KCET 2013 Chemistry Question Paper with Answer and Solution

(C) Isoelectronic species are those which have the same number of electrons.
For option $C$:
$K^{+} = 19 - 1 = 18 \text{ electrons}$
$Ca^{2+} = 20 - 2 = 18 \text{ electrons}$
$Sc^{3+} = 21 - 3 = 18 \text{ electrons}$
$Cl^{-} = 17 + 1 = 18 \text{ electrons}$
Since all these ions have $18$ electrons,they are isoelectronic.

(B) ligand is defined as an atom,molecule,or ion that can donate at least one pair of electrons to a central metal atom or ion to form a coordinate covalent bond. Therefore,the essential requirement for a species to act as a ligand is the presence of at least one unshared (lone) pair of electrons.

(B) Let $z = \frac{1 + 2i}{1 - (1 - i)^2}$.
First,simplify the denominator: $(1 - i)^2 = 1^2 + i^2 - 2i = 1 - 1 - 2i = -2i$.
So,$z = \frac{1 + 2i}{1 - (-2i)} = \frac{1 + 2i}{1 + 2i} = 1$.
We can write $z$ as $1 + 0i$.
The modulus $|z| = \sqrt{1^2 + 0^2} = 1$.
The amplitude $\theta = \tan^{-1}(\frac{0}{1}) = 0$.

(B) Given equation: $\sin x - \cos x = \sqrt{2}$
Divide both sides by $\sqrt{2}$:
$\frac{1}{\sqrt{2}} \sin x - \frac{1}{\sqrt{2}} \cos x = 1$
Using the identity $\sin(x - \frac{\pi}{4}) = \sin x \cos \frac{\pi}{4} - \cos x \sin \frac{\pi}{4}$:
$\sin(x - \frac{\pi}{4}) = 1$
We know that $\sin \theta = 1$ implies $\theta = 2n\pi + \frac{\pi}{2}$ for any integer $n$.
So,$x - \frac{\pi}{4} = 2n\pi + \frac{\pi}{2}$
$x = 2n\pi + \frac{\pi}{2} + \frac{\pi}{4}$
$x = 2n\pi + \frac{3\pi}{4}$

(C) The system consists of two capacitors connected in parallel between plate $Q$ and the grounded plates $P$ and $R$.
The capacitance of the first capacitor (between $Q$ and $P$) is $C_1 = \frac{\epsilon_0 A}{d}$.
The capacitance of the second capacitor (between $Q$ and $R$) is $C_2 = \frac{\epsilon_0 A}{2d}$.
Since both capacitors are connected in parallel,the equivalent capacitance is $C_{eff} = C_1 + C_2 = \frac{\epsilon_0 A}{d} + \frac{\epsilon_0 A}{2d} = \frac{3 \epsilon_0 A}{2d}$.
Given $A = 2.0\ m^2$,$d = 2 \times 10^{-3}\ m$,and $\epsilon_0 = 8.85 \times 10^{-12}\ F/m$,we have:
$C_{eff} = \frac{3 \times 8.85 \times 10^{-12} \times 2.0}{2 \times 2 \times 10^{-3}} = \frac{3 \times 8.85 \times 10^{-12}}{2 \times 10^{-3}} = 1.5 \times 8.85 \times 10^{-9}\ F$.
The potential $V$ of plate $Q$ is given by $V = \frac{q}{C_{eff}}$.
Substituting the values: $V = \frac{8.85 \times 10^{-8}}{1.5 \times 8.85 \times 10^{-9}} = \frac{10^{-8}}{1.5 \times 10^{-9}} = \frac{10}{1.5} = 6.67\ V$.

(D) The magnetic field at the centre $O$ due to the circular loop of radius $R$ carrying current $I_1$ is given by $B_1 = \frac{\mu_0 I_1}{2R}$ (directed outwards).
The magnetic field at point $O$ due to the infinitely long straight wire carrying current $I_2$ at a distance $d = OA + AB = R + R = 2R$ is given by $B_2 = \frac{\mu_0 I_2}{2\pi d} = \frac{\mu_0 I_2}{2\pi (2R)} = \frac{\mu_0 I_2}{4\pi R}$ (directed inwards).
Since the net magnetic field at $O$ is zero,the magnitudes of the magnetic fields must be equal:
$B_1 = B_2$
$\frac{\mu_0 I_1}{2R} = \frac{\mu_0 I_2}{4\pi R}$
Canceling common terms $\mu_0$ and $R$ from both sides:
$\frac{I_1}{2} = \frac{I_2}{4\pi}$
Rearranging to find the ratio $I_1/I_2$:
$\frac{I_1}{I_2} = \frac{2}{4\pi} = \frac{1}{2\pi}$
Therefore,the correct option is $D$.

