KCET 2007 Chemistry Question Paper with Answer and Solution

(D) The optical rotation $\theta$ is directly proportional to the length $l$ of the solution,i.e.,$\theta \propto l$.
For the first solution,the rotation per unit length is $\frac{38^\circ}{20 \, cm} = 1.9^\circ/cm$ (right-handed).
For the second solution,the rotation per unit length is $\frac{24^\circ}{30 \, cm} = 0.8^\circ/cm$ (left-handed,denoted as $-0.8^\circ/cm$).
In a mixture with a volume ratio of $1:2$ in a $30 \, cm$ tube,the lengths occupied are $10 \, cm$ of the first solution and $20 \, cm$ of the second solution.
Rotation by the first part: $\theta_1 = 1.9^\circ/cm \times 10 \, cm = +19^\circ$.
Rotation by the second part: $\theta_2 = -0.8^\circ/cm \times 20 \, cm = -16^\circ$.
Total rotation: $\theta_{total} = \theta_1 + \theta_2 = 19^\circ - 16^\circ = +3^\circ$.
$A$ positive value indicates a right-handed rotation.

(C) The given equation of the parabola is $y = 2x^2 + x$.
Divide by $2$: $x^2 + \frac{x}{2} = \frac{y}{2}$.
Complete the square: $(x + \frac{1}{4})^2 = \frac{y}{2} + \frac{1}{16}$.
$(x + \frac{1}{4})^2 = \frac{1}{2}(y + \frac{1}{8})$.
Comparing this with the standard form $X^2 = 4AY$,where $X = x + \frac{1}{4}$,$Y = y + \frac{1}{8}$,and $4A = \frac{1}{2} \Rightarrow A = \frac{1}{8}$.
The focus of the standard parabola $X^2 = 4AY$ is $(0, A) = (0, \frac{1}{8})$.
Thus,$X = 0$ and $Y = \frac{1}{8}$.
Substituting back: $x + \frac{1}{4} = 0 \Rightarrow x = -\frac{1}{4}$ and $y + \frac{1}{8} = \frac{1}{8} \Rightarrow y = 0$.
Therefore,the focus is $(-\frac{1}{4}, 0)$.

(B) The area of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is given by $A_{e} = \pi ab$.
The auxiliary circle of this ellipse is $x^{2} + y^{2} = a^{2}$,and its area is $A_{c} = \pi a^{2}$.
According to the problem,$A_{c} = 2 A_{e}$.
Substituting the values,we get $\pi a^{2} = 2 \pi ab$,which simplifies to $a = 2b$.
The eccentricity $e$ of an ellipse is given by $e = \sqrt{1 - \frac{b^{2}}{a^{2}}}$.
Substituting $a = 2b$,we get $e = \sqrt{1 - \frac{b^{2}}{(2b)^{2}}} = \sqrt{1 - \frac{b^{2}}{4b^{2}}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

(B) The area of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $A_{e} = \pi ab$.
The auxiliary circle of the ellipse is $x^2 + y^2 = a^2$,and its area is $A_{c} = \pi a^2$.
According to the problem,$A_{c} = 2 A_{e}$,so $\pi a^2 = 2 \pi ab$.
Dividing by $\pi a$ (since $a \neq 0$),we get $a = 2b$.
The eccentricity $e$ of the ellipse is given by $e = \sqrt{1 - \frac{b^2}{a^2}}$.
Substituting $a = 2b$,we get $e = \sqrt{1 - \frac{b^2}{(2b)^2}} = \sqrt{1 - \frac{b^2}{4b^2}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

(C) The velocity of sound in an ideal gas is given by the formula $\nu = \sqrt{\frac{\gamma R T}{M}}$.
Since the temperature $T$ is the same for both gases,the velocity is proportional to $\sqrt{\frac{\gamma}{M}}$,where $\gamma$ is the adiabatic index and $M$ is the molar mass.
For hydrogen $(H_2)$,$M_{H_2} = 2 \times 10^{-3} \ kg/mol$ and $\gamma_{H_2} = 7/5$.
For helium $(He)$,$M_{He} = 4 \times 10^{-3} \ kg/mol$ and $\gamma_{He} = 5/3$.
The ratio of the velocity of sound in hydrogen to that in helium is:
$\frac{\nu_{H_2}}{\nu_{He}} = \sqrt{\frac{\gamma_{H_2}}{\gamma_{He}} \times \frac{M_{He}}{M_{H_2}}}$
Substituting the values:
$\frac{\nu_{H_2}}{\nu_{He}} = \sqrt{\frac{7/5}{5/3} \times \frac{4}{2}} = \sqrt{\frac{7}{5} \times \frac{3}{5} \times 2} = \sqrt{\frac{42}{25}} = \frac{\sqrt{42}}{5}$.

