AIEEE 2004 Chemistry Question Paper with Answer and Solution

(A) The gravitational potential energy $U$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GMm}{r}$.
At the surface of the Earth,$r = R$,so $U_{surface} = -\frac{GMm}{R}$.
At a height $h = R$ above the surface,the distance from the center is $r = R + h = R + R = 2R$.
Thus,the potential energy at height $h$ is $U_{height} = -\frac{GMm}{2R}$.
The gain in potential energy is $\Delta U = U_{height} - U_{surface} = -\frac{GMm}{2R} - (-\frac{GMm}{R}) = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R}$.
Since the acceleration due to gravity $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting this into the expression for $\Delta U$,we get $\Delta U = \frac{(gR^2)m}{2R} = \frac{1}{2}mgR$.

(B) The gravitational potential energy $U$ at a distance $r$ from the center of the earth is given by $U = -\frac{GMm}{r}$.
At the surface of the earth,$r = R$,so $U_i = -\frac{GMm}{R}$.
At a height $h = R$ above the surface,the distance from the center is $r = R + h = R + R = 2R$.
Thus,the potential energy at height $h$ is $U_f = -\frac{GMm}{2R}$.
The gain in potential energy is $\Delta U = U_f - U_i = -\frac{GMm}{2R} - (-\frac{GMm}{R}) = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R}$.
Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.
Substituting this into the expression for $\Delta U$,we get $\Delta U = \frac{(gR^2)m}{2R} = \frac{1}{2}mgR$.

(B) For the system to be in equilibrium,the net force on every charge must be zero.
Consider a charge $-Q$ at corner $B$ of the square with side length $a$.
The forces acting on this charge due to the other three charges at the corners are:
$1$. Force due to charge at $A$: $F_A = \frac{kQ^2}{a^2}$ (along $AB$)
$2$. Force due to charge at $C$: $F_C = \frac{kQ^2}{a^2}$ (along $CB$)
$3$. Force due to charge at $D$: $F_D = \frac{kQ^2}{(a\sqrt{2})^2} = \frac{kQ^2}{2a^2}$ (along $DB$)
The resultant force of $F_A$ and $F_C$ is $F_{AC} = \sqrt{F_A^2 + F_C^2} = \sqrt{2} \frac{kQ^2}{a^2}$.
The total force away from the center is $F_{net} = F_{AC} + F_D = \sqrt{2} \frac{kQ^2}{a^2} + \frac{kQ^2}{2a^2} = \frac{kQ^2}{a^2} (\sqrt{2} + 0.5)$.
For equilibrium,the force due to the central charge $q$ must balance this force:
$F_O = \frac{k |Q| |q|}{(a/\sqrt{2})^2} = \frac{2kQq}{a^2}$.
Equating the forces: $\frac{2kQq}{a^2} = \frac{kQ^2}{a^2} (\sqrt{2} + 0.5)$.
$2q = Q (\frac{2\sqrt{2} + 1}{2})$.
$q = \frac{Q}{4} (1 + 2\sqrt{2})$.
Since the corner charges are negative,the central charge $q$ must be positive to provide an attractive force.

(B) Let the length of the wire be $L$ and the current be $i$.
For a single turn loop of radius $r$,the circumference is $2\pi r = L$,so $r = \frac{L}{2\pi}$.
The magnetic field at the center is $B = \frac{\mu_0 i}{2r} = \frac{\mu_0 i}{2(L/2\pi)} = \frac{\mu_0 i \pi}{L}$.
When the wire is bent into $n$ turns,the new radius $r'$ satisfies $n(2\pi r') = L$,so $r' = \frac{L}{2\pi n} = \frac{r}{n}$.
The magnetic field at the center for $n$ turns is $B_n = n \times \frac{\mu_0 i}{2r'} = n \times \frac{\mu_0 i}{2(r/n)} = n^2 \times \frac{\mu_0 i}{2r}$.
Since $B = \frac{\mu_0 i}{2r}$,we have $B_n = n^2 B$.

(C) The Rydberg formula for the hydrogen atom is given by: $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$
Here,$n_1 = 1$ (stationary state) and $n_2 = \infty$ (infinity).
Substituting the values: $\frac{1}{\lambda} = 1.097 \times 10^7 \ m^{-1} \left[ \frac{1}{1^2} - \frac{1}{\infty^2} \right]$
Since $\frac{1}{\infty} = 0$,we have: $\frac{1}{\lambda} = 1.097 \times 10^7 \ m^{-1} \times 1$
$\lambda = \frac{1}{1.097 \times 10^7} \ m \approx 9.115 \times 10^{-8} \ m$
Converting to nanometers $(nm)$: $\lambda = 9.115 \times 10^{-8} \times 10^9 \ nm = 91.15 \ nm \approx 91 \ nm$.

(A) For a $4f$ orbital,the principal quantum number $n = 4$.
The azimuthal quantum number $l$ for an $f$ orbital is $3$ (since $l = 0, 1, 2, 3$ correspond to $s, p, d, f$ orbitals respectively).
The magnetic quantum number $m$ can take values from $-l$ to $+l$,i.e.,$m = \{-3, -2, -1, 0, +1, +2, +3\}$.
The spin quantum number $s$ can be either $+\frac{1}{2}$ or $-\frac{1}{2}$.
Comparing these with the given options,option $A$ $(n = 4, l = 3, m = +1, s = +\frac{1}{2})$ satisfies all conditions.

(B) The electronic configuration of $Cr$ $(Z = 24)$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^5, 4s^1$.
For $l = 1$ (which corresponds to $p$-orbitals),the electrons are in $2p^6$ and $3p^6$. Total electrons $= 6 + 6 = 12$.
For $l = 2$ (which corresponds to $d$-orbitals),the electrons are in $3d^5$. Total electrons $= 5$.
Therefore,the numbers of electrons with $l = 1$ and $l = 2$ are $12$ and $5$ respectively.

(C) To determine the bond angles,we analyze the hybridization and molecular geometry of each molecule:
$1$. $H_2S$: The central atom $S$ has two bond pairs and two lone pairs. Due to the presence of lone pairs,the bond angle is significantly reduced from the tetrahedral angle,approximately $92.6^o$.
$2$. $NH_3$: The central atom $N$ has three bond pairs and one lone pair. The lone pair-bond pair repulsion reduces the bond angle to approximately $107^o$.
$3$. $SiH_4$: The central atom $Si$ is $sp^3$ hybridized with four bond pairs and no lone pairs,resulting in a perfect tetrahedral geometry with a bond angle of $109^o 28'$.
$4$. $BF_3$: The central atom $B$ is $sp^2$ hybridized with three bond pairs and no lone pairs,resulting in a trigonal planar geometry with a bond angle of $120^o$.
Thus,the correct order of bond angles (smallest first) is $H_2S < NH_3 < SiH_4 < BF_3$ $(92.6^o < 107^o < 109^o 28' < 120^o)$.

