KCET 2008 Chemistry Question Paper with Answer and Solution

(A) In a steady state,the rate of heat flow through the copper rod must equal the rate of heat flow through the steel rod.
Let $\theta$ be the temperature at the junction.
Let $K_{Cu}$ and $K_S$ be the thermal conductivities,$A$ be the cross-sectional area,and $L_{Cu}$ and $L_S$ be the lengths of the copper and steel rods respectively.
Given: $K_{Cu} = 9K_S$,$L_{Cu} = 18 \text{ cm}$,$L_S = 6 \text{ cm}$,$T_{Cu} = 100^\circ C$,$T_S = 0^\circ C$.
The rate of heat flow is given by $H = \frac{KA(T_1 - T_2)}{L}$.
Equating the heat flow rates: $\frac{K_{Cu} A (100 - \theta)}{L_{Cu}} = \frac{K_S A (\theta - 0)}{L_S}$.
Substituting the values: $\frac{9K_S (100 - \theta)}{18} = \frac{K_S (\theta - 0)}{6}$.
Simplifying: $\frac{9(100 - \theta)}{18} = \frac{\theta}{6}$.
$\frac{100 - \theta}{2} = \frac{\theta}{6}$.
$3(100 - \theta) = \theta$.
$300 - 3\theta = \theta$.
$4\theta = 300$.
$\theta = 75^\circ C$.

(C) When carbon monoxide $(CO)$ is passed over solid caustic soda $(NaOH)$ at $200^\circ C$ and $6-10 \text{ atmospheric pressure}$,it undergoes a chemical reaction to form sodium formate $(HCOONa)$.
The chemical equation is: $CO + NaOH \xrightarrow{200^\circ C} HCOONa$.

(B) Let $OP = h$ be the height of the tower. Let $O$ be the foot of the tower.
In $\triangle OAP$,$\angle OAP = \alpha$. So,$OA = h \cot \alpha$.
In $\triangle OBP$,$\angle OBP = 2\alpha$. So,$OB = h \cot 2\alpha$.
Given $AB = a$,we have $OA - OB = a$.
Therefore,$h(\cot \alpha - \cot 2\alpha) = a$.
$h \left( \frac{\cos \alpha}{\sin \alpha} - \frac{\cos 2\alpha}{\sin 2\alpha} \right) = a$
$h \left( \frac{\sin 2\alpha \cos \alpha - \cos 2\alpha \sin \alpha}{\sin \alpha \sin 2\alpha} \right) = a$
$h \left( \frac{\sin(2\alpha - \alpha)}{\sin \alpha \sin 2\alpha} \right) = a$
$h \left( \frac{\sin \alpha}{\sin \alpha \sin 2\alpha} \right) = a$
$h \left( \frac{1}{\sin 2\alpha} \right) = a$
$h = a \sin 2\alpha$.

(A) The conductor consists of three parts: two semi-infinite straight wires ($A$ and $C$) and a three-quarter circular arc $(B)$.
$1$. For the straight wire $A$,the point $O$ lies on its axis,so the magnetic field $B_A = 0$.
$2$. For the straight wire $C$,the distance from $O$ is $r$. The magnetic field due to a semi-infinite wire at a perpendicular distance $r$ is $B_C = \frac{\mu_0 I}{4\pi r}$. Using the right-hand rule,the direction is into the page $(\otimes)$.
$3$. For the three-quarter circular arc $B$,the angle subtended is $\theta = \frac{3\pi}{2}$. The magnetic field is $B_B = \frac{\mu_0 I \theta}{4\pi r} = \frac{\mu_0 I (3\pi/2)}{4\pi r}$. The direction is also into the page $(\otimes)$.
$4$. The net magnetic field at $O$ is $B_{net} = B_A + B_B + B_C = 0 + \frac{\mu_0 I}{4\pi r} \left( \frac{3\pi}{2} \right) + \frac{\mu_0 I}{4\pi r} = \frac{\mu_0 I}{4\pi r} \left( \frac{3\pi}{2} + 1 \right)$.

(C) The magnetic field at the centre of a circular current-carrying coil is given by:
$B_c = \frac{\mu_0 I}{2r} \quad \dots(i)$
The magnetic field on the axis of a circular coil at a distance $d$ from the centre is given by:
$B_a = \frac{\mu_0 I r^2}{2(r^2 + d^2)^{3/2}}$
Given $d = r$,we substitute this into the formula:
$B_a = \frac{\mu_0 I r^2}{2(r^2 + r^2)^{3/2}} = \frac{\mu_0 I r^2}{2(2r^2)^{3/2}} = \frac{\mu_0 I r^2}{2(2\sqrt{2} r^3)} = \frac{\mu_0 I}{4\sqrt{2} r} \quad \dots(ii)$
Now,calculating the ratio $B_c : B_a$:
$\frac{B_c}{B_a} = \left( \frac{\mu_0 I}{2r} \right) \div \left( \frac{\mu_0 I}{4\sqrt{2} r} \right) = \frac{4\sqrt{2} r}{2r} = 2\sqrt{2}$
Therefore,$B_c : B_a = 2\sqrt{2} : 1$.