(B) According to Brewster's Law,when the reflected ray is completely polarized,the angle of incidence is equal to the Brewster's angle $(i_{p})$.
Given $i_{p} = 60^{\circ}$.
The refractive index $\mu$ is given by $\mu = \tan(i_{p})$.
$\mu = \tan(60^{\circ}) = \sqrt{3}$.
The velocity of light in a medium is given by $v = \frac{c}{\mu}$,where $c$ is the speed of light in vacuum $(3 \times 10^{8} \, m/s)$.
$v = \frac{3 \times 10^{8}}{\sqrt{3}} = \sqrt{3} \times 10^{8} \, m/s$.

(D) Glycine $(NH_2-CH_2-COOH)$ is the only amino acid that is not optically active because it does not have a chiral carbon atom.
For a carbon atom to be chiral,it must be bonded to $4$ different groups.
Glycine has two identical hydrogen atoms attached to the $\alpha$-carbon,which means it lacks a chiral center and is therefore achiral and optically inactive.

(C) The condition for the $n^{th}$ secondary maximum in a single slit diffraction pattern is given by $x = \frac{(2n+1) \lambda D}{2a}$,where $n$ is the order of the secondary maximum.
For the second secondary maximum $(n=2)$ of red light $(\lambda_1 = 6500 \ \mathring{A})$,the position is:
$x = \frac{(2 \times 2 + 1) \lambda_1 D}{2a} = \frac{5 \lambda_1 D}{2a}$.
For the third secondary maximum $(n=3)$ of light with wavelength $\lambda_2$,the position is:
$x = \frac{(2 \times 3 + 1) \lambda_2 D}{2a} = \frac{7 \lambda_2 D}{2a}$.
Since the positions coincide,we equate them:
$\frac{5 \lambda_1 D}{2a} = \frac{7 \lambda_2 D}{2a}$.
This simplifies to:
$5 \lambda_1 = 7 \lambda_2$.
Substituting $\lambda_1 = 6500 \ \mathring{A}$:
$\lambda_2 = \frac{5 \times 6500}{7} = \frac{32500}{7} \approx 4642.8 \ \mathring{A}$.

(B) The phase angle $\theta$ in an $LCR$ series circuit is given by $\tan \theta = \frac{X_L - X_C}{R}$.
First,calculate the inductive reactance: $X_L = 2 \pi f L = 2 \pi \times 50 \times \left(\frac{200}{\pi} \times 10^{-3}\right) = 20 \ \Omega$.
Next,calculate the capacitive reactance: $X_C = \frac{1}{2 \pi f C} = \frac{1}{2 \pi \times 50 \times \left(\frac{10^{-3}}{\pi}\right)} = \frac{1}{100 \times 10^{-3}} = 10 \ \Omega$.
Given resistance $R = 10 \ \Omega$.
Substituting these values into the formula: $\tan \theta = \frac{20 - 10}{10} = \frac{10}{10} = 1$.
Therefore,$\theta = \tan^{-1}(1) = \frac{\pi}{4}$.

(D) During the formation of a chemical bond,the system moves to a state of lower potential energy to achieve stability. Therefore,the energy of the system decreases as the atoms approach each other and form a bond.

(B) The chemical formula of acetylene is $C_2H_2$,and its structural formula is $H-C \equiv C-H$.
In the triple bond between the two carbon atoms,there is one $\sigma$ bond (formed by the head-on overlap of $sp$ hybrid orbitals) and two $\pi$ bonds (formed by the lateral overlap of unhybridized $2p$ orbitals).

(B) The total number of electrons in $O_{2}^{2-}$ is $8 + 8 + 2 = 18$.
The molecular orbital configuration is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$.
The antibonding orbitals are $\sigma^* 1s$,$\sigma^* 2s$,$\pi^* 2p_x$,and $\pi^* 2p_y$.
Each of these contains $2$ electrons,forming $4$ antibonding electron pairs.

(C) The gram molecular volume is defined as the volume occupied by $1 \ \text{mole}$ of any gas at $STP$ $(Standard \ Temperature \ and \ Pressure)$.
According to Avogadro's hypothesis,$1 \ \text{mole}$ of any ideal gas occupies $22.4 \ L$ or $22400 \ cm^{3}$ at $STP$.
Therefore,the gram molecular volume of oxygen at $STP$ is $22400 \ cm^{3}$.