(A) To determine the structure,we calculate the hybridization and number of lone pairs for each species:
$1$. $BF_4^-$: Boron has $3$ valence electrons. It forms $4$ bonds with $F$ atoms and gains $1$ electron from the negative charge. Total electron pairs = $(3+4+1)/2 = 4$. Hybridization is $sp^3$,which corresponds to a regular tetrahedral geometry.
$2$. $SF_4$: Sulfur has $6$ valence electrons. It forms $4$ bonds with $F$ atoms,leaving $1$ lone pair. Total electron pairs = $5$ ($sp^3d$ hybridization). Due to the lone pair,it has a see-saw shape.
$3$. $XeF_4$: Xenon has $8$ valence electrons. It forms $4$ bonds with $F$ atoms,leaving $2$ lone pairs. Total electron pairs = $6$ ($sp^3d^2$ hybridization). Due to the two lone pairs,it has a square planar geometry.
$4$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ has a $d^8$ configuration. $CN^-$ is a strong field ligand,causing pairing of electrons. Hybridization is $dsp^2$,resulting in a square planar geometry.
Thus,only $BF_4^-$ has a regular tetrahedral structure.

(A) In boric acid $(H_3BO_3)$,the boron atom is bonded to three oxygen atoms via single bonds. Boron has $3$ valence electrons,all of which are involved in bonding,resulting in a trigonal planar geometry with $sp^2$ hybridization.
Each oxygen atom is bonded to one boron atom and one hydrogen atom,and it also possesses two lone pairs of electrons. Thus,each oxygen atom has $4$ electron domains (two bond pairs and two lone pairs),resulting in $sp^3$ hybridization.

(D) The number of $90^{\circ}$ bond pair-bond pair angles for different geometries are as follows:
$1$. $dsp^2$ hybridization (Square planar): There are $4$ angles of $90^{\circ}$ between adjacent bond pairs.
$2$. $sp^3d$ hybridization (Trigonal bipyramidal): There are $6$ angles of $90^{\circ}$ (three between equatorial and axial bonds,and three more between the other equatorial and axial bonds).
$3$. $sp^3d^2$ hybridization (Octahedral): There are $12$ angles of $90^{\circ}$ between adjacent bond pairs.
Comparing these,$sp^3d^2$ hybridization has the maximum number of $90^{\circ}$ angles. Therefore,the correct option is $D$.

(B) The bond length is inversely proportional to the bond order.
Since the bond order of $NO^{+}$ $(3)$ is greater than the bond order of $NO$ $(2.5)$,the bond length of $NO^{+}$ will be shorter than that of $NO$.
Therefore,the bond length in $NO$ is greater than in $NO^{+}$.

(A) The correct option is $A$. In $Vander$ $Waal$'s equation,$(P + \frac{an^2}{V^2})(V - nb) = nRT$,the constant '$b$' is known as the co-volume or excluded volume. It represents the effective volume occupied by the gas molecules themselves.

(A) The average kinetic energy $(K.E.)$ of an ideal gas is directly proportional to its absolute temperature $(T)$ in Kelvin: $K.E. \propto T$.
Given temperatures:
$T_1 = 20 + 273 = 293 \ K$
$T_2 = 40 + 273 = 313 \ K$
The factor by which the kinetic energy changes is the ratio of the final kinetic energy to the initial kinetic energy:
$\frac{K.E._2}{K.E._1} = \frac{T_2}{T_1} = \frac{313}{293}$.

(B) For the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$,the equilibrium constant is $K_c = \frac{[NO]^2}{[N_2][O_2]} = 4 \times 10^{-4}$.
For the reaction $NO_{(g)} \rightleftharpoons \frac{1}{2}N_{2(g)} + \frac{1}{2}O_{2(g)}$,the equilibrium constant $K'_c$ is given by $K'_c = \frac{[N_2]^{1/2}[O_2]^{1/2}}{[NO]}$.
Comparing the two expressions,we see that $K'_c = \frac{1}{\sqrt{K_c}}$.
Substituting the value of $K_c$,we get $K'_c = \frac{1}{\sqrt{4 \times 10^{-4}}} = \frac{1}{2 \times 10^{-2}} = \frac{100}{2} = 50$.

(A) For the reaction $P_{4(s)} + 5O_{2(g)} \rightleftharpoons P_4O_{10(s)}$,the equilibrium constant expression is given by the ratio of the product of concentrations of products to the reactants,each raised to the power of their stoichiometric coefficients.
$K_c = \frac{[P_4O_{10(s)}]}{[P_{4(s)}][O_{2(g)}]^5}$
Since $P_{4(s)}$ and $P_4O_{10(s)}$ are pure solids,their concentrations are taken as unity $(1)$.
Therefore,$K_c = \frac{1}{[O_2]^5}$.

(C) The relationship between $K_p$ and $K_c$ is given by the equation: $K_p = K_c(RT)^{\Delta n}$.
For the reaction $CO_{(g)} + Cl_{2(g)} \rightleftharpoons COCl_{2(g)}$,the change in the number of moles of gaseous products and reactants is $\Delta n = n_p - n_r = 1 - (1 + 1) = 1 - 2 = -1$.
Substituting $\Delta n = -1$ into the equation,we get: $K_p = K_c(RT)^{-1}$.
Therefore,the ratio $\frac{K_p}{K_c} = (RT)^{-1} = \frac{1}{RT}$.

(D) . The conjugate base of an acid is formed by the removal of one proton $(H^+)$ from the acid molecule.
For the acid $H_2PO_4^-$,the reaction is:
$H_2PO_4^- \to H^+ + HPO_4^{2-}$
Thus,the conjugate base of $H_2PO_4^-$ is $HPO_4^{2-}$.

(D) For a sparingly soluble salt $MX_4$,the dissociation equilibrium is:
$MX_4(s) \rightleftharpoons M^{4+}(aq) + 4X^-(aq)$
Let the molar solubility be $s$.
At equilibrium,$[M^{4+}] = s$ and $[X^-] = 4s$.
The solubility product $K_{sp}$ is defined as:
$K_{sp} = [M^{4+}][X^-]^4$
$K_{sp} = (s)(4s)^4$
$K_{sp} = s \times 256s^4 = 256s^5$
Solving for $s$:
$s^5 = K_{sp} / 256$
$s = (K_{sp} / 256)^{1/5}$
Therefore,the correct option is $D$.