(A) Let $V_A$ be the potential of the outer sphere (radius $R$,charge $Q$) and $V_B$ be the potential of the inner sphere (radius $r$,charge $q$).
The potential at the surface of the outer sphere $(A)$ is due to its own charge $Q$ and the charge $q$ on the inner sphere:
$V_A = \frac{1}{4\pi \epsilon_0} \frac{Q}{R} + \frac{1}{4\pi \epsilon_0} \frac{q}{R} = \frac{1}{4\pi \epsilon_0} \left( \frac{Q+q}{R} \right)$
The potential at the surface of the inner sphere $(B)$ is due to its own charge $q$ and the potential due to the outer sphere (which is constant inside it):
$V_B = \frac{1}{4\pi \epsilon_0} \frac{q}{r} + \frac{1}{4\pi \epsilon_0} \frac{Q}{R} = \frac{1}{4\pi \epsilon_0} \left( \frac{q}{r} + \frac{Q}{R} \right)$
The potential difference between the spheres is $V_B - V_A$:
$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left( \frac{q}{r} + \frac{Q}{R} \right) - \frac{1}{4\pi \epsilon_0} \left( \frac{Q+q}{R} \right)$
$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left( \frac{q}{r} + \frac{Q}{R} - \frac{Q}{R} - \frac{q}{R} \right)$
$V_B - V_A = \frac{1}{4\pi \epsilon_0} \left( \frac{q}{r} - \frac{q}{R} \right)$

(C) The magnetic field at the centre of a circular current-carrying coil of radius $r$ is given by:
$B_c = \frac{\mu_0 I}{2r} \quad \dots(i)$
The magnetic field on the axis of a circular coil at a distance $x$ from the centre is given by:
$B_a = \frac{\mu_0 I r^2}{2(r^2 + x^2)^{3/2}}$
Given $x = r$,we substitute this into the formula:
$B_a = \frac{\mu_0 I r^2}{2(r^2 + r^2)^{3/2}} = \frac{\mu_0 I r^2}{2(2r^2)^{3/2}} = \frac{\mu_0 I r^2}{2(2\sqrt{2}r^3)} = \frac{\mu_0 I}{4\sqrt{2}r} \quad \dots(ii)$
Now,taking the ratio $B_c : B_a$:
$\frac{B_c}{B_a} = \frac{\frac{\mu_0 I}{2r}}{\frac{\mu_0 I}{4\sqrt{2}r}} = \frac{4\sqrt{2}r}{2r} = 2\sqrt{2}$
Therefore,$B_c : B_a = 2\sqrt{2} : 1$.

(C) Given: Primary voltage $V_{p} = 220 ~V$,Primary current $I_{p} = 5 ~A$,Secondary voltage $V_{s} = 2200 ~V$,Power loss $= 50 \%$.
Efficiency $\eta = 100 \% - 50 \% = 50 \%$.
The efficiency of a transformer is defined as the ratio of output power to input power:
$\eta = \frac{P_{out}}{P_{in}} = \frac{V_{s} I_{s}}{V_{p} I_{p}}$
Substituting the given values:
$0.5 = \frac{2200 \times I_{s}}{220 \times 5}$
$0.5 = \frac{2200 \times I_{s}}{1100}$
$0.5 = 2 \times I_{s}$
$I_{s} = \frac{0.5}{2} = 0.25 ~A$.
Therefore,the current in the secondary coil is $0.25 ~A$.

(B) The wavelength of spectral lines in the hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the minimum wavelength,the transition must be from $n_2 = \infty$ to $n_1$.
For the Lyman series,$n_1 = 1$,so $\frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R$,which gives $\lambda_L = \frac{1}{R}$.
For the Balmer series,$n_1 = 2$,so $\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = \frac{R}{4}$,which gives $\lambda_B = \frac{4}{R}$.
The ratio of the minimum wavelengths is $\frac{\lambda_L}{\lambda_B} = \frac{1/R}{4/R} = \frac{1}{4} = 0.25$.

