two moles of methyl chloride $(CH_3Cl)$ to form $N, N$-dimethylphenylmethanamine ($C_6H_5CH_2N(CH_3)_2$,$B$).

(A) The reactivity of alkyl halides in $S_N2$ reactions is primarily governed by steric hindrance. The nucleophile attacks from the side opposite to the leaving group. As the number of bulky groups attached to the electrophilic carbon increases,the steric hindrance increases,making the $S_N2$ reaction slower.
The compounds are:
$1$. $C_6H_5CH_2Br$ (Primary,least hindered)
$2$. $C_6H_5CH(CH_3)Br$ (Secondary,more hindered than primary)
$3$. $C_6H_5CH(C_6H_5)Br$ (Secondary,more hindered than $C_6H_5CH(CH_3)Br$ due to the bulky phenyl group)
$4$. $C_6H_5C(CH_3)(C_6H_5)Br$ (Tertiary,most hindered)
Thus,the order of reactivity in $S_N2$ is: $C_6H_5C(CH_3)(C_6H_5)Br < C_6H_5CH(C_6H_5)Br < C_6H_5CH(CH_3)Br < C_6H_5CH_2Br$.

(A) The reaction proceeds in two steps as $HBr$ is in excess:
$1$. The $-OH$ group is substituted by $-Br$ to form allyl bromide: $CH_{2}=CH-CH_{2}-OH + HBr \rightarrow CH_{2}=CH-CH_{2}Br + H_{2}O$.
$2$. The second equivalent of $HBr$ adds across the double bond following Markownikoff's rule to form $1,2-\text{dibromopropane}$: $CH_{2}=CH-CH_{2}Br + HBr \rightarrow CH_{3}-CH(Br)-CH_{2}Br$.

(D) The reaction sequence is as follows:
$1$. $CH_3COOH + SOCl_2 \rightarrow CH_3COCl$ ($A$ is acetyl chloride).
$2$. $CH_3COCl + C_6H_6 \xrightarrow{Anhy. AlCl_3} C_6H_5COCH_3$ ($B$ is acetophenone,a Friedel-Crafts acylation reaction).
$3$. $C_6H_5COCH_3 + HCN \rightarrow C_6H_5C(OH)(CH_3)CN$ ($C$ is acetophenone cyanohydrin).
$4$. $C_6H_5C(OH)(CH_3)CN + H_2O \rightarrow C_6H_5C(OH)(CH_3)COOH$ ($D$ is $2-$hydroxy$-2-$phenylpropanoic acid,formed by the hydrolysis of the cyanohydrin group).

(D) Aryl halides,such as chlorobenzene,do not undergo nucleophilic substitution reactions with potassium phthalimide under ordinary conditions because the $C-Cl$ bond in chlorobenzene has partial double bond character and the electron-rich benzene ring repels the nucleophile.
Therefore,the Gabriel phthalimide synthesis cannot be used for the preparation of aniline.
Thus,method $(d)$ is incorrect for the preparation of aniline.

(C) The reaction of $p$-nitroethylbenzene with $Br_2$ in the presence of $UV$ light proceeds via a free radical mechanism. The bromine radical abstracts a hydrogen atom from the benzylic position because the resulting benzylic radical is stabilized by resonance with the benzene ring. Therefore,the major product is $1$-bromo-$1$-($4$-nitrophenyl)ethane.

(B) The thermal stability of the hydrides of group $15$ elements decreases down the group as $NH_{3} > PH_{3} > AsH_{3} > SbH_{3} > BiH_{3}$.
As we move down the group,the atomic size of the central atom increases (from $N$ to $Bi$).
This leads to an increase in the bond length and a decrease in the bond dissociation enthalpy (bond strength).
Consequently,the thermal stability decreases.

(C) The reactions are as follows:
$1$. $2NO(g) + O_2(g) \rightarrow 2NO_2(g)$ (Reddish brown gas $Q$)
$2$. $2NO_2(g) \rightleftharpoons N_2O_4(g)$ (Colourless gas $R$ on cooling)
$3$. $NO(g) + NO_2(g) \rightarrow N_2O_3(s)$ (Blue solid $S$)
Thus,$P = NO$,$Q = NO_2$,$R = N_2O_4$,and $S = N_2O_3$.

(C) $XeF_{6}$ on partial hydrolysis with water gives $XeOF_{4}$.
The reaction involved is as follows:
$XeF_{6} + H_{2}O \rightarrow XeOF_{4} + 2HF$
The structure of $XeOF_{4}$ is square pyramidal.

(B) $NH_{4}NO_{3}$ on heating gives nitrous oxide $(N_{2}O)$.
The reaction is as follows:
$NH_{4}NO_{3} \stackrel{\Delta}{\longrightarrow} N_{2}O_{(g)} + 2H_{2}O_{(g)}$
Nitrous oxide is also known as laughing gas.

(A) When a salt containing chloride $(Cl^-)$ is treated with concentrated sulphuric acid $(H_2SO_4)$,it undergoes a displacement reaction to produce hydrogen chloride gas $(HCl)$.
The reaction is: $NaCl(s) + H_2SO_4(conc.) \rightarrow NaHSO_4(s) + HCl(g) \uparrow$.
$HCl$ is a colourless gas with a pungent odour.

(A) Let the number of oxide ions $(O^{2-})$ in the $ccp$ structure be $N$.
Number of octahedral voids = $N$.
Number of tetrahedral voids = $2N$.
Given that $Al^{3+}$ ions occupy $\frac{1}{4}$ of the octahedral voids,so number of $Al^{3+} = \frac{1}{4} \times N = \frac{N}{4}$.
Given that $Be^{2+}$ ions occupy $\frac{1}{8}$ of the tetrahedral voids,so number of $Be^{2+} = \frac{1}{8} \times 2N = \frac{N}{4}$.
The ratio of $Be : Al : O$ is $\frac{N}{4} : \frac{N}{4} : N$.
Multiplying by $4$,we get the ratio $1 : 1 : 4$.
Therefore,the formula of the compound is $BeAlO_{4}$.