(C) The work done during expansion against a constant external pressure is given by the formula: $W = -P_{ext} \Delta V$.
Here,$P_{ext} = 1 \times 10^5 \ N \ m^{-2}$,$V_1 = 1 \times 10^{-3} \ m^3$,and $V_2 = 1 \times 10^{-2} \ m^3$.
$\Delta V = V_2 - V_1 = (1 \times 10^{-2} - 1 \times 10^{-3}) \ m^3 = (10 \times 10^{-3} - 1 \times 10^{-3}) \ m^3 = 9 \times 10^{-3} \ m^3$.
Substituting the values: $W = -(1 \times 10^5 \ N \ m^{-2}) \times (9 \times 10^{-3} \ m^3) = -900 \ J$.
Thus,the work done is $-900 \ J$.

(D) $C_{(s)} + O_{2(g)} \to CO_{2(g)}$; $\Delta H = -393.5 \, kJ \, mol^{-1}$ ..... $(I)$
$CO_{(g)} + 1/2 O_{2(g)} \to CO_{2(g)}$; $\Delta H = -283 \, kJ \, mol^{-1}$ ..... $(II)$
Subtracting equation $(II)$ from equation $(I)$:
$(C_{(s)} + O_{2(g)}) - (CO_{(g)} + 1/2 O_{2(g)}) = CO_{2(g)} - CO_{2(g)}$
$C_{(s)} + 1/2 O_{2(g)} \to CO_{(g)}$
$\Delta H_f = (-393.5) - (-283) = -110.5 \, kJ \, mol^{-1}$

(A) The ionic radius is inversely proportional to the $Z/e$ ratio (where $Z$ is the nuclear charge and $e$ is the number of electrons).
For the given ions:
$O^{2-}$: $Z=8, e=10$,$Z/e = 0.8$
$F^{-}$: $Z=9, e=10$,$Z/e = 0.9$
$Li^{+}$: $Z=3, e=2$,$Z/e = 1.5$
$B^{3+}$: $Z=5, e=2$,$Z/e = 2.5$
Since $O^{2-}$ has the lowest $Z/e$ ratio,it has the largest ionic radius.

(A) The formation of $O_{(g)}^{2-}$ involves the addition of an electron to a negatively charged $O_{(g)}^{-}$ ion.
Since both the incoming electron and the $O_{(g)}^{-}$ ion are negatively charged,there is a strong electrostatic repulsion between them.
To overcome this repulsion and force the electron into the $O_{(g)}^{-}$ ion,energy must be supplied to the system,making the process endothermic.
Therefore,the $O^{-}$ ion resists the addition of another electron.

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

(D) The acidic strength of oxides increases as we move from left to right across a period in the periodic table due to the increase in electronegativity and decrease in metallic character.
$Al_2O_3$ is amphoteric.
$SiO_2$ is weakly acidic.
$P_2O_3$ is acidic (anhydride of $H_3PO_3$).
$SO_2$ is more acidic (anhydride of $H_2SO_3$).
Therefore,the correct order of acidic strength is $Al_2O_3 < SiO_2 < P_2O_3 < SO_2$.

(C) Beryllium $(Be)$ and aluminium $(Al)$ show a diagonal relationship due to similar ionic potential.
However,they differ in their maximum covalency.
Beryllium has a maximum covalency of $4$ because it has only $2s$ and $2p$ orbitals available for bonding.
Aluminium has a maximum covalency of $6$ because it has vacant $3d$ orbitals available for bonding.

(A) Aluminium chloride $(Al_2Cl_6)$ is a covalent dimer in non-polar solvents.
When it dissolves in water,it undergoes hydration to form the stable octahedral complex ion,$[Al(H_2O)_6]^{3+}$,and releases chloride ions.
The reaction is: $Al_2Cl_6 + 12H_2O \rightarrow 2[Al(H_2O)_6]^{3+} + 6Cl^-$.
Therefore,the correct species formed are $[Al(H_2O)_6]^{3+}$ and $Cl^-$ ions.

(A) Smog is a type of air pollution. Photochemical smog is primarily caused by the presence of oxides of nitrogen $(NO_x)$ and volatile organic compounds $(VOCs)$ in the atmosphere,which react in the presence of sunlight. Classical smog (London smog) is caused by a mixture of smoke,fog,and sulphur dioxide $(SO_2)$. Therefore,the presence of oxides of sulphur and nitrogen is the primary cause of smog.

(C) The correct answer is $(C)$.
Helium is actually twice as heavy as hydrogen,not lighter. While it is non-inflammable,the statement that it is lighter than hydrogen is scientifically incorrect.
Helium has the lowest melting and boiling points of any element,making liquid helium an ideal coolant for superconducting magnets and cryogenic research.
It is also used in gas-cooled nuclear reactors as a heat transfer medium.

(A) The balanced chemical equation for the reaction of magnesium nitride with water is:
$Mg_3N_2 + 6H_2O \to 3Mg(OH)_2 + 2NH_3$
From the stoichiometry of the reaction,$1 \text{ mole}$ of $Mg_3N_2$ reacts with $6 \text{ moles}$ of $H_2O$ to produce $3 \text{ moles}$ of $Mg(OH)_2$ and $2 \text{ moles}$ of $NH_3$ (ammonia).
Therefore,the correct option is $A$.

(C) In the Lassaigne test for nitrogen,the organic compound is fused with sodium metal to form sodium cyanide $(NaCN)$.
This $NaCN$ is then treated with ferrous sulfate $(FeSO_4)$ and ferric chloride $(FeCl_3)$.
The final product formed is ferric ferrocyanide,which is $Fe_4[Fe(CN)_6]_3$,also known as Prussian blue.

(A) compound is chiral if it contains at least one chiral carbon atom (a carbon atom bonded to four different groups).
$1-$chloropentane: $CH_2(Cl)-CH_2-CH_2-CH_2-CH_3$. The carbon atom bonded to chlorine is attached to two identical hydrogen atoms,so it is achiral.
$2-$chloropentane: $CH_3-CH(Cl)-CH_2-CH_2-CH_3$. The $C-2$ atom is bonded to $-H, -Cl, -CH_3, \text{ and } -CH_2CH_2CH_3$. Since all four groups are different,it is chiral.
$1-$chloro$-2-$methylpentane: $CH_2(Cl)-CH(CH_3)-CH_2-CH_2-CH_3$. The $C-2$ atom is bonded to $-H, -CH_3, -CH_2Cl, \text{ and } -CH_2CH_2CH_3$. Since all four groups are different,it is chiral.
$3-$chloro$-2-$methylpentane: $CH_3-CH(CH_3)-CH(Cl)-CH_2-CH_3$. The $C-3$ atom is bonded to $-H, -Cl, -CH(CH_3)_2, \text{ and } -CH_2CH_3$. Since all four groups are different,it is chiral.
Therefore,$1-$chloropentane is the only achiral compound.

