AIPMT 1995 Chemistry Question Paper with Answer and Solution

(C) Pressure is defined as force per unit area: $P = \frac{F}{A} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$.
Energy per unit volume is defined as energy divided by volume: $\frac{E}{V} = \frac{[ML^2T^{-2}]}{[L^3]} = [ML^{-1}T^{-2}]$.
Since both have the same dimensional formula $[ML^{-1}T^{-2}]$,the dimensions of pressure are the same as that of energy per unit volume.

(D) Volume of air = $1 \ L = 1000 \ mL$.
Volume of $O_2$ in $1 \ L$ of air = $21 \%$ of $1000 \ mL = 210 \ mL$.
At $STP$,$1 \ mole$ of any gas occupies $22400 \ mL$.
Number of moles of $O_2 = \frac{\text{Volume of } O_2 \text{ in } mL}{22400 \ mL/mol} = \frac{210}{22400} \approx 0.009375 \ mol$.
Rounding to the given options,the correct answer is $0.0093 \ mol$.

(C) The bond length is inversely proportional to the bond order.
Bond orders for the given species are:
$O_2$: Bond order = $2.0$
$O_3$: Bond order = $1.5$
$H_2O_2$: Bond order = $1.0$
Since bond length increases as bond order decreases,the order of bond length is $H_2O_2$ $(1.48 \ \mathring{A})$ > $O_3$ $(1.28 \ \mathring{A})$ > $O_2$ $(1.21 \ \mathring{A})$.
Thus,the correct option is $C$.

(A) According to $VSEPR$ theory,the shape of a molecule is determined by the number of bond pairs and lone pairs around the central atom.
$BCl_3$: The central Boron atom has $3$ bond pairs and $0$ lone pairs,resulting in $sp^2$ hybridization and a trigonal planar geometry.
$NCl_3$: The central Nitrogen atom has $3$ bond pairs and $1$ lone pair,resulting in $sp^3$ hybridization and a pyramidal geometry due to the repulsion between the lone pair and bond pairs.

(C) According to Molecular Orbital Theory $(MOT)$,a species is paramagnetic if it contains one or more unpaired electrons.
$1$. $O_2^-$ has $17$ electrons. Its electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. Since it has an unpaired electron,it is paramagnetic.
$2$. $NO$ has $15$ electrons. Its electronic configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Since it has an unpaired electron,it is paramagnetic.
$3$. $CN^-$ has $14$ electrons. All electrons are paired,so it is diamagnetic.
Therefore,both $(a)$ and $(b)$ are paramagnetic.

(C) The dimension of pressure $(P)$ is defined as force per unit area,which is $[M L^{-1} T^{-2}]$.
Energy $(E)$ has dimensions $[M L^2 T^{-2}]$.
Volume $(V)$ has dimensions $[L^3]$.
Therefore,the dimension of energy per unit volume is $\frac{[M L^2 T^{-2}]}{[L^3]} = [M L^{-1} T^{-2}]$.
Since the dimensions of pressure and energy per unit volume are identical,the correct option is $C$.

(D) The correct answer is $(D)$.
According to the kinetic theory of gases,an ideal gas is defined by the absence of inter-molecular forces of attraction.
Liquefaction of a gas requires the presence of inter-molecular forces to bring molecules together to form a liquid phase.
Since an ideal gas lacks these forces,it cannot be liquefied at any value of pressure $(P)$ and temperature $(T)$.

(A) According to Le Chatelier's principle,if the concentration of a product $(trans-2-pentene)$ is increased in a system at equilibrium,the system will shift in the direction that consumes the added substance to counteract the change.
Therefore,the equilibrium will shift in the backward direction to form more $cis-2-pentene$.

(C) The solubility of a sparingly soluble salt like $AgCl$ decreases in the presence of a common ion due to the common ion effect.
$AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$
$K_{sp} = [Ag^+][Cl^-]$
Comparing the concentration of common ions:
$(A)$ $0.001 \ M \ AgNO_3$ provides $0.001 \ M \ Ag^+$.
$(B)$ Pure water has no common ions.
$(C)$ $0.01 \ M \ CaCl_2$ provides $2 \times 0.01 = 0.02 \ M \ Cl^-$.
$(D)$ $0.01 \ M \ NaCl$ provides $0.01 \ M \ Cl^-$.
Since $0.01 \ M \ CaCl_2$ provides the highest concentration of common ion $(Cl^-)$,the equilibrium shifts furthest to the left,resulting in the minimum solubility of $AgCl$.

(A) The correct answer is $(A)$.
Blood acts as a buffer solution because it contains a mixture of weak acids and their conjugate bases (such as the bicarbonate buffer system and serum proteins).
Buffer solutions are resistant to changes in $pH$ upon the addition of small amounts of acid or base.