(B) The minimum wavelength of a spectral series corresponds to the transition from $n_{i}=\infty$ to the final orbit $n_{f}$.
For the Lyman series,$n_{f}=1$:
$\frac{1}{(\lambda_{\min})_{L}} = R \left[ \frac{1}{1^2} - \frac{1}{\infty^2} \right] = R$
$\Rightarrow (\lambda_{\min})_{L} = \frac{1}{R}$ --- $(i)$
For the Balmer series,$n_{f}=2$:
$\frac{1}{(\lambda_{\min})_{B}} = R \left[ \frac{1}{2^2} - \frac{1}{\infty^2} \right] = \frac{R}{4}$
$\Rightarrow (\lambda_{\min})_{B} = \frac{4}{R}$ --- (ii)
Dividing equation $(i)$ by (ii):
$\frac{(\lambda_{\min})_{L}}{(\lambda_{\min})_{B}} = \frac{1/R}{4/R} = \frac{1}{4} = 0.25$

(C) The molecular orbital picture of benzene shows that all six carbon atoms are $sp^{2}$ hybridized.
Out of these three $sp^{2}$ hybrid orbitals of each $C$ atom,two orbitals overlap with $sp^{2}$ hybrid orbitals of adjacent $C$ atoms to form six $C-C$ single bonds.
The remaining $sp^{2}$ orbital of each $C$ atom overlaps with the $s$-orbital of each hydrogen atom to form six $C-H$ single sigma bonds.
Each $C$ atom is now left with one unhybridized $p$-orbital perpendicular to the plane of the ring.

(A) In an antibonding molecular orbital,such as $\sigma^{*}s$,the wave functions of the atomic orbitals interfere destructively.
This results in a region of zero electron density between the two nuclei,which is known as a nodal plane.
As shown in the diagram,there is exactly $1$ nodal plane perpendicular to the internuclear axis for the $\sigma^{*}s$ antibonding molecular orbital.

(C) The molecular orbital $(MO)$ configuration of the superoxide ion,$O_{2}^{-} (17 \ e^{-})$,is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$.
Bond order is calculated as: $\text{Bond order} = \frac{1}{2} (N_b - N_a) = \frac{1}{2} (10 - 7) = \frac{3}{2} = 1.5$.

(D) single water molecule $(H_2O)$ has two hydrogen atoms that can act as donors for hydrogen bonding and two lone pairs on the oxygen atom that can act as acceptors for hydrogen bonding.
Thus,one water molecule can form a maximum of four hydrogen bonds with surrounding water molecules.
This is represented by the structure where the central oxygen atom is bonded to two hydrogen atoms via covalent bonds and interacts with two other hydrogen atoms from neighboring water molecules via hydrogen bonds,while the two hydrogen atoms of the central molecule interact with the oxygen atoms of two other neighboring water molecules.

(A) The given reaction is $2 \ HI_{(g)} \rightleftharpoons H_{2(g)} + I_{2(g)} - Q \ kJ$.
For this reaction,the change in the number of moles of gaseous species is $\Delta n_g = n_p - n_R = 2 - 2 = 0$.
Since $\Delta n_g = 0$,the equilibrium constant is not affected by changes in pressure or volume.
$A$ catalyst only speeds up the attainment of equilibrium and does not alter the value of the equilibrium constant.
The equilibrium constant ($K_c$ or $K_p$) is a function of temperature only,as described by the van't Hoff equation. Therefore,for this reaction,the equilibrium constant depends only upon the temperature.

(A) $N_{2} + 3H_{2} \rightleftharpoons 2NH_{3} + \text{heat}$
According to Le Chatelier's principle,for an exothermic reaction,an increase in temperature shifts the equilibrium in the direction that absorbs heat.
Since the forward reaction is exothermic,increasing the temperature will favour the backward reaction.
Therefore,the equilibrium shifts to the left.

(C) In metallic bonds,the valence shell electrons are delocalised and shared between many atoms.
These delocalised electrons allow the metal atoms to slide past one another without being subjected to strong repulsive forces.
The malleability and ductility of metals is due to this sliding capacity of the layers of metal ions.

(D) The tendency to form a metal oxide is related to the stability of the oxide formed and the reactivity of the metal.
According to the Ellingham diagram,metals that are more reactive and have a more negative Gibbs free energy of formation $(\Delta G_f^\circ)$ for their oxides have a greater tendency to form oxides.
Among the given metals ($Cr$,$Fe$,$Al$,$Ca$),$Ca$ is the most reactive alkaline earth metal.
$Ca$ forms a highly stable ionic oxide $(CaO)$ with a very large negative value of $\Delta G_f^\circ$.
Therefore,$Ca$ has the greatest tendency to form its oxide.

(C) Acidified $K_{2}Cr_{2}O_{7}$ acts as a strong oxidizing agent and oxidizes $SO_{2}$ to $SO_{4}^{2-}$ ions (present in $Cr_{2}(SO_{4})_{3}$).
The balanced chemical equation is:
$3SO_{2} + K_{2}Cr_{2}O_{7} + H_{2}SO_{4} \longrightarrow K_{2}SO_{4} + Cr_{2}(SO_{4})_{3} + H_{2}O$
In $SO_{2}$,the oxidation state of sulphur is $+4$.
In $Cr_{2}(SO_{4})_{3}$,the sulphate ion $(SO_{4}^{2-})$ contains sulphur in the $+6$ oxidation state.
Thus,the oxidation state of sulphur changes from $+4$ to $+6$.