(B) In the given compound,the hydroxyl group $(-OH)$ is the principal functional group and has higher priority than the methyl groups $(-CH_3)$.
Therefore,the carbon atom attached to the $-OH$ group is assigned number $1$.
Numbering the cyclohexane ring to give the substituents (methyl groups) the lowest possible locants results in them being at position $3$.
Thus,the $IUPAC$ name is $3, 3-$dimethylcyclohexan$-1-$ol.

(A) The hybridization of carbon atoms in the given compounds is as follows:
$1$. $\text{Acetonitrile} (CH_3CN)$: The structure is $CH_3-C \equiv N$. The methyl carbon is $sp^3$ and the nitrile carbon is $sp$ hybridized.
$2$. $\text{Acetic acid} (CH_3COOH)$: The structure is $CH_3-C(=O)OH$. The carbonyl carbon is $sp^2$ hybridized.
$3$. $\text{Acetone} (CH_3COCH_3)$: The structure is $CH_3-C(=O)-CH_3$. The carbonyl carbon is $sp^2$ hybridized.
$4$. $\text{Acetamide} (CH_3CONH_2)$: The structure is $CH_3-C(=O)NH_2$. The carbonyl carbon is $sp^2$ hybridized.
Therefore,$\text{Acetonitrile}$ does not contain any $sp^2$ hybridized carbon.

(B) meso compound is an optically inactive molecule that contains chiral centers but also possesses an internal plane of symmetry or a center of inversion.
$2, 3-$Dichlorobutane $(CH_3-CHCl-CHCl-CH_3)$ has two identical chiral centers at $C2$ and $C3$.
In its eclipsed conformation,it possesses a plane of symmetry,making it a meso isomer.
Therefore,the correct option is $(B)$.

(D) The structural isomers $C_2H_5OH$ (ethanol) and $CH_3OCH_3$ (dimethyl ether) have the same molecular formula,$C_2H_6O$,and therefore the same molar mass $(M = 46.07 \ g/mol)$.
According to the ideal gas law,$PV = nRT = (m/M)RT$,which can be rearranged to $d = (PM)/(RT)$,where $d$ is the density.
Since both compounds have the same molar mass $(M)$,they will have the same gaseous density $(d)$ at the same temperature $(T)$ and pressure $(P)$.
Properties like boiling point,vapour pressure,and heat of vaporization depend on intermolecular forces.
Ethanol exhibits strong hydrogen bonding,whereas dimethyl ether does not,leading to significant differences in these physical properties.

(A) The Anti-Markownikoff's rule (also known as the peroxide effect or Kharasch effect) is applicable only to unsymmetrical alkenes.
$2-$butene is a symmetrical alkene because the double bond is located between the $2^{nd}$ and $3^{rd}$ carbon atoms,resulting in identical groups on both sides of the double bond $(CH_3-CH=CH-CH_3)$.
Therefore,$2-$butene does not show Anti-Markownikoff's addition.
$1-$butene,$2-$pentene,and $2-$hexene are unsymmetrical alkenes and can follow the rule.

(B) The number of moles of urea is calculated as: $n = \frac{6.02 \times 10^{20}}{6.02 \times 10^{23}} = 10^{-3} \ mol$.
The volume of the solution is $100 \ mL = 0.1 \ L$.
The molarity $(M)$ is defined as the number of moles of solute per liter of solution: $M = \frac{n}{V(L)} = \frac{10^{-3} \ mol}{0.1 \ L} = 0.01 \ M$.

(A) $H_3PO_3$ is a dibasic acid,meaning it provides $2$ moles of $H^+$ ions per mole of acid.
Using the principle of equivalence: $n_{acid} \times \text{basicity} = n_{base} \times \text{acidity}$.
$M_1 \times V_1 \times \text{basicity} = M_2 \times V_2 \times \text{acidity}$.
$0.1 \times 20 \times 2 = 0.1 \times V_2 \times 1$.
$4 = 0.1 \times V_2$.
$V_2 = \frac{4}{0.1} = 40 \ mL$.

(C) The correct answer is $(C)$. Chlorophyll is a green pigment in plants and contains magnesium $(Mg^{2+})$ instead of calcium.

(C) The map distance between two genes on a chromosome is measured in centimorgans $(cM)$ or map units. This distance is directly proportional to the frequency of recombination between the genes. The percentage of crossing over is equivalent to the recombination frequency,which is calculated as the number of recombinant offspring divided by the total number of offspring,multiplied by $100$. Therefore,to determine the map distance,one must know the cross over percentage.

(D) The conjugate base of an acid is formed by the removal of one proton $(H^+)$ from the acid molecule.
For the species $H_2PO_4^-$,the removal of one $H^+$ ion results in the formation of $HPO_4^{2-}$.
The reaction is: $H_2PO_4^- \rightarrow H^+ + HPO_4^{2-}$.

(A) The total number of observations is $2n$.
There are $n$ observations equal to $a$ and $n$ observations equal to $-a$.
The mean $\bar{x} = \frac{n(a) + n(-a)}{2n} = \frac{0}{2n} = 0$.
The standard deviation $\sigma$ is given by $\sigma = \sqrt{\frac{\sum (x_i - \bar{x})^2}{2n}}$.
Since $\bar{x} = 0$,$\sigma = \sqrt{\frac{\sum x_i^2}{2n}}$.
Here,$\sum x_i^2 = n(a^2) + n(-a)^2 = 2na^2$.
Substituting the values,$2 = \sqrt{\frac{2na^2}{2n}} = \sqrt{a^2} = |a|$.
Therefore,$|a| = 2$.

(C) The van der Waals equation is $(P + \frac{an^2}{V^2})(V - nb) = nRT$.
In this equation,the constant $a$ represents the magnitude of intermolecular forces of attraction,while the constant $b$ represents the excluded volume or the volume occupied by the gas molecules themselves.

(C) To determine the bond angles,we analyze the hybridization and the number of lone pairs on the central atom:

Compound	Hybridization	Lone Pairs	Bond Angle
$H_2S$	$sp^3$	$2$	$\approx 92.2^o$
$NH_3$	$sp^3$	$1$	$\approx 107^o$
$SiH_4$	$sp^3$	$0$	$109.5^o$
$BF_3$	$sp^2$	$0$	$120^o$

As the number of lone pairs increases,the bond angle decreases due to increased lone pair-bond pair repulsion.
Therefore,the correct order is $H_2S < NH_3 < SiH_4 < BF_3$.

(D) The bond order is inversely proportional to the bond length,i.e.,$\text{Bond Order} \propto \frac{1}{\text{Bond Length}}$.
Since the bond order of $NO$ $(2.5)$ is less than the bond order of $NO^+$ $(3.0)$,the bond length of $NO$ must be greater than the bond length of $NO^+$.
Therefore,the bond length of $NO > NO^+$.