(A) The $pH$ of a solution is defined as $pH = -\log[H^+]$.
For a strong acid like $HCl$,the concentration of $H^+$ ions is equal to the concentration of the acid.
Given $[H^+] = 10 \ M$.
$pH = -\log(10) = -1$.
Since the $pH$ scale is typically defined for dilute solutions,for highly concentrated solutions,the $pH$ can be negative.
Therefore,the $pH$ of a $10 \ M$ $HCl$ solution is $-1$,which is less than $0$.

(B) The combustion reaction for $C_2H_{4(g)}$ is: $C_2H_{4(g)} + 3O_{2(g)} \to 2CO_{2(g)} + 2H_2O_{(l)}$
$\Delta H_{combustion} = \sum \Delta H_f^o(\text{products}) - \sum \Delta H_f^o(\text{reactants})$
$\Delta H_{combustion} = [2 \times \Delta H_f^o(CO_2) + 2 \times \Delta H_f^o(H_2O)] - [\Delta H_f^o(C_2H_4) + 3 \times \Delta H_f^o(O_2)]$
Given: $\Delta H_f^o(C_2H_4) = 52 \ kJ \ mol^{-1}$,$\Delta H_f^o(CO_2) = -394 \ kJ \ mol^{-1}$,$\Delta H_f^o(H_2O) = -286 \ kJ \ mol^{-1}$,and $\Delta H_f^o(O_2) = 0 \ kJ \ mol^{-1}$.
$\Delta H_{combustion} = [2(-394) + 2(-286)] - [52 + 3(0)]$
$\Delta H_{combustion} = [-788 - 572] - 52 = -1360 - 52 = -1412 \ kJ \ mol^{-1}$.

(A) For a reaction to be spontaneous,the change in Gibbs free energy,$\Delta G$,must be negative.
Since $\Delta G = \Delta H - T\Delta S$,the expression $(\Delta H - T\Delta S)$ must be negative.

(A) The given electronic configuration $1s^2, 2s^2 2p^6, 3s^2 3p^3$ corresponds to the element Phosphorus $(P)$ with an atomic number of $15$.
Phosphorus belongs to Group $15$ of the periodic table.
The element located immediately below Phosphorus in Group $15$ is Arsenic $(As)$.
To find the atomic number of the element below,we add the magic number $18$ to the atomic number of Phosphorus $(15 + 18 = 33)$.
Thus,the atomic number of Arsenic is $33$.

(A) Gypsum is $CaSO_4 \cdot 2H_2O$ and Plaster of Paris is $CaSO_4 \cdot \frac{1}{2}H_2O$.
When Plaster of Paris is mixed with water,it rehydrates to form gypsum: $CaSO_4 \cdot \frac{1}{2}H_2O + \frac{3}{2}H_2O \rightarrow CaSO_4 \cdot 2H_2O$.
When gypsum is heated at $373 \ K$,it loses water to form Plaster of Paris: $CaSO_4 \cdot 2H_2O \xrightarrow{373 \ K} CaSO_4 \cdot \frac{1}{2}H_2O + \frac{3}{2}H_2O$.
Comparing the molar masses,the percentage of $Ca$ in gypsum is lower than in Plaster of Paris because of the higher water content in gypsum.

(A) $AlCl_3$ forms a dimer and exists as $Al_2Cl_6$ to achieve a higher coordination number of $6$ for the aluminium atom.
In $AlCl_3$,the aluminium atom is electron-deficient with an incomplete octet.
By forming a dimer,each aluminium atom accepts a lone pair of electrons from a chlorine atom of another $AlCl_3$ molecule,thereby completing its octet and increasing its coordination number.

(A) The hydration energy decreases from $Be^{2+}$ to $Ba^{2+}$.
Since the hydration energy of the smaller $Be^{2+}$ ion is significantly higher than the lattice energy of $BeSO_4$,it is highly soluble.
As the size of the cation increases down the group,the hydration energy decreases more rapidly than the lattice energy,leading to a decrease in solubility.

(C) The parent carbon chain is selected by including the principal functional group (carboxylic acid) and the double bond.
Numbering starts from the $COOH$ group as $C-1$.
The chain is $CH_3(7)-CH(6)=CH(5)-CH_2(4)-CH(NH_2)(3)-CH_2(2)-COOH(1)$.
There is an amino group at position $3$ and a double bond starting at position $5$.
Thus,the name is $3-$amino$-5-$heptenoic acid.

