AIEEE 2007 Chemistry Question Paper with Answer and Solution

(A) The energy of a photon emitted during a transition between two energy levels $n_i$ and $n_f$ is given by $\Delta E = E_{n_i} - E_{n_f} = h\nu$,where $\nu$ is the frequency of the emitted photon.
For emission to occur,the electron must transition from a higher energy level to a lower energy level $(n_i > n_f)$.
Among the given options,$n = 2$ to $n = 1$ and $n = 6$ to $n = 2$ are emission transitions.
The energy difference for $n = 2$ to $n = 1$ is $\Delta E_{2 \to 1} = E_2 - E_1 = -3.4 \text{ eV} - (-13.6 \text{ eV}) = 10.2 \text{ eV}$.
The energy difference for $n = 6$ to $n = 2$ is $\Delta E_{6 \to 2} = E_6 - E_2 = -0.38 \text{ eV} - (-3.4 \text{ eV}) = 3.02 \text{ eV}$.
Since $\Delta E_{2 \to 1} > \Delta E_{6 \to 2}$,the frequency $\nu = \frac{\Delta E}{h}$ is highest for the transition $n = 2$ to $n = 1$.

(D) Given the equation: $\sin^{-1}(\frac{x}{5}) + \csc^{-1}(\frac{5}{4}) = \frac{\pi}{2}$
We know that $\csc^{-1}(\frac{1}{y}) = \sin^{-1}(y)$,so $\csc^{-1}(\frac{5}{4}) = \sin^{-1}(\frac{4}{5})$.
Substituting this into the equation: $\sin^{-1}(\frac{x}{5}) + \sin^{-1}(\frac{4}{5}) = \frac{\pi}{2}$
Rearranging the terms: $\sin^{-1}(\frac{x}{5}) = \frac{\pi}{2} - \sin^{-1}(\frac{4}{5})$
Using the identity $\sin^{-1}(z) + \cos^{-1}(z) = \frac{\pi}{2}$,we have $\frac{\pi}{2} - \sin^{-1}(\frac{4}{5}) = \cos^{-1}(\frac{4}{5})$.
Thus,$\sin^{-1}(\frac{x}{5}) = \cos^{-1}(\frac{4}{5})$.
Since $\cos^{-1}(\frac{4}{5}) = \sin^{-1}(\sqrt{1 - (\frac{4}{5})^2}) = \sin^{-1}(\sqrt{1 - \frac{16}{25}}) = \sin^{-1}(\sqrt{\frac{9}{25}}) = \sin^{-1}(\frac{3}{5})$.
Therefore,$\sin^{-1}(\frac{x}{5}) = \sin^{-1}(\frac{3}{5})$,which implies $\frac{x}{5} = \frac{3}{5}$,so $x = 3$.

(B) According to the first law of thermodynamics,for any path,$Q = \Delta U + W$,where $\Delta U$ is the change in internal energy.
For path $iaf$:
$Q_{iaf} = \Delta U_{iaf} + W_{iaf}$
$50 = \Delta U_{iaf} + 20$
$\Delta U_{iaf} = 50 - 20 = 30 \, cal$
Since internal energy is a state function,the change in internal energy depends only on the initial and final states,not on the path taken.
Therefore,$\Delta U_{ibf} = \Delta U_{iaf} = 30 \, cal$.
For path $ibf$:
$Q_{ibf} = \Delta U_{ibf} + W_{ibf}$
$36 = 30 + W_{ibf}$
$W_{ibf} = 36 - 30 = 6 \, cal$.

(D) When the dielectric slab is removed from the capacitor,the capacitance changes from $KC$ to $C$. Since the battery is disconnected,the charge $Q = KCV$ remains constant.
Work done by an external agent to remove the slab is $W_{ext} = U_{final} - U_{initial} = \frac{Q^2}{2C} - \frac{Q^2}{2KC} = \frac{Q^2}{2C}(1 - \frac{1}{K}) = \frac{K^2C^2V^2}{2C}(\frac{K-1}{K}) = \frac{1}{2}KC(K-1)V^2$.
When the slab is reinserted,the external agent does negative work of the same magnitude to restore the initial state.
Since the process is conservative and the system returns to its initial state,the net work done by the system (or external agent) over the complete cycle is zero.

(D) Let the roots of the equation $x^2 + ax + 1 = 0$ be $\alpha$ and $\beta$.
Given that the difference between the roots is $|\alpha - \beta| < \sqrt{5}$.
We know that $|\alpha - \beta| = \frac{\sqrt{D}}{|a_{coeff}|}$,where $D$ is the discriminant $b^2 - 4ac$.
Here,$D = a^2 - 4(1)(1) = a^2 - 4$.
Since the roots must be real for the difference to be defined in the real number system,we require $D \ge 0$,which means $a^2 \ge 4$,so $a \in (-\infty, -2] \cup [2, \infty)$.
The condition $|\alpha - \beta| < \sqrt{5}$ implies $\sqrt{a^2 - 4} < \sqrt{5}$.
Squaring both sides,we get $a^2 - 4 < 5$,which simplifies to $a^2 < 9$.
This implies $-3 < a < 3$,or $a \in (-3, 3)$.
Combining the conditions $a \in (-\infty, -2] \cup [2, \infty)$ and $a \in (-3, 3)$,we get $a \in (-3, -2] \cup [2, 3)$.