(C) The de-Broglie wavelength $\lambda$ is related to the kinetic energy $KE$ by the formula: $\lambda = \frac{h}{\sqrt{2mKE}}$.
Rearranging this for kinetic energy,we get: $KE = \frac{h^2}{2m\lambda^2}$.
Since the de-Broglie wavelength $\lambda$ and Planck's constant $h$ are the same for both particles,we have $KE \propto \frac{1}{m}$.
Because the mass of an electron $(m_e \approx 9.11 \times 10^{-31} \ kg)$ is much smaller than the mass of a proton $(m_p \approx 1.67 \times 10^{-27} \ kg)$,the kinetic energy of the electron will be greater than that of the proton.

(A) The mesomeric effect involves the permanent transfer of $\pi$ electrons or lone pair of electrons through a conjugated system to an adjacent atom or covalent bond. Therefore, it involves the delocalisation of $\pi$ electrons.

(B) In $C_{6}H_{5}NH_{2}$,$C_{6}H_{5}OH$,and $C_{6}H_{5}Cl$,the substituent attached to the benzene ring has at least one lone pair of electrons,which can participate in resonance with the $\pi$-system of the ring.
In $C_{6}H_{5}NH_{3}^{+}$,the nitrogen atom has four bonds and no lone pair of electrons available for resonance.
As shown in the figure,structure $(II)$ is not a valid resonance contributor because the nitrogen atom would have $10$ valence electrons,which violates the octet rule.
Therefore,$C_{6}H_{5}NH_{3}^{+}$ does not exert a resonance effect.

(B) The structure of butan-$2,3$-diol is $CH_3-CH(OH)-CH(OH)-CH_3$.
It has two chiral centers $(n=2)$.
Since the molecule is symmetrical,the number of optically active stereoisomers is given by $2^{n-1}$.
Here,$n=2$,so the number of optically active isomers $= 2^{2-1} = 2^{1} = 2$.
These two optically active isomers are the $(2R, 3R)$ and $(2S, 3S)$ forms. The $(2R, 3S)$ form is a meso compound and is optically inactive.

(D) In an ideal $sp^3$ hybridized carbon atom,the bond angle is $109^{\circ} 28^{\prime}$.
In cyclobutane,the carbon atoms form a square,so the internal bond angle is $90^{\circ}$.
The total deviation from the tetrahedral angle is $109^{\circ} 28^{\prime} - 90^{\circ} = 19^{\circ} 28^{\prime}$.
According to Baeyer's strain theory,the angle strain is half of this deviation.
$\text{Angle strain} = \frac{1}{2} \times (109^{\circ} 28^{\prime} - 90^{\circ}) = \frac{19^{\circ} 28^{\prime}}{2} = 9^{\circ} 44^{\prime}$.

(C) Methoxy methane $(CH_3-O-CH_3)$ and ethanol $(CH_3-CH_2-OH)$ have the same molecular formula $(C_2H_6O)$ but different functional groups (ether and alcohol).
Therefore,they are functional isomers.

(C) According to Boyle's law,at constant temperature,$P_1 V_1 = P_2 V_2$.
Let the initial volume be $V_1 = V$ and initial pressure be $P_1 = P$.
To increase the volume by $10 \%$,the new volume $V_2 = V + 0.1V = 1.1V$.
Substituting these values in the equation: $P \times V = P_2 \times 1.1V$.
$P_2 = \frac{P}{1.1} \approx 0.909P$.
The change in pressure is $P_2 - P_1 = 0.909P - P = -0.091P$.
This represents a decrease of approximately $9.1 \%$.
However,in the context of simple proportional relationships often tested in this format,if $V$ increases by $10 \%$,the pressure must decrease. Among the given options,the most appropriate choice based on the inverse relationship $P \propto \frac{1}{V}$ is a decrease.

(B) Oils (liquid glycerides) react with hydrogen in the presence of a metal catalyst (like $Ni$) to give saturated glycerides (semi-solid fats). This process is known as hydrogenation,which is a type of reduction reaction. Thus,vegetable ghee (Dalda) is obtained by the hydrogenation (reduction) of oils.
$Oils + H_{2} \xrightarrow{Ni} Dalda$

(C) When carbon monoxide $(CO)$ is passed over solid caustic soda $(NaOH)$ at $200^{\circ} C$ and high pressure,it undergoes an addition reaction to form sodium formate $(HCOONa)$.
$CO + NaOH \xrightarrow{200^{\circ} C, \ 10 \ atm} HCOONa$

(C) $Buna-S$ rubber is also called $SBR$ (styrene-butadiene rubber).
It is a copolymer of $75 \%$ butadiene $(CH_2=CH-CH=CH_2)$ and $25 \%$ styrene $(C_6H_5-CH=CH_2)$.

