AIPMT 2006 Chemistry Question Paper with Answer and Solution

(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$.
When the battery is disconnected,the charge $Q$ on the plates remains constant.
The potential difference $V$ across the plates is given by $V = \frac{Q}{C}$.
Substituting the expression for $C$,we get $V = \frac{Q d}{\varepsilon_0 A}$.
Since $Q$,$\varepsilon_0$,and $A$ are constant,we have $V \propto d$.
When the distance $d$ between the plates is increased,the potential difference $V$ increases.

(C) The energy of the electron in the ground state $(n=1)$ of a hydrogen atom is $E_1 = -13.6 \ eV$.
When the atom absorbs a photon of energy $12.1 \ eV$,the final energy of the electron becomes $E_f = E_1 + 12.1 \ eV = -13.6 \ eV + 12.1 \ eV = -1.51 \ eV$.
Using the formula $E_n = -\frac{13.6}{n^2} \ eV$,we find the excited state $n$:
$-1.51 = -\frac{13.6}{n^2} \implies n^2 = \frac{13.6}{1.51} \approx 9 \implies n = 3$.
The number of spectral lines emitted when an electron transitions from an excited state $n$ to lower energy states is given by the formula $N = \frac{n(n-1)}{2}$.
For $n=3$,the number of spectral lines is $N = \frac{3(3-1)}{2} = \frac{3 \times 2}{2} = 3$.

(A) Sertoli cells are located in the seminiferous tubules of the testes.
They are regulated by the Follicle Stimulating Hormone $(FSH)$,which is secreted by the anterior pituitary gland.
$FSH$ acts on the Sertoli cells and stimulates the secretion of some factors which help in the process of spermiogenesis.
Therefore,the correct option is $A$.

(C) In $1953$,Stanley Miller and Harold Urey conducted an experiment to simulate the conditions of the primitive Earth's atmosphere. They used a mixture of $CH_4$,$NH_3$,$H_2$,and water vapor in a closed flask and subjected it to electric discharges. The experiment resulted in the formation of simple organic compounds,including amino acids like glycine,alanine,and aspartic acid. Glutamic acid was not among the amino acids synthesized in this specific experiment.

(A) The evolutionary history and relationship of a species or group of organisms is called $Phylogeny$.
$Ontogeny$ refers to the developmental history of an individual organism.
$Paleontology$ is the study of fossils.
$Ancestry$ refers to the lineage or descent of an organism.
Therefore,the correct term for evolutionary history is $Phylogeny$.

(B) In angiosperms,the synergid cells are haploid $(n)$.
Given that the number of chromosomes in the synergid cell is $n = 8$.
The endosperm in angiosperms is formed by the fusion of two polar nuclei and one male gamete,resulting in a triploid $(3n)$ structure.
Therefore,the number of chromosomes in the endosperm = $3 \times n = 3 \times 8 = 24$.

(A) The formula for freezing point depression is $\Delta T_f = K_f \times m$,where $m$ is the molality of the solution.
Molality $m = \frac{W_1 \times 1000}{M_1 \times W_2}$,where $W_1$ is the mass of solute,$M_1$ is the molar mass of solute,and $W_2$ is the mass of solvent in grams.
Given: $W_1 = 1.00 \ g$,$M_1 = 250 \ g \ mol^{-1}$,$W_2 = 51.2 \ g$,and $K_f = 5.12 \ K \ kg \ mol^{-1}$.
Substituting the values: $\Delta T_f = 5.12 \times \frac{1.00 \times 1000}{250 \times 51.2}$.
$\Delta T_f = 5.12 \times \frac{1000}{12800} = 5.12 \times 0.078125 = 0.4 \ K$.

(C) Let $r_{1}$ and $r_{2}$ be the radii of coil $1$ and $2$ respectively. Given $r_{1} = 2r_{2}$.
The magnetic field at the center of a circular coil is given by $B = \frac{\mu_{0} I}{2r}$.
For the fields to be equal,$B_{1} = B_{2} \implies \frac{\mu_{0} I_{1}}{2r_{1}} = \frac{\mu_{0} I_{2}}{2r_{2}}$.
Substituting $r_{1} = 2r_{2}$,we get $\frac{I_{1}}{2r_{2}} = \frac{I_{2}}{r_{2}}$,which simplifies to $I_{1} = 2I_{2}$.
The resistance of a wire is proportional to its length,$R \propto L = 2\pi r$. Since both coils are made of the same wire,$R_{1} = 2R_{2}$.
The potential difference is $V = IR$. Therefore,the ratio of potential differences is $\frac{V_{1}}{V_{2}} = \frac{I_{1} R_{1}}{I_{2} R_{2}}$.
Substituting the values,$\frac{V_{1}}{V_{2}} = \frac{(2I_{2})(2R_{2})}{I_{2} R_{2}} = 4$.