(B) The current $I$ in a tangent galvanometer is given by $I = \frac{2 r H}{\mu_{0} N} \tan \theta$, where $r$ is the radius, $H$ is the horizontal component of the Earth's magnetic field, $N$ is the number of turns, and $\theta$ is the deflection.
Since the galvanometers are connected in parallel to a $4 \, V$ source, the voltage across each is $4 \, V$. The current in each galvanometer is $I_A = I_B = \frac{V}{R} = \frac{4 \, V}{8 \, \Omega} = 0.5 \, A$.
From the formula, we have $\frac{N I}{r \tan \theta} = \frac{2 H}{\mu_{0}} = \text{constant}$.
Therefore, $\frac{N_A I_A}{r_A \tan \theta_A} = \frac{N_B I_B}{r_B \tan \theta_B}$.
Since $I_A = I_B$, we have $\frac{N_A}{r_A \tan \theta_A} = \frac{N_B}{r_B \tan \theta_B}$.
Substituting the values: $\frac{2}{8 \cdot \tan 30°} = \frac{N_B}{16 \cdot \tan 60°}$.
$\frac{2}{8 \cdot (1/\sqrt{3})} = \frac{N_B}{16 \cdot \sqrt{3}}$.
$\frac{2 \sqrt{3}}{8} = \frac{N_B}{16 \sqrt{3}}$.
$N_B = \frac{2 \sqrt{3} \cdot 16 \sqrt{3}}{8} = \frac{32 \cdot 3}{8} = 4 \cdot 3 = 12$ turns.

(B) The area of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is given by $A_{e} = \pi ab$.
The auxiliary circle of the ellipse is $x^{2}+y^{2}=a^{2}$,and its area is $A_{c} = \pi a^{2}$.
According to the problem,the area of the auxiliary circle is twice the area of the ellipse:
$\pi a^{2} = 2(\pi ab)$
$a^{2} = 2ab$
Since $a \neq 0$,we have $a = 2b$.
The eccentricity $e$ of the ellipse is given by $e = \sqrt{1-\frac{b^{2}}{a^{2}}}$.
Substituting $a = 2b$ into the formula:
$e = \sqrt{1-\frac{b^{2}}{(2b)^{2}}} = \sqrt{1-\frac{b^{2}}{4b^{2}}} = \sqrt{1-\frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

(B) Cycloalkanes are saturated cyclic hydrocarbons.
They contain one ring and no double or triple bonds.
The general formula for an acyclic alkane is $C_n H_{2n+2}$.
Forming a ring requires the removal of two hydrogen atoms from the ends of the chain,resulting in the general formula $C_n H_{2n}$ for cycloalkanes.

(A) The inductive effect ($I$-effect) is the permanent displacement of sigma electrons along a carbon chain due to the presence of an electron-withdrawing or electron-donating group.
Groups that donate electron density towards the carbon chain exhibit the $+I$ effect (e.g.,alkyl groups like $-CH_3$).
Groups that withdraw electron density from the carbon chain exhibit the $-I$ effect (e.g.,halogens like $-Br$,$-Cl$,and electron-withdrawing groups like $-NO_2$).
Therefore,$-CH_3$ shows the $+I$ effect.

(C) The stability of a molecule is directly proportional to the bond order,represented as $\text{Bond order} \propto \text{Stability}$.
For a molecule to be stable,it must have a positive bond order.
If the bond order is $0$ or negative,the molecule is unstable and will not exist.

(C) The structure of $2,3,4$-trichloropentane is $CH_3-CHCl-CHCl-CHCl-CH_3$.
In this molecule,the carbon atoms at positions $2$ and $4$ are bonded to four different groups: a hydrogen atom $(-H)$,a chlorine atom $(-Cl)$,a methyl group $(-CH_3)$,and a $-CHCl-CHCl_2$ group (for $C-2$) or a $-CHCl-CH_3$ group (for $C-4$).
The carbon atom at position $3$ is not chiral because it is bonded to two identical $-CHCl-CH_3$ groups.
Therefore,there are $2$ chiral carbon atoms present in $2,3,4$-trichloropentane.