(D) To determine the hybridization of carbon atoms,we count the number of sigma bonds and lone pairs (if any) attached to each carbon atom.
$(i)$ $CH_3-CH_2-CH_2-CH_3$: All carbon atoms are bonded to four other atoms via single bonds,so all are $sp^3$ hybridized.
$(ii)$ $CH_3-CH=CH-CH_3$: The terminal carbons are $sp^3$ hybridized,while the carbons involved in the double bond are $sp^2$ hybridized. Thus,it contains both $sp^2$ and $sp^3$ hybridized carbons.
$(iii)$ $CH_2=CH-CH=CH_2$: All carbon atoms are involved in double bonds,so all are $sp^2$ hybridized.
$(iv)$ $HC\equiv CH$: Both carbon atoms are involved in a triple bond,so both are $sp$ hybridized.
Therefore,only compound $(ii)$ has more than one kind of hybridization for carbon.

(A) To determine if a system is planar,we look at the hybridization of the atoms involved in the structure.
$(i)$ Phenyl group: The benzene ring is planar because all carbon atoms are $sp^2$ hybridized.
$(ii)$ Cyclohexyl group: This is a saturated ring where carbons are $sp^3$ hybridized,adopting a non-planar chair conformation.
$(iii)$ Cyclopentyl group: This is a saturated ring where carbons are $sp^3$ hybridized,adopting a non-planar envelope conformation.
$(iv)$ Butyl group: This is an open-chain alkyl group with $sp^3$ hybridized carbons,which is non-planar.
$(v)$ Ethenyl (vinyl) group: The two carbons are $sp^2$ hybridized,making the system planar.
Therefore,systems $(i)$ and $(v)$ have essentially planar geometry.

(B) The hydroboration-oxidation of alkenes follows anti-Markovnikov regioselectivity to yield primary alcohols.
The reaction proceeds as follows:
$3R-CH=CH_2 + 1/2 B_2H_6 \rightarrow (R-CH_2-CH_2)_3B$
$(R-CH_2-CH_2)_3B + 3H_2O_2 \xrightarrow{OH^-} 3R-CH_2-CH_2-OH + H_3BO_3$
Thus,the final product is a primary alcohol,$R-CH_2-CH_2-OH$.

(B) Applying the Clausius-Clapeyron equation:
$\log \frac{P_2}{P_1} = \frac{\Delta H_{vap}}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$
Given: $P_2 = 760 \ mm$,$T_2 = 373 \ K$,$P_1 = 23 \ mm$,$\Delta H_{vap} = 40656 \ J/mol$,$R = 8.314 \ J/K \cdot mol$.
Substituting the values:
$\log \frac{760}{23} = \frac{40656}{2.303 \times 8.314} \left[ \frac{373 - T_1}{373 T_1} \right]$
$1.519 = 2123.5 \times \left[ \frac{373 - T_1}{373 T_1} \right]$
$0.000715 = \frac{373 - T_1}{373 T_1}$
$0.2667 T_1 = 373 - T_1$
$1.2667 T_1 = 373$
$T_1 \approx 294.4 \ K$.

(B) Sodium is obtained by the electrolytic reduction of its chloride. The melting point of pure $NaCl$ is very high $(1074 \ K)$. To lower the melting point of the electrolyte to a manageable range (around $873 \ K$),$CaCl_2$ is added to the mixture.

(A) Metals contain a large number of free electrons. When one end of the metal is heated,these electrons gain kinetic energy and move rapidly towards the cooler end,transferring thermal energy through collisions with other atoms and electrons. This process is known as thermal conduction.

(A) The correct answer is $(A)$.
Mercury $(Hg)$ has a very high ionisation energy,which makes it difficult for its valence electrons to participate in metallic bonding.
Consequently,the metallic bonds between $Hg$ atoms are very weak,resulting in a low melting point that makes it liquid at room temperature ($0 \, ^oC$ is below its melting point of $-38.8 \, ^oC$).

(A) The $Pronephros$ kidney is the functional kidney during the embryonic stage of cyclostomes,fishes,and amphibians.
$A$ frog tadpole is the larval stage of a frog,and during this stage,the $Pronephros$ serves as the functional excretory organ.

(A) The thyroid gland specifically accumulates iodine for the synthesis of thyroid hormones ($T_3$ and $T_4$).
Because of this physiological property,the radioactive isotope Iodine-$131$ is used as a tracer to detect thyroid cancer and to treat hyperthyroidism.

(B) Nicotine acts as a stimulant because it mimics the effect of the neurotransmitter $Acetylcholine$.
It binds to nicotinic $Acetylcholine$ receptors at the autonomic ganglia and the neuromuscular junction.
Initially,it stimulates these receptors,leading to increased neural activity,but in high doses,it can cause inhibition of neural impulses.