(A) The radius of a nucleus is given by the formula $R = R_0 A^{1/3}$,where $R_0$ is a constant and $A$ is the mass number (number of nucleons).
Given that the radius of $Ge$ is twice the radius of ${}_4^9Be$,we have:
$R_{Ge} = 2 R_{Be}$
Using the formula $R \propto A^{1/3}$:
$\frac{R_{Ge}}{R_{Be}} = \left( \frac{A_{Ge}}{A_{Be}} \right)^{1/3}$
Substituting the given values:
$2 = \left( \frac{A_{Ge}}{9} \right)^{1/3}$
Cubing both sides:
$2^3 = \frac{A_{Ge}}{9}$
$8 = \frac{A_{Ge}}{9}$
$A_{Ge} = 8 \times 9 = 72$
Therefore,the number of nucleons in $Ge$ is $72$.

(A) The correct formula for exponential population growth is $dN/dt = rN$.
In this equation:
$N$ represents the population size.
$t$ represents time.
$r$ represents the intrinsic rate of natural increase.
$dN/dt$ represents the rate of change in population size over time.
When resources are unlimited,populations tend to grow exponentially,which is described by this differential equation.

(A) $AIDS$ (Acquired Immuno Deficiency Syndrome) is a syndrome caused by the retrovirus $HIV$ (Human Immunodeficiency Virus).
Upon entering the body,the virus targets and destroys a specific subgroup of white blood cells known as helper $T$-cells (also called $CD4$ lymphocytes).
These cells are crucial for coordinating the immune response; their destruction leads to a severe suppression of the body's immune system,making the individual susceptible to opportunistic infections.
$HIV$ is transmitted through infected blood,semen,and vaginal fluids.
While antiviral therapy can delay the progression of the disease,it cannot completely cure $AIDS$.

(C) : Hybridisation,heterosis,or hybrid vigour is defined as the superiority of the hybrid over the parents. It has been commercially exploited in different commercial crops like maize,sorghum,bajra,etc. The main steps include: selection of parents,selfing of parents,emasculation,bagging,crossing of desired and selected parents,and finally seed setting and harvesting.

(B) : Hotspots are areas with high density of biodiversity or megadiversity which are also the most threatened ones.
Ecologically,hotspots are determined by four factors: number of species/species diversity,degree of endemism,degree of threat to habitat due to its degradation and fragmentation,and degree of exploitation.
India has three biodiversity hotspots: Indo-Burma,Himalayas,and Western Ghats & Sri Lanka.
India is a country of megadiversity with $2.4\%$ of the global land area and $8.1\%$ of the global species diversity.
Major centers of biodiversity within the Western Ghats include Agasthyamalai hills,Silent Valley,and Amambalam reserve.
There is a high degree of endemism as well as richness of species of flowering plants,amphibians,reptiles,some mammals,and butterflies.

(D) The magnetic quantum number is denoted by $m_l$.
It provides information about the spatial orientation of the orbital with respect to the standard set of coordinate axes.
Explanation for incorrect options:
- The azimuthal quantum number $(l)$ determines the shape of the orbital.
- The spin quantum number $(m_s)$ denotes the orientation of the spin of the electron.
- The principal quantum number $(n)$ determines the size and,to a large extent,the energy of the orbital.
Therefore,options $(A)$,$(B)$,and $(C)$ are incorrect.
Hence,option $(D)$ is correct.