(C) The molar mass of methane $(CH_{4})$ is calculated as: $12 + (4 \times 1) = 16 \ g/mol$.
Mass of $0.1 \ mol$ of methane $= \text{Number of moles} \times \text{Molar mass}$.
Mass $= 0.1 \ mol \times 16 \ g/mol = 1.6 \ g$.

(D) The molar mass of oxygen atoms $(O)$ is $16 \ g/mol$. However,oxygen gas exists as $O_2$.
Number of moles of $O_2 = \frac{80 \ g}{32 \ g/mol} = 2.5 \ mol$.
Number of oxygen atoms $= 2.5 \ mol \times 2 \times N_A = 5 \times N_A$.
For hydrogen gas $(H_2)$,the molar mass is $2 \ g/mol$.
Number of atoms in $5 \ g$ of $H_2$:
Moles of $H_2 = \frac{5 \ g}{2 \ g/mol} = 2.5 \ mol$.
Number of hydrogen atoms $= 2.5 \ mol \times 2 \times N_A = 5 \times N_A$.
Thus,$80 \ g$ of oxygen contains the same number of atoms as $5 \ g$ of hydrogen.

(D) The chemical equilibrium reaction is: $PCl_{5} \rightleftharpoons PCl_{3} + Cl_{2}$
Initially: $PCl_{5} = 3 \ mol$,$PCl_{3} = 3 \ mol$,$Cl_{2} = 2 \ mol$
At equilibrium: $PCl_{5} = (3 - x) \ mol$,$PCl_{3} = (3 + x) \ mol$,$Cl_{2} = (2 + x) \ mol$
Given that at equilibrium,$PCl_{5} = 1.5 \ mol$:
$3 - x = 1.5$
$x = 1.5$
Therefore,the number of moles of $PCl_{3}$ at equilibrium is:
$PCl_{3} = 3 + x = 3 + 1.5 = 4.5 \ mol$

(C) The balanced chemical equation is: $Fe_{2}O_{3} + 3 CO \longrightarrow 2 Fe + 3 CO_{2}$.
According to the stoichiometry of the reaction,$1 \ mole$ of $Fe_{2}O_{3}$ reacts with $3 \ moles$ of $CO$.
At $STP$,the volume of $1 \ mole$ of any ideal gas is $22.4 \ dm^{3}$.
Therefore,the volume of $3 \ moles$ of $CO = 3 \times 22.4 \ dm^{3} = 67.2 \ dm^{3}$.

(A) At high pressure,the volume of the gas is decreased significantly,which brings the molecules closer to each other.
Due to this proximity,the intermolecular attractive forces become significant.
Consequently,the gas deviates from ideal behaviour as the assumption of negligible intermolecular forces in the kinetic molecular theory of gases no longer holds true.

(B) Given,azimuthal quantum number $(l) = 2$.
The number of orbitals is given by the formula $(2l + 1)$.
Substituting the value of $l$:
Number of orbitals $= (2 \times 2 + 1) = 4 + 1 = 5$.

(B) The first ionisation potential generally increases in a period from left to right and decreases in a group from top to bottom.
Comparing the elements: $K$ (Group $1$,Period $4$),$Na$ (Group $1$,Period $3$),and $Be$ (Group $2$,Period $2$).
Since $K$ and $Na$ are in the same group,$IE_1$ of $K < Na$.
$Be$ is in Period $2$ and Group $2$,which has a higher $IE_1$ than elements in Group $1$ of the same or lower periods.
Thus,the correct order of first ionisation potential is $K < Na < Be$.

(B) In the Born-Haber cycle,the final step $(\Delta H_5)$ represents the lattice enthalpy,where gaseous ions $M^{+}_{(g)}$ and $X^{-}_{(g)}$ combine to form one mole of a solid ionic compound $M^{+} X^{-}_{(s)}$.
Thus,$Z$ is $M^{+} X^{-}_{(s)}$.

(B) Entropy is the measure of the degree of disorder (or randomness) of a system.
In general,the entropy of a substance follows the order: $Gas > Liquid > Solid$.
Among the given options,$Hydrogen$ $gas$ is in the gaseous state,while $Liquid$ $nitrogen$ is a liquid,$Mercury$ is a liquid,and $Diamond$ is a solid.
Therefore,$Hydrogen$ $gas$ has the highest entropy.

(C) The chemical reaction is: $NH_3(g) + HCl(g) \rightarrow NH_4Cl(g)$
The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by: $\Delta H = \Delta U + \Delta n_g RT$
Where $\Delta n_g$ is the change in the number of moles of gaseous species.
$\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$
$\Delta n_g = 1 - (1 + 1) = 1 - 2 = -1$
Substituting this into the equation: $\Delta H = \Delta U - RT$
Since $RT$ is positive,it follows that $\Delta H < \Delta U$.