(B) Initially,the nucleus is at rest,so the total initial momentum is $0$.
By the law of conservation of linear momentum,the final momentum must also be $0$.
Let $M_d$ be the mass of the daughter nucleus and $M_{\alpha}$ be the mass of the $\alpha$-particle.
The mass of the $\alpha$-particle is approximately $4$ units,and the mass of the daughter nucleus is $(A - 4)$ units.
Let $v_1$ be the recoil velocity of the daughter nucleus.
According to the conservation of momentum:
$M_d v_1 = M_{\alpha} v$
$(A - 4) v_1 = 4v$
$v_1 = \frac{4v}{A - 4}$

(A) The impedance of the circuit is $Z = \sqrt{R^2 + X_C^2}$.
When the angular frequency is changed to $\omega' = \omega / 3$,the capacitive reactance becomes $X_C' = \frac{1}{\omega' C} = \frac{1}{(\omega/3) C} = 3 X_C$.
The new impedance is $Z' = \sqrt{R^2 + (3X_C)^2} = \sqrt{R^2 + 9X_C^2}$.
Given that the current becomes half,$I' = I/2$,which implies $Z' = 2Z$.
Substituting the expressions,we get $2 \sqrt{R^2 + X_C^2} = \sqrt{R^2 + 9X_C^2}$.
Squaring both sides,$4(R^2 + X_C^2) = R^2 + 9X_C^2$.
$4R^2 + 4X_C^2 = R^2 + 9X_C^2$.
$3R^2 = 5X_C^2$.
Therefore,$\frac{X_C^2}{R^2} = \frac{3}{5} = 0.6$.
Thus,the ratio $\frac{X_C}{R} = \sqrt{0.6}$.

(B) Insulin is a peptide hormone composed of two polypeptide chains,referred to as the $A$ chain and $B$ chain.
These two chains are linked together by two inter-chain disulphide bonds.
Additionally,there is one intra-chain disulphide bond present within the $A$ chain.
Therefore,the total number of disulphide linkages in insulin is $2 + 1 = 3$.

(D) The total number of electrons in $O_{2}^{-}$ is $8 + 8 + 1 = 17$.
The molecular orbital configuration of $O_{2}^{-}$ is: $\sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \sigma 2p_{z}^{2}, \pi 2p_{x}^{2} = \pi 2p_{y}^{2}, \pi^{*} 2p_{x}^{2} = \pi^{*} 2p_{y}^{1}$.
Anti-bonding electrons are those in orbitals denoted with an asterisk $(*)$.
Number of anti-bonding electrons $= 2(\sigma^{*} 1s) + 2(\sigma^{*} 2s) + 2(\pi^{*} 2p_{x}) + 1(\pi^{*} 2p_{y}) = 7$.

(B) Intramolecular hydrogen bonding occurs within a single molecule. It typically happens in compounds where a hydrogen atom is bonded to a highly electronegative atom (like $O$,$N$,or $F$) and is positioned close to another electronegative atom within the same molecule,allowing for the formation of a stable ring structure. In salicylaldehyde,the hydrogen atom of the hydroxyl $(-OH)$ group is close to the oxygen atom of the aldehyde $(-CHO)$ group,leading to the formation of an intramolecular hydrogen bond. $H_2O$ and $NH_3$ exhibit intermolecular hydrogen bonding,while benzophenone lacks the necessary donor group ($-OH$,$-NH_2$,etc.) for hydrogen bonding.

(C) The given reaction is $A_{2(g)} + 2B_{(g)} \rightleftharpoons C_{(g)} + Q \ kJ$.
Since the reaction releases heat $(+Q \ kJ)$,it is an exothermic reaction.
According to Le Chatelier's principle,for an exothermic reaction,a decrease in temperature favors the forward reaction,thereby increasing the yield of the products.
Furthermore,the number of moles of gaseous reactants is $1 + 2 = 3$,while the number of moles of gaseous products is $1$.
Since the forward reaction proceeds with a decrease in the number of gaseous molecules,an increase in pressure will shift the equilibrium toward the product side.
Therefore,low temperature and high pressure favor the formation of products.

(B) The atomic number of copper $(Cu)$ is $29$.
The ground state electronic configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$.
To form the $Cu^{2+}$ ion,two electrons are removed,one from the $4s$ orbital and one from the $3d$ orbital.
Therefore,the electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9 4s^0$.

(C) Standard reduction potential is a measure of the tendency of an element to lose electrons in an aqueous solution.
More negative the reduction potential,higher is the ability of the element to lose electrons,and hence stronger is the reducing character.
Therefore,alkali metals behave as reducing agents.

(A) $1$. Identify the principal functional group: The $-COOH$ group is the principal functional group,so the parent chain is named as a pentanoic acid.
$2$. Number the chain: Start numbering from the carbon of the $-COOH$ group as $C-1$. The chain is numbered to give the lowest possible locants to the substituents.
$3$. Identify substituents: There is a methyl group at $C-2$ and a hydroxy group at $C-4$.
$4$. Alphabetical order: 'Hydroxy' comes before 'methyl'.
$5$. Final name: Combining these,the $IUPAC$ name is $4-$hydroxy$-2-$methylpentanoic acid.