(A) Given, metal crystallises in $bcc$ lattice, therefore $Z = 2$.
Edge length $a = 300 \ pm = 300 \times 10^{-10} \ cm$.
Density $d = 6.15 \ g \ cm^{-3}$.
Using the formula for density: $d = \frac{Z \times M}{a^3 \times N_A}$.
Rearranging for molar mass $M$: $M = \frac{d \times a^3 \times N_A}{Z}$.
Substituting the values: $M = \frac{6.15 \times (300 \times 10^{-10})^3 \times 6.022 \times 10^{23}}{2}$.
$M = \frac{6.15 \times 27 \times 10^{-24} \times 6.022 \times 10^{23}}{2} = \frac{99.965}{2} \approx 49.98 \ g \ mol^{-1}$.
Thus, the molar mass is approximately $50 \ g \ mol^{-1}$.

(C) The correct statement is that the trapping of an electron in the lattice leads to the formation of an $F$-centre.
Frenkel defect is a dislocation defect where an ion leaves its lattice site and occupies an interstitial site.
Schottky defect is a vacancy defect where equal numbers of cations and anions are missing from the lattice,which leads to a decrease in the density of the solid.

(A) Given,vapour pressure of pure liquid $A$,$p_{A}^{\circ} = 450 \ mm \ Hg$.
Vapour pressure of pure liquid $B$,$p_{B}^{\circ} = 700 \ mm \ Hg$.
Total vapour pressure,$p_{\text{Total}} = 600 \ mm \ Hg$.
From Raoult's law,$p_{\text{Total}} = p_{A}^{\circ}\chi_{A} + p_{B}^{\circ}\chi_{B}$.
Since $\chi_{B} = 1 - \chi_{A}$,we have:
$600 = 450\chi_{A} + 700(1 - \chi_{A})$
$600 = 450\chi_{A} + 700 - 700\chi_{A}$
$600 = 700 - 250\chi_{A}$
$250\chi_{A} = 100$
$\chi_{A} = \frac{100}{250} = 0.4$
$\therefore \chi_{B} = 1 - 0.4 = 0.6$.

(A) Given,Henry's law constant $(K_H)$ for the solubility of $N_2$ gas in water at $298 \ K = 1 \times 10^5 \ atm$.
Mole fraction of $N_2$ in air $(\chi_{N_2, \text{air}}) = 0.8$.
Total pressure $(P_{\text{total}}) = 5 \ atm$.
Partial pressure of nitrogen $(p_{N_2}) = P_{\text{total}} \times \chi_{N_2, \text{air}} = 5 \times 0.8 = 4 \ atm$.
According to Henry's law,$p_{N_2} = K_H \times \chi_{N_2, \text{water}}$.
$4 = 10^5 \times \chi_{N_2, \text{water}} \Rightarrow \chi_{N_2, \text{water}} = 4 \times 10^{-5}$.
We know that $\chi_{N_2, \text{water}} = \frac{n_{N_2}}{n_{N_2} + n_{H_2O}}$.
Since $n_{N_2} \ll n_{H_2O}$,we can approximate $\chi_{N_2, \text{water}} \approx \frac{n_{N_2}}{n_{H_2O}}$.
$4 \times 10^{-5} = \frac{n_{N_2}}{10}$.
$n_{N_2} = 4 \times 10^{-4} \ moles$.

(C) Among the given statements,only statement $(c)$ is correct regarding the Henry's law constant $(K_{H})$.
$(a)$ $K_{H}$ is a function of the nature of the gas and the solvent used; therefore,the $K_{H}$ value is not the same for a gas in different solutions.
$(b)$ According to Henry's law,$p = K_{H} \cdot x$,which implies $x = p / K_{H}$. Thus,a higher $K_{H}$ value corresponds to lower solubility of the gas.
$(c)$ $K_{H}$ is temperature-dependent and generally increases with an increase in temperature,which explains why gases are less soluble in warm water.
$(d)$ Easily liquefiable gases have stronger intermolecular forces,which leads to lower $K_{H}$ values.

(C) According to Henry's law,the solubility of a gas in a liquid is inversely proportional to its Henry's law constant $(K_H)$.
Mathematically,$\text{Solubility} \propto \frac{1}{K_H}$.
Given $K_H$ values at $298 \ K$:
Argon $(I)$: $40.3 \ k\text{bar}$
Carbon dioxide $(II)$: $1.67 \ k\text{bar}$
Methane $(IV)$: $0.413 \ k\text{bar}$
Formaldehyde $(III)$: $1.83 \times 10^{-5} \ k\text{bar}$
Comparing the $K_H$ values: $1.83 \times 10^{-5} < 0.413 < 1.67 < 40.3$.
Therefore,the order of $K_H$ is $III < IV < II < I$.
Since solubility is inversely proportional to $K_H$,the increasing order of solubility is $I < II < IV < III$.

(A) Since the colloidal particles move towards the anode,they are negatively charged.
According to the Hardy-Schulze rule,the coagulating power of an electrolyte increases with the increase in the magnitude of the charge on the cation (for negatively charged colloids).
The coagulating power order is $Al^{3+} > Ba^{2+} > Na^{+}$.
Since the coagulation value is inversely proportional to the coagulating power,the amount of electrolyte required follows the order: $NaCl > BaCl_{2} > AlCl_{3}$.

(C) Zeta potential,also known as electrokinetic potential,is the potential difference between the fixed charged layer and the diffused layer of opposite charges surrounding the colloidal particle.