(C) $C_2H_5OH$ shows functional isomerism with dimethyl ether.
$\text{CH}_3\text{CH}_2\text{OH} \text{ (Ethanol)} \rightleftharpoons \text{CH}_3\text{OCH}_3 \text{ (Dimethyl ether)}$

(A) The bond angle in cyclopropane is calculated as $\theta = \frac{180(n-2)}{n} = \frac{180(3-2)}{3} = 60^{\circ}$.
Angle strain is defined as $\alpha = \frac{1}{2} (109^{\circ} 28^{\prime} - \theta)$.
Substituting the value of $\theta = 60^{\circ}$:
$\alpha = \frac{1}{2} (109^{\circ} 28^{\prime} - 60^{\circ}) = \frac{1}{2} (49^{\circ} 28^{\prime}) = 24^{\circ} 44^{\prime}$.

(B) When both nitrogen $(N)$ and sulfur $(S)$ are present in an organic compound,sodium fusion produces sodium thiocyanate $(NaCNS)$.
Upon adding acidified ferric chloride $(FeCl_3)$ solution to the extract,the ferric ions react with thiocyanate ions to form ferric thiocyanide,which exhibits a blood-red colouration.
The chemical reaction is: $3 \ NaCNS + FeCl_3 \longrightarrow Fe(CNS)_3 + 3 \ NaCl$ (blood-red colour).

(D) The presence of halogen in an organic compound can be detected by $Beilstein's$ test. In this test,a copper wire is heated in a flame until it shows no green color,then it is dipped in the organic compound and heated again. The appearance of a green flame indicates the presence of a halogen.

(B) Benzene reacts with chlorine in the presence of sunlight via a free radical addition reaction to give benzene hexachloride $(C_{6}H_{6}Cl_{6})$,also known as gammexane or lindane.
$C_{6}H_{6} + 3Cl_{2} \xrightarrow{\text{Sunlight}} C_{6}H_{6}Cl_{6}$

(B) For a strong acid like $HCl$,the concentration of hydrogen ions is $[H^+] = 10^{-2} \ M$.
Using the ionic product of water,$K_w = [H^+][OH^-] = 10^{-14}$ at $298 \ K$.
Substituting the values: $[10^{-2}][OH^-] = 10^{-14}$.
Therefore,$[OH^-] = \frac{10^{-14}}{10^{-2}} = 10^{-12} \ mol \ dm^{-3}$.

(A) Geometrical isomerism is shown by compounds containing a $C=C$ double bond,provided that each carbon atom of the double bond is attached to two different groups.
The general condition for geometrical isomerism is $abC=Cab$,where $a \neq b$.

(A) $KCN$ is a salt of a weak acid $(HCN)$ and a strong base $(KOH)$.
When dissolved in water,the $CN^-$ ion undergoes hydrolysis to form $OH^-$ ions,resulting in a basic solution with $pH > 7$ at $25^{\circ} C$.
$KNO_3$ is a salt of a strong acid and a strong base $(pH = 7)$.
$NH_4Cl$ is a salt of a weak base and a strong acid $(pH < 7)$.
$NH_4CN$ is a salt of a weak acid and a weak base $(pH \approx 7)$.

(C) Graphite is a soft solid lubricant that is extremely difficult to melt.
This is because graphite has a layered structure where carbon atoms are arranged in large hexagonal plates (rings) held together by strong covalent bonds within the layers,while the layers themselves are held together by weak van der Waals forces (interplate bonds).
These weak interplate bonds allow the layers to slide over each other,making it a lubricant,while the strong covalent bonds within the layers require very high temperatures to break,making it difficult to melt.

(B) To find the value of $x$,we must balance the charge on both sides of the equation.
In the reactant side,the total charge is: $(-1) + 6(+1) + x(-1) = 5 - x$.
In the product side,the total charge is: $(+3) + 3(0) = +3$.
Equating the charges: $5 - x = 3$.
Solving for $x$: $x = 5 - 3 = 2$.
Thus,the balanced equation is $BiO_{3}^{-} + 6H^{+} + 2e^{-} \longrightarrow Bi^{3+} + 3H_{2}O$.

(A) The given reaction is: $SnCl_{2} + Cl_{2} \longrightarrow SnCl_{4}$.
In this reaction,the oxidation state of $Sn$ changes from $+2$ in $SnCl_{2}$ to $+4$ in $SnCl_{4}$.
The $n$-factor is the change in oxidation state per molecule,which is $|4 - 2| = 2$.
The molar mass of $SnCl_{2} = 118.7 + 2 \times 35.5 = 118.7 + 71 = 189.7 \approx 190 \ g/mol$.
The equivalent weight is calculated as: $\text{Equivalent weight} = \frac{\text{Molar mass}}{n\text{-factor}} = \frac{190}{2} = 95$.

