IIT JEE 2001 Chemistry Question Paper with Answer and Solution

(C) The electric field lines for a system of positive charges must originate from the charges and extend to infinity.
$1$. Since all three charges are positive,the electric field lines will repel each other and cannot terminate on any of the charges.
$2$. Option $A$ shows lines terminating on other charges,which is incorrect for like charges.
$3$. Option $B$ and $D$ show closed loops,which is impossible for electrostatic fields as they are conservative.
$4$. Option $C$ correctly depicts the field lines originating from each positive charge and moving outwards,repelling each other,which is the characteristic behavior of a system of identical positive charges.
Therefore,the correct representation is option $C$.

(A) The total number of electrons in each species is as follows:
$CN^{-} = 6 + 7 + 1 = 14$ electrons.
$CO = 6 + 8 = 14$ electrons.
$NO^{+} = 7 + 8 - 1 = 14$ electrons.
Since all species have $14$ electrons,they are isoelectronic.
Using the molecular orbital theory,the bond order is calculated as:
$B.O. = \frac{1}{2} [N_b - N_a] = \frac{1}{2} [10 - 4] = 3$.
Thus,they all have a bond order of $3$ and are isoelectronic.

(D) The structure of the $SO_3$ trimer $(S_3O_9)$ consists of a cyclic ring of alternating $S$ and $O$ atoms,where each $S$ atom is also bonded to two terminal oxygen atoms via double bonds.
In this structure,there are no direct $S-S$ bonds.
Therefore,the number of $S-S$ bonds is $0$.

(D) The equilibrium constant $(K_p)$ for a given reaction depends only on temperature.
It is independent of the initial pressure $(P)$ and the extent of decomposition $(x)$.
Therefore,$K_p$ remains constant even if $P$ or $x$ changes at a constant temperature.

(B) The first ionisation potential $(I.P.)$ generally decreases as we move down a group due to the increase in atomic size and shielding effect.
In Group $2$ $(Be, Mg, Ca)$,the order of $I.P.$ is $Be > Mg > Ca$.
In Group $1$ $(Li, Na, K)$,the order is $Li > Na > K$.
In Group $14$ $(C, Si, Ge)$,the order is $C > Si > Ge$.
In Group $13$ $(B, Al, Ga)$,the order is $B > Al > Ga$.
Therefore,the correct set is $Be > Mg > Ca$.

(A) The molar mass of oxalic acid dihydrate $(H_2C_2O_4 \cdot 2H_2O)$ is $126 \ g/mol$.
The $n$-factor for oxalic acid is $2$.
Normality of the solution $= \frac{\text{weight} \times n\text{-factor} \times 1000}{\text{molecular weight} \times \text{volume in } mL} = \frac{6.3 \times 2 \times 1000}{126 \times 250} = 0.4 \ N$.
Using the law of equivalence,$N_1 V_1 = N_2 V_2$ for neutralization:
$0.4 \times 10 = 0.1 \times V_2$.
$V_2 = \frac{4}{0.1} = 40 \ mL$.

(D) The molecular formula $C_2BrClFI$ corresponds to a haloalkene where each carbon of the double bond is substituted with two different halogens.
For a general alkene $C(ab)=C(cd)$,if $a \neq b$ and $c \neq d$,the molecule exhibits geometrical isomerism ($E$ and $Z$ forms).
In the case of $C_2BrClFI$,we can arrange the four different halogens $(Br, Cl, F, I)$ on the two carbons in different ways.
There are $3$ possible structural skeletons based on the relative positions of the halogens:
$1$. $Br-C(Cl)=C(F)-I$
$2$. $Br-C(F)=C(Cl)-I$
$3$. $Br-C(I)=C(Cl)-F$
Each of these $3$ structural isomers can exist as a pair of geometrical isomers ($E$ and $Z$).
Therefore,the total number of stereoisomers is $3 \times 2 = 6$.