(C) The term $Florigen$ was proposed by the Russian scientist $M.K. Chailakhyan$ in $1936$ while studying photoperiodism.
He hypothesized that $Florigen$ is a hypothetical floral hormone synthesized in the leaves and transported to the shoot apices to induce flowering.
According to his theory,$Florigen$ consists of two groups of substances:
$(i)$ $Gibberellins$,which are required for stem elongation.
$(ii)$ $Anthesins$,which are required for the actual formation of flowers.
Together,these substances constitute the flowering hormone known as $Florigen$.

(D) For stable equilibrium,the net force on the system must be zero,which implies that the derivative of the potential energy with respect to distance must be zero.
$F = -\frac{dU}{dx} = 0$
Given $U(x) = ax^{-12} - bx^{-6}$,we differentiate with respect to $x$:
$\frac{dU}{dx} = -12ax^{-13} + 6bx^{-7} = 0$
Rearranging the terms:
$12ax^{-13} = 6bx^{-7}$
$\frac{12a}{x^{13}} = \frac{6b}{x^7}$
$\frac{2a}{b} = \frac{x^{13}}{x^7} = x^6$
$x = \sqrt[6]{\frac{2a}{b}}$
Thus,the atoms are in equilibrium at $x = \sqrt[6]{\frac{2a}{b}}$.

(C) The moment of inertia $I$ of a body about an axis is given by $I = Mk^2$,where $M$ is the mass and $k$ is the radius of gyration.
For a planar object,the moment of inertia about an axis is proportional to the square of the average distance of the mass from the axis.
In the given right-angled triangle $ABC$ with sides $AB=4$,$BC=3$,and $AC=5$:
$1$. The axis $BC$ is at a distance from the mass distributed along $AB$. The average distance of the mass from axis $BC$ is larger than the average distance of the mass from axis $AB$.
$2$. Specifically,the mass is more spread out relative to the axis $BC$ compared to axis $AB$.
$3$. Therefore,the radius of gyration $k_{BC} > k_{AB}$.
$4$. Consequently,$I_{BC} > I_{AB}$.
$5$. The axis $CA$ (the hypotenuse) is closest to the center of mass of the triangle,resulting in the smallest moment of inertia.
Thus,the correct relation is $I_{BC} > I_{AB} > I_{CA}$.

(A) According to Heisenberg's uncertainty principle: $\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$.
Given: $m = 9.1 \times 10^{-28} \ g$,$v = 3.0 \times 10^4 \ cm \ s^{-1}$,and accuracy is $0.001\%$.
First,calculate the uncertainty in velocity $(\Delta v)$:
$\Delta v = v \times \frac{0.001}{100} = 3.0 \times 10^4 \times 10^{-5} = 0.3 \ cm \ s^{-1}$.
Now,calculate the uncertainty in momentum $(\Delta p)$:
$\Delta p = m \times \Delta v = 9.1 \times 10^{-28} \times 0.3 = 2.73 \times 10^{-28} \ g \ cm \ s^{-1}$.
Using the uncertainty principle formula:
$\Delta x = \frac{h}{4 \pi \cdot m \cdot \Delta v} = \frac{6.626 \times 10^{-27}}{4 \times 3.1416 \times 2.73 \times 10^{-28}}$.
$\Delta x = \frac{6.626 \times 10^{-27}}{34.287 \times 10^{-28}} = \frac{66.26}{34.287} \approx 1.93 \ cm$.
Rounding to the nearest provided option,the correct answer is $1.92 \ cm$.

(C) reducing agent is a substance that can be oxidized to a higher oxidation state.
In $SO_2$,the oxidation state of $S$ is $+4$,which can be oxidized to $+6$.
In $NO_2$,the oxidation state of $N$ is $+4$,which can be oxidized to $+5$.
In $ClO_2$,the oxidation state of $Cl$ is $+4$,which can be oxidized to $+7$.
In $CO_2$,the oxidation state of $C$ is $+4$,which is its maximum oxidation state (group valence).
Therefore,$CO_2$ cannot be further oxidized and cannot act as a reducing agent.

(A) Given the partial fraction decomposition: $\frac{1}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$
Multiply both sides by $x(x^2 + 1)$: $1 = A(x^2 + 1) + (Bx + C)x$
$1 = Ax^2 + A + Bx^2 + Cx$
$1 = (A + B)x^2 + Cx + A$
Comparing the coefficients of $x^2$,$x$,and the constant term on both sides:
$A + B = 0$
$C = 0$
$A = 1$
Substituting $A = 1$ into $A + B = 0$,we get $1 + B = 0$,which implies $B = -1$.
Thus,$(A, B, C) = (1, -1, 0)$.