(C) Let the total weight of the oxide be $100 \ g$.
Since the oxide contains $20 \%$ oxygen by weight,the weight of oxygen is $20 \ g$ and the weight of the element is $100 - 20 = 80 \ g$.
The equivalent weight of oxygen is $8 \ g$.
According to the law of equivalence,the number of gram equivalents of the element must be equal to the number of gram equivalents of oxygen.
$\frac{\text{Weight of element}}{\text{Equivalent weight of element}} = \frac{\text{Weight of oxygen}}{\text{Equivalent weight of oxygen}}$
$\frac{80}{E} = \frac{20}{8}$
$E = \frac{80 \times 8}{20} = 32$
Therefore,the equivalent weight of the element is $32$.

(A) When $KBr$ is dissolved in water,it dissociates into $K^{+}$ and $Br^{-}$ ions.
These ions interact with water molecules through ion-dipole interactions,a process known as hydration.

(A) Orthophosphoric acid $(H_{3}PO_{4})$ is a tribasic acid,meaning its basicity is $3$.
The relationship between normality and molarity is given by the formula: $\text{Normality} = \text{Molarity} \times \text{basicity}$.
Substituting the given values: $\text{Normality} = 3 \ M \times 3 = 9 \ N$.

(B) The chemical formula for hydrogen chloride is $HCl$.
One molecule of $HCl$ contains one atom of chlorine.
Therefore,$6.02 \times 10^{25}$ molecules of $HCl$ contain $6.02 \times 10^{25}$ atoms of chlorine.
The number of moles (gram molecules) of chlorine atoms is calculated as:
$\text{Number of moles} = \frac{\text{Total number of atoms}}{\text{Avogadro's number}} = \frac{6.02 \times 10^{25}}{6.02 \times 10^{23}} = 100$.

(D) The number of oxygen atoms is calculated as: $\text{Number of atoms} = \frac{\text{mass}}{\text{molar mass}} \times N_A \times \text{number of oxygen atoms in one molecule}$.
For $2 \ g$ of $CO$ $(M = 28 \ g/mol)$: $\frac{2}{28} \times 1 = 0.071 \ N_A$.
For $2 \ g$ of $CO_2$ $(M = 44 \ g/mol)$: $\frac{2}{44} \times 2 = 0.091 \ N_A$.
For $2 \ g$ of $SO_2$ $(M = 64 \ g/mol)$: $\frac{2}{64} \times 2 = 0.0625 \ N_A$.
For $2 \ g$ of $H_2O$ $(M = 18 \ g/mol)$: $\frac{2}{18} \times 1 = 0.111 \ N_A$.
Comparing the values,$2 \ g$ of $H_2O$ contains the maximum number of oxygen atoms.

(D) Dalton's law of partial pressure is applicable only to a mixture of non-reacting gases.
In the given options,$NH_{3} + HCl$ react to form $NH_{4}Cl$,$NO + O_{2}$ react to form $NO_{2}$,and $H_{2} + Cl_{2}$ react to form $HCl$.
However,$CO$ and $H_{2}$ do not react with each other under normal conditions.
Therefore,Dalton's law of partial pressure is applicable to the system $CO + H_{2}$.

(B) The average kinetic energy $(KE)$ of one mole of an ideal gas is given by the formula: $KE = \frac{3}{2} RT$.
Since $R$ is a constant,$KE \propto T$.
For oxygen $(O_{2})$: $T_{O_{2}} = 273 \ K$.
For sulphur dioxide $(SO_{2})$: $T_{SO_{2}} = 546 \ K$.
Comparing the kinetic energies: $\frac{KE_{O_{2}}}{KE_{SO_{2}}} = \frac{T_{O_{2}}}{T_{SO_{2}}} = \frac{273}{546} = \frac{1}{2}$.
Therefore,$KE_{SO_{2}} = 2 \times KE_{O_{2}}$,which implies $KE_{SO_{2}} > KE_{O_{2}}$ or $KE_{O_{2}} < KE_{SO_{2}}$.

(C) Isoelectronic species are those that have the same number of electrons.
$Mg$ has an atomic number of $12$,so $Mg^{2+}$ has $12 - 2 = 10$ electrons.
$Na$ has an atomic number of $11$,so $Na^{+}$ has $11 - 1 = 10$ electrons.
Since both $Mg^{2+}$ and $Na^{+}$ have $10$ electrons,they are isoelectronic.