(B) The given compound is $CH_3-C \equiv C-CH(CH_3)-CH=CH_2$.
Hydrogenation with a poisoned palladium catalyst (Lindlar's catalyst) results in $syn$-addition of hydrogen to the triple bond,converting it into a $cis$-double bond.
The resulting product is $CH_3-CH=CH-CH(CH_3)-CH=CH_2$.
Due to the specific geometry and the presence of internal symmetry elements or the formation of a meso-like structure depending on the configuration,the product formed is optically inactive.

(C) The anti-Markovnikov addition of $HX$ to alkenes proceeds via a free radical mechanism.
For $HCl$,the bond dissociation energy is very high,making the homolytic cleavage of the $H-Cl$ bond endothermic.
For $HI$,the addition of the iodine radical to the alkene is an endothermic step.
Since one of the propagation steps is endothermic for both $HCl$ and $HI$,the overall reaction does not proceed via the anti-Markovnikov pathway in the presence of peroxide.

(B) Alkenes undergo electrophilic addition reactions.
$HOCl$ acts as a source of the electrophile $Cl^{+}$.
In the first step,the electrophile $Cl^{+}$ attacks the double bond of propene to form a cyclic chloronium ion intermediate.
Subsequently,the nucleophile $OH^{-}$ attacks the more substituted carbon atom to form the final product,chlorohydrin.

(A) Let $\angle PRQ = \theta$. Since $PQ$ is a tangent at $P$ and $PR$ is the diameter,$\angle RPQ = 90^{\circ}$.
In $\triangle RPQ$,$\tan \theta = \frac{PQ}{PR} = \frac{PQ}{2r}$.
Since $RS$ is a tangent at $R$ and $PR$ is the diameter,$\angle PRS = 90^{\circ}$.
Given $X$ is on the circumference,$\angle RXP = 90^{\circ}$ (angle in a semicircle).
In $\triangle RXP$,$\angle XRP = 90^{\circ} - \theta$.
In $\triangle PRS$,$\tan(90^{\circ} - \theta) = \frac{RS}{PR} = \frac{RS}{2r}$.
Thus,$\cot \theta = \frac{RS}{2r}$.
Multiplying the two expressions: $\tan \theta \cdot \cot \theta = \left(\frac{PQ}{2r}\right) \cdot \left(\frac{RS}{2r}\right) = 1$.
$1 = \frac{PQ \cdot RS}{4r^2} \implies 4r^2 = PQ \cdot RS \implies 2r = \sqrt{PQ \cdot RS}$.

(B) Let the temperature of the junction be $\theta$.
Since the rods are made of the same material and have the same length and cross-section,they have the same thermal resistance $R$.
According to the principle of conservation of energy (steady state),the heat flowing into the junction must equal the heat flowing out of the junction.
Let the heat flow from the $90^{\circ}C$ ends towards the junction and then towards the $0^{\circ}C$ end.
$Q_{in} = Q_{out}$
$\frac{90 - \theta}{R} + \frac{90 - \theta}{R} = \frac{\theta - 0}{R}$
$2(90 - \theta) = \theta$
$180 - 2\theta = \theta$
$3\theta = 180$
$\theta = 60^{\circ}C$.

(A) The molar mass of oxalic acid dihydrate $(H_2C_2O_4 \cdot 2H_2O)$ is $126 \ g/mol$.
Since it is a dibasic acid,its equivalent mass is $126 / 2 = 63 \ g/eq$.
First,calculate the normality $(N_1)$ of the oxalic acid solution:
$N_1 = \frac{\text{mass}}{\text{equivalent mass}} \times \frac{1000}{V(mL)} = \frac{6.3}{63} \times \frac{1000}{250} = 0.1 \times 4 = 0.4 \ N$.
Using the law of equivalence for neutralization:
$N_1 V_1 = N_2 V_2$
$(0.4 \ N) \times (10 \ mL) = (0.1 \ N) \times V_2$
$V_2 = \frac{0.4 \times 10}{0.1} = 40 \ mL$.