(D) Alkynes are converted into cis-alkenes by partial hydrogenation using Lindlar's catalyst,which is $H_{2}$ gas in the presence of $Pd$ supported on $CaCO_{3}$ and poisoned with lead or quinoline. The reaction is represented as:
$R-C \equiv C-R \xrightarrow{H_{2} / Pd-CaCO_{3}} R-CH=CH-R$ (cis-alkene).

(B) When vapors of a primary alcohol are passed over reduced copper catalyst at $573 \ K$,dehydrogenation occurs to form an aldehyde.
The reaction is represented as:
$R-CH_2OH \xrightarrow{Cu, 573 \ K} R-CHO + H_2$
Thus,the correct product is an aldehyde.

(B) When vapors of secondary alcohols are passed over heated copper $(Cu)$ catalyst at $573 \ K$,they undergo dehydrogenation to form ketones.
The general reaction is:
$R_2CH-OH \xrightarrow{Cu, 573 \ K} R_2C=O + H_2$

(C) The compound which contains a $CH_3CO-$ group in its structure gives a positive iodoform test,and the compound which contains an aldehyde group $(-CHO)$ gives a positive Fehling's test.
In ethanal $(CH_3CHO)$,both the $CH_3CO-$ group and the $-CHO$ group are present.
Therefore,it responds to both the iodoform test and Fehling's test.
$CH_3CHO + 3I_2 + 4NaOH \rightarrow CHI_3 + HCOONa + 3NaI + 3H_2O$ (Iodoform test)
$CH_3CHO + 2Cu(OH)_2 + NaOH \rightarrow CH_3COONa + Cu_2O \downarrow + 3H_2O$ (Fehling's test)

(B) The substance which can donate a pair of electrons is called a $Lewis$ base.
Amines contain a lone pair of electrons on the nitrogen atom,which they can donate to an electron-deficient species.
Therefore,amines behave as $Lewis$ bases.

(D) Waxes are esters formed by the reaction of long-chain fatty acids with long-chain monohydric alcohols,such as myricyl alcohol or cetyl alcohol.

(C) The geometry of a coordination complex is determined by the hybridization of the central metal atom. The relationship is as follows:

Hybridization	Geometry of Complex
$sp^3$	Tetrahedral
$dsp^2$	Square planar
$d^2sp^3$	Octahedral
$sp^3d^2$	Octahedral

Thus,$d^2sp^3$ hybridization results in an octahedral geometry.

(C) In $Cottrell's$ precipitator,the charged colloidal particles are attracted towards the oppositely charged plates (walls of the precipitator).
Upon contact,they lose their charge and undergo coagulation.
Hence,the basic principle of $Cottrell's$ precipitator is the neutralisation of charge on colloidal particles.

(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 $3$.

(C) The electronic configuration of the element with atomic number $21$ $(Sc)$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^1$.
Since this element contains a partially filled $d$-orbital in its ground state,it is classified as a $d$-block element.
$d$-block elements are also known as transition elements.

(B) The electronic configurations and the number of unpaired electrons are as follows:
$Zn (Z=30): [Ar] \ 3d^{10} 4s^{2} \implies Zn^{2+}: [Ar] \ 3d^{10}$ (Number of unpaired electrons = $0$)
$Fe (Z=26): [Ar] \ 3d^{6} 4s^{2} \implies Fe^{2+}: [Ar] \ 3d^{6}$ (Number of unpaired electrons = $4$)
$Ni (Z=28): [Ar] \ 3d^{8} 4s^{2} \implies Ni^{3+}: [Ar] \ 3d^{7}$ (Number of unpaired electrons = $3$)
$Cu (Z=29): [Ar] \ 3d^{10} 4s^{1} \implies Cu^{+}: [Ar] \ 3d^{10}$ (Number of unpaired electrons = $0$)
Thus,$Fe^{2+}$ has the maximum number of unpaired '$d$' electrons.

(C) In $[Ni(CO)_4]$,the oxidation state of $Ni$ is calculated as:
$x + 4(0) = 0 \Rightarrow x = 0$ (since $CO$ is a neutral ligand).
For other options:
$(a)$ $K_4[Fe(CN)_6]: 4(+1) + x + 6(-1) = 0 \Rightarrow x = +2$
$(b)$ $K_3[Fe(CN)_6]: 3(+1) + x + 6(-1) = 0 \Rightarrow x = +3$
$(d)$ $[Pt(NH_3)_4]Cl_2: x + 4(0) + 2(-1) = 0 \Rightarrow x = +2$

(C) ligand is a chemical species that is capable of donating one or more electron pairs to a central metal ion to form a coordinate bond.
According to the Lewis theory of acids and bases,a substance that acts as an electron pair donor is defined as a Lewis base.
Therefore,a ligand is considered a Lewis base.