(C) Given: Mass $m = 10 \ mg = 10 \times 10^{-6} \ kg = 10^{-5} \ kg$.
Velocity $v = 100 \ ms^{-1} = 10^2 \ ms^{-1}$.
Planck's constant $h = 6.63 \times 10^{-34} \ Js$.
Using the de-Broglie wavelength formula: $\lambda = \frac{h}{mv}$.
Substituting the values: $\lambda = \frac{6.63 \times 10^{-34}}{10^{-5} \times 10^2} = \frac{6.63 \times 10^{-34}}{10^{-3}} = 6.63 \times 10^{-31} \ m$.

(D) The properties of a system whose values are independent of the amount of substance present in the system are called intensive properties.
Examples include $temperature$,$viscosity$,$surface \ tension$,$pressure$,$density$,$molar \ heat \ capacity$,etc.
Since all the given options are independent of the amount of matter,they are all intensive properties.

(C) The enthalpy change $(\Delta H)$ is related to the internal energy change $(\Delta U)$ by the equation:
$\Delta H = \Delta U + \Delta(PV)$
For gaseous reactions,this is expressed as:
$\Delta H = \Delta U + \Delta n_{g}RT$
where $\Delta n_{g}$ is the change in the number of moles of gaseous products and reactants.

(D) According to the second law of thermodynamics,the entropy of the universe always increases in the course of every spontaneous (natural) change.

(B) The order of entropy for states of matter is $Gas > Liquid > Solid$.
Sublimation is the process where a solid directly converts into a gas.
Since the transition from solid to gas involves the greatest increase in disorder,the sublimation of naphthalene shows the maximum increase in entropy.

(C) $1$. $\Delta H$ is negative $(-)$ because the reaction is exothermic,as indicated by the release of $393.5 \ kJ$ of heat.
$2$. $\Delta S$ is negative $(-)$ because the number of moles of gaseous species decreases from $1 \ mol$ of $O_{2(g)}$ on the reactant side to $0 \ mol$ of gas on the product side (since $CO_2$ is produced,but the total moles of gas decrease from $1$ to $1$,wait,let's re-evaluate: $C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$. Here,$1 \ mol$ of gas reacts to form $1 \ mol$ of gas. However,the entropy change is generally considered negative or near zero due to the loss of the solid phase and the nature of the gas phase,but in standard textbook contexts for this specific reaction,$\Delta S$ is often considered negative due to the decrease in disorder).
$3$. $\Delta G$ is negative $(-)$ because the combustion of carbon is a spontaneous process at standard conditions.

(A) For a spontaneous process,$\Delta G < 0$. For a non-spontaneous process,$\Delta G > 0$. At equilibrium,the system is at its minimum Gibbs free energy,and the change in Gibbs free energy is $\Delta G = 0$.

(C) cycloalkane is a saturated hydrocarbon that contains one ring structure.
For an open-chain alkane,the general formula is $C_n H_{2n+2}$.
When a ring is formed,two hydrogen atoms are removed from the chain to connect the ends,resulting in the loss of $2$ hydrogen atoms.
Therefore,the general formula for a cycloalkane becomes $C_n H_{2n+2-2} = C_n H_{2n}$.

(B) Denatured alcohol is ethyl alcohol that has been made unfit for drinking by adding poisonous substances such as methanol,naphtha,pyridine,or rubber. This process is done to prevent the misuse of industrial alcohol for consumption.

(C) Hofmann's bromamide reaction is a degradation reaction used to convert a primary amide into a primary amine with one carbon atom less than the original amide.
The general chemical equation is:
$RCONH_{2} + Br_{2} + 4KOH \longrightarrow RNH_{2} + K_{2}CO_{3} + 2KBr + 2H_{2}O$
Thus,the reaction converts $\text{Amide} \longrightarrow \text{Amine}$.

(D) When an alkyl halide reacts with alcoholic ammonia in a sealed tube,a mixture of primary,secondary,and tertiary amines is formed due to successive alkylation reactions.
$R-X + NH_{3} \rightarrow R-NH_{2} + HX$ (Primary amine)
$R-NH_{2} + R-X \rightarrow R_{2}NH + HX$ (Secondary amine)
$R_{2}NH + R-X \rightarrow R_{3}N + HX$ (Tertiary amine)
$R_{3}N + R-X \rightarrow R_{4}N^{+}X^{-}$ (Quaternary ammonium salt)

(A) $CH_3NH_2$ reacts with $CH_3I$ through a process called exhaustive alkylation (Hofmann's alkylation).
The reaction proceeds as follows:
$1. CH_3NH_2 + CH_3I \rightarrow (CH_3)_2NH + HI$ (Secondary amine)
$2. (CH_3)_2NH + CH_3I \rightarrow (CH_3)_3N + HI$ (Tertiary amine)
$3. (CH_3)_3N + CH_3I \rightarrow (CH_3)_4N^+I^-$ (Quaternary ammonium salt)
Thus,a total of $3$ molecules of $CH_3I$ react with one molecule of $CH_3NH_2$ to reach the final stable product.