(D) The spin quantum number $(m_s)$ describes the intrinsic angular momentum of an electron.
Historically,it was visualized as the electron spinning on its axis,where $+1/2$ corresponds to clockwise rotation and $-1/2$ corresponds to anticlockwise rotation.
Alternatively,it represents the orientation of the electron's magnetic moment,where $+1/2$ indicates the magnetic moment is pointing up and $-1/2$ indicates it is pointing down.
Therefore,both statements $A$ and $C$ are accepted interpretations.

(A) The dissociation of the sparingly soluble salt $A_pB_q$ is given by:
$A_pB_q(s) \rightleftharpoons pA^{q+}(aq) + qB^{p-}(aq)$
If $S$ is the solubility of the salt,then the concentration of ions at equilibrium is:
$[A^{q+}] = p \cdot S$
$[B^{p-}] = q \cdot S$
The solubility product $(L_S)$ is defined as:
$L_S = [A^{q+}]^p [B^{p-}]^q$
Substituting the concentrations:
$L_S = (p \cdot S)^p \cdot (q \cdot S)^q$
$L_S = p^p \cdot S^p \cdot q^q \cdot S^q$
$L_S = S^{p + q} \cdot p^p \cdot q^q$

(C) Given $\alpha + \beta = \frac{\pi}{2}$,we have $\beta = \frac{\pi}{2} - \alpha$,so $\tan \beta = \cot \alpha = \frac{1}{\tan \alpha}$.
Also,$\beta + \gamma = \alpha$,so $\tan(\beta + \gamma) = \tan \alpha$.
Using the formula $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we get $\tan \alpha = \frac{\tan \beta + \tan \gamma}{1 - \tan \beta \tan \gamma}$.
Substituting $\tan \beta = \cot \alpha$:
$\tan \alpha = \frac{\cot \alpha + \tan \gamma}{1 - \cot \alpha \tan \gamma}$.
Multiply both sides by $(1 - \cot \alpha \tan \gamma)$:
$\tan \alpha (1 - \cot \alpha \tan \gamma) = \cot \alpha + \tan \gamma$.
$\tan \alpha - (\tan \alpha \cot \alpha) \tan \gamma = \cot \alpha + \tan \gamma$.
Since $\tan \alpha \cot \alpha = 1$,we have $\tan \alpha - \tan \gamma = \cot \alpha + \tan \gamma$.
Since $\cot \alpha = \tan \beta$,we get $\tan \alpha = \tan \beta + 2\tan \gamma$.

(D) Given equation: ${y^2} + 4y + 4x + 2 = 0$
Rewrite as: ${y^2} + 4y + 4 = -4x + 2$
${(y + 2)^2} = -4x + 2$
${(y + 2)^2} = -4(x - \frac{1}{2})$
Comparing with standard form ${Y^2} = -4aX$,where $Y = y + 2$,$X = x - \frac{1}{2}$,and $4a = 4 \implies a = 1$.
The directrix of ${Y^2} = -4aX$ is $X = a$.
Substituting the values: $x - \frac{1}{2} = 1$
$x = 1 + \frac{1}{2} = \frac{3}{2}$.

(B) Let $E$ be the $EMF$ of each cell and $R$ be the resistance of the wire of length $L$. The mass of the wire is $m = \rho A L$,where $\rho$ is density and $A$ is cross-sectional area.
In the first case,the heat produced is equal to the heat absorbed by the wire:
$\frac{(3E)^2}{R} t = m s \Delta T$ ....................... $(i)$
When the length is doubled to $2L$,the new resistance becomes $2R$ and the new mass becomes $2m$.
In the second case,with $N$ cells in series:
$\frac{(NE)^2}{2R} t = (2m) s \Delta T$ ....................... $(ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{(NE)^2 / 2R}{(3E)^2 / R} = \frac{2m s \Delta T}{m s \Delta T}$
$\frac{N^2}{2 \times 9} = 2$
$N^2 = 36$
$N = 6$