(A) Resonance or mesomeric effect is defined as the polarity produced in a molecule by the interaction of two $\pi$-bonds or between a $\pi$-bond and a lone pair of electrons present on an adjacent atom.
It involves the complete delocalisation of $\pi$-electrons through the conjugated system.
There are two types of mesomeric effect:
$1$. $+M$ effect: Observed when the direction of electron displacement is away from an atom or substituent group attached to the conjugated system,e.g.,$-halogen$,$-OH$,$-NH_2$.
$2$. $-M$ effect: Observed when the transfer of electrons is towards the atom or substituent group attached to the conjugated system,e.g.,$-NO_2$,$-COOH$.

(D) The given compound is $CH_{3}CHBrCHBrCOOH$. It contains two chiral carbon atoms $(n=2)$.
Since the molecule is unsymmetrical (the two ends are different: $-CH_{3}$ and $-COOH$),it cannot be divided into two equal halves.
For an unsymmetrical molecule,the number of optical isomers is given by $2^{n}$,where $n$ is the number of chiral centers.
Here,$n=2$,so the number of optical isomers $= 2^{2} = 4$.
These $4$ isomers consist of $2$ pairs of enantiomers,and there are no meso forms $(m=0)$.
Thus,the total number of optical isomers is $4$.

(C) The correct answer is $C$. Meso compounds are those compounds whose molecules are superimposable on their mirror images despite the presence of asymmetric carbon atoms.
This is due to internal compensation.
The two halves of the molecule rotate the plane of polarised light in opposite directions and hence cancel the effect of each other,making the molecule optically inactive.
Thus,the optical inactivity in meso compounds is due to the presence of a plane of symmetry or molecular symmetry,which leads to internal compensation.

(A) Cyclohexane exists in various conformations,primarily the chair and boat forms.
In the chair conformation,all $C-H$ bonds on adjacent carbon atoms are in a staggered (skew) position,which minimizes torsional strain.
In the boat conformation,some $C-H$ bonds are eclipsed,leading to torsional strain. Additionally,there is steric repulsion between the hydrogen atoms at the $1$ and $4$ positions (flagpole interactions).
Because the boat conformation has higher total strain than the chair conformation,it is less stable.
Therefore,the chair conformation is the most stable and least energetic conformation of cyclohexane.

(A) The molecular formula for both acetone $(CH_3COCH_3)$ and propanal $(CH_3CH_2CHO)$ is $C_3H_6O$.
Acetone contains a ketonic functional group $(>C=O)$,while propanal contains an aldehydic functional group $(-CHO)$.
Since they have the same molecular formula but different functional groups,they are classified as functional isomers.

(A) When $1,4$-dibromopentane is treated with sodium metal,an intramolecular Wurtz reaction occurs.
This leads to the formation of a four-membered ring with a methyl substituent.
The reaction is represented as:
$CH_3-CH(Br)-CH_2-CH_2-CH_2-Br + 2Na \rightarrow \text{Methylcyclobutane} + 2NaBr$.
Thus,the product formed is methyl cyclobutane.

(C) The chlorination of $methane$ is carried out by treating it with chlorine in the presence of ultraviolet light or by heating the reaction mixture to $520-670 \ K$.
$CH_{4} + Cl_{2} \xrightarrow{hv \ or \ \Delta} CH_{3}Cl + HCl$
This $methyl \ chloride$ reacts with metallic $Na$ in the presence of dry ether to form a symmetrical alkane with a higher number of carbon atoms $(ethane)$.
$2CH_{3}Cl + 2Na \xrightarrow{\text{dry ether}} CH_{3}-CH_{3} + 2NaCl$
This reaction is called the $Wurtz$ reaction.

(B) conjugate acid-base pair differs by a single proton $(H^{+})$.
To find the conjugate base of an acid,we remove one proton $(H^{+})$ from the species:
$H_{2}PO_{4}^{-} \rightarrow H^{+} + HPO_{4}^{2-}$
Therefore,the conjugate base of $H_{2}PO_{4}^{-}$ is $HPO_{4}^{2-}$.

(A) Bronsted acid is defined as a substance that has a tendency to donate a proton $(H^{+})$.
Conversely,a Lewis acid is defined as a substance that is capable of accepting a lone pair of electrons.
$H_{3}O^{+}$,$NH_{3}$ (as a conjugate acid),and $HCl$ contain $H^{+}$ ions and can act as Bronsted acids.
$BF_{3}$ is an electron-deficient molecule with an incomplete octet on the boron atom,making it an electron pair acceptor,which classifies it as a Lewis acid.