(C) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{1}{10}$.
From this,we find $\frac{T_2}{T_1} = 1 - \frac{1}{10} = \frac{9}{10}$.
When the same engine is used as a refrigerator,its coefficient of performance $\beta$ is given by $\beta = \frac{T_2}{T_1 - T_2}$.
Dividing numerator and denominator by $T_1$,we get $\beta = \frac{T_2/T_1}{1 - T_2/T_1}$.
Substituting the value $\frac{T_2}{T_1} = \frac{9}{10}$,we get $\beta = \frac{9/10}{1 - 9/10} = \frac{9/10}{1/10} = 9$.
The coefficient of performance is also defined as $\beta = \frac{Q_2}{W}$,where $Q_2$ is the heat absorbed from the cold reservoir and $W$ is the work done on the system.
Given $W = 10 \, J$,we have $Q_2 = \beta \times W = 9 \times 10 = 90 \, J$.

(B) Let the mass per unit area be $\sigma$.
The mass of the complete disc of radius $2R$ is $M_1 = \sigma \pi (2R)^2 = 4\pi \sigma R^2$. Its center of mass is at the origin $O$.
The mass of the removed disc of radius $R$ is $M_2 = \sigma \pi R^2$. Its center of mass is at a distance $R$ from the origin $O$ along the positive x-axis.
Using the concept of negative mass for the cavity,the center of mass of the remaining part is given by:
$x_{cm} = \frac{M_1 x_1 - M_2 x_2}{M_1 - M_2}$
$x_{cm} = \frac{(4\pi \sigma R^2)(0) - (\pi \sigma R^2)(R)}{4\pi \sigma R^2 - \pi \sigma R^2}$
$x_{cm} = \frac{-\pi \sigma R^3}{3\pi \sigma R^2} = -\frac{R}{3}$
The magnitude of the distance is $\frac{R}{3}$,so $\alpha R = \frac{R}{3}$,which gives $\alpha = \frac{1}{3}$.

(D) The relationship between Molarity $(M)$,percentage by mass $(\% \; w/w)$,and density $(d)$ is given by the formula:
$M = \frac{10 \times \% \; w/w \times d}{M_{solute}}$
Given values are:
$M = 3.60 \; mol \; L^{-1}$
$\% \; w/w = 29 \%$
$M_{solute} = 98 \; g \; mol^{-1}$
Substituting these values into the formula:
$3.60 = \frac{10 \times 29 \times d}{98}$
$d = \frac{3.60 \times 98}{10 \times 29}$
$d = \frac{352.8}{290} \approx 1.2165 \; g \; mL^{-1}$
Rounding to two decimal places,we get $d = 1.22 \; g \; mL^{-1}$.

(D) From the balanced chemical equation: $2Al_{(s)} + 6HCl_{(aq)} \rightarrow 2Al^{3+}_{(aq)} + 6Cl^{-}_{(aq)} + 3H_{2(g)}$
$6 \ moles$ of $HCl_{(aq)}$ produces $3 \ moles$ of $H_{2(g)}$.
Therefore,$1 \ mole$ of $HCl_{(aq)}$ produces $\frac{3}{6} = 0.5 \ moles$ of $H_{2(g)}$.
At $STP$,$1 \ mole$ of any gas occupies $22.4 \ L$.
Thus,$0.5 \ moles$ of $H_{2(g)}$ occupies $0.5 \times 22.4 \ L = 11.2 \ L$ at $STP$.
Hence,option $D$ is correct.

(C) According to the $(n+l)$ rule,the energy of an orbital increases as the sum of the principal quantum number $(n)$ and the azimuthal quantum number $(l)$ increases.
For option $A$: $n+l = 3+0 = 3$
For option $B$: $n+l = 3+1 = 4$
For option $C$: $n+l = 3+2 = 5$
For option $D$: $n+l = 4+0 = 4$
Since option $C$ has the highest $(n+l)$ value of $5$,it represents the highest energy state.

(B) To determine the magnetic behavior,we analyze the molecular orbital configuration of each species.
$1$. $NO$ has $15$ electrons: $\sigma_{1s}^2, \sigma_{1s}^{*2}, \sigma_{2s}^2, \sigma_{2s}^{*2}, \sigma_{2p_z}^2, \pi_{2p_x}^2 = \pi_{2p_y}^2, \pi_{2p_x}^{*1}$. It is paramagnetic due to the unpaired electron.
$2$. $O_2^{2-}$ has $18$ electrons. Its configuration is: $\sigma_{1s}^2, \sigma_{1s}^{*2}, \sigma_{2s}^2, \sigma_{2s}^{*2}, \sigma_{2p_z}^2, \pi_{2p_x}^2 = \pi_{2p_y}^2, \pi_{2p_x}^{*2} = \pi_{2p_y}^{*2}$. Since all electrons are paired,it is diamagnetic.
$3$. $O_2^{+}$ has $15$ electrons and is paramagnetic.
$4$. $O_2$ has $16$ electrons and is paramagnetic due to two unpaired electrons in the $\pi^*$ orbitals.

(D) The polarizing power of a cation is directly proportional to its charge density,which is defined by the ratio of charge to size.
As the charge on the cation increases and its ionic radius decreases,the polarizing power increases.
For the given ions,the ionic radii follow the order: $K^{+} > Ca^{2+} > Mg^{2+} > Be^{2+}$.
Since $Be^{2+}$ has the highest charge-to-size ratio and $K^{+}$ has the lowest,the increasing order of polarizing power is $K^{+} < Ca^{2+} < Mg^{2+} < Be^{2+}$.

(C) The bond order $(BO)$ is calculated as $\frac{1}{2}(N_b - N_a)$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
For $NO$ ($15$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. $BO = \frac{1}{2}(10 - 5) = 2.5$. It is paramagnetic.
For $NO^+$ ($14$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2$. $BO = \frac{1}{2}(10 - 4) = 3.0$. It is diamagnetic.
In the process $NO \rightarrow NO^+$,the electron is removed from the antibonding $\pi^*$ orbital,which increases the bond order from $2.5$ to $3.0$ and changes the magnetic behavior from paramagnetic to diamagnetic.