(B) According to Heisenberg's uncertainty principle:
$\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}$
Since $\Delta p = m \cdot \Delta v$,the formula becomes:
$\Delta x \cdot m \cdot \Delta v \geq \frac{h}{4 \pi}$
$\Delta v \geq \frac{h}{4 \pi \cdot m \cdot \Delta x}$
Given:
$\Delta x = 0.1 \ \mathring{A} = 0.1 \times 10^{-10} \ m = 10^{-11} \ m$
$m = 9.11 \times 10^{-31} \ kg$
$h = 6.626 \times 10^{-34} \ J \ s$
$\pi = 3.14$
Substituting the values:
$\Delta v = \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.11 \times 10^{-31} \times 10^{-11}}$
$\Delta v = \frac{6.626 \times 10^{-34}}{114.46 \times 10^{-42}}$
$\Delta v \approx 0.05789 \times 10^8 \ m \ s^{-1} = 5.79 \times 10^6 \ m \ s^{-1}$

(A) The correct order of bond dissociation energy for halogens is $Cl_2 > Br_2 > F_2 > I_2$.
The given order $F_2 > Cl_2 > Br_2 > I_2$ is incorrect because the $F-F$ bond is weaker than the $Cl-Cl$ bond due to the small size of the $F$ atom,which causes significant inter-electronic repulsion between the lone pairs of the two fluorine atoms.

(D) Generally,$AB_5$ molecules have a trigonal bipyramidal structure due to $sp^3d$ hybridization.
For example,$PCl_5$ has a trigonal bipyramidal geometry.
Therefore,the statement that every $AB_5$ molecule has a square pyramidal structure is incorrect.

(D) $NO_2^+$ has a linear shape due to $sp$ hybridization of the $N$ atom.
$O_3$,$NO_2^-$,and $SO_2$ all have an angular (bent) shape due to the presence of a lone pair on the central atom.

(B) $SiCl_4$ has $sp^3$ hybridization and a tetrahedral geometry.
$NH_4^+$,$SO_4^{2-}$,and $PO_4^{3-}$ also exhibit $sp^3$ hybridization and a tetrahedral geometry,making them isostructural with $SiCl_4$.
$SCl_4$ has $sp^3d$ hybridization with one lone pair on the $S$ atom,resulting in a see-saw geometry.
Therefore,$SCl_4$ is not isostructural with $SiCl_4$.

(C) For a process occurring at constant temperature and pressure:
$1$. If $\Delta G_{system} < 0,$ the process is spontaneous.
$2$. If $\Delta G_{system} > 0,$ the process is non-spontaneous.
$3$. If $\Delta G_{system} = 0,$ the system is in a state of equilibrium.

(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by the equation: $\Delta H = \Delta E + \Delta n_{g} RT$.
For $\Delta H = \Delta E$ to hold true,the change in the number of moles of gaseous species $(\Delta n_{g})$ must be $0$.
Calculating $\Delta n_{g}$ for option $B$: $H_{2_{(g)}} + Br_{2_{(g)}} \rightarrow 2HBr_{(g)}$.
$\Delta n_{g} = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) = 2 - (1 + 1) = 0$.
Since $\Delta n_{g} = 0$,it follows that $\Delta H = \Delta E$.

(B) For the reaction $Br_{2(l)} + Cl_{2(g)} \rightarrow 2BrCl_{(g)}$,we have:
$\Delta H = 30 \ kJ \ mol^{-1} = 30000 \ J \ mol^{-1}$
$\Delta S = 105 \ J \ K^{-1} \ mol^{-1}$
At equilibrium,the Gibbs free energy change $\Delta G$ is $0$.
Using the relation $\Delta G = \Delta H - T \Delta S$,we set $\Delta G = 0$:
$0 = \Delta H - T \Delta S$
$\Delta H = T \Delta S$
$T = \frac{\Delta H}{\Delta S} = \frac{30000 \ J \ mol^{-1}}{105 \ J \ K^{-1} \ mol^{-1}} \approx 285.7 \ K$.

(B) The enthalpy of hydrogenation of cyclohexene is $-119.5 \ kJ \ mol^{-1},$ which corresponds to the hydrogenation of one $C=C$ double bond.
Benzene contains three $C=C$ double bonds. Therefore,the calculated enthalpy of hydrogenation for benzene (assuming no resonance) is:
$\Delta H_{Cal} = 3 \times (-119.5 \ kJ \ mol^{-1}) = -358.5 \ kJ \ mol^{-1}.$
Resonance energy is defined as the difference between the experimental enthalpy of hydrogenation and the calculated enthalpy of hydrogenation:
$\text{Resonance Energy} = \Delta H_{Exp} - \Delta H_{Cal}.$
Given that the resonance energy is $-150.4 \ kJ \ mol^{-1},$ we have:
$-150.4 = \Delta H_{Exp} - (-358.5).$
Solving for $\Delta H_{Exp}:$
$\Delta H_{Exp} = -150.4 - 358.5 = -508.9 \ kJ \ mol^{-1}.$