(B) Since $HCl$ is an acid,its $pH$ must be less than $7$.
For very dilute solutions of strong acids,the contribution of $H^+$ ions from the auto-ionization of water cannot be neglected.
From $HCl$,$[H^+]_{acid} = 10^{-8} \ M$.
From water,$[H^+]_{water} = 10^{-7} \ M$.
Total $[H^+] = 10^{-8} + 10^{-7} = 10^{-8}(1 + 10) = 11 \times 10^{-8} \ M$.
$pH = -\log[H^+] = -\log(11 \times 10^{-8}) = -(\log 11 + \log 10^{-8}) = -(1.0414 - 8) = 6.9586$.

(C) buffer solution is classified based on the $pH$ of the solution.
$1$. Acidic buffers consist of a weak acid and its salt with a strong base (e.g.,$CH_3COOH$ and $CH_3COONa$).
$2$. Basic buffers consist of a weak base and its salt with a strong acid (e.g.,$NH_4OH$ and $NH_4Cl$).
$3$. $A$ neutral buffer is typically formed by the mixture of a weak acid and a weak base,such as $CH_3COOH$ and $NH_4OH$,which results in a solution with a $pH$ near $7$.

(D) In graphite,each carbon atom is $sp^{2}$ hybridised and linked to three other carbon atoms in a hexagonal planar structure.
In diamond,each carbon atom is $sp^{3}$ hybridised and linked to four other carbon atoms in a tetrahedral geometry via strong $C-C$ $\sigma$-bonds.
Therefore,the hybridisation states for graphite and diamond are $sp^{2}$ and $sp^{3}$ respectively.

(C) In the given reaction,the oxidation state of iron $(Fe)$ increases from $+2$ in $FeSO_{4}$ to $+3$ in $Fe_{2}(SO_{4})_{3}$.
This indicates that $FeSO_{4}$ is being oxidised.
Hydrogen peroxide $(H_{2} O_{2})$ is a powerful oxidising agent that accepts electrons and gets reduced to water $(H_{2} O)$.
Therefore,$H_{2} O_{2}$ acts as the oxidising agent in this reaction.

(C) The lustre of metals is due to the presence of free or mobile electrons in the metallic lattice.
When light strikes the surface of the metal,these mobile electrons absorb the energy of the incident light and then re-emit it.
This reflection of light by the mobile electrons gives the metal its characteristic shiny appearance or lustre.

(D) To determine the magnetic property,we look at the Molecular Orbital $(MO)$ configuration of each species:
$H_{2}^{+}$ ($1$ electron): $\sigma 1s^{1}$ (Paramagnetic due to $1$ unpaired electron).
$He_{2}^{+}$ ($3$ electrons): $\sigma 1s^{2}, \sigma^{*} 1s^{1}$ (Paramagnetic due to $1$ unpaired electron).
$O_{2}$ ($16$ electrons): $\sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \sigma 2p_{x}^{2}, \pi 2p_{y}^{2} = \pi 2p_{z}^{2}, \pi^{*} 2p_{y}^{1} = \pi^{*} 2p_{z}^{1}$ (Paramagnetic due to $2$ unpaired electrons).
$N_{2}$ ($14$ electrons): $\sigma 1s^{2}, \sigma^{*} 1s^{2}, \sigma 2s^{2}, \sigma^{*} 2s^{2}, \pi 2p_{y}^{2} = \pi 2p_{z}^{2}, \sigma 2p_{x}^{2}$ (Diamagnetic as all electrons are paired).
Therefore,$N_{2}$ is the correct answer.

(D) Colligative properties are those properties of ideal solutions that depend only on the number of solute particles (molecules or ions) dissolved in a definite amount of solvent,and not on the nature of the solute.
The four main colligative properties are:
$(i)$ Relative lowering of vapour pressure.
$(ii)$ Osmotic pressure.
$(iii)$ Elevation in boiling point.
$(iv)$ Depression in freezing point.
Note that 'Lowering of vapour pressure' is not a colligative property; rather,the 'Relative lowering of vapour pressure' is. Therefore,option $D$ is the correct answer.

(C) The elevation in boiling point is a colligative property,which depends on the van't Hoff factor $(i)$ and the molar concentration $(M)$. The formula is $\Delta T_{b} = i \times K_{b} \times m$.
$1.$ $0.01 \ M$ urea: $i = 1$,effective concentration = $0.01 \times 1 = 0.01 \ M$.
$2.$ $0.01 \ M \ KNO_{3}$: $i = 2$ $(K^{+} + NO_{3}^{-})$,effective concentration = $0.01 \times 2 = 0.02 \ M$.
$3.$ $0.01 \ M \ Na_{2}SO_{4}$: $i = 3$ $(2Na^{+} + SO_{4}^{2-})$,effective concentration = $0.01 \times 3 = 0.03 \ M$.
$4.$ $0.015 \ M \ C_{6}H_{12}O_{6}$: $i = 1$,effective concentration = $0.015 \times 1 = 0.015 \ M$.
Since $0.01 \ M \ Na_{2}SO_{4}$ has the highest effective concentration of particles,it will exhibit the highest boiling point.