(C) The total momentum of the system at time $t = 0$ is $\vec{P}_i = m_1 \vec{v}_1 + m_2 \vec{v}_2$.
The collision between the two particles is an internal interaction and does not change the total momentum of the system.
The only external force acting on the system throughout the motion is the gravitational force,which is $\vec{F}_{ext} = (m_1 + m_2)g$ acting downwards.
According to the impulse-momentum theorem,the change in momentum is equal to the impulse of the external force: $\Delta \vec{P} = \int_{t_1}^{t_2} \vec{F}_{ext} dt$.
Here,the time interval is from $t = 0$ to $t = 2t_0$,so $\Delta t = 2t_0$.
Since the force is constant,the impulse is $\vec{J} = \vec{F}_{ext} \times \Delta t = (m_1 + m_2)g \times 2t_0$.
Therefore,the magnitude of the change in momentum is $|(m_1 \vec{v}_1' + m_2 \vec{v}_2') - (m_1 \vec{v}_1 + m_2 \vec{v}_2)| = 2(m_1 + m_2)gt_0$.

(D) The electron spin quantum number $(m_s)$ describes the intrinsic angular momentum of an electron.
While it is often visualized as the electron spinning on its axis,this is a simplified model.
In quantum mechanics,the values $+1/2$ and $-1/2$ represent two distinct quantum mechanical spin states that do not have a direct classical analogue.

(D) The electron spin quantum number $(m_s)$ describes the intrinsic angular momentum of an electron.
While it is often visualized as the electron spinning about its axis,this is a semi-classical model.
In quantum mechanics,these values ($+1/2$ and $-1/2$) represent two distinct quantum mechanical spin states that have no true classical analogue.

(D) To determine the number of $d$ electrons,we calculate the oxidation state of the central metal atom:
$1$. In $[Co(NH_3)_6]^{3+}$,$Co$ is in $+3$ state. $Co$ $(Z=27)$ is $[Ar] 3d^7 4s^2$,so $Co^{3+}$ is $3d^6$.
$2$. In $[Fe(CN)_6]^{3-}$,$Fe$ is in $+3$ state. $Fe$ $(Z=26)$ is $[Ar] 3d^6 4s^2$,so $Fe^{3+}$ is $3d^5$.
$3$. In $[Cr(H_2O)_6]^{3+}$,$Cr$ is in $+3$ state. $Cr$ $(Z=24)$ is $[Ar] 3d^5 4s^1$,so $Cr^{3+}$ is $3d^3$.
$4$. In $[MnO_4]^-$,$Mn$ is in $+7$ state. $Mn$ $(Z=25)$ is $[Ar] 3d^5 4s^2$,so $Mn^{7+}$ is $3d^0$.
Therefore,$[MnO_4]^-$ has no $d$ electrons.

(B) Given that $\vec{a}, \vec{b}, \text{ and } \vec{c}$ are unit vectors,we have $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$.
We know that $|\vec{x} - \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2(\vec{x} \cdot \vec{y})$.
Expanding the expression:
$|\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 = (|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}) + (|\vec{b}|^2 + |vec{c}|^2 - 2\vec{b} \cdot \vec{c}) + (|\vec{c}|^2 + |\vec{a}|^2 - 2\vec{c} \cdot \vec{a})$.
Since $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$,this simplifies to:
$2(1+1+1) - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 6 - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$.
We also know that $|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 3 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$.
Thus,$2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = |\vec{a} + \vec{b} + \vec{c}|^2 - 3$.
Substituting this back into the expression:
$6 - (|\vec{a} + \vec{b} + \vec{c}|^2 - 3) = 9 - |\vec{a} + \vec{b} + \vec{c}|^2$.
Since $|\vec{a} + \vec{b} + \vec{c}|^2 \geq 0$,the maximum value of the expression is $9$.

(D) The electron spin quantum number $(m_s)$ describes the intrinsic angular momentum of an electron.
While it is often visualized as the electron spinning about its axis,this is a quantum mechanical property.
These two values,$+\frac{1}{2}$ and $-\frac{1}{2}$,represent two distinct quantum mechanical spin states that do not have a direct classical analogue.