(C) The moment of inertia of a body about an axis is given by $I = MK^2$,where $M$ is the mass and $K$ is the radius of gyration. The radius of gyration $K$ depends on the distribution of mass relative to the axis of rotation. The farther the mass is distributed from the axis,the larger the radius of gyration.
In the given right-angled triangle $ABC$ with sides $AB=4$,$BC=3$,and $AC=5$:
$1$. For axis $AB$,the mass is distributed within a distance of $3$ units.
$2$. For axis $BC$,the mass is distributed within a distance of $4$ units.
$3$. For axis $AC$,the mass is distributed within a smaller perpendicular distance compared to the other two axes.
Since the mass is spread further away from the axis $BC$ compared to axis $AB$,the radius of gyration $K_{BC} > K_{AB}$.
Consequently,$I_{BC} > I_{AB} > I_{CA}$.
Therefore,the correct relation is $I_{BC} > I_{AB}$.

(C) When the key $K$ is pressed at $t = 0$,the capacitor acts as a short circuit (uncharged). Thus,the entire voltage $2 \; V$ appears across the $1000 \; \Omega$ resistor connected between $A$ and $B$.
At $t = 0$,the current $I = \frac{V}{R} = \frac{2 \; V}{1000 \; \Omega} = 2 \times 10^{-3} \; A = 2 \; mA$.
As time passes,the capacitor charges. Once the capacitor is fully charged,it acts as an open circuit. The current then flows through both $1000 \; \Omega$ resistors in series.
After a long time $(t \to \infty)$,the equivalent resistance is $R_{eq} = 1000 \; \Omega + 1000 \; \Omega = 2000 \; \Omega$.
The steady-state current is $I = \frac{V}{R_{eq}} = \frac{2 \; V}{2000 \; \Omega} = 1 \times 10^{-3} \; A = 1 \; mA$.
Therefore,at $t = 0, I = 2 \; mA$ and as time increases,it decreases to $1 \; mA$.

(B) The de-Broglie wavelength $\lambda$ of a particle of mass $m$ accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2mqV}}$.
Since both the electron and the proton have the same magnitude of charge $q = e$ and are accelerated through the same potential difference $V$,we have $\lambda \propto \frac{1}{\sqrt{m}}$.
For an electron: $\lambda_e = \lambda = \frac{h}{\sqrt{2meV}}$.
For a proton: $\lambda_p = \frac{h}{\sqrt{2MeV}}$.
Dividing the two equations: $\frac{\lambda_p}{\lambda_e} = \sqrt{\frac{m}{M}}$.
Therefore,$\lambda_p = \lambda \sqrt{\frac{m}{M}}$.

(C) Gibberellins are plant hormones that promote stem elongation by stimulating cell division and cell elongation in the internodal regions of plants. This is particularly evident in 'rosette' plants like cabbage and sugar beet,where they induce 'bolting' (internode elongation just prior to flowering). Therefore,the correct option is $C$.

(D) The stability of an ecosystem is often determined by its size,complexity,and the ability to maintain homeostasis despite environmental fluctuations.
Among the given options,the ocean is considered the most stable ecosystem.
This is because the ocean covers a vast area,has a high heat capacity (which regulates temperature),and possesses a complex food web that provides resilience against disturbances.
While forests are also stable,they are more susceptible to localized changes like fire or deforestation compared to the vast,interconnected marine environment.

(C) In good conductors of electricity,such as metals,the atoms are held together by metallic bonding.
In this type of bonding,the valence electrons are not bound to any specific atom but are free to move throughout the crystal lattice,forming a 'sea of electrons'.
This mobility of electrons is the fundamental reason why metals are excellent conductors of electricity.

(D) The rate of heat flow through a rod is given by the formula $\frac{Q}{t} = \frac{KA \Delta T}{L}$,where $K$ is the thermal conductivity,$A$ is the cross-sectional area,$\Delta T$ is the temperature difference,and $L$ is the length.
Since the material is the same,$K_1 = K_2$. Given $\Delta T_1 = \Delta T_2$,the ratio of the rates of heat flow is $\frac{(Q/t)_1}{(Q/t)_2} = \frac{A_1}{A_2} \times \frac{L_2}{L_1}$.
The area $A = \pi r^2 = \pi (d/2)^2$,so the ratio of areas is $\frac{A_1}{A_2} = \left(\frac{d_1}{d_2}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.
Given the ratio of lengths $\frac{L_1}{L_2} = \frac{2}{1}$,we have $\frac{L_2}{L_1} = \frac{1}{2}$.
Substituting these values: $\frac{(Q/t)_1}{(Q/t)_2} = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$.