(D) The molar mass of $H_{2}O$ is $18 \ g/mol$.
$18 \ g$ of $H_{2}O$ contains $6.022 \times 10^{23}$ molecules.
Therefore,$0.018 \ g$ of $H_{2}O$ contains:
$\text{Number of molecules} = \frac{\text{Given mass}}{\text{Molar mass}} \times N_{A}$
$= \frac{0.018 \ g}{18 \ g/mol} \times 6.022 \times 10^{23} \text{ molecules/mol}$
$= 0.001 \times 6.022 \times 10^{23}$
$= 6.022 \times 10^{20} \text{ molecules}$.

(A) $1 \ mol$ of $NH_3 = 17 \ g = 22400 \ cm^3$ at $STP$.
Since $22400 \ cm^3$ of $NH_3$ at $STP$ has a mass of $17 \ g$.
Therefore,the mass of $112 \ cm^3$ of $NH_3$ at $STP$ is calculated as:
$\text{Mass} = \frac{17 \ g}{22400 \ cm^3} \times 112 \ cm^3 = 0.085 \ g$.

(A) The empirical formula of the compound is $CH_2O$.
Empirical formula mass $= (1 \times 12) + (2 \times 1) + (1 \times 16) = 30 \ g/mol$.
Molecular mass $= 90 \ g/mol$.
Calculate the value of $n$ using the formula: $n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{90}{30} = 3$.
The molecular formula is given by $n \times (\text{Empirical formula}) = 3 \times CH_2O = C_3H_6O_3$.

(A) The root mean square $(RMS)$ velocity of a gas is given by the formula $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$.
Since $R$ and $T$ are constant for all gases at a given temperature,the $RMS$ velocity is inversely proportional to the square root of the molar mass,i.e.,$v_{\text{rms}} \propto \frac{1}{\sqrt{M}}$.
Comparing the molar masses $(M)$:
$CH_{4} = 16 \ g/mol$
$CO = 28 \ g/mol$
$Cl_{2} = 71 \ g/mol$
$CO_{2} = 44 \ g/mol$
Since $CH_{4}$ has the lowest molar mass,it will have the highest $RMS$ velocity.

(C) Isoelectronic species are those that have the same number of electrons.
Calculating the number of electrons for each ion:
$K^{+} = 19 - 1 = 18$
$Ca^{2+} = 20 - 2 = 18$
$Sc^{3+} = 21 - 3 = 18$
$Cl^{-} = 17 + 1 = 18$
Since all ions in option $C$ have $18$ electrons,they are isoelectronic.

(D) The given thermochemical equation is:
$2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(l)}$; $\Delta H = -571.6 \ kJ$
The decomposition of water is represented by the reverse reaction:
$H_2O_{(l)} \rightarrow H_{2(g)} + \frac{1}{2} O_{2(g)}$
To obtain this,we reverse the given equation and divide it by $2$:
$\Delta H_{\text{decomposition}} = -(\frac{\Delta H}{2}) = -(\frac{-571.6 \ kJ}{2}) = +285.8 \ kJ$
Thus,the heat of decomposition of water is $+285.8 \ kJ$.

(B) According to Gibbs' energy equation,$ \Delta G = \Delta H - T \Delta S $.
For a process to be spontaneous,$ \Delta G $ must be negative.
When $ \Delta H $ is negative (exothermic) and $ \Delta S $ is positive (increase in entropy),the term $ -T \Delta S $ becomes negative.
Thus,$ \Delta G = (\text{negative}) - (\text{positive}) = \text{negative}$.
Therefore,the process is spontaneous at all temperatures when $ \Delta H < 0 $ and $ \Delta S > 0 $.

(A) According to the second law of thermodynamics,all spontaneous processes are thermodynamically irreversible and are accompanied by a net increase in entropy.
Therefore,for all spontaneous processes,the total entropy change (sum of the entropy changes of the system and the surroundings) is positive.
This implies that the entropy of the universe is continuously increasing.

(C) The elevation in boiling point is given by the formula: $\Delta T_b = i \cdot K_b \cdot m$.
Since $K_b$ is constant for the same solvent (water) and assuming molality $m \approx M$,the boiling point depends on the product of the van't Hoff factor $(i)$ and molarity $(M)$.
Calculating $i \times M$ for each:
$A$: Urea is a non-electrolyte,$i = 1$. $1 \times 0.01 = 0.01$.
$B$: $KNO_3 \rightarrow K^+ + NO_3^-$,$i = 2$. $2 \times 0.01 = 0.02$.
$C$: $Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-}$,$i = 3$. $3 \times 0.01 = 0.03$.
$D$: Glucose $(C_6H_{12}O_6)$ is a non-electrolyte,$i = 1$. $1 \times 0.015 = 0.015$.
Since the product $i \times M$ is highest for $Na_2SO_4$,it will exhibit the highest boiling point.