(C) Cerium $(Z = 58)$ has the electronic configuration $[Xe] 4f^1 5d^1 6s^2$.
It exhibits $+3$ and $+4$ oxidation states.
The $+4$ oxidation state is achieved by losing all four valence electrons,resulting in a stable $[Xe]$ configuration.
However,in aqueous solution,$Ce^{4+}$ is a strong oxidizing agent because it tends to revert to the more stable $Ce^{3+}$ state.
Therefore,the statement that the $+4$ oxidation state is not known in solution is incorrect,as $Ce^{4+}$ compounds are well-known in analytical chemistry.

(A) The given parabolas are $y^2 = 4ax$ $(i)$ and $x^2 = 4ay$ $(ii)$.
Substituting $y = \frac{x^2}{4a}$ from $(ii)$ into $(i)$,we get $(\frac{x^2}{4a})^2 = 4ax$,which simplifies to $x^4 = 64a^3x$.
This gives $x(x^3 - 64a^3) = 0$,so $x = 0$ or $x = 4a$.
For $x = 0$,$y = 0$. For $x = 4a$,$y = 4a$. The intersection points are $A(0, 0)$ and $B(4a, 4a)$.
The line $2bx + 3cy + 4d = 0$ passes through $A(0, 0)$,so $2b(0) + 3c(0) + 4d = 0$,which implies $d = 0$.
The line also passes through $B(4a, 4a)$,so $2b(4a) + 3c(4a) + 4d = 0$.
Since $d = 0$,this becomes $8ab + 12ac = 0$. Dividing by $4a$ (given $a \neq 0$),we get $2b + 3c = 0$.
Thus,$d = 0$ and $2b + 3c = 0$,which implies $d^2 + (2b + 3c)^2 = 0$.

(D) Given $B = A^{-1}$,we have $AB = I$,where $I$ is the identity matrix.
$B = \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix} = \begin{bmatrix} 0.4 & 0.2 & 0.2 \\ -0.5 & 0 & \alpha/10 \\ 0.1 & -0.2 & 0.3 \end{bmatrix}$.
For $AB = I$,the product of the second row of $A$ and the third column of $B$ must be $0$ (since it is the element at position $(2,3)$ of the identity matrix).
The second row of $A$ is $(2, 1, -3)$ and the third column of $B$ is $\begin{bmatrix} 0.2 \\ \alpha/10 \\ 0.3 \end{bmatrix}$.
Thus,$(2)(0.2) + (1)(\alpha/10) + (-3)(0.3) = 0$.
$0.4 + \frac{\alpha}{10} - 0.9 = 0$.
$\frac{\alpha}{10} - 0.5 = 0$.
$\frac{\alpha}{10} = 0.5$.
$\alpha = 5$.

(A) The gravitational force is given by $F \propto \frac{1}{R^n} = R^{-n}$.
For a planet in a circular orbit of radius $R$,the gravitational force provides the necessary centripetal force:
$F = \frac{M v^2}{R} = M R \omega^2$,where $\omega = \frac{2\pi}{T}$.
Equating the forces: $M R \omega^2 \propto R^{-n}$.
$\omega^2 \propto R^{-n-1} \Rightarrow \omega \propto R^{-\frac{n+1}{2}}$.
Since $\omega = \frac{2\pi}{T}$,we have $\frac{1}{T} \propto R^{-\frac{n+1}{2}}$.
Therefore,$T \propto R^{\frac{n+1}{2}}$.

(A) According to Faraday's first law of electrolysis,the mass $(m)$ of the substance liberated is given by the formula:
$m = Z \times I \times t$
Where:
$Z$ (electrochemical equivalent) = $3.3 \times 10^{-7} \ kg/C$
$I$ (current) = $3 \ A$
$t$ (time) = $2 \ s$
Substituting the values:
$m = 3.3 \times 10^{-7} \times 3 \times 2$
$m = 19.8 \times 10^{-7} \ kg$

(D) The correct answer is $D$. The standard electrode potential of a halogen depends on three factors: enthalpy of dissociation,ionization enthalpy,and hydration enthalpy.
Fluorine has a very low $F-F$ bond dissociation energy due to the high inter-electronic repulsion between the lone pairs of the small $F$ atoms.
This low bond dissociation energy,combined with high hydration enthalpy,makes fluorine the strongest oxidizing agent among the halogens.

(B) The correct answer is $B$.
Grey tin ($\alpha$-tin) is very brittle and easily crumbles into a powder in very cold climates.
The transformation is: $\text{Grey tin (Cubic)} \rightleftharpoons \text{White tin (Tetragonal)}$.
The change of white tin to grey tin is accompanied by an increase in volume,which causes the buttons to crumble. This phenomenon is known as $\text{tin disease}$ or $\text{tin plague}$.

(A) $1$. Calculate the milliequivalents of $H_2SO_4$ taken: $100 \, mL \times 0.1 \, M \times 2 \, (\text{basicity}) = 20 \, meq$.
$2$. Calculate the milliequivalents of $NaOH$ used for back titration: $20 \, mL \times 0.5 \, M \times 1 \, (\text{acidity}) = 10 \, meq$.
$3$. Milliequivalents of $NH_3$ evolved = $20 - 10 = 10 \, meq = 0.01 \, eq$.
$4$. Percentage of nitrogen = $\frac{1.4 \times \text{meq of } NH_3}{\text{mass of compound}} = \frac{1.4 \times 10}{0.30} = 46.66 \%$.
$5$. Nitrogen percentage in Urea $(NH_2)_2CO$: $\frac{28}{60} \times 100 = 46.66 \%$.
$6$. Since the percentage matches,the compound is Urea.

(B) The dehydrohalogenation of $2-$bromobutane follows Saytzeff's rule.
According to this rule,the more substituted alkene is the major product.
$2-$butene $(CH_3-CH=CH-CH_3)$ is a disubstituted alkene,while $1-$butene $(CH_2=CH-CH_2-CH_3)$ is a monosubstituted alkene.
Therefore,$2-$butene is formed as the major product.

(A) The dehydration of alcohols follows the order: $3^\circ > 2^\circ > 1^\circ$ alcohol,because the rate-determining step involves the formation of a carbocation intermediate.
$CH_3-CH_2-C(OH)(CH_3)-CH_2-CH_3$ is a $3^\circ$ alcohol,which forms a stable $3^\circ$ carbocation upon dehydration.
$CH_3-CH_2-CH_2-CH(OH)-CH_3$ is a $2^\circ$ alcohol.
$CH_3-CH_2-CH_2-CH_2-CH_2-OH$ and $CH_3-CH_2-CH(CH_3)-CH_2-CH_2-OH$ are $1^\circ$ alcohols.
Since the $3^\circ$ carbocation is the most stable,the $3^\circ$ alcohol undergoes dehydration most easily.