(C) The moment of inertia $(I)$ of a body depends on the distribution of mass relative to the axis of rotation. The farther the mass is from the axis,the greater the moment of inertia.
For a triangular plate,the moment of inertia about an axis is proportional to the square of the distance of the center of mass from that axis.
Let the height of the triangle with respect to base $BC$ be $h_A = AB = 4$ units.
Let the height of the triangle with respect to base $AB$ be $h_C = BC = 3$ units.
Let the height of the triangle with respect to hypotenuse $AC$ be $h_B$. Since the area is $\frac{1}{2} \times 3 \times 4 = 6$,we have $\frac{1}{2} \times 5 \times h_B = 6$,so $h_B = 2.4$ units.
The moment of inertia is proportional to the square of the perpendicular distance of the center of mass from the axis. The distance of the centroid from any side is $\frac{1}{3}$ of the altitude to that side.
Thus,$I \propto h^2$.
$I_{AB} \propto (BC)^2 = 3^2 = 9$
$I_{BC} \propto (AB)^2 = 4^2 = 16$
$I_{CA} \propto (h_B)^2 = (2.4)^2 = 5.76$
Comparing these values,we get $I_{BC} > I_{AB} > I_{CA}$.
Therefore,the relation $I_{BC} > I_{AB}$ is correct.

(D) To determine the bond length,we look at the bond order of the $O-O$ bond in each species:
$1$. In $O_2$,the bond order is $2.0$ (double bond).
$2$. In $O_3$ (ozone),the resonance hybrid has a bond order of $1.5$ for the $O-O$ bond.
$3$. In $H_2O_2$,the $O-O$ bond is a single bond with a bond order of $1.0$.
Since bond length is inversely proportional to bond order,the order of bond length is $H_2O_2 (1.0) > O_3 (1.5) > O_2 (2.0)$.

(D) The number of $\alpha$ particles emitted is given by the change in mass number divided by $4$:
$n_{\alpha} = \frac{A - A'}{4} = \frac{200 - 168}{4} = \frac{32}{4} = 8$.
The number of $\beta$ particles emitted is given by the formula:
$n_{\beta} = 2n_{\alpha} - (Z - Z')$.
Substituting the values:
$n_{\beta} = 2(8) - (90 - 80) = 16 - 10 = 6$.
Therefore,$8$ $\alpha$ particles and $6$ $\beta$ particles are emitted.

(C) The wave number $\bar{\nu}$ is given by the Rydberg formula:
$\bar{\nu} = \frac{1}{\lambda} = R \left[ \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right]$
Here,the transition is from $n_{2} = 4$ to $n_{1} = 2$.
Substituting the values:
$\bar{\nu} = R \left[ \frac{1}{2^{2}} - \frac{1}{4^{2}} \right]$
$\bar{\nu} = R \left[ \frac{1}{4} - \frac{1}{16} \right]$
$\bar{\nu} = R \left[ \frac{4 - 1}{16} \right] = \frac{3R}{16}$

(B) According to Newton's Law of Cooling,the rate of cooling $\frac{dT}{dt}$ is directly proportional to the temperature difference between the body and its surroundings: $\frac{dT}{dt} = -k(T - T_s)$.
As the temperature of the water decreases,the temperature difference $(T - T_s)$ between the water and the room decreases.
Since the rate of cooling decreases as the temperature difference decreases,the time taken to cool by the same amount $(5\,^{\circ}C)$ will increase.
Therefore,$t_1 < t_2 < t_3$.

(D) The bond length is inversely proportional to the bond order.
$1$. In $O_2$,the bond order is $2.0$ (double bond).
$2$. In $O_3$,the bond order is $1.5$ (resonance hybrid).
$3$. In $H_2O_2$,the bond order is $1.0$ (single bond).
Since the bond order follows the order $O_2 > O_3 > H_2O_2$,the bond length follows the order $H_2O_2 > O_3 > O_2$.

(A) An amplifier increases the magnitude of an externally applied $AC$ input voltage at the output.
An oscillator is a device that produces an output signal at a desired frequency without requiring any external $AC$ input voltage.
This self-sustaining oscillation is achieved by providing a positive feedback from a portion of the amplifier's output back to its input circuit in the correct phase.
Therefore,an oscillator is essentially an amplifier with positive feedback.

(A) Blood maintains a constant $pH$ because it contains serum proteins and bicarbonate ions which act as a buffer system.
This buffer system resists changes in $pH$ upon the addition of small amounts of acid or base.

(B) Enterokinase (also known as enteropeptidase) is an enzyme secreted by the intestinal mucosa.
It acts on the inactive enzyme trypsinogen,which is secreted by the pancreas into the duodenum.
Enterokinase converts inactive trypsinogen into its active form,trypsin.
This activation is a critical step in the process of protein digestion,as trypsin then activates other pancreatic zymogens like chymotrypsinogen and procarboxypeptidase.