(C) The strength of a hydrogen bond depends on the electronegativity difference between the atoms involved in the bond.
Greater the electronegativity difference,the stronger the hydrogen bond.
Fluorine $(F)$ is the most electronegative element.
Therefore,the $F-H \cdots F$ hydrogen bond is the strongest among the given options.

(A) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G^{\circ}$ must be less than $0$.
Using the relation $\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}$,we set $\Delta G^{\circ} < 0$:
$\Delta H^{\circ} - T \Delta S^{\circ} < 0$
$\Delta H^{\circ} < T \Delta S^{\circ}$
$T > \frac{\Delta H^{\circ}}{\Delta S^{\circ}}$
Given $\Delta H^{\circ} = 179.1 \ kJ \ mol^{-1} = 179100 \ J \ mol^{-1}$ and $\Delta S^{\circ} = 160.2 \ J \ K^{-1} \ mol^{-1}$:
$T > \frac{179100}{160.2} \ K$
$T > 1117.97 \ K$
Rounding to the nearest whole number,$T > 1118 \ K$.

(D) The process is: $H_2O(l) \rightarrow H_2O(g)$.
The change in the number of moles of gas is $\Delta n_g = 1 - 0 = 1$.
Given: $\Delta H = 41\, kJ\, mol^{-1} = 41000\, J\, mol^{-1}$,$T = 373\, K$,and $R = 8.3\, J\, mol^{-1}\, K^{-1}$.
Using the relation $\Delta H = \Delta U + \Delta n_g RT$,we get $\Delta U = \Delta H - \Delta n_g RT$.
$\Delta U = 41000\, J\, mol^{-1} - (1 \times 8.3\, J\, mol^{-1}\, K^{-1} \times 373\, K)$.
$\Delta U = 41000 - 3095.9 = 37904.1\, J\, mol^{-1}$.
Converting to $kJ\, mol^{-1}$,we get $\Delta U = 37.9041\, kJ\, mol^{-1}$.

(B) The spontaneity of a process depends on the tendency to acquire a minimum energy state and maximum randomness (entropy).
For a spontaneous process in an isolated system,the total change in entropy $(\Delta S_{total})$ must be greater than zero,i.e.,$\Delta S > 0$.

(D) For a polyprotic acid $H_2A$,the dissociation steps are:
$H_2A \rightleftharpoons H^{+} + HA^{-}$; $K_1 = \frac{[H^{+}][HA^{-}]}{[H_2A]} = 1.0 \times 10^{-5}$
$HA^{-} \rightleftharpoons H^{+} + A^{2-}$; $K_2 = \frac{[H^{+}][A^{2-}]}{[HA^{-}]} = 5.0 \times 10^{-10}$
The overall dissociation reaction is the sum of these two steps:
$H_2A \rightleftharpoons 2H^{+} + A^{2-}$
The overall dissociation constant $K$ is the product of the individual dissociation constants:
$K = K_1 \times K_2$
$K = (1.0 \times 10^{-5}) \times (5.0 \times 10^{-10})$
$K = 5.0 \times 10^{-15}$

(D) For a buffer solution,the Henderson-Hasselbalch equation is given by: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$
Given that $50\%$ of the acid is ionized,the concentration of the salt (conjugate base) is equal to the concentration of the remaining undissociated acid,i.e.,$[Salt] = [Acid]$.
Substituting this into the equation: $pH = 4.5 + \log(1) = 4.5 + 0 = 4.5$.
We know that at $25^{\circ}C$,$pH + pOH = 14$.
Therefore,$pOH = 14 - pH = 14 - 4.5 = 9.5$.

(C) The equilibrium is $AgIO_{3(s)} \rightleftharpoons Ag^+_{(aq)} + IO^-_{3(aq)}$.
Let $S$ be the solubility in $mol/L$.
$K_{sp} = [Ag^+][IO_3^-] = S \times S = S^2$.
Given $K_{sp} = 1.0 \times 10^{-8}$,so $S = \sqrt{1.0 \times 10^{-8}} = 1.0 \times 10^{-4} \ mol/L$.
The molar mass of $AgIO_3$ is $283 \ g/mol$.
Mass in $1 \ L$ (or $1000 \ mL$) $= 1.0 \times 10^{-4} \ mol/L \times 283 \ g/mol = 2.83 \times 10^{-2} \ g/L$.
Mass in $100 \ mL = \frac{2.83 \times 10^{-2} \ g}{1000 \ mL} \times 100 \ mL = 2.83 \times 10^{-3} \ g$.

(A) $1$. Identify the longest carbon chain in the given structure. The longest chain contains $7$ carbon atoms,so the parent alkane is heptane.
$2$. Number the chain from the end that gives the lowest locants to the substituents. Numbering from right to left gives the substituents at positions $3$ and $4$.
$3$. At position $3$,there is an ethyl group $(-CH_2CH_3)$.
$4$. At position $4$,there are two methyl groups $(-CH_3)$.
$5$. Combining these,the $IUPAC$ name is $3-$ethyl$-4,4-$dimethylheptane.