(C) The acidity of phenols is due to the greater resonance stabilization of the phenoxide ion relative to phenol. Electron withdrawing groups $(EWG)$ like $-NO_{2}$ stabilize the phenoxide ion by dispersing the negative charge,which increases the acidity of phenols.
Conversely,electron donating groups $(EDG)$ like alkyl groups destabilize the phenoxide ion by intensifying the negative charge,which decreases the acidic strength of phenols.
Since the methyl group has a $+I$ effect,which is stronger at the $o$-position than at the $p$-position (as the $+I$ effect decreases with distance),$o$-cresol is a weaker acid than $p$-cresol.
Therefore,the order of acidic strength is: $p$-nitrophenol $>$ phenol $>$ $p$-cresol $>$ $o$-cresol.

(B) Acetaldehyde is an aliphatic aldehyde,while benzaldehyde is an aromatic aldehyde.
Fehling's solution is a mild oxidizing agent that can oxidize aliphatic aldehydes (like acetaldehyde) to their corresponding carboxylic acids,resulting in a red precipitate of cuprous oxide $(Cu_2O)$.
However,aromatic aldehydes (like benzaldehyde) are not strong enough to be oxidized by Fehling's solution.
Therefore,Fehling's solution can be used to distinguish between acetaldehyde and benzaldehyde.
Tollen's reagent reacts with both aliphatic and aromatic aldehydes,so it cannot distinguish between them.

(A) The reaction described is the Hofmann bromamide degradation reaction.
In this reaction,an amide is treated with bromine $(Br_{2})$ and an aqueous solution of a strong base like potassium hydroxide $(KOH)$ to produce a primary amine.
The reaction is as follows:
$R-CONH_{2} + Br_{2} + 4KOH \rightarrow R-NH_{2} + K_{2}CO_{3} + 2KBr + 2H_{2}O$
This reaction is used to decrease the carbon chain length by one carbon atom,which is removed as a carbonate ion $(CO_{3}^{2-})$.

(A) When glucose is heated with $HI$ and red phosphorus,it undergoes complete reduction to form $n$-hexane. The reaction is as follows:
$CHO-(CHOH)_4-CH_2OH \xrightarrow{HI/\text{red } P} CH_3-(CH_2)_4-CH_3$ ($n$-hexane).
This reaction confirms that glucose contains a straight chain of six carbon atoms.

(B) Formic acid $(HCOOH)$ undergoes dehydration when heated with concentrated sulfuric acid $(H_{2}SO_{4})$.
$HCOOH \xrightarrow{\text{conc. } H_{2}SO_{4}} CO + H_{2}O$
The concentrated $H_{2}SO_{4}$ acts as a dehydrating agent,removing a water molecule from the formic acid to produce carbon monoxide $(CO)$ gas.

(D) When $1$ $s$,$3$ $p$ and $2$ $d$-orbitals belonging to the same shell of an atom mix together to form six new equivalent orbitals,the type of hybridisation is called $d^{2}sp^{3}$ or octahedral hybridisation.
The new orbitals are called $d^{2}sp^{3}$ or octahedral orbitals.
These orbitals are directed towards the corners of an octahedron,resulting in an octahedral geometry.

(D) For a collision to be effective,the colliding molecules must possess energy equal to or greater than a specific value known as the threshold energy. At room temperature,most reactant molecules have energy less than this threshold value. When the temperature increases,the kinetic energy of the reactant molecules increases,leading to a higher fraction of molecules possessing energy equal to or greater than the threshold energy. Consequently,the number of effective collisions per unit time increases,which results in an increase in the rate of reaction.

(B) The temperature coefficient $(n)$ is defined as the ratio of rate constants at temperatures differing by $10^{\circ} C$:
$n = \frac{k_{T+10}}{k_T} = 2$.
This means for every $10^{\circ} C$ rise in temperature,the rate of reaction doubles.
The total rise in temperature is $\Delta T = 90^{\circ} C - 30^{\circ} C = 60^{\circ} C$.
The number of $10^{\circ} C$ intervals is $x = \frac{60}{10} = 6$.
The rate of reaction increases by a factor of $n^x = 2^6$.
$2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$.
Therefore,the rate of reaction increases by $64$ times.

(A) For a first order reaction,the half-life $t_{1/2} = 25 \ min$.
The number of half-lives $n$ in $100 \ min$ is calculated as $n = \frac{100 \ min}{25 \ min} = 4$.
The amount of reactant remaining after $n$ half-lives is given by $\frac{A_0}{2^n}$.
Amount remaining $= \frac{100}{2^4} = \frac{100}{16} = 6.25 \%$.
The amount of reactant converted into product $= 100 \% - 6.25 \% = 93.75 \%$.