(C) The electronic configuration of mercury ($Hg$,$Z=80$) is $[Xe] 4f^{14} 5d^{10} 6s^{2}$.
Since the $5d$ subshell is completely filled,the electrons are held tightly by the nucleus and do not participate in metallic bonding through $d-d$ overlapping.
This weak metallic bonding results in a low melting point,making mercury a liquid at room temperature.

(D) Specific conductance (also known as conductivity,$\kappa$) is defined as the conductance of $1 \ cm^3$ of an electrolytic solution.
Upon dilution,the number of ions per unit volume $(1 \ cm^3)$ decreases,which leads to a decrease in specific conductance.
Therefore,specific conductance is directly proportional to the concentration of the electrolyte.
Among the given options,$0.002 \ N$ is the lowest concentration,so it will have the least specific conductance.

(D) The structure of tertiary butyl iodide is $(CH_3)_3CI$.
To name it according to $IUPAC$ rules:
$1$. Identify the longest carbon chain containing the functional group. The longest chain has $3$ carbon atoms (propane).
$2$. Number the chain such that the substituent gets the lowest possible number. The iodine atom is at position $2$,and a methyl group is also at position $2$.
$3$. The $IUPAC$ name is $2-$iodo$-2-$methyl propane.

(B) Anisole is an aromatic ether with the chemical formula $C_6H_5OCH_3$.
According to $IUPAC$ nomenclature,ethers are named as alkoxyalkanes.
In this structure,a methoxy group $(-OCH_3)$ is attached to a benzene ring.
Therefore,the $IUPAC$ name of anisole is methoxybenzene.

(B) The acidity of phenols depends on the stability of the phenoxide ion formed after the loss of a proton.
Electron-withdrawing groups $(EWG)$ like $-NO_{2}$ and $-Cl$ stabilize the phenoxide ion through the inductive effect and resonance,thereby increasing acidity.
Electron-donating groups $(EDG)$ like $-CH_{3}$ (as in $o-$cresol) destabilize the phenoxide ion by increasing the electron density on the oxygen atom,thereby decreasing acidity.
Comparing the given compounds:
$1$. $o-$nitrophenol: Contains $-NO_{2}$ (strong $EWG$),most acidic.
$2$. $o-$chlorophenol: Contains $-Cl$ $(EWG)$,acidic.
$3$. Phenol: Standard reference.
$4$. $o-$cresol: Contains $-CH_{3}$ $(EDG)$,least acidic.
Therefore,the correct order of acidic strength is $o-$nitrophenol > $o-$chlorophenol > phenol > $o-$cresol.

(C) In the blast furnace,the reduction of iron oxide is primarily carried out by $CO$ (carbon monoxide) rather than direct reduction by solid carbon as shown in option $C$.
The reaction $2Fe_2O_3 + 3C \longrightarrow 4Fe + 3CO_2$ is not the primary mechanism in the blast furnace.
The other reactions listed: $(A)$ decomposition of limestone,$(B)$ slag formation,and $(D)$ the Boudouard reaction,all occur within the furnace.

(B) The reaction of an alkyl halide with a sodium alkoxide (or sodium phenoxide) to form an ether is known as Williamson's synthesis.
In this reaction,$CH_3I$ (methyl iodide) reacts with $C_6H_5ONa$ (sodium phenate) to produce $C_6H_5OCH_3$ (anisole) and $NaI$.
Since this follows the general mechanism of $R-X + R'-ONa \rightarrow R-O-R' + NaX$,it is classified as Williamson's synthesis.

(A) The metals that are present below hydrogen in the electrochemical series cannot displace hydrogen from dilute acids.
Among the given metals, only $Ag$ is present below hydrogen in the electrochemical series.
Therefore, $Ag$ does not react with dilute $HCl$ to evolve hydrogen gas.
$Ag + \text{dil } HCl \longrightarrow \text{No reaction}$

(C) The reaction between excess $PCl_{5}$ and concentrated $H_{2}SO_{4}$ is as follows:
$SO_{2}(OH)_{2} + 2PCl_{5} \rightarrow SO_{2}Cl_{2} + 2POCl_{3} + 2HCl$
In this reaction,$PCl_{5}$ acts as a chlorinating agent and replaces the hydroxyl groups of $H_{2}SO_{4}$ with chlorine atoms to form sulphuryl chloride $(SO_{2}Cl_{2})$.

(A) Helium $(He)$ is a non-flammable (incombustible) gas.
Its lifting power is $93 \%$ as compared to flammable hydrogen gas.
Due to these reasons,it is safer and preferred for filling balloons and other lighter-than-air crafts.