(B) When a carbonyl compound (aldehyde or ketone) reacts with hydroxylamine $(NH_2OH)$,it undergoes a condensation reaction to form an oxime $(>C=N-OH)$.
The general reaction is:
$R_2C=O + NH_2OH \rightarrow R_2C=N-OH + H_2O$.

(D) An amino acid is optically active if it contains at least one chiral carbon atom (an asymmetric carbon atom bonded to four different groups).
Glycine has the structure $NH_2-CH_2-COOH$.
In glycine,the central carbon atom is bonded to two identical hydrogen atoms,meaning it does not have a chiral center.
Therefore,glycine is the only naturally occurring amino acid that is not optically active.

(D) Vitamin $B_{12}$ contains the metal cobalt.
The chemical name of vitamin $B_{12}$ is cyanocobalamin.

(C) $Sc^{3+}$ has completely empty $d$-orbitals ($d^0$ configuration); hence,it forms a colourless solution in aqueous medium.
$Sc \ (21): [Ar] \ 3d^1 \ 4s^2$
$Sc^{3+}: [Ar] \ 3d^0 \ 4s^0$
Since there are no unpaired electrons to undergo $d-d$ transitions,the solution is colourless.

(D) reaction is said to be of second order if its rate depends on the concentration of two reactants or the square of the concentration of one reactant.
The saponification of an ester,such as the reaction between methyl acetate and sodium hydroxide,follows second-order kinetics.
The rate law for this reaction is $Rate = k[CH_{3}COOCH_{3}][NaOH]$.
Thus,$CH_{3}COOCH_{3} + NaOH \longrightarrow CH_{3}COONa + CH_{3}OH$ is a second-order reaction.

(C) Paracetamol acts as both an antipyretic and an analgesic.
An antipyretic is a drug that reduces fever,while an analgesic is a drug that relieves pain.
Therefore,it is used to lower body temperature during high fever and also to provide relief from pain.

(C) The coordination number of a central metal ion in a complex is defined as the number of ligand donor atoms to which the metal is directly bonded. For unidentate ligands,the coordination number is equal to the number of ligands attached.

(D) Let the oxidation state of iron in $K_{4}[Fe(CN)_{6}]$ be $x$.
$4(+1) + x + 6(-1) = 0$
$4 + x - 6 = 0$
$x = +2$
Therefore,the oxidation state of iron is $+2$.

(A) complex ionises in solution only if it has counter-ions outside the coordination sphere.
In the complex $[CoCl_{3}(NH_{3})_{3}]$,all three $Cl^{-}$ ions are directly bonded to the central $Co^{3+}$ metal ion as ligands (satisfying both primary and secondary valencies).
Since there are no ions outside the square brackets,it does not dissociate into ions in an aqueous solution.
Therefore,it will not be precipitated by the addition of $AgNO_{3}$.

(B) $1$. Identify the cation: $Na^+$ is sodium.
$2$. Identify the coordination entity: $[Co(NO_{2})_{6}]^{3-}$.
$3$. The ligand is $NO_{2}^-$,which is named 'nitro' when bonded through nitrogen.
$4$. There are $6$ such ligands,so the prefix is 'hexa'.
$5$. The central metal is cobalt,and since the complex is anionic,it is named 'cobaltate'.
$6$. Calculate the oxidation state of $Co$: $x + 6(-1) = -3$,so $x = +3$.
$7$. Combining these,the $IUPAC$ name is sodium hexanitrocobaltate $(III)$.

(C) The formation of a coloured solution is possible when the metal ion in the compound contains unpaired electrons,which allow for $d-d$ transitions.
For example:
$Cu^{+}: 3d^{10} 4s^{0}$ (no unpaired electrons,colourless)
$Cu^{2+}: 3d^{9} 4s^{0}$ (one unpaired electron,blue)

(C) The reaction is $2 SO_{2(g)} + O_{2(g)} \xrightarrow{V_{2}O_{5(s)}} 2 SO_{3(g)}$.
In this reaction,the reactants ($SO_2$ and $O_2$) are in the gaseous phase,while the catalyst $(V_2O_5)$ is in the solid phase.
Since the reactants and the catalyst are in different phases,this is an example of heterogeneous catalysis.