(B) The correct answer is $(b)$.
In a mixture of benzene and methanol,methanol molecules are initially held together by strong hydrogen bonds.
When benzene is added,it disrupts these hydrogen bonds,resulting in weaker intermolecular forces in the mixture compared to the pure components.
This reduction in intermolecular attraction leads to a higher vapor pressure than predicted by Raoult's law,which is defined as a positive deviation.

(D) The correct answer is $(d)$.
Freezing point depression $(\Delta T_f)$ is a colligative property given by $\Delta T_f = K_f \times m$,where $K_f$ is the cryoscopic constant of the solvent and $m$ is the molality.
Since $K_f$ is a characteristic property of the solvent,it varies from one solvent to another.
Therefore,even if the molality $(m)$ is the same,the freezing point depression will differ if the solvents are different.

(C) The elevation in boiling point $(\Delta T_b)$ is a colligative property,which is directly proportional to the van't Hoff factor $(i)$ and the molarity $(M)$ of the solution: $\Delta T_b = i \times K_b \times M$.
For non-electrolytes like urea and glucose,$i = 1$.
For $KNO_3$,$i = 2$ $(K^{+} + NO_3^{-})$.
For $Na_2SO_4$,$i = 3$ $(2Na^{+} + SO_4^{2-})$.
Calculating the effective concentration $(i \times M)$:
$A: 1 \times 0.015 = 0.015 \ M$
$B: 2 \times 0.01 = 0.02 \ M$
$C: 3 \times 0.01 = 0.03 \ M$
$D: 1 \times 0.015 = 0.015 \ M$
Since $0.01 \ M \ Na_2SO_4$ has the highest effective concentration of solute particles,it will exhibit the highest boiling point.

(B) In the provided diagram,an equal number of cations $(Na^{+})$ and anions $(Cl^{-})$ are missing from their respective lattice sites,creating vacancies $(Box)$.
This is the characteristic feature of a Schottky defect,which is a type of stoichiometric point defect in ionic crystals.

(B) In the first reaction:
$_{92}^{238}M \to _{y}^{x}N + 2_{2}^{4}He$
Applying conservation of mass number: $x = 238 - (2 \times 4) = 230$
Applying conservation of atomic number: $y = 92 - (2 \times 2) = 88$
So,the element is $_{88}^{230}N$.
In the second reaction:
$_{88}^{230}N \to _{B}^{A}L + 2\beta^{+}$
For $\beta^{+}$ decay (positron emission),the mass number $A$ remains the same and the atomic number decreases by $1$ per particle.
$A = 230$
$B = 88 - (2 \times 1) = 86$
So,the element is $_{86}^{230}L$.
Number of neutrons in $_{86}^{230}L = A - B = 230 - 86 = 144$.

(A) The formula for radioactive decay is $N = N_0 \times (\frac{1}{2})^n$,where $n$ is the number of half-lives.
Number of half-lives $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{24 \ h}{4 \ h} = 6$.
Initial mass $N_0 = 200 \ g$.
Remaining mass $N = 200 \times (\frac{1}{2})^6$.
$N = 200 \times \frac{1}{64} = 3.125 \ g$.

(D) For the reaction $2A + B \to C$,the rate law is given as $\text{rate} = k[A][B]$.
According to the Arrhenius equation,$k = Ae^{-E_a/RT}$.
This shows that the rate constant $k$ depends only on temperature and the activation energy $(E_a)$,not on the initial concentrations of the reactants $A$ and $B$.
Therefore,the value of $k$ is independent of the initial concentrations of $A$ and $B$.

(C) For a first order reaction,the half-life period $(T_{1/2})$ is the time required for the concentration to become half of its initial value.
Given that the concentration decreases from $0.8 \ M$ to $0.4 \ M$ in $15 \ min$,this represents one half-life,so $T_{1/2} = 15 \ min$.
To find the time taken for the concentration to change from $0.1 \ M$ to $0.025 \ M$,we observe the number of half-lives:
$0.1 \ M$ $\xrightarrow{T_{1/2}} 0.05 \ M$ $\xrightarrow{T_{1/2}} 0.025 \ M$.
This process involves $2$ half-lives.
Therefore,the total time taken $= 2 \times T_{1/2} = 2 \times 15 \ min = 30 \ min$.

(B) In a hydrogen-oxygen fuel cell,the following reactions take place to create a potential difference between the two electrodes:
Anode: $2H_{2(g)} + 4OH^{-}_{(aq)} \to 4H_2O_{(l)} + 4e^{-}$
Cathode: $O_{2(g)} + 2H_2O_{(l)} + 4e^{-} \to 4OH^{-}_{(aq)}$
Overall reaction: $2H_{2(g)} + O_{2(g)} \to 2H_2O_{(l)}$
The net reaction is equivalent to the combustion of hydrogen to form water,which facilitates the flow of electrons and creates a potential difference.

(A) The cell reaction is $Zn_{(s)} + 2H^{+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + H_{2_{(g)}}$.
According to the Nernst equation: $E_{cell} = E^0_{cell} - \frac{0.059}{2} \log \frac{[Zn^{2+}]}{[H^{+}]^2}$.
When $H_2SO_4$ is added to the cathode compartment,the concentration of $H^{+}$ ions increases.
As $[H^{+}]$ increases,the term $\log \frac{[Zn^{2+}]}{[H^{+}]^2}$ decreases,which causes the $E_{cell}$ to increase.
According to Le Chatelier's principle,an increase in the concentration of reactants $(H^{+})$ shifts the equilibrium to the right.

(A) The relationship between standard cell potential $(E_{cell}^0)$ and the equilibrium constant $(K_c)$ is given by the Nernst equation at equilibrium: $E_{cell}^0 = \frac{0.0591}{n} \log K_c$ at $298 \ K$.
Given: $E_{cell}^0 = 0.591 \ V$,$n = 1$.
Substituting the values: $0.591 = \frac{0.0591}{1} \log K_c$.
$\log K_c = \frac{0.591}{0.0591} = 10$.
$K_c = 10^{10} = 1.0 \times 10^{10}$.

(A) The cell reaction is represented as: $\frac{1}{2} H_2 | H^{+} || Ag^{+} | Ag$.
The standard cell potential is calculated using the formula: $E_{Cell}^{0} = E_{Cathode}^{0} - E_{Anode}^{0}$.
Here,the cathode is the $Ag^{+}/Ag$ electrode $(E^{0} = 0.80 \ V)$ and the anode is the standard hydrogen electrode $(E^{0} = 0.00 \ V)$.
Therefore,$E_{Cell}^{0} = 0.80 \ V - 0.00 \ V = 0.80 \ V$.