(C) The given reaction is $:$
$CH_{4(g)} + 2O_{2(g)} \rightleftharpoons CO_{2(g)} + 2H_2O_{(l)}$
The equilibrium constant expression for this reaction is $K_c = \frac{[CO_2]}{[CH_4][O_2]^2}$ because $H_2O$ is a pure liquid and its activity is taken as $1$.
Option $A$ is true because $\Delta H_r$ is negative,indicating an exothermic reaction.
Option $B$ is true because there is no requirement for the concentrations of products to be equal at equilibrium.
Option $C$ is incorrect because $K_p$ is defined in terms of partial pressures,not concentrations. The expression given is for $K_c$,and it is also missing the pressure terms for the gaseous species.
Option $D$ is true according to Le Chatelier's principle; adding reactants shifts the equilibrium to the right.

(B) buffer solution is formed by a mixture of a weak acid and its conjugate base (salt with a strong base) or a weak base and its conjugate acid (salt with a strong acid).
$HNO_2$ is a weak acid and $NaNO_2$ is its salt with a strong base $(NaOH)$.
Therefore,the pair $HNO_2$ and $NaNO_2$ constitutes an acidic buffer solution.

(C) In an aqueous solution of $10^{-8} \ M \ HCl$,the total $[H^{+}]$ is the sum of $[H^{+}]$ from $HCl$ and $[H^{+}]$ from the dissociation of water.
Let $[H^{+}] = x$.
From $HCl$,$[H^{+}] = 10^{-8} \ M$.
From water,$[H^{+}] = [OH^{-}] = x - 10^{-8} \ M$.
Using the ionic product of water,$K_w = [H^{+}][OH^{-}] = 10^{-14}$.
$x(x - 10^{-8}) = 10^{-14}$
$x^{2} - 10^{-8}x - 10^{-14} = 0$
Solving this quadratic equation using the quadratic formula $x = \frac{-b + \sqrt{b^{2} - 4ac}}{2a}$:
$x = \frac{10^{-8} + \sqrt{(10^{-8})^{2} - 4(1)(-10^{-14})}}{2}$
$x = \frac{10^{-8} + \sqrt{10^{-16} + 4 \times 10^{-14}}}{2}$
$x = \frac{10^{-8} + \sqrt{10^{-16} + 400 \times 10^{-16}}}{2}$
$x = \frac{10^{-8} + \sqrt{401 \times 10^{-16}}}{2}$
$x = \frac{10^{-8} + 20.025 \times 10^{-8}}{2}$
$x = \frac{21.025 \times 10^{-8}}{2} = 1.05125 \times 10^{-7} \ M \approx 1.0525 \times 10^{-7} \ M$.

(A) In an aqueous solution,alkali metal ions get hydrated. The extent of hydration depends on the size of the cation; smaller cations have higher charge density and thus higher hydration energy.
The order of hydration energy is $Li^{+} > Na^{+} > K^{+} > Rb^{+}$.
Due to higher hydration,the effective size of the hydrated $Li^{+}$ ion is the largest,which makes it move the slowest in an electric field.
Therefore,the mobility of the ions in an aqueous solution is inversely proportional to their hydrated size,resulting in the order: $Rb^{+} > K^{+} > Na^{+} > Li^{+}$.

(D) In the periodic table,the acidic nature of oxides increases from left to right across a period,while the basic character increases from top to bottom down a group.
$SeO_2$ is an acidic oxide (non-metal oxide).
$Al_2O_3$ is an amphoteric oxide.
$Sb_2O_3$ is amphoteric (with more basic character than $Al_2O_3$ due to its position in the group).
$Bi_2O_3$ is the most basic oxide among the given options because $Bi$ is the most metallic element in the group.
Therefore,the order of basic nature is $Bi_2O_3 > Sb_2O_3 > Al_2O_3 > SeO_2$.