(D) The reaction $CH_3CHO + HCN \to CH_3CH(OH)CN$ involves the nucleophilic addition of cyanide to the carbonyl carbon of acetaldehyde.
Since the carbonyl carbon is planar,the cyanide ion can attack from either side with equal probability.
This leads to the formation of both $d$ (dextro) and $l$ (laevo) enantiomers in equal amounts.
Therefore,the resulting product is a racemic mixture.

(C) The dehydration of $3, 3-$dimethyl$-2-$butanol with ${H_2SO_4}$ proceeds via the formation of a carbocation.
$1.$ Protonation of the alcohol group leads to the formation of a $2^o$ carbocation: $CH_3-CH^+-C(CH_3)_2-CH_3$.
$2.$ This $2^o$ carbocation undergoes a $1, 2-$methyl shift to form a more stable $3^o$ carbocation: $CH_3-CH(CH_3)-C^+(CH_3)-CH_3$.
$3.$ Loss of a proton from the $3^o$ carbocation yields the most stable alkene,which is $2, 3-$dimethyl$-2-$butene,according to Saytzeff's rule.

(B) Molality is defined as the number of moles of solute per $1 \ kg$ of solvent.
Since mass does not change with temperature,molality is independent of temperature.
In contrast,molarity,formality,and normality involve volume,which changes with temperature.

(B) According to Raoult's law,for a solution containing a non-volatile solute,the relative lowering of vapour pressure is given by the expression: $\frac{P^{\circ} - P_s}{P^{\circ}} = X_2$.
Here,$\frac{P^{\circ} - P_s}{P^{\circ}}$ represents the relative lowering of vapour pressure,and $X_2$ is the mole fraction of the solute.
Thus,the relative lowering of vapour pressure is equal to the mole fraction of the solute.

(D) The initial nucleus is $_{92}X^{238}$.
After the emission of one $\alpha$-particle $(_{2}He^{4})$,the nucleus becomes $_{90}Y^{234}$.
After the subsequent emission of one $\beta$-particle $(_{-1}e^{0})$,the nucleus becomes $_{91}Z^{234}$.
The number of neutrons is calculated as $A - Z = 234 - 91 = 143$.

(A) The $E.M.F.$ of a concentration cell is given by $E_{cell} = \frac{0.0591}{n} \log \frac{[H^+]_{cathode}}{[H^+]_{anode}}$.
Since $HCl$ is a strong acid,it dissociates completely,whereas $CH_3COOH$ is a weak acid and dissociates partially.
Therefore,the concentration of $H^+$ ions in both compartments is different,leading to different $pH$ values.
As a result,the $E.M.F.$ of the cell is not zero.

(D) To determine the colourless species,we check for the presence of unpaired electrons in the metal ions:
$1$. In $TiF_6^{2-}$,$Ti$ is in the $+4$ oxidation state. The electronic configuration of $Ti^{4+}$ is $[Ar] 3d^0$. Since there are no $d$-electrons,it is colourless.
$2$. In $CoF_6^{3-}$,$Co$ is in the $+3$ oxidation state. The configuration of $Co^{3+}$ is $[Ar] 3d^6$. It has unpaired electrons,so it is coloured.
$3$. In $Cu_2Cl_2$,$Cu$ is in the $+1$ oxidation state. The configuration of $Cu^+$ is $[Ar] 3d^{10}$. Since the $d$-subshell is completely filled,there are no unpaired electrons,making it colourless.
$4$. In $NiCl_4^{2-}$,$Ni$ is in the $+2$ oxidation state. The configuration of $Ni^{2+}$ is $[Ar] 3d^8$. It has unpaired electrons,so it is coloured.
Therefore,$TiF_6^{2-}$ and $Cu_2Cl_2$ are the colourless species.

(A) The coordination complex $[Pt(NH_3)_2Cl_2]$ is a square planar complex of the type $[MA_2B_2]$.
It exhibits $2$ geometrical isomers:
$1$. $\text{Cis-isomer}$: Where similar ligands ($NH_3$ or $Cl$) are adjacent to each other.
$2$. $\text{Trans-isomer}$: Where similar ligands ($NH_3$ or $Cl$) are opposite to each other.

(B) The complex is $K_3[Cr(C_2O_4)_3]$.
$1$. Coordination Number: The ligand oxalate $(C_2O_4)^{2-}$ is a bidentate ligand. Since there are $3$ such ligands,the coordination number is $3 \times 2 = 6$.
$2$. Oxidation State: Let the oxidation state of $Cr$ be $x$. The charge on the oxalate ion is $-2$ and the charge on the potassium ion is $+1$. The total charge of the complex is $0$.
$3(+1) + x + 3(-2) = 0$
$3 + x - 6 = 0$
$x - 3 = 0$
$x = +3$
Therefore,the coordination number is $6$ and the oxidation state is $+3$.