(C) Insulin is composed of two peptide chains referred to as the $A$ chain and $B$ chain.
The $A$ chain consists of $21$ amino acid residues and the $B$ chain consists of $30$ amino acid residues.
These two chains are linked together by two inter-chain disulphide bridges.
Additionally,there is one intra-chain disulphide bridge within the $A$ chain.
Therefore,the total number of disulphide linkages in insulin is $3$.

(A) Oils are liquids at room temperature because they contain a high proportion of unsaturated fatty acid residues,such as oleic acid,linoleic acid,and linolenic acid. These residues form esters known as oleates. In contrast,fats are solids at room temperature and contain a high proportion of saturated fatty acid residues like myristic,palmitic,and stearic acids. For example,glyceryl trioleate (triolein) is an oil,while glyceryl tristearate (tristearin) is a fat.

(B) ligand is defined as an atom,molecule,or ion that donates at least one pair of electrons to the central metal atom or ion to form a coordinate bond.
Since ligands donate electron pairs,they act as $Lewis$ bases.
Therefore,the essential requirement for a species to act as a ligand is the presence of at least one unshared electron pair (lone pair) that can be donated to the metal center.

(A) The magnetic moment $(\mu)$ is related to the number of unpaired electrons $(n)$ by the formula: $\mu = \sqrt{n(n+2)} \ BM$.
Greater the number of unpaired electrons,higher is the magnetic moment.
Let us determine the number of unpaired electrons for each ion:
$Cr^{3+}$ $(3d^3)$: $n = 3$
$Fe^{2+}$ $(3d^6)$: $n = 4$
$Co^{2+}$ $(3d^7)$: $n = 3$
$Ni^{2+}$ $(3d^8)$: $n = 2$
Since $Fe^{2+}$ has the maximum number of unpaired electrons $(n=4)$,it exhibits the highest magnetic moment.

(C) The paramagnetism of transition elements in a given series depends on the number of unpaired electrons in the $d$-orbitals.
As we move from left to right across a transition series,the number of unpaired electrons increases until the $d^5$ configuration is reached.
Consequently,the paramagnetism increases to a maximum value at the $d^5$ configuration.
Beyond $d^5$,the electrons begin to pair up in the $d$-orbitals ($d^6$ to $d^{10}$),which leads to a decrease in the number of unpaired electrons and thus a decrease in paramagnetism.
Therefore,the paramagnetism first increases to a maximum and then decreases.

(C) The standard reduction potentials are $E^o_{Zn^{2+}/Zn} = -0.76 \ V$ and $E^o_{Fe^{2+}/Fe} = -0.41 \ V$.
In a galvanic cell,the electrode with the higher reduction potential acts as the cathode and the one with the lower reduction potential acts as the anode.
$E^o_{cell} = E^o_{cathode} - E^o_{anode}$
$E^o_{cell} = (-0.41 \ V) - (-0.76 \ V) = -0.41 \ V + 0.76 \ V = +0.35 \ V$.

(C) Calcination is the process of converting an ore into its oxide by heating it strongly below its melting point in the absence or limited supply of air. Carbonate ores are converted into their respective oxides by the loss of carbon dioxide.
$CaCO_{3} \stackrel{\Delta}{\longrightarrow} CaO + CO_{2} \uparrow$

(B) Zone refining is a method used to obtain metals of very high purity.
It is based on the principle that the impurities are more soluble in the melt than in the solid state of the metal.
Metals like $Si$,$Ge$,and $Ga$,which are used in semiconductor devices,are purified by this method.

(C) Let us analyze each reaction:
$I.$ $RX + AgOH(aq) \longrightarrow ROH + AgX$. The product is an alcohol $(ROH)$,not an alkane $(RH)$. So,$I$ is incorrect.
$II.$ $RX + AgCN(alc) \longrightarrow RNC + AgX$. This is a correct reaction where alkyl isocyanide is formed.
$III.$ $RX + KCN(alc) \longrightarrow RCN + KX$. The product is alkyl cyanide $(RCN)$,not isocyanide $(RNC)$. So,$III$ is incorrect.
$IV.$ $RX + Na(ether) \longrightarrow R-R + 2NaX$. This is the Wurtz reaction,which correctly forms an alkane $(R-R)$.
Thus,pairs $II$ and $IV$ are correctly matched.

(B) The dry distillation of a mixture of calcium acetate $(CH_3COO)_2Ca$ and calcium formate $(HCOO)_2Ca$ results in the formation of acetaldehyde $(CH_3CHO)$ and calcium carbonate $(CaCO_3)$.
The chemical reaction is as follows:
$(CH_3COO)_2Ca + (HCOO)_2Ca \xrightarrow{\text{dry distillation}} 2CH_3CHO + 2CaCO_3$

(D) The $Wurtz$ reaction involves the coupling of two alkyl halide molecules in the presence of metallic $Na$ and dry ether to form a symmetrical alkane.
The product $4,5-$diethyl octane has the structure $CH_{3}CH_{2}CH_{2}CH(CH_{2}CH_{3})CH(CH_{2}CH_{3})CH_{2}CH_{2}CH_{3}$.
This molecule is formed by the coupling of two $3-$bromohexane units at the $C-3$ position.
Therefore,the alkyl bromide $(X)$ is $3-$bromohexane,which is $CH_{3}CH_{2}CH_{2}CH(Br)CH_{2}CH_{3}$.