(B) molecule is expected to rotate the plane of plane-polarised light if it is optically active.
Optically active molecules must be chiral,meaning they lack a plane of symmetry or center of inversion.
$1$. Glycine $(H_2N-CH_2-COOH)$ has a plane of symmetry and is achiral.
$2$. Glyceraldehyde $(HOCH_2-CH(OH)-CHO)$ has a chiral carbon atom bonded to four different groups $(-H, -OH, -CHO, -CH_2OH)$ and is optically active.
$3$. Butane-$2$-thiol $(CH_3-CH(SH)-CH_2-CH_3)$ has a chiral carbon atom bonded to four different groups $(-H, -SH, -CH_3, -CH_2CH_3)$ and is optically active.
$4$. $1,2$-Diphenylethane-$1,2$-diamine can exist as a meso compound (optically inactive) or as chiral enantiomers depending on the configuration.
In the context of standard chemistry questions,both Glyceraldehyde and Butane-$2$-thiol are classic examples of chiral molecules. However,Glyceraldehyde is the most fundamental example of a chiral molecule in carbohydrates. Given the options,both $B$ and $C$ are chiral. Assuming a single choice is required,Glyceraldehyde is the standard textbook example.

(B) molecule is chiral if it lacks a plane of symmetry or a center of inversion.
The $chair$ and $boat$ conformations of cyclohexane possess planes of symmetry,making them achiral.
The $twist-boat$ conformation of cyclohexane lacks a plane of symmetry,making it chiral.

(B) The reaction of propyne $(CH_3-C \equiv CH)$ with two equivalents of $HBr$ follows Markovnikov's rule.
In the first step,$H^+$ adds to the terminal carbon and $Br^-$ to the middle carbon to form $2-$bromopropene $(CH_3-C(Br)=CH_2)$.
In the second step,another $HBr$ adds following the same rule to form $2, 2-$dibromopropane $(CH_3-C(Br)_2-CH_3)$.
Reaction:
$CH_3-C \equiv CH + HBr$ $\rightarrow CH_3-C(Br)=CH_2$ $\xrightarrow{HBr} CH_3-C(Br)_2-CH_3$

(D) Let the mass of methane $(CH_4)$ and oxygen $(O_2)$ be $m \ g$.
Moles of $CH_4 = \frac{m}{16}$
Moles of $O_2 = \frac{m}{32}$
According to Dalton's Law of Partial Pressures,the partial pressure of a gas is equal to its mole fraction multiplied by the total pressure.
Mole fraction of $O_2 (x_{O_2}) = \frac{\text{Moles of } O_2}{\text{Moles of } O_2 + \text{Moles of } CH_4}$
$x_{O_2} = \frac{m/32}{m/32 + m/16} = \frac{m/32}{m/32 + 2m/32} = \frac{m/32}{3m/32} = 1/3$
Therefore,the fraction of the total pressure exerted by oxygen is $1/3$.

(A) The hydrolysis of ammonium sulphate,$(NH_4)_2SO_4$,in the soil produces sulphuric acid as follows:
$(NH_4)_2SO_4 + 2H_2O \longrightarrow H_2SO_4 + 2NH_4OH$
Since $H_2SO_4$ is a strong acid,its accumulation in the soil leads to an increase in soil acidity.

(A) The reluctance of valence shell $ns^2$ electrons to participate in bonding due to poor shielding of $d$ and $f$ orbitals is known as the inert pair effect.
As we move down group $14$,the stability of the $+2$ oxidation state increases relative to the $+4$ oxidation state.
Therefore,the stability of dihalides $(MX_2)$ increases in the order: $SiX_2 < GeX_2 < SnX_2 < PbX_2$.

(C) Let $C$ be the capacitance of the capacitor and $e$ be the electromotive force $(emf)$ of the battery.
When the capacitor is fully charged,the potential difference across the plates $V$ is equal to the $emf$ of the battery,i.e.,$V = e$.
The energy stored in the capacitor is given by $U = \frac{1}{2} C V^2 = \frac{1}{2} C e^2$.
The work done by the battery in charging the capacitor is given by $W = qV = (Ce)e = C e^2$.
The ratio of the energy stored in the capacitor to the work done by the battery is $\frac{U}{W} = \frac{\frac{1}{2} C e^2}{C e^2} = \frac{1}{2}$.

(D) When a charged particle enters a magnetic field perpendicular to its direction of motion,it experiences a magnetic Lorentz force $F = q(v \times B)$.
Since the force is always perpendicular to the velocity,it acts as a centripetal force,causing the particle to move in a circular path.
In uniform circular motion,the magnitude of velocity (speed) remains constant,so the kinetic energy $K = \frac{1}{2}mv^2$ remains constant.
However,the direction of the velocity vector changes continuously at every point along the circular path.
Since momentum $p = mv$ is a vector quantity,a change in the direction of velocity implies a change in the momentum vector.
Therefore,the momentum changes while the kinetic energy remains constant.

(C) According to the question,the half-life of $X$ is equal to the mean life of $Y$.
$T_{1/2}(X) = \tau_{av}(Y)$
Using the relations $T_{1/2} = \frac{0.693}{\lambda}$ and $\tau_{av} = \frac{1}{\lambda}$,we get:
$\frac{0.693}{\lambda_X} = \frac{1}{\lambda_Y}$
$\lambda_X = 0.693 \cdot \lambda_Y$
Since $0.693 < 1$,it follows that $\lambda_X < \lambda_Y$.
The rate of decay is given by $R = \lambda N$.
Since both elements have the same number of atoms $N$ initially,the element with the larger decay constant $\lambda$ will have a higher decay rate.
Therefore,$Y$ will decay faster than $X$.

(C) Step $1$: $CH_3CH_2OH + P + I_2 \rightarrow CH_3CH_2I$ $(A)$
Step $2$: $CH_3CH_2I + Mg \xrightarrow{\text{Ether}} CH_3CH_2MgI$ ($B$,Grignard reagent)
Step $3$: $CH_3CH_2MgI + HCHO \rightarrow CH_3CH_2CH_2OMgI$ $(C)$
Step $4$: $CH_3CH_2CH_2OMgI + H_2O \rightarrow CH_3CH_2CH_2OH + Mg(OH)I$ $(D)$
Compound $D$ is $n-$propyl alcohol $(CH_3CH_2CH_2OH)$.