(A) The given reaction is: $Sn_{(s)} + 2Fe^{3+}_{(aq)} \to 2Fe^{2+}_{(aq)} + Sn^{2+}_{(aq)}$
Here,$Sn$ is oxidized to $Sn^{2+}$ (anode) and $Fe^{3+}$ is reduced to $Fe^{2+}$ (cathode).
$E^0_{cell} = E^0_{cathode} - E^0_{anode}$
$E^0_{cell} = E^0_{Fe^{3+}/Fe^{2+}} - E^0_{Sn^{2+}/Sn}$
$E^0_{cell} = (+0.77 \ V) - (-0.14 \ V)$
$E^0_{cell} = 0.77 \ V + 0.14 \ V = 0.91 \ V$

(C) The ease of oxidation $(M^{2+} \to M^{3+} + e^-)$ is determined by the oxidation potential,which is the negative of the reduction potential $({E^0}_{M^{2+}/M^{3+}} = -{E^0}_{M^{3+}/M^{2+}})$. $A$ higher (more positive) oxidation potential indicates that the oxidation process is easier.

Metal	Oxidation Potential $({E^0}_{M^{2+}/M^{3+}})$
$Cr$	$+0.41 \ V$
$Mn$	$-1.57 \ V$
$Fe$	$-0.77 \ V$
$Co$	$-1.97 \ V$

Since $Cr$ has the highest oxidation potential $(+0.41 \ V)$,the change in oxidation state from $+2$ to $+3$ is easiest for $Cr$.

(C) According to Kohlrausch's law of independent migration of ions:
$\wedge _{NaCl}^0 = \lambda _{Na^{+}}^0 + \lambda _{Cl^{-}}^0 = 126 \ S \ cm^2 \ mol^{-1}$ $(1)$
$\wedge _{KBr}^0 = \lambda _{K^{+}}^0 + \lambda _{Br^{-}}^0 = 152 \ S \ cm^2 \ mol^{-1}$ $(2)$
$\wedge _{KCl}^0 = \lambda _{K^{+}}^0 + \lambda _{Cl^{-}}^0 = 150 \ S \ cm^2 \ mol^{-1}$ $(3)$
To find $\wedge _{NaBr}^0 = \lambda _{Na^{+}}^0 + \lambda _{Br^{-}}^0$,we perform the operation: $(1)$ + $(2)$ - $(3)$
$\wedge _{NaBr}^0 = (\lambda _{Na^{+}}^0 + \lambda _{Cl^{-}}^0) + (\lambda _{K^{+}}^0 + \lambda _{Br^{-}}^0) - (\lambda _{K^{+}}^0 + \lambda _{Cl^{-}}^0)$
$\wedge _{NaBr}^0 = 126 + 152 - 150 = 128 \ S \ cm^2 \ mol^{-1}$

(A) The correct answer is $(A)$.
Froth-flotation is a process specifically used for the concentration of sulphide ores.
Galena $(PbS)$ is a sulphide ore.
Cassiterite $(SnO_2)$ is an oxide ore.
Magnetite $(Fe_3O_4)$ is an oxide ore.
Malachite $(Cu(OH)_2 \cdot CuCO_3)$ is a carbonate ore.
Therefore,only Galena is concentrated by the froth-flotation method,which relies on the preferential wetting of sulphide particles by oil (frothing agent) and gangue particles by water.

(A) Cerium $(Z = 58)$ has an electronic configuration of $[Xe] 4f^1 5d^1 6s^2$.
It exhibits both $+3$ and $+4$ oxidation states.
The $+4$ oxidation state of cerium is well-known and stable in aqueous solutions,where it acts as a strong oxidizing agent.
Therefore,the statement that the $+4$ oxidation state is not known in solutions is incorrect.

(D) The maximum oxidation state of a transition metal is determined by the total number of electrons in the $(n-1)d$ and $ns$ orbitals.
For $(n-1)d^3ns^2$,the total number of valence electrons is $3 + 2 = 5$.
For $(n-1)d^5ns^1$,the total number of valence electrons is $5 + 1 = 6$.
For $(n-1)d^8ns^2$,the total number of valence electrons is $8 + 2 = 10$,but only the unpaired electrons or those in the outer shell are typically involved in bonding,limiting the oxidation state.
For $(n-1)d^5ns^2$,the total number of valence electrons is $5 + 2 = 7$.
Therefore,the configuration $(n - 1)d^5ns^2$ can achieve the maximum oxidation state of $+7$ (e.g.,in $Mn$).

(C) The magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.
$1$. For $[MnCl_4]^{2-}$: $Mn^{2+}$ is $3d^5$. Since $Cl^-$ is a weak field ligand,electrons do not pair up. $n = 5$.
$2$. For $[CoCl_4]^{2-}$: $Co^{2+}$ is $3d^7$. Since $Cl^-$ is a weak field ligand,electrons do not pair up. $n = 3$.
$3$. For $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $3d^6$. Since $CN^-$ is a strong field ligand,electrons pair up. $n = 0$.
Comparing the number of unpaired electrons: $5 > 3 > 0$.
Therefore,the order of magnetic moments is $[MnCl_4]^{2-} > [CoCl_4]^{2-} > [Fe(CN)_6]^{4-}$.

(B) The reaction between $KI$ and $CuSO_4$ is: $2CuSO_4 + 4KI \to Cu_2I_2 + I_2 + 2K_2SO_4$.
In this reaction,$Cu^{2+}$ is reduced to $Cu^+$ (forming $Cu_2I_2$) and $I^-$ is oxidized to $I_2$.
Then,$I_2$ reacts with $Na_2S_2O_3$: $I_2 + 2Na_2S_2O_3 \to Na_2S_4O_6 + 2NaI$.
Here,$Na_2S_2O_3$ is oxidized to $Na_2S_4O_6$ and $I_2$ is reduced to $I^-$.
$CuI_2$ is not formed because $Cu^{2+}$ oxidizes $I^-$ to $I_2$ immediately,resulting in the formation of $Cu_2I_2$ (cuprous iodide).

(C) The coordination number of a central metal atom or ion in a complex is defined as the total number of ligand donor atoms that are directly bonded to the central metal atom or ion through coordinate covalent bonds (which are essentially $\sigma$-bonds).
Therefore,the coordination number of a metal = number of $\sigma$-bonds formed by the metal with the ligands.

(A) $CN^{-}$ ion acts as a reducing agent because it can be oxidised to $(CN)_2$.
It also acts as a strong complexing agent due to the presence of a lone pair on the carbon atom,which allows it to form stable coordinate bonds with metal ions.
Therefore,the properties shown by $CN^{-}$ are $(a)$ reducing and $(c)$ complexing.