(B) The width of the segment on the screen is given by $x = n \beta$,where $\beta = \frac{\lambda D}{d}$ is the fringe width.
Since the segment length $x$ remains constant,we have $n_1 \beta_1 = n_2 \beta_2$.
Substituting the formula for fringe width: $n_1 \left( \frac{\lambda_1 D}{d} \right) = n_2 \left( \frac{\lambda_2 D}{d} \right)$.
This simplifies to $n_1 \lambda_1 = n_2 \lambda_2$.
Given $n_1 = 12$,$\lambda_1 = 600 \, nm$,and $\lambda_2 = 400 \, nm$,we find $n_2 = \frac{n_1 \lambda_1}{\lambda_2}$.
$n_2 = \frac{12 \times 600}{400} = 12 \times 1.5 = 18$.
Therefore,the number of fringes observed is $18$.

(A) The mutual inductance $M$ between two coils depends on the magnetic flux linkage between them.
$M$ is directly proportional to the amount of magnetic flux produced by one coil that passes through the area of the second coil.
In situation $(A)$,the two coils are placed coaxially with their planes parallel to each other. This configuration allows the maximum number of magnetic field lines produced by the first coil to pass through the second coil,resulting in the maximum flux linkage.
In situations $(B)$ and $(C)$,the coils are either perpendicular or at an angle,which significantly reduces the effective flux linkage.
Therefore,the mutual inductance is maximum in situation $(A)$.

(C) The energy $E$ of a standing wave is proportional to the square of the amplitude $(A)$ and the square of the angular frequency $(\omega)$.
Mathematically,$E \propto A^2 \omega^2$.
In the first case,the displacement is $y_1 = A \sin(\pi x/L) \sin(\omega t)$,so the amplitude is $A$ and the angular frequency is $\omega$. Thus,$E_1 = k A^2 \omega^2$ (where $k$ is a constant).
In the second case,the displacement is $y_2 = A \sin(2\pi x/L) \sin(2\omega t)$,so the amplitude is $A$ and the angular frequency is $2\omega$.
Thus,$E_2 = k A^2 (2\omega)^2 = k A^2 (4\omega^2) = 4(k A^2 \omega^2)$.
Comparing the two,we get $E_2 = 4E_1$.

(A) The mutual inductance $M$ between two coils depends on the magnetic flux linkage between them. The flux linkage is maximum when the magnetic field lines produced by one coil pass through the area of the other coil most effectively.
In situation $(A)$,the planes of the two coils are parallel to each other. This orientation allows the maximum number of magnetic field lines produced by the larger coil to pass through the area of the smaller coil,resulting in the maximum magnetic flux linkage.
In situation $(B)$,the smaller coil is placed perpendicular to the plane of the larger coil. In this case,the magnetic field lines from the larger coil are mostly parallel to the plane of the smaller coil,resulting in minimal flux linkage.
In situation $(C)$,the smaller coil is placed in the same plane as the larger coil but shifted to the side. This also results in a lower flux linkage compared to situation $(A)$.
Therefore,the mutual inductance is maximum in situation $(A)$.

(D) In a steady state,the capacitor acts as an open circuit,so no current flows through the middle branch.
The current $I$ flows through the outer loop containing the $V$ and $2V$ batteries and the $R$ and $2R$ resistors.
Applying Kirchhoff's voltage law to the outer loop:
$2V - V = I(R + 2R)$
$V = I(3R)$
$I = \frac{V}{3R}$
The potential difference across the capacitor $C$ is equal to the potential difference between the two nodes to which it is connected.
Let the left node be $A$ and the right node be $B$. The potential difference $V_{AB}$ across the capacitor branch is:
$V_{AB} = V - I \times R$ (or $V_{AB} = 2V - I \times 2R$)
$V_{AB} = V - (\frac{V}{3R}) \times R = V - \frac{V}{3} = \frac{2V}{3}$
Thus,the potential drop across the capacitor is $\frac{2V}{3}$.