(C) The electronegativity of hybrid orbitals depends on the $s$-character present in the orbital.
As the $s$-character increases,the electron-attracting tendency (electronegativity) increases.
The $s$-character in different hybrid orbitals is as follows:
$sp$: $50\%$
$sp^2$: $33.3\%$
$sp^3$: $25\%$
Therefore,the correct order of electronegativity is $sp > sp^2 > sp^3$.

(C) $1$. Identify the principal functional group: The compound contains an acyl chloride group $(-COCl)$,which takes priority in numbering.
$2$. Select the longest carbon chain containing the functional group: The longest chain starting from the carbonyl carbon has $5$ carbons,making it a pentanoyl chloride derivative.
$3$. Number the chain: The carbonyl carbon of the $-COCl$ group is assigned position $1$.
$4$. Identify substituents: There are methyl groups at positions $2$ and $3$.
$5$. Combine the parts: The name is $2,3-$dimethylpentanoyl chloride.

(C) For isotonic solutions,the osmotic pressures are equal,so $\pi_1 = \pi_2$.
Since the temperature is constant,the molar concentrations are equal: $C_1 = C_2$.
For urea: Concentration $C_1 = \frac{10 \ g}{60 \ g \ mol^{-1} \times 1 \ L} = \frac{1}{6} \ mol \ L^{-1}$.
For the non-volatile solute: $A$ $5 \%$ solution means $5 \ g$ of solute in $100 \ mL$ of solution,which is $50 \ g \ L^{-1}$.
So,$C_2 = \frac{50}{M_w} \ mol \ L^{-1}$,where $M_w$ is the molecular mass.
Equating the concentrations: $\frac{1}{6} = \frac{50}{M_w}$.
Solving for $M_w$: $M_w = 50 \times 6 = 300 \ g \ mol^{-1}$.

(C) The electronegativity of $N$ $(3.04)$ is greater than that of $H$ $(2.20)$,so the bond dipoles $(N-H)$ and the lone pair dipole on $N$ point in the same direction,reinforcing each other.
In $NF_3$,the electronegativity of $F$ $(3.98)$ is greater than that of $N$ $(3.04)$,so the bond dipoles $(N-F)$ point away from $N$,which is opposite to the direction of the lone pair dipole on $N$. This causes the bond dipoles to partially cancel the lone pair dipole,resulting in a much smaller net dipole moment.

(B) $BF_3$ has a trigonal planar structure with $sp^2$ hybridization,where all $B-F$ bonds are equivalent.
$AlF_3$ is an ionic compound with a crystal lattice structure where all $Al-F$ bonds are equivalent.
$NF_3$ has a trigonal pyramidal geometry with $sp^3$ hybridization,where all $N-F$ bonds are equivalent.
$ClF_3$ has a $T$-shaped geometry due to $sp^3d$ hybridization with two lone pairs in the equatorial positions. This results in two different bond lengths (axial and equatorial),meaning all bonds are not equal.
Therefore,the correct option is $B$.

(D) molecule is chiral if it contains at least one chiral carbon atom,which is a carbon atom bonded to four different groups.
$1$. $2-$Hydroxypropanoic acid: $CH_3-CH(OH)-COOH$. The $C-2$ atom is bonded to $-H$,$-OH$,$-CH_3$,and $-COOH$. It is chiral.
$2$. $2-$Butanol: $CH_3-CH(OH)-CH_2-CH_3$. The $C-2$ atom is bonded to $-H$,$-OH$,$-CH_3$,and $-CH_2CH_3$. It is chiral.
$3$. $2,3-$Dibromopentane: $CH_3-CH(Br)-CH(Br)-CH_2-CH_3$. Both $C-2$ and $C-3$ are bonded to four different groups. It is chiral.
$4$. $3-$Bromopentane: $CH_3-CH_2-CH(Br)-CH_2-CH_3$. The $C-3$ atom is bonded to $-H$,$-Br$,and two identical $-CH_2CH_3$ groups. Since it is bonded to two identical groups,it is achiral.
Therefore,$3-$Bromopentane is not chiral.

(C) Organic compounds,other than carbohydrates and proteins,which are required for normal growth and nutrition in the human body but are not synthesized by the human body are called vitamins. They must be supplied through the diet.

(B) By observing the given waveforms:
At $t_1$,$A=0, B=0 \implies C=0$.
At $t_2$,$A=1, B=1 \implies C=1$.
At $t_3$,$A=0, B=1 \implies C=0$.
At $t_4$,$A=1, B=0 \implies C=0$.
This behavior corresponds to the truth table of an $AND$ gate,where the output is $1$ only when both inputs are $1$.