(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{1}{10}$.
From this,we get $\frac{T_2}{T_1} = 1 - \frac{1}{10} = \frac{9}{10}$,which implies $\frac{T_1}{T_2} = \frac{10}{9}$.
The Coefficient of Performance $(COP)$ of a refrigerator is given by $COP = \frac{Q_2}{W} = \frac{T_2}{T_1 - T_2} = \frac{1}{\frac{T_1}{T_2} - 1}$.
Substituting the values,$COP = \frac{1}{\frac{10}{9} - 1} = \frac{1}{1/9} = 9$.
Since $COP = \frac{Q_2}{W}$,where $Q_2$ is the heat absorbed from the cold reservoir and $W$ is the work done,we have $Q_2 = COP \times W$.
Given $W = 10 \ J$,we get $Q_2 = 9 \times 10 \ J = 90 \ J$.

(B) The distance of point $A(\sqrt{2}, \sqrt{2})$ from the origin $(0, 0)$ is given by $r_A = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2 \text{ units}$.
The distance of point $B(2, 0)$ from the origin $(0, 0)$ is given by $r_B = \sqrt{2^2 + 0^2} = \sqrt{4} = 2 \text{ units}$.
The electric potential $V$ at a distance $r$ from a point charge $Q$ is given by $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$.
Since both points $A$ and $B$ are at the same distance $r = 2 \text{ units}$ from the charge $Q = 10^{-3} \ \mu C$,the potential at both points is equal:
$V_A = \frac{kQ}{r_A} = \frac{kQ}{2}$ and $V_B = \frac{kQ}{r_B} = \frac{kQ}{2}$.
Therefore,the potential difference between points $A$ and $B$ is $\Delta V = V_A - V_B = 0 \text{ V}$.

(D) The electric potential $V$ at the centre of the square is the algebraic sum of potentials due to individual charges: $V = k \sum \frac{q_i}{r}$. Since the distance $r$ from each vertex to the centre is the same,$V = \frac{k}{r} (q_A + q_B + q_C + q_D)$. Interchanging charges does not change the sum $(q_A + q_B + q_C + q_D)$,so $V$ remains unchanged.
The electric field $\overrightarrow{E}$ at the centre is a vector sum: $\overrightarrow{E} = \sum \overrightarrow{E_i}$. Initially,the field vectors from $A$ and $C$ (both $q$) and $B$ and $D$ (both $-q$) result in a net field. When charges are interchanged,the positions of the charges change relative to the centre,which alters the direction and magnitude of the individual field vectors $\overrightarrow{E_i}$. Thus,the resultant vector $\overrightarrow{E}$ changes.

(A) The change in sound level $\Delta L$ in decibels $(dB)$ is given by the formula: $\Delta L = 10 \log_{10} \left( \frac{I_1}{I_2} \right)$.
Given that the sound level is attenuated by $20\, dB$,we have $\Delta L = 20\, dB$.
Substituting the values into the formula: $20 = 10 \log_{10} \left( \frac{I_1}{I_2} \right)$.
Dividing by $10$,we get: $2 = \log_{10} \left( \frac{I_1}{I_2} \right)$.
Converting the logarithmic form to exponential form: $10^2 = \frac{I_1}{I_2}$.
Therefore,$\frac{I_1}{I_2} = 100$,which means the intensity decreases by a factor of $100$.

(A) Let $\sigma$ be the surface mass density of the discs.
Mass of the bigger disc $M = \sigma \pi (2R)^2 = 4 \sigma \pi R^2$.
Mass of the removed smaller disc $m = \sigma \pi R^2$.
Let the origin be at the centre of the bigger disc. The centre of the removed disc is at a distance $d = R$ from the origin.
The centre of mass of the remaining part is given by $X_{cm} = \frac{M(0) - m(d)}{M - m}$.
$X_{cm} = \frac{0 - (\sigma \pi R^2)(R)}{4 \sigma \pi R^2 - \sigma \pi R^2} = \frac{-\sigma \pi R^3}{3 \sigma \pi R^2} = -\frac{R}{3}$.
The magnitude of the distance is $\frac{R}{3}$,so $\alpha R = \frac{R}{3}$,which gives $\alpha = \frac{1}{3}$.

(C) The Henderson-Hasselbalch equation for a weak acid is given by: $pH = pK_a + \log \frac{[A^-]}{[HA]}$
Given that $50\%$ of the acid is ionized,the concentration of the conjugate base $[A^-]$ is equal to the concentration of the remaining unionized acid $[HA]$.
Thus,$\frac{[A^-]}{[HA]} = 1$.
Substituting the values: $pH = 4.5 + \log(1) = 4.5 + 0 = 4.5$.
We know that at $25^{\circ}C$,$pH + pOH = 14$.
Therefore,$pOH = 14 - pH = 14 - 4.5 = 9.5$.

(A) According to the first law of thermodynamics,the heat supplied to a system is equal to the sum of the change in internal energy and the work done by the system: $Q = \Delta U + W$.
For the path $iaf$:
$Q_{iaf} = 50 \, cal$
$W_{iaf} = 20 \, cal$
Using the first law: $50 = \Delta U + 20$
Therefore,the change in internal energy between states $i$ and $f$ is $\Delta U = 30 \, cal$.
Since internal energy is a state function,it depends only on the initial and final states and not on the path taken. Thus,for path $ibf$,the change in internal energy remains $\Delta U = 30 \, cal$.
For the path $ibf$:
$Q_{ibf} = 36 \, cal$
Using the first law: $Q_{ibf} = \Delta U + W_{ibf}$
$36 = 30 + W_{ibf}$
$W_{ibf} = 36 - 30 = 6 \, cal$.