(D) The conversion of benzene to acetophenone is achieved by reacting benzene with acetyl chloride $(CH_3COCl)$ in the presence of an anhydrous Lewis acid catalyst like $AlCl_3$.
This reaction is a specific type of electrophilic aromatic substitution known as Friedel-Crafts acylation,where an acyl group $(-COCH_3)$ is introduced into the benzene ring.
The reaction is represented as:
$C_6H_6 + CH_3COCl \xrightarrow{AlCl_3} C_6H_5COCH_3 + HCl$

(C) When excess $PCl_{5}$ reacts with concentrated $H_{2}SO_{4}$,the hydroxyl groups of the acid are replaced by chlorine atoms to form sulphuryl chloride $(SO_{2}Cl_{2})$.
The chemical reaction is as follows:
$SO_{2}(OH)_{2} + 2PCl_{5} \rightarrow SO_{2}Cl_{2} + 2POCl_{3} + 2HCl$

(D) Xenon $(Xe)$ is used in discharge tubes for producing a high-speed flash of bluish light,which is used in high-speed photography.

(D) When a mixture of $1,3-$butadiene and styrene in the ratio of $3:1$ undergoes copolymerisation in the presence of sodium,it forms a styrene-butadiene copolymer,commonly known as styrene-butadiene rubber $(SBR)$ or Buna-$S$.
In the name Buna-$S$,'$Bu$' stands for butadiene,'$Na$' stands for sodium (which acts as the polymerising agent),and '$S$' stands for styrene.

(A) In a crystal lattice,a face is common to two adjacent unit cells.
Therefore,each face of a unit cell is shared equally by $2$ unit cells.

(B) Colligative properties are properties of solutions that depend only on the number of solute particles present in a given amount of solvent,not on their identity.
Common colligative properties include:
$1$. Relative lowering of vapour pressure
$2$. Elevation in boiling point
$3$. Depression in freezing point
$4$. Osmotic pressure
Optical activity is a property related to the interaction of a substance with plane-polarized light and is not dependent on the number of solute particles.
Therefore,optical activity is not a colligative property.

(C) The relative lowering of vapour pressure is given by the formula: $\frac{p^{\circ}-p_s}{p^{\circ}} = \frac{n_2}{n_1+n_2}$
where $p^{\circ}$ is the vapour pressure of the pure solvent,$p_s$ is the vapour pressure of the solution,$n_2$ is the number of moles of the solute (urea),and $n_1$ is the number of moles of the solvent $(H_2O)$.
Molar mass of urea $(NH_2CONH_2)$ = $60 \ g/mol$.
$n_2 = \frac{w_2}{M_2} = \frac{3 \ g}{60 \ g/mol} = 0.05 \ mol$.
Molar mass of $H_2O = 18 \ g/mol$.
$n_1 = \frac{w_1}{M_1} = \frac{45 \ g}{18 \ g/mol} = 2.5 \ mol$.
Substituting the values: $\frac{p^{\circ}-p_s}{p^{\circ}} = \frac{0.05}{2.5 + 0.05} = \frac{0.05}{2.55} \approx 0.0196 \approx 0.02$.

(B) Adsorption theory is applicable for solid catalysts which show heterogeneous catalysis.
According to this theory,the gaseous reactants are adsorbed on the surface of the solid catalyst.
As a result,the concentration of the reactants increases on the surface and hence the rate of reaction increases.

(C) The stability of a lyophobic colloidal solution is primarily due to the presence of electrical charges on the colloidal particles.
Because all particles carry the same type of charge (either positive or negative),they repel each other.
This electrostatic repulsion prevents the particles from coming close enough to coalesce and settle down,thereby maintaining the stability of the colloid.

(D) The minimum amount of an electrolyte that must be added to one litre of a colloidal solution to bring about complete coagulation is called the coagulation or flocculation value of the electrolyte.
According to the Hardy-Schulze rule,the greater the valency of the coagulating or flocculating ion,the smaller the quantity of the electrolyte required to coagulate a definite amount of the colloidal sol.
Arsenic sulphide sol is a negatively charged sol.
Therefore,it is coagulated by cations.
The effectiveness of cations follows the order: $Al^{3+} > Ca^{2+} = Mg^{2+} > Na^{+}$.
Since the flocculation value is inversely proportional to the coagulating power,the order of flocculation value is: $Na^{+} > Ca^{2+} = Mg^{2+} > Al^{3+}$.
Thus,$Al^{3+}$ has the minimum flocculation value.