$A$	$B$	$C$
$0$	$0$	$0$
$0$	$1$	$0$
$1$	$0$	$0$
$1$	$1$	$1$

(B) In the mechanism of hormone action,hormones that cannot enter the target cell (like peptide or protein hormones) bind to specific receptors on the cell membrane. This binding triggers the production of intracellular molecules known as second messengers,which regulate cellular metabolism. Common second messengers include $cAMP$,$IP_3$ (Inositol triphosphate),and $Ca^{2+}$ (Calcium ions). Sodium $(Na^+)$ ions are involved in membrane potential and nerve impulse conduction but do not function as second messengers in hormone signaling pathways.

(C) Bowman's glands (also known as olfactory glands) are specialized glands located in the olfactory epithelium of the nasal cavity.
These glands secrete mucus,which helps in dissolving odorants,thereby facilitating the sense of smell (olfaction).
Therefore,the correct option is $C$.

(B) In $1953$, Stanley Miller and Harold Urey conducted an experiment to simulate the conditions of the primitive Earth's atmosphere. They used a mixture of $CH_4$, $NH_3$, $H_2$, and water vapor in a closed flask and subjected it to electric discharges. The experiment resulted in the formation of several organic compounds, including amino acids like $Glycine$, $Alanine$, and $Aspartic$ acid. $Glutamic$ acid was not among the amino acids synthesized in this specific experiment.

(B) Benzyl amine,$C_6H_5CH_2NH_2$,is more basic than aniline because the $C_6H_5CH_2-$ group is an electron-donating group due to the $+I$ effect.
This increases the electron density on the nitrogen atom of the $-NH_2$ group.
Consequently,the availability of the lone pair of electrons for donation increases,making it more basic.
In contrast,phenyl and nitro groups are electron-withdrawing groups,which decrease the electron density on the nitrogen atom,making them less basic than aniline.

(A) The torque $\tau$ acting on the rod about point $A$ is due to the weight of the rod acting at its center of mass,which is at a distance of $\frac{l}{2}$ from $A$.
$\tau = mg \left(\frac{l}{2}\right)$
Using the rotational form of Newton's second law,$\tau = I\alpha$,where $I$ is the moment of inertia about $A$ and $\alpha$ is the angular acceleration.
Given $I = \frac{ml^2}{3}$,we have:
$mg \left(\frac{l}{2}\right) = \left(\frac{ml^2}{3}\right) \alpha$
Solving for $\alpha$:
$\alpha = \frac{mg(l/2)}{ml^2/3} = \frac{mg l}{2} \cdot \frac{3}{ml^2} = \frac{3g}{2l}$

(A) The relationship between enthalpy change $(\Delta H)$ and internal energy change ($\Delta E$ or $\Delta U$) is given by the equation: $\Delta H = \Delta E + \Delta n_g RT$.
Here,$\Delta n_g$ is the change in the number of moles of gaseous species,calculated as: $\Delta n_g = \sum n_g(\text{products}) - \sum n_g(\text{reactants})$.
For $\Delta H = \Delta E$,the term $\Delta n_g RT$ must be equal to $0$,which implies $\Delta n_g = 0$.
Evaluating the options:
$A$: $H_{2(g)} + Br_{2(g)} \to 2HBr_{(g)}$,$\Delta n_g = 2 - (1 + 1) = 0$.
$B$: $C_{(s)} + 2H_2O_{(g)} \to 2H_{2(g)} + CO_{2(g)}$,$\Delta n_g = (2 + 1) - 2 = 1$.
$C$: $PCl_{5(g)} \to PCl_{3(g)} + Cl_{2(g)}$,$\Delta n_g = (1 + 1) - 1 = 1$.
$D$: $2CO_{(g)} + O_{2(g)} \to 2CO_{2(g)}$,$\Delta n_g = 2 - (2 + 1) = -1$.
Thus,the condition $\Delta H = \Delta E$ is satisfied for reaction $A$.