(C) Consider a small elemental ring of radius $r$ and thickness $dr$ within the spiral coil.
The number of turns in this elemental ring is $dn = \frac{N}{b - a} dr$.
The magnetic field at the centre due to this elemental ring is $dB = \frac{\mu_0 (dn) I}{2r} = \frac{\mu_0 N I}{2(b - a)} \frac{dr}{r}$.
To find the total magnetic field $B$ at the centre,we integrate $dB$ from $r = a$ to $r = b$:
$B = \int_{a}^{b} \frac{\mu_0 N I}{2(b - a)} \frac{dr}{r} = \frac{\mu_0 N I}{2(b - a)} [\ln r]_{a}^{b}$.
$B = \frac{\mu_0 N I}{2(b - a)} \ln \left( \frac{b}{a} \right)$.

(B) The resultant intensity $I_{net}$ for two interfering beams is given by the formula: $I_{net} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta \phi)$.
At point $A$,the phase difference $\Delta \phi = \pi/2$. Substituting the values $I_1 = I$ and $I_2 = 4I$:
$I_A = I + 4I + 2\sqrt{I \cdot 4I} \cos(\pi/2) = 5I + 2(2I)(0) = 5I$.
At point $B$,the phase difference $\Delta \phi = \pi$. Substituting the values:
$I_B = I + 4I + 2\sqrt{I \cdot 4I} \cos(\pi) = 5I + 2(2I)(-1) = 5I - 4I = I$.
The difference between the resultant intensities at $A$ and $B$ is:
$|I_A - I_B| = |5I - I| = 4I$.

(A) The mutual inductance $M$ between two coils depends on the magnetic flux linkage between them.
Mutual inductance is directly proportional to the amount of magnetic flux produced by one coil that passes through the area of the other coil.
In situation $(A)$,the planes of the two circular coils are parallel to each other. This orientation allows the maximum number of magnetic field lines produced by one coil to pass through the area of the second coil,resulting in maximum flux linkage.
In situations $(B)$ and $(C)$,the coils are placed perpendicular to each other,which results in minimum or zero flux linkage.
Therefore,the mutual inductance is maximum in situation $(A)$.

(D) When a conductor moves in a magnetic field,a motional electromotive force $(EMF)$ is induced across its ends,which is given by $\varepsilon = \int (\vec{v} \times \vec{B}) \cdot d\vec{l}$.
This induced $EMF$ corresponds to an induced electric field $\vec{E} = \vec{v} \times \vec{B}$.
In this problem,the square loop $ABCD$ moves with velocity $\vec{v}$ in a uniform magnetic field $\vec{B}$ perpendicular to the plane of the loop.
For any segment of the loop,if the velocity $\vec{v}$ is perpendicular to the magnetic field $\vec{B}$,an electric field $\vec{E} = \vec{v} \times \vec{B}$ is induced along the conductor.
Since the segments $AD$ and $BC$ are perpendicular to the velocity $\vec{v}$ and the magnetic field $\vec{B}$ is uniform and perpendicular to the plane,the cross product $\vec{v} \times \vec{B}$ is non-zero for both segments $AD$ and $BC$.
Therefore,an electric field is induced in both $AD$ and $BC$.

(B) The resultant intensity $I_R$ for two interfering beams is given by the formula: $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$,where $\phi$ is the phase difference.
Given $I_1 = I$ and $I_2 = 4I$.
At point $A$,the phase difference $\phi_A = \pi / 2$. Therefore,$I_A = I + 4I + 2\sqrt{I \cdot 4I} \cos(\pi / 2) = 5I + 4I(0) = 5I$.
At point $B$,the phase difference $\phi_B = \pi$. Therefore,$I_B = I + 4I + 2\sqrt{I \cdot 4I} \cos(\pi) = 5I + 4I(-1) = 5I - 4I = I$.
The difference between the resultant intensities at $A$ and $B$ is $I_A - I_B = 5I - I = 4I$.