(A) The sound level $L$ in decibels $(dB)$ is given by $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$,where $I$ is the intensity and $I_0$ is the reference intensity.
Let $L_1$ and $L_2$ be the initial and final sound levels,and $I_1$ and $I_2$ be the corresponding intensities.
The change in sound level is given by $\Delta L = L_1 - L_2 = 20\, dB$.
Using the formula: $\Delta L = 10 \log_{10} \left( \frac{I_1}{I_0} \right) - 10 \log_{10} \left( \frac{I_2}{I_0} \right) = 10 \log_{10} \left( \frac{I_1}{I_2} \right)$.
Substituting the given value: $20 = 10 \log_{10} \left( \frac{I_1}{I_2} \right)$.
Dividing by $10$: $2 = \log_{10} \left( \frac{I_1}{I_2} \right)$.
Taking the antilog: $\frac{I_1}{I_2} = 10^2 = 100$.
Therefore,the intensity decreases by a factor of $100$.

(A) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G^{\circ}$ must be less than $0$.
Given the equation: $\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}$.
At equilibrium,$\Delta G^{\circ} = 0$,so $T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}}$.
Given values: $\Delta H^{\circ} = 179.1 \ kJ \ mol^{-1} = 179100 \ J \ mol^{-1}$ and $\Delta S^{\circ} = 160.2 \ J \ K^{-1} \ mol^{-1}$.
Substituting these values into the equation:
$T = \frac{179100 \ J \ mol^{-1}}{160.2 \ J \ K^{-1} \ mol^{-1}} \approx 1117.97 \ K$.
Since the reaction is endothermic and entropy increases,the reaction becomes spontaneous at temperatures above $1118 \ K$.

(B) Let the centre of the bigger disc be the origin $O(0,0)$.
Let the mass of the original disc be $M$ and its radius be $2R$.
The mass of the removed smaller disc of radius $R$ is $m = \frac{\pi R^2}{\pi (2R)^2} M = \frac{M}{4}$.
The centre of the removed disc is at a distance $R$ from the origin along the $x$-axis,so its coordinates are $(R, 0)$.
The centre of mass of the remaining part $(x_{CM})$ is given by the formula:
$x_{CM} = \frac{M_1 x_1 - m x_2}{M_1 - m}$
Here,$M_1 = M$,$x_1 = 0$,$m = M/4$,and $x_2 = R$.
$x_{CM} = \frac{M(0) - (M/4)(R)}{M - M/4} = \frac{-MR/4}{3M/4} = -\frac{R}{3}$.
The distance from the centre is $|x_{CM}| = \frac{R}{3}$.
Comparing this with $\alpha R$,we get $\alpha = \frac{1}{3}$.

(B) Let the centre of the bigger disc of radius $2R$ be at the origin $(0,0)$.
Let $M$ be the mass of the bigger disc. The mass per unit area is $\sigma = \frac{M}{\pi(2R)^2} = \frac{M}{4\pi R^2}$.
The mass of the removed smaller disc of radius $R$ is $m = \sigma \cdot \pi R^2 = \frac{M}{4\pi R^2} \cdot \pi R^2 = \frac{M}{4}$.
The centre of mass of the bigger disc is at $x_1 = 0$. The centre of mass of the removed smaller disc is at $x_2 = R$ (since the circumferences touch).
The centre of mass of the remaining part is given by:
$x_{CM} = \frac{M x_1 - m x_2}{M - m}$
$x_{CM} = \frac{M(0) - (M/4)(R)}{M - M/4} = \frac{-MR/4}{3M/4} = -\frac{R}{3}$.
The distance from the centre of the bigger disc is $|x_{CM}| = \frac{R}{3}$.
Comparing this with $\alpha R$,we get $\alpha = \frac{1}{3}$.

(A) The nitro group $(-NO_2)$ is a strong electron-withdrawing group due to its $-I$ and $-M$ effects.
It withdraws electron density from the benzene ring,thereby reducing the electron density available for electrophilic attack.
Consequently,it deactivates the benzene ring towards electrophilic substitution reactions.

(D) When alkyl benzenes (like ethyl benzene) are treated with strong oxidizing agents such as alkaline $KMnO_4$,the alkyl side chain is oxidized to a carboxylic acid group $(-COOH)$ attached to the benzene ring,regardless of the length of the alkyl chain,provided that the benzylic carbon has at least one hydrogen atom.
In the case of ethyl benzene $(C_6H_5-CH_2-CH_3)$,the benzylic carbon has two hydrogen atoms,so it undergoes oxidation to form benzoic acid $(C_6H_5-COOH)$.

(D) The reaction of toluene with $Cl_2$ in the presence of a Lewis acid catalyst like $FeCl_3$ is an electrophilic aromatic substitution reaction (chlorination).
The methyl group $(-CH_3)$ attached to the benzene ring is an electron-donating group due to the inductive effect and hyperconjugation.
This group activates the benzene ring and directs the incoming electrophile $(Cl^+)$ to the ortho $(o-)$ and para $(p-)$ positions.
Therefore,the reaction yields a mixture of $o-$chlorotoluene and $p-$chlorotoluene.

(B) According to Raoult's law,the total vapour pressure of the solution is given by $P = P_A^\circ x_A + P_B^\circ x_B$.
Given:
$P = 290 \, mm$
$P_B^\circ = 200 \, mm$ (vapour pressure of propyl alcohol)
$x_A = 0.6$ (mole fraction of ethyl alcohol)
$x_B = 1 - x_A = 1 - 0.6 = 0.4$ (mole fraction of propyl alcohol)
Substituting the values in the equation:
$290 = P_A^\circ \times 0.6 + 200 \times 0.4$
$290 = P_A^\circ \times 0.6 + 80$
$290 - 80 = P_A^\circ \times 0.6$
$210 = P_A^\circ \times 0.6$
$P_A^\circ = \frac{210}{0.6} = 350 \, mm$.