(B) The nuclear fusion reaction is given by: ${}_1^2H + {}_1^2H \rightarrow {}_2^4He + Q$.
Binding energy of one deuteron is $2.2 \, MeV$. Therefore,the total binding energy of two deuterons is $2 \times 2.2 \, MeV = 4.4 \, MeV$.
The binding energy of one ${}_2^4He$ nucleus is $28 \, MeV$.
The energy released $(Q)$ in the fusion process is equal to the difference between the binding energy of the product nucleus and the total binding energy of the reactant nuclei.
$Q = BE({}_2^4He) - 2 \times BE({}_1^2H)$
$Q = 28 \, MeV - 4.4 \, MeV = 23.6 \, MeV$.

(B) When $CuSO_4$ reacts with $KCN$,initially $Cu(CN)_2$ is formed,which is unstable and decomposes to form $CuCN$ and $(CN)_2$ gas.
$2Cu^{2+} + 4CN^- \longrightarrow 2CuCN + (CN)_2$
In the presence of excess $KCN$,$CuCN$ reacts to form a stable complex:
$CuCN + 3KCN \longrightarrow K_3[Cu(CN)_4]$
Thus,the final product is $[Cu(CN)_4]^{3-}$.

(B) The basic character of oxides increases as the metallic character of the central atom increases down a group.
$SeO_2$ is an acidic oxide because $Se$ is a non-metal.
$Al_2O_3$ is amphoteric.
Between $Sb_2O_3$ and $Bi_2O_3$,both are oxides of Group $15$ elements.
As we move down the group from $Sb$ to $Bi$,the metallic character increases,which leads to an increase in the basic nature of the oxides.
Therefore,$Bi_2O_3$ is the most basic oxide among the given options.

(D) The frequency of an $L-C$ resonant circuit is given by the formula: $f = \frac{1}{2 \pi \sqrt{LC}}$.
Let the initial frequency be $f_1 = f$ with inductance $L_1 = L$ and capacitance $C_1 = C$.
The new frequency $f_2$ with $L_2 = 2L$ and $C_2 = 4C$ is given by:
$f_2 = \frac{1}{2 \pi \sqrt{L_2 C_2}} = \frac{1}{2 \pi \sqrt{(2L)(4C)}}$.
Comparing the two frequencies:
$\frac{f_2}{f_1} = \frac{\frac{1}{2 \pi \sqrt{8LC}}}{\frac{1}{2 \pi \sqrt{LC}}} = \sqrt{\frac{LC}{8LC}} = \frac{1}{\sqrt{8}} = \frac{1}{2\sqrt{2}}$.
Therefore,$f_2 = \frac{f}{2\sqrt{2}}$.

(B) The nuclear fusion reaction is given by: $2(_1^2H) \rightarrow _2^4He + Q$.
The energy released $(Q)$ in a nuclear reaction is equal to the difference between the total binding energy of the products and the total binding energy of the reactants.
Total binding energy of the reactants (two deuterons) $= 2 \times 2.2\,MeV = 4.4\,MeV$.
Total binding energy of the product ($^4He$ nucleus) $= 28\,MeV$.
Energy released $(Q)$ $= (\text{Binding energy of product}) - (\text{Binding energy of reactants})$.
$Q = 28\,MeV - 4.4\,MeV = 23.6\,MeV$.

(D) The radius of a nucleus is given by the formula $R = R_{0}A^{1/3}$,where $R_{0}$ is a constant and $A$ is the mass number (number of nucleons).
Given that the radius of $Ge$ is twice the radius of $_{4}^{9}\text{Be}$,we have $R_{Ge} = 2R_{Be}$.
Using the formula,we can write the ratio as $\frac{R_{Ge}}{R_{Be}} = \frac{A_{Ge}^{1/3}}{A_{Be}^{1/3}}$.
Substituting the given values: $2 = \left(\frac{A_{Ge}}{9}\right)^{1/3}$.
Cubing both sides,we get $2^{3} = \frac{A_{Ge}}{9}$.
$8 = \frac{A_{Ge}}{9}$.
$A_{Ge} = 8 \times 9 = 72$.
Therefore,the number of nucleons in $Ge$ is $72$.

(C) The complex $[Co(NH_3)_4(NO_2)_2]Cl$ exhibits the following isomerisms:
$1$. Linkage Isomerism: Due to the presence of the ambidentate ligand $NO_2^-$,which can coordinate through either $N$ or $O$.
$2$. Ionization Isomerism: Due to the exchange of the $Cl^-$ ion inside the coordination sphere with the $NO_2^-$ ligand.
$3$. Geometrical Isomerism: The complex exists in $cis$ and $trans$ forms due to the arrangement of the $NO_2^-$ ligands.