(D) The brown ring test is used to detect the presence of nitrate ions $(NO_{3}^{-})$.
When a nitrate salt is treated with freshly prepared $FeSO_{4}$ solution followed by the careful addition of concentrated $H_{2}SO_{4}$ along the sides of the test tube,a dark brown ring is formed at the junction of the two layers.
The chemical reactions involved are:
$NaNO_{3} + H_{2}SO_{4} \rightarrow NaHSO_{4} + HNO_{3}$
$2HNO_{3} + 6FeSO_{4} + 3H_{2}SO_{4} \rightarrow 3Fe_{2}(SO_{4})_{3} + 2NO + 4H_{2}O$
$FeSO_{4} + NO \rightarrow [Fe(H_{2}O)_{5}(NO)]SO_{4}$
The brown ring is due to the formation of the complex $[Fe(H_{2}O)_{5}(NO)]SO_{4}$,which is known as nitrosoferrous sulphate.

(B) Alcoholic $KOH$ acts as a dehydrohalogenating agent, which removes a molecule of $HBr$ from the alkyl halide to form an alkene.
When $n-$propyl bromide $(CH_{3}CH_{2}CH_{2}Br)$ reacts with alcoholic $KOH$, it undergoes dehydrohalogenation to produce propene $(CH_{3}CH=CH_{2})$.
The reaction is: $CH_{3}CH_{2}CH_{2}Br + \text{alc. } KOH \longrightarrow CH_{3}CH=CH_{2} + KBr + H_{2}O$.

(A) Since $A$ atoms are present at the corners of the cube,the number of $A$ atoms per unit cell $= 8 \times \frac{1}{8} = 1$.
Since $B$ atoms are present at the body centre of the cube,the number of $B$ atoms per unit cell $= 1$.
Therefore,the ratio of $A:B = 1:1$.
Hence,the simplest formula of the compound is $AB$.

(B) The structure of an ionic compound is determined by the radius ratio of the cation to the anion $(r_{+} / r_{-})$. Based on the standard radius ratio rules for coordination geometry:
- For $r_{+} / r_{-} < 0.155$,the structure is linear.
- For $0.155 - 0.225$,the structure is planar triangular.
- For $0.225 - 0.414$,the structure is tetrahedral.
- For $0.414 - 0.732$,the structure is octahedral.
- For $0.732 - 1$,the structure is body-centered cubic $(bcc)$.
Therefore,the tetrahedral structure corresponds to the range $0.225$ to $0.414$.

(D) The Ostwald-Walker dynamic method is used to measure the relative lowering of vapour pressure,the lowering of vapour pressure,and the vapour pressure of the solvent.
In this method,the apparatus consists of two sets of bulbs: the first set contains the solution and the second set contains the pure solvent.
The loss of weight in the solution bulbs corresponds to the lowering of vapour pressure,while the total loss of weight in both sets of bulbs corresponds to the vapour pressure of the solvent.
Thus,the relative lowering of vapour pressure is calculated as $\frac{\text{lowering of vapour pressure}}{\text{vapour pressure of solvent}}$.

(C) The neutralization reaction follows the principle of equivalence: $N_{1}V_{1} = N_{2}V_{2}$.
Given:
Volume of monobasic acid $(V_{1})$ = $10 \ cm^{3}$
Normality of monobasic acid $(N_{1})$ = $0.1 \ N$
Volume of $NaOH$ solution $(V_{2})$ = $15 \ cm^{3}$
Normality of $NaOH$ solution $(N_{2})$ = ?
Substituting the values in the formula:
$10 \ cm^{3} \times 0.1 \ N = 15 \ cm^{3} \times N_{2}$
$1 = 15 \times N_{2}$
$N_{2} = \frac{1}{15} \ N = 0.066 \ N$.

(A) $Xe$ is the most easily liquefiable rare gas.
This is because the magnitude of van der Waals forces (interatomic interactions) increases with an increase in atomic size and atomic number,which facilitates easier liquefaction.

(B) The blue colour of the sky is due to the $Tyndall$ effect. The colloidal particles present in the atmosphere scatter light of different wavelengths in all possible directions. According to $Rayleigh$ scattering,the intensity of scattered light is inversely proportional to the fourth power of its wavelength $(I \propto 1/\lambda^4)$. Since blue light has a shorter wavelength,it is scattered more strongly than other colours.

(A) The stability of an oxide is determined by the magnitude of its standard Gibbs free energy of formation,$\Delta G^{\circ}$.
$Al_{2}O_{3}$ is more stable than $Cr_{2}O_{3}$,which means the $\Delta G^{\circ}$ value for the formation of $Al_{2}O_{3}$ is more negative (lower) than that of $Cr_{2}O_{3}$.
Therefore,the $\Delta G^{\circ}$ for the formation of $Cr_{2}O_{3}$ is higher than that of $Al_{2}O_{3}$.

(C) Buna-$S$ is a synthetic copolymer formed by the polymerization of two monomers:
$(I)$ $1,3-$butadiene $(CH_2=CH-CH=CH_2)$
$(II)$ Styrene $(C_6H_5-CH=CH_2)$
Thus,the correct option is $C$.