(A) For isotonic solutions,the osmotic pressure $(\pi)$ is equal,which implies that their molar concentrations $(C)$ are equal at the same temperature.
Given that the solutions are $5.25\%$ and $1.5\%$ by mass,and assuming the density is $1.0\ g\ cm^{-3}$,the concentration in $g\ L^{-1}$ is the same as the percentage mass multiplied by $10$.
Concentration of substance $(C_1) = \frac{5.25 \times 10}{M} = \frac{52.5}{M} \ mol\ L^{-1}$.
Concentration of urea $(C_2) = \frac{1.5 \times 10}{60} = \frac{15}{60} = 0.25 \ mol\ L^{-1}$.
Since the solutions are isotonic,$C_1 = C_2$.
$\frac{52.5}{M} = 0.25$.
$M = \frac{52.5}{0.25} = 210 \ g\ mol^{-1}$.

(D) At $1\, atm$ atmospheric pressure,the boiling point of the mixture is $80\,^oC.$
At the boiling point,the total vapour pressure of the mixture,$P_{T} = 1\, atm = 760\, mm\,Hg.$
Using Raoult's law,$P_{T} = P_{A}^{\circ} X_{A} + P_{B}^{\circ} X_{B}.$
Given $P_{A}^{\circ} = 520\, mm\,Hg,$ $P_{B}^{\circ} = 1000\, mm\,Hg,$ and $X_{A} + X_{B} = 1,$
$760 = 520 X_{A} + 1000(1 - X_{A}).$
$760 = 520 X_{A} + 1000 - 1000 X_{A}.$
$480 X_{A} = 240.$
$X_{A} = \frac{240}{480} = 0.5.$
Therefore,the amount of $'A'$ in the mixture is $50\, mol\, percent.$

(B) According to Kohlrausch's law of independent migration of ions, the molar conductivity of a weak electrolyte at infinite dilution can be determined using the molar conductivities of strong electrolytes.
For acetic acid $(CH_3COOH)$, the expression is:
$\Lambda_{CH_3COOH}^o = \Lambda_{CH_3COONa}^o + \Lambda_{HCl}^o - \Lambda_{NaCl}^o$
Given the values for $\Lambda_{CH_3COONa}^o$ and $\Lambda_{HCl}^o$, we require the value of $\Lambda_{NaCl}^o$ to complete the calculation.
Thus, the correct option is $(B)$.

(D) When the cell is completely discharged,the cell potential $E_{cell} = 0 \ V.$
The Nernst equation for the cell reaction is given by:
$E_{cell} = E_{cell}^{\circ} - \frac{0.059}{n} \log \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right)$
Here,$n = 2$ (number of electrons transferred) and $E_{cell}^{\circ} = 1.10 \ V.$
Substituting the values:
$0 = 1.10 - \frac{0.059}{2} \log \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right)$
Rearranging the equation:
$\log \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right) = \frac{1.10 \times 2}{0.059} \approx 37.288 \approx 37.3$
Therefore,$\frac{[Zn^{2+}]}{[Cu^{2+}]} = 10^{37.3}$.

$(A)$ The enthalpy change of a reaction $(\Delta H)$ is defined as the difference between the activation energy of the forward reaction $(E_{af})$ and the activation energy of the reverse reaction $(E_{ab})$.
$\Delta H = E_{af} - E_{ab} = 180 \, kJ \, mol^{-1} - 200 \, kJ \, mol^{-1} = -20 \, kJ \, mol^{-1}$.
$A$ catalyst provides an alternative reaction pathway with lower activation energy, but it does not change the energy of the reactants or the products.
Since $\Delta H$ is a state function, it depends only on the initial and final states of the system.
Therefore, the enthalpy change of the reaction remains unchanged in the presence of a catalyst, which is $-20 \, kJ \, mol^{-1}$.

(B) For a reaction,the rate law can be expressed as $Rate = k[A]^x[B]^y$.
$1$. When the concentration of $B$ is doubled,the half-life does not change. This indicates that the reaction is of first order with respect to $B$ $(y=1)$,as the half-life of a first-order reaction is independent of the initial concentration of the reactant.
$2$. When the concentration of $A$ is doubled,the rate increases by two times. This indicates that the reaction is of first order with respect to $A$ $(x=1)$.
$3$. The overall order of the reaction is $n = x + y = 1 + 1 = 2$.
$4$. The unit of the rate constant for a reaction of order $n$ is $(mol \ L^{-1})^{1-n} \ s^{-1}$. For $n=2$,the unit is $(mol \ L^{-1})^{1-2} \ s^{-1} = (mol \ L^{-1})^{-1} \ s^{-1} = L \ mol^{-1} \ s^{-1}$.

(A) The radioactive decay follows first-order kinetics,where the activity $A$ at time $t$ is given by $A = A_0 e^{-\lambda t}$.
Given that the initial activity $A_0 = 10A$,where $A$ is the permissible safe activity level.
The decay constant $\lambda$ is related to the half-life $t_{1/2}$ by $\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{30 \text{ days}}$.
Using the first-order integrated rate equation: $t = \frac{2.303}{\lambda} \log_{10} \left( \frac{A_0}{A} \right)$.
Substituting the values: $t = \frac{2.303}{(0.693 / 30)} \log_{10} (10)$.
Since $\frac{2.303}{0.693} \approx \frac{1}{0.434} \approx 3.32$ (or more precisely $\frac{2.303}{0.693} \approx 3.32$ and $3.32 \times 30 \approx 100$),we get $t = 30 \times \frac{2.303}{0.693} \times 1 = 30 \times 3.325 \approx 100 \text{ days}$.