(A) Let the diameter be $PR = 2r$. Since $PQ$ and $RS$ are tangents at $P$ and $R$ respectively,$PQ \perp PR$ and $RS \perp PR$.
In $\triangle PXR$,$\angle PXR = 90^{\circ}$ because it is an angle in a semicircle.
Let $\angle RPX = \theta$. Then $\angle PRX = 90^{\circ} - \theta$.
In $\triangle PQR$,$\angle PQR = 90^{\circ} - \theta$ and $\angle PRQ = 90^{\circ}$.
Thus,$\tan(\angle RPX) = \tan \theta = \frac{RS}{PR} = \frac{RS}{2r}$.
Also,in $\triangle PQR$,$\tan(\angle PRQ) = \tan(90^{\circ} - \theta) = \cot \theta = \frac{PQ}{PR} = \frac{PQ}{2r}$.
Therefore,$\tan \theta = \frac{2r}{PQ}$.
Equating the two expressions for $\tan \theta$:
$\frac{RS}{2r} = \frac{2r}{PQ} \implies (2r)^2 = PQ \cdot RS \implies 2r = \sqrt{PQ \cdot RS}$.

(B) The hybridization of the central atom in each species is determined as follows:
$1$. In $NH_3$,the central atom $N$ has $3$ bond pairs and $1$ lone pair,resulting in $sp^3$ hybridization.
$2$. In $[PtCl_4]^{2-}$,$Pt^{2+}$ has a $d^8$ configuration,which undergoes $dsp^2$ hybridization to form a square planar geometry.
$3$. In $PCl_5$,the central atom $P$ has $5$ bond pairs,resulting in $sp^3d$ hybridization.
$4$. In $BCl_3$,the central atom $B$ has $3$ bond pairs,resulting in $sp^2$ hybridization.
Thus,the correct order is $sp^3$,$dsp^2$,$sp^3d$,$sp^2$.

(B) In the iodometric titration,$K_2Cr_2O_7$ acts as an oxidizing agent.
The reduction half-reaction is: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$.
Here,the change in oxidation state of $Cr$ is from $+6$ to $+3$ for each $Cr$ atom. Since there are two $Cr$ atoms in $K_2Cr_2O_7$,the total change in oxidation state is $2 \times (6 - 3) = 6$.
Therefore,the $n$-factor for $K_2Cr_2O_7$ is $6$.
Equivalent weight = $\frac{Molecular \text{ } weight}{n\text{-}factor} = \frac{Molecular \text{ } weight}{6}$.

(D) In an $NaCl$ structure,$A$ atoms (cations) occupy the corners and face centers,while $B$ atoms (anions) occupy the edge centers and body center.
Initial number of $A$ atoms per unit cell $= 8 \times \frac{1}{8} (\text{corners}) + 6 \times \frac{1}{2} (\text{face-centers}) = 1 + 3 = 4$.
Initial number of $B$ atoms per unit cell $= 12 \times \frac{1}{4} (\text{edge-centers}) + 1 (\text{body-center}) = 3 + 1 = 4$.
Removing face-centered atoms along one of the axes means removing $2$ face-centered $A$ atoms.
New number of $A$ atoms $= 4 - 2 = 2$.
Wait,the question states $A$ atoms occupy corners. In $NaCl$,$A$ is typically the cation. If $A$ is at corners and face centers,and $B$ is at edge centers and body center,removing $2$ face-centered $A$ atoms leaves $A = 8 \times \frac{1}{8} + 4 \times \frac{1}{2} = 1 + 2 = 3$.
Number of $B$ atoms remains $4$.
Therefore,the stoichiometry is $A_3B_4$.

(B) In a photochemical reaction,the rate of formation of the excited species $AB^*$ is governed by the primary photochemical process.
According to the Stark-Einstein law of photochemical equivalence,the rate of a primary photochemical process is directly proportional to the intensity of the absorbed light $(I)$.