(D) An outer orbital complex uses the outer $d$-orbitals (e.g.,$4d$) for hybridization,typically $sp^3d^2$.
$1$. $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $3d^6$. $CN^-$ is a strong field ligand,causing pairing. Hybridization is $d^2sp^3$ (inner orbital).
$2$. $[Mn(CN)_6]^{4-}$: $Mn^{2+}$ is $3d^5$. $CN^-$ is a strong field ligand,causing pairing. Hybridization is $d^2sp^3$ (inner orbital).
$3$. $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ is $3d^6$. $NH_3$ is a strong field ligand for $Co^{3+}$,causing pairing. Hybridization is $d^2sp^3$ (inner orbital).
$4$. $[Ni(NH_3)_6]^{2+}$: $Ni^{2+}$ is $3d^8$. $NH_3$ is a weak field ligand relative to $Ni^{2+}$. Hybridization is $sp^3d^2$ (outer orbital).
Thus,$[Ni(NH_3)_6]^{2+}$ is an outer orbital complex.

(D) $[Co(en)_2Cl_2]^+$ has the largest number of isomers.
It exhibits geometrical isomerism (cis and trans forms).
The cis-isomer is optically active and exists as two enantiomers ($d$ and $l$ forms).
Thus,it shows both geometrical and optical isomerism,resulting in a total of $3$ isomers.

(B) The reaction between chlorobenzene and chloral $(CCl_3CHO)$ in the presence of concentrated $H_2SO_4$ leads to the formation of $DDT$ ($1,1,1$-trichloro-$2,2$-bis($p$-chlorophenyl)ethane).
The chemical reaction is:
$2C_6H_5Cl + CCl_3CHO \xrightarrow{conc. H_2SO_4} (ClC_6H_4)_2CHCCl_3 + H_2O$

(A) Acetyl bromide $(CH_3COBr)$ reacts with the first equivalent of $CH_3MgI$ to form acetone $(CH_3COCH_3)$.
Since $CH_3MgI$ is in excess,it further reacts with the formed acetone to produce a tertiary alkoxide intermediate $(CH_3-C(CH_3)_2-OMgI)$.
Finally,hydrolysis of this intermediate with a saturated solution of $NH_4Cl$ yields $2$-methyl-$2$-propanol $(CH_3-C(CH_3)_2-OH)$.
Reaction: $CH_3COBr + 2CH_3MgI$ $\rightarrow CH_3-C(CH_3)_2-OMgI$ $\xrightarrow{NH_4Cl} CH_3-C(CH_3)_2-OH$.

(B) The reaction of an aldehyde with no $\alpha$-hydrogen with concentrated alkali $(50\% \, NaOH)$ is known as the Cannizzaro reaction.
In this reaction,one molecule of the aldehyde is oxidized to the corresponding carboxylic acid (as a salt),and another molecule is reduced to the corresponding alcohol.
Benzaldehyde $(C_6H_5CHO)$ does not contain any $\alpha$-hydrogen atom.
Therefore,it undergoes the Cannizzaro reaction to produce benzyl alcohol $(C_6H_5CH_2OH)$ and sodium benzoate $(C_6H_5COONa)$.
The reaction is:
$2C_6H_5CHO + NaOH (50\%) \rightarrow C_6H_5CH_2OH + C_6H_5COONa$
Thus,the correct option is $(B)$.

(D) Butan-$2$-one is a ketone. It undergoes Clemmensen reduction when treated with zinc amalgam and hydrochloric acid $(Zn-Hg/HCl)$ to give the corresponding hydrocarbon,butane.
$CH_3-CO-CH_2-CH_3 \xrightarrow{Zn-Hg/HCl} CH_3-CH_2-CH_2-CH_3$
Amides,carboxylic acids,and esters are not reduced to hydrocarbons by this reagent.

(B) The acidity of substituted benzoic acids depends on the electronic effects of the substituents.
$1$. The $-NO_2$ group is a strong electron-withdrawing group ($-I$ and $-M$ effects),which increases the acidity of benzoic acid compared to $PhCOOH$ $(A)$.
$2$. Due to the ortho effect,the ortho-substituted isomer $(B)$ is the most acidic among the nitrobenzoic acids.
$3$. For the meta and para isomers,the $-NO_2$ group at the para position $(C)$ exerts both $-I$ and $-M$ effects,whereas at the meta position $(D)$,it exerts only the $-I$ effect. However,the $-M$ effect is stronger at the para position,making $p-NO_2C_6H_4COOH$ more acidic than $m-NO_2C_6H_4COOH$.
$4$. Therefore,the correct order of acidity is: $o-NO_2C_6H_4COOH (B) > p-NO_2C_6H_4COOH (C) > m-NO_2C_6H_4COOH (D) > PhCOOH (A)$.
Thus,the correct option is $B$.

(C) . Ethyl acetate $(CH_3COOC_2H_5)$ is an ester. Sodium chloride $(NaCl)$ is a salt that dissociates into $Na^+$ and $Cl^-$ ions in water.
There is no chemical reaction between an ester and a neutral salt like $NaCl$ in an aqueous medium.
Therefore,the composition of the resultant solution remains the same as the mixture of the two reactants: $CH_3COOC_2H_5 + NaCl$.

(D) The basicity of amines depends on the availability of the lone pair of electrons on the nitrogen atom for protonation.
In aniline $(C_6H_5NH_2)$,$N$-methylaniline $(C_6H_5NHCH_3)$,and $o$-toluidine $(2-CH_3C_6H_4NH_2)$,the lone pair on the nitrogen atom is in conjugation with the benzene ring,which decreases its availability for protonation.
In benzylamine $(C_6H_5CH_2NH_2)$,the nitrogen atom is attached to a $CH_2$ group,not directly to the benzene ring. Therefore,the lone pair on the nitrogen atom is not involved in resonance with the benzene ring,making it more available for protonation.
Thus,benzylamine is the strongest base among the given options.

(D) . Enzymes are shape-selective,specific biological catalysts that possess well-defined active sites and function optimally at physiological temperatures.

(B) $DNA$ contains $2$-deoxyribose sugar and the nitrogenous base thymine $(T)$.
$RNA$ contains ribose sugar and the nitrogenous base uracil $(U)$ instead of thymine.
Therefore,$RNA$ is characterized by the presence of ribose sugar and uracil.

(B) Insulin is a proteinaceous hormone secreted by $\beta$-cells of the Islets of Langerhans in the pancreas of our body.

(C) Helium $(He)$ is twice as heavy as hydrogen $(H_2)$. While it is non-inflammable,it is not lighter than hydrogen. Therefore,the statement that it is lighter than hydrogen is incorrect.
Helium has the lowest melting and boiling point of any element,which makes liquid helium an ideal coolant for many extremely low-temperature applications,such as superconducting magnets and cryogenic research where temperatures close to absolute zero are needed.
$He$ is also used in gas-cooled nuclear reactors as a heat transfer agent.

(C) Chlorophyll is a green pigment found in plants that plays a crucial role in photosynthesis. It contains a porphyrin ring with a central magnesium $(Mg^{2+})$ ion,not calcium $(Ca^{2+})$. Therefore,the statement that chlorophyll contains calcium is incorrect.