(B) Isotopes are atoms of the same element having the same atomic number but different atomic masses.
Neutron has an atomic number of $0$ and an atomic mass of $1$.
Therefore,the loss or gain of a neutron changes the mass number while keeping the atomic number constant,thus generating an isotope.
For example: $_{92}U^{238} + _{0}n^{1} \rightarrow _{92}U^{239}$

(D) Let us analyze each reaction:
$1$. $3Br_2 6NaOH (\text{hot and conc.}) \rightarrow 5NaBr NaBrO_3 3H_2O$. Thus,option $A$ is incorrect as it does not mention $NaBrO_3$.
$2$. $SO_2 O_3 \rightarrow SO_3 O_2$. This is correct.
$3$. $Si 2NaOH H_2O \rightarrow Na_2SiO_3 2H_2$. This is correct.
$4$. With excess $NH_3$,$Cl_2$ reacts as: $3Cl_2 8NH_3 \rightarrow N_2 6NH_4Cl$. The statement in option $D$ claims $HCl$ is formed,but $HCl$ reacts with excess $NH_3$ to form $NH_4Cl$. Therefore,both $A$ and $D$ are technically incorrect based on standard chemical descriptions,but $D$ is the most commonly cited incorrect statement in this context due to the product being $NH_4Cl$ instead of $HCl$.

(A) The shielding effect of $f-$orbitals is poor. Specifically,the $4f$ orbitals are closer to the nucleus than the $5f$ orbitals,meaning the shielding provided by $4f$ electrons is greater than that provided by $5f$ electrons. Therefore,the statement that $4f$ and $5f$ orbitals are equally shielded is incorrect.

(A) In $[PtCl_4]^{2-}$,the central metal ion is $Pt^{2 }$ ($5d^8$ configuration).
$Pt$ is a $5d$ series element,and for $5d$ elements,the crystal field splitting energy $(\Delta_o)$ is very high,even with weak ligands like $Cl^-$.
This high splitting energy forces the pairing of electrons in the $5d$ orbitals,resulting in $dsp^2$ hybridization.
Therefore,$[PtCl_4]^{2-}$ exhibits a square planar geometry.
In contrast,$[NiCl_4]^{2-}$,$[CoCl_4]^{2-}$,and $[FeCl_4]^{2-}$ involve $3d$ metals with weak ligands,leading to $sp^3$ hybridization and tetrahedral geometry.

(A) The $5f$ orbitals extend further from the nucleus compared to the $4f$ orbitals.
Due to this,the $5f$ electrons are less tightly held by the nucleus and are more available for bonding.
Consequently,the actinoids exhibit a greater variety of oxidation states compared to the lanthanoids.

(C) In the $S_{N^{2}}$ mechanism,the transition state is pentavalent.
Steric hindrance plays a crucial role in $S_{N^{2}}$ reactions.
Bulky alkyl groups hinder the approach of the nucleophile,while smaller alkyl groups facilitate it.
Therefore,the reactivity order for $S_{N^{2}}$ is primary $(RCH_2X)$ > secondary $(R_2CHX)$ > tertiary $(R_3CX)$.
The correct order is $RCH_2X > R_2CHX > R_3CX$.

(D) The reaction sequence is as follows:
$CH_3CH_2OH \xrightarrow{P + I_2} CH_3CH_2I (A)$
$CH_3CH_2I \xrightarrow{Mg, \text{ether}} CH_3CH_2MgI (B)$
$CH_3CH_2MgI + HCHO \rightarrow CH_3CH_2CH_2OMgI (C)$
$CH_3CH_2CH_2OMgI \xrightarrow{H_2O} CH_3CH_2CH_2OH (D) + Mg(OH)I$
Compound $D$ is $n-$propyl alcohol (propan$-1-$ol).

(D) Aromatic amines are less basic than aliphatic amines.
Among aliphatic amines,the order of basicity in aqueous solution is $2^{o} > 1^{o} > 3^{o}$ due to a combination of inductive effect,solvation effect,and steric hindrance.
In $3^{o}$ amines (Trimethylamine),the steric hindrance of three alkyl groups makes the approach and bonding of a proton relatively difficult.
Aniline is much less basic than aliphatic amines because the lone pair of electrons on the nitrogen atom is involved in resonance with the benzene ring.
Therefore,Dimethylamine ($2^{o}$ aliphatic amine) is the strongest base among the given choices.
$\therefore$ The correct order of basic strength is $\text{Dimethylamine} > \text{Methylamine} > \text{Trimethylamine} > \text{Aniline}$.

(A) The given reaction is the carbylamine reaction,which is a characteristic test for primary amines.
In this reaction,a primary amine reacts with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ to form an isocyanide (carbylamine) and potassium chloride $(KCl)$.
The balanced chemical equation is:
$CH_3CH_2NH_2 + CHCl_3 + 3KOH \rightarrow C_2H_5NC + 3KCl + 3H_2O$
Here,$(A)$ is $C_2H_5NC$ (ethyl isocyanide) and $(B)$ is $3KCl$.

(A) The primary structure of a protein refers to the specific sequence of $\alpha -$ amino acids in the polypeptide chain.
Secondary structure refers to the regular folding patterns of the polypeptide backbone,such as the $\alpha -$ helix and $\beta -$ pleated sheet structures.
These structures are stabilized by hydrogen bonding between the carbonyl oxygen $(C=O)$ and the amide hydrogen $(N-H)$ of the peptide backbone.