(B) During the extraction of copper from copper pyrites $(CuFeS_2)$,the iron impurity $(FeO)$ is removed by adding silica $(SiO_2)$ as a flux.
The reaction is: $FeO (s) + SiO_2 (s) \to FeSiO_3 (l)$ (slag).
Thus,the slag formed is iron silicate $(FeSiO_3)$.

(A) The complex ion $[MnO_4]^-$ has no $d$-electron in the central metal atom.
The electronic configuration of $Mn$ is $[Ar] 3d^5 4s^2$. In $[MnO_4]^-$,the oxidation number of $Mn$ is $+7$,meaning all the $3d$ and $4s$ electrons are lost to form the $[MnO_4]^-$ complex.
Thus,the electronic configuration of $Mn(VII)$ is $[Ar] 3d^0 4s^0$,confirming it has no $d$-electrons.
In contrast,the central metal atoms in $[Co(NH_3)_6]^{3+}$,$[Fe(CN)_6]^{3-}$,and $[Cr(H_2O)_6]^{3+}$ have $6$,$5$,and $3$ $d$-electrons respectively.

(A) The reaction involves a nucleophilic substitution $(S_N2)$ where $OH^-$ acts as the nucleophile.
In $(CH_3)_4N^+I^-$,the quaternary ammonium salt acts as a good leaving group (trimethylamine,$(CH_3)_3N$) because the positive charge on the nitrogen atom makes the attached methyl group highly electrophilic.
$(CH_3)_4N^+I^- + OH^- \rightarrow CH_3OH + (CH_3)_3N + I^-$.
This reaction proceeds readily due to the high reactivity of the methyl group attached to the positively charged nitrogen atom.

(A) The reaction between benzaldehyde $(C_6H_5CHO)$ and formaldehyde $(HCHO)$ in the presence of aqueous $NaOH$ is a $Crossed \ Cannizzaro \ reaction$.
In this reaction,the more reactive aldehyde (formaldehyde) undergoes oxidation to form a formate salt,while the less reactive aldehyde (benzaldehyde) undergoes reduction to form an alcohol.
$C_6H_5CHO + HCHO \xrightarrow{NaOH_{(aq)}} C_6H_5CH_2OH + HCOONa$
Thus,the products are benzyl alcohol $(C_6H_5CH_2OH)$ and sodium formate $(HCOONa)$.

(A) The basicity of the given compounds is determined by the availability of the lone pair on the nitrogen atom:
$1$. $CH_3-C(=NH)NH_2$ (Acetamidine): It is a very strong base because the conjugate acid is stabilized by resonance across two equivalent nitrogen atoms.
$2$. $(CH_3)_2NH$ (Dimethylamine): $A$ secondary aliphatic amine,which is a strong base due to the inductive effect $(+I)$ of two methyl groups.
$3$. $CH_3-CH_2-NH_2$ (Ethylamine): $A$ primary aliphatic amine,which is a strong base,but less basic than the secondary amine due to fewer $+I$ groups.
$4$. $CH_3-CONH_2$ (Acetamide): The lone pair on the nitrogen is involved in resonance with the carbonyl group $(C=O)$,making it very weakly basic.
Thus,the correct order of basicity is $1 > 2 > 3 > 4$.

(B) The molar mass of oxalic acid dihydrate $(H_2C_2O_4 \cdot 2H_2O)$ is $126 \ g/mol$.
Since it is a dibasic acid,its equivalent mass is $\frac{126}{2} = 63 \ g/eq$.
Normality of the oxalic acid solution = $\frac{\text{mass}}{\text{equivalent mass} \times \text{volume in L}} = \frac{6.3}{63 \times 0.250} = 0.4 \ N$.
Using the titration formula $N_1 V_1 = N_2 V_2$ for $10 \ mL$ of the solution:
$0.4 \ N \times 10 \ mL = 0.1 \ N \times V_2$.
$V_2 = \frac{4}{0.1} = 40 \ mL$.