(C) The acidity of phenols depends on the stability of the phenoxide ion formed after the loss of a proton. Electron-withdrawing groups $(EWG)$ like $-NO_2$ increase acidity by stabilizing the phenoxide ion through $-I$ and $-M$ effects,while electron-donating groups $(EDG)$ like $-CH_3$ decrease acidity through $+I$ and hyperconjugation effects.
$1$. $p-$methylphenol: The $-CH_3$ group is an $EDG$,which destabilizes the phenoxide ion,making it the least acidic.
$2$. Phenol: Acts as the reference compound.
$3$. $m-$nitrophenol: The $-NO_2$ group exerts only a $-I$ effect from the meta position,increasing acidity more than phenol.
$4$. $p-$nitrophenol: The $-NO_2$ group exerts both $-I$ and $-M$ effects from the para position,providing the strongest stabilization to the phenoxide ion,making it the most acidic.
Thus,the increasing order of acidity is: $p-$methylphenol < phenol < $m-$nitrophenol < $p-$nitrophenol.

(C) The oxidation of toluene with $CrO_3$ in the presence of acetic anhydride $(CH_3CO)_2O$ forms a gem-diacetate intermediate,$C_6H_5CH(OCOCH_3)_2$,which is product '$A$'.
Upon hydrolysis with aqueous $NaOH$,this intermediate yields benzaldehyde $(C_6H_5CHO)$.
Benzaldehyde,lacking $\alpha$-hydrogens,undergoes the Cannizzaro reaction in the presence of concentrated aqueous $NaOH$ to produce sodium benzoate $(C_6H_5COONa)$ and benzyl alcohol $(C_6H_5CH_2OH)$.
Therefore,the final product obtained is $C_6H_5COONa$.

(C) $\alpha-D$-glucose and $\beta-D$-glucose are anomers of each other.
Anomers are a type of stereoisomer that differ in configuration only at the anomeric carbon (the $C-1$ carbon in glucose).
Since they differ in the spatial arrangement of the $-OH$ group at the $C-1$ position,they are said to differ in their configuration.

(A) The complete oxidation of one molecule of glucose $(C_6H_{12}O_6)$ in a living cell follows the overall reaction:
$C_6H_{12}O_6 + 6O_2 \to 6CO_2 + 6H_2O + 38 \, ATP$.
Thus,the total number of $ATP$ molecules generated is $38$.

(D) Secondary structure refers to the regular folding patterns of continuous portions of the polypeptide chain,such as $\alpha$-helices and $\beta$-pleated sheets.
These structures are stabilized by hydrogen bonds formed between the carbonyl oxygen $(C=O)$ of one amino acid residue and the amino hydrogen $(N-H)$ of another residue in the polypeptide backbone.

(C) $i$. False: Enzymes contain various nucleophilic groups (e.g.,$-OH$,$-SH$,$-NH_2$) in their active sites that participate in catalysis.
$ii$. True: Enzymes exhibit high stereospecificity and substrate specificity.
$iii$. True: Enzymes function by providing an alternative pathway with a lower activation energy $(E_a)$.
$iv$. True: Pepsin is a digestive enzyme that breaks down proteins into peptides.
Therefore,statements $ii$,$iii$,and $iv$ are correct.

(C) catalyst provides an alternative pathway for the reaction with lower activation energy. By lowering the activation energy,a larger fraction of molecules can cross the energy barrier,which effectively increases the rate of the reaction. While the frequency of collisions is determined by temperature and concentration,the catalyst's primary role is to alter the reaction mechanism to a lower energy path. However,in the context of standard multiple-choice questions regarding the effect of catalysts on reaction rates,option $C$ is the most scientifically accurate description of how a catalyst functions.

(A) . The reactivity of carbonyl compounds towards nucleophilic addition is governed by two factors: steric hindrance and electronic effects.
$1$. Steric hindrance: As the number and size of alkyl or aryl groups attached to the carbonyl carbon increase,the approach of the nucleophile becomes more difficult.
$2$. Electronic effects: Alkyl groups are electron-donating ($+I$ effect),which decreases the electrophilicity of the carbonyl carbon.
Formaldehyde $(H_2C = O)$ has the least steric hindrance and no electron-donating groups,making it the most reactive.
Aldehydes $(RCHO)$ are more reactive than ketones $(R_2C = O)$ due to less steric hindrance.
Aryl groups $(Ar)$ provide resonance stabilization to the carbonyl carbon,further reducing its electrophilicity compared to alkyl groups.
Thus,the correct order is $H_2C = O > RCHO > ArCHO > R_2C = O > Ar_2C = O$.