(B) In the periodic table,transition metals are commonly used as catalysts because they provide a large surface area and have variable oxidation states.
These metals belong to the $d$-block.
Examples include $Ni$,$Pt$,$V_2O_5$,and $Fe$.

(B) The atomic number of Chromium $(Cr)$ is $24$.
Its ground state electronic configuration is $[Ar] 3d^{5} 4s^{1}$.
To form the $Cr^{3+}$ ion,we remove $3$ electrons: one from the $4s$ orbital and two from the $3d$ orbital.
Therefore,the electronic configuration of $Cr^{3+}$ is $[Ar] 3d^{3} 4s^{0}$.

(C) The chemical formulas for the given ores are:
$1$. Limonite: $Fe_{2}O_{3} \cdot 3H_{2}O$ (Contains iron)
$2$. Siderite: $FeCO_{3}$ (Contains iron)
$3$. Carnallite: $KCl \cdot MgCl_{2} \cdot 6H_{2}O$ (Does not contain iron)
$4$. Chalcopyrites: $CuFeS_{2}$ (Contains iron)
Therefore,Carnallite is the ore that does not contain iron.

(D) The extraction of gold involves the leaching of gold with a cyanide solution in the presence of air $(O_2)$:
$4Au + 8CN^{-} + 2H_2O + O_2 \longrightarrow 4[Au(CN)_2]^{-} + 4OH^{-}$
In this reaction,the gold is oxidized to the complex $[Au(CN)_2]^{-}$.
Next,the gold is recovered from the complex by displacement with zinc:
$2[Au(CN)_2]^{-} + Zn \longrightarrow 2Au + [Zn(CN)_4]^{2-}$
Comparing these with the given reactions,$X$ is $[Au(CN)_2]^{-}$ and $Y$ is $[Zn(CN)_4]^{2-}$.

(B) In Clemmensen's reduction,$Zn-Hg$ amalgam with concentrated $HCl$ is used as the reagent.
This reaction reduces carbonyl groups $(>C=O)$ to methylene groups $(-CH_2-)$.
The general reaction is: $>C=O + 4[H] \xrightarrow{Zn-Hg / \text{conc. } HCl} >CH_2 + H_2O$.
This method is specifically used to convert aldehydes and ketones into their corresponding alkanes.

(B) The reaction of benzene diazonium chloride with $Cu_{2}Cl_{2} / HCl$ to produce chlorobenzene is a classic example of Sandmeyer's reaction.
In this reaction,the diazonium group $(-N_{2}^{+}Cl^{-})$ is replaced by a chlorine atom.

(A) $P_{2}O_{5}$ is a powerful dehydrating agent. When concentrated $H_{2}SO_{4}$ is heated with $P_{2}O_{5}$,it undergoes dehydration to form sulphur trioxide $(SO_{3})$.
The chemical reaction is as follows:
$2H_{2}SO_{4} + P_{2}O_{5} \rightarrow 2SO_{3} + 2H_{3}PO_{4}$ (or $2H_{2}SO_{4} + 2P_{2}O_{5} \rightarrow 2SO_{3} + 4HPO_{3}$).

(D) The separation of noble gases is based on the principle of selective adsorption on coconut charcoal at different temperatures.
At $173 \ K$,$Ar$,$Kr$,and $Xe$ are adsorbed on the surface of the coconut charcoal.
However,$He$ and $Ne$ have very low boiling points and weak van der Waals forces,so they are not adsorbed at this temperature.
Therefore,the gases that remain in the gaseous phase are $He$ and $Ne$.

(A) Let $V \ L$ of $10 \ N$ $HCl$ be mixed with $(1-V) \ L$ of $4 \ N$ $HCl$ to obtain $1 \ L$ of $7 \ N$ $HCl$.
Using the mixture formula $N_1 V_1 + N_2 V_2 = N_3 V_3$:
$10V + 4(1-V) = 7 \times 1$
$10V + 4 - 4V = 7$
$6V = 3$
$V = 0.50 \ L$
Thus,the volume of $10 \ N$ $HCl$ is $0.50 \ L$ and the volume of $4 \ N$ $HCl$ is $1 - 0.50 = 0.50 \ L$.

(C) Hydrophobic sols are irreversible in nature.
They have no affinity between the dispersed phase and the dispersion medium $(H_{2}O)$.
Once precipitated,they do not form the colloidal sol by simple addition of water.