(A) Let the potential of the left junction be $V_L = 0$ and the right junction be $V_R = V$.
The potential at point $A$ is determined by the voltage divider of the top two resistors ($4 \, \Omega$ and $4 \, \Omega$):
$V_A = V_L + \frac{4}{4+4} \times (V_R - V_L) = 0 + \frac{4}{8} \times V = \frac{V}{2}$.
The potential at point $B$ is determined by the voltage divider of the bottom two resistors ($1 \, \Omega$ and $3 \, \Omega$):
$V_B = V_L + \frac{1}{1+3} \times (V_R - V_L) = 0 + \frac{1}{4} \times V = \frac{V}{4}$.
Comparing the potentials, $V_A = \frac{V}{2}$ and $V_B = \frac{V}{4}$.
Since $V_A > V_B$, the current will flow from the higher potential to the lower potential, i.e., from $A$ to $B$.

(C) Alkanols are derivatives of alkanes,formed by replacing one hydrogen atom $(-H)$ of an alkane with a hydroxyl group $(-OH)$.
The general formula for alkanes is $C_nH_{2n+2}$.
Replacing one $H$ with $OH$ gives $C_nH_{2n+1}OH$,which simplifies to $C_nH_{2n+2}O$.
Therefore,the general molecular formula for the homologous series of alkanols is $C_nH_{2n+2}O$.

(B) $F$-centers are the sites where anions are missing and instead electrons are present. The appearance of colour in solid alkali metal halides is generally due to these $F$-centers.

(A) The density of a unit cell is given by the formula: $\rho = \frac{Z \times M}{a^{3} \times N_{A}}$
For $CsBr$ in a body-centred cubic (bcc) lattice,the number of formula units per unit cell $Z = 1$.
The molar mass $M = 133 + 80 = 213 \, g/mol$.
The edge length $a = 436.6 \, pm = 436.6 \times 10^{-10} \, cm$.
Substituting these values into the formula:
$\rho = \frac{1 \times 213}{(436.6 \times 10^{-10})^{3} \times 6.02 \times 10^{23}}$
$\rho = \frac{213}{83.25 \times 10^{-24} \times 6.02 \times 10^{23}}$
$\rho = \frac{213}{50.11} \approx 4.25 \, g/cm^{3}$.

(D) Osmosis is the phenomenon of the flow of solvent molecules through a semipermeable membrane.
In reality,solvent molecules move in both directions across the membrane.
However,the net flow occurs from the region of lower solute concentration (higher solvent concentration) to the region of higher solute concentration (lower solvent concentration).
Since no membrane is perfectly semipermeable,the flow of water occurs from both sides but at unequal rates,resulting in a net flow.

(C) solution of acetone in ethanol shows a positive deviation from Raoult's law.
In pure ethanol,molecules are held together by strong hydrogen bonding.
When acetone is added,it disrupts these hydrogen bonds,leading to weaker solute-solvent interactions compared to the original solvent-solvent interactions.
This results in an increase in vapor pressure,which is characteristic of a positive deviation from Raoult's law.

(B) The formula for freezing point depression is $\Delta T_f = K_f \times m$,where $m$ is the molality of the solution.
Molality $m = \frac{W_1 \times 1000}{M_1 \times W_2}$,where $W_1$ is the mass of solute,$M_1$ is the molar mass of solute,and $W_2$ is the mass of solvent in grams.
Given: $W_1 = 1.00 \ g$,$M_1 = 250 \ g \ mol^{-1}$,$W_2 = 51.2 \ g$,$K_f = 5.12 \ K \ kg \ mol^{-1}$.
Substituting the values: $\Delta T_f = 5.12 \times \frac{1.00 \times 1000}{250 \times 51.2}$.
$\Delta T_f = 5.12 \times \frac{1000}{12800} = 5.12 \times 0.078125 = 0.4 \ K$.

(A) In an electrochemical cell representation,the left side represents the anode (oxidation) and the right side represents the cathode (reduction).
Given cell: $A \,|\, A^{+} \,(x \ M) \, ||\, B^{+} \,(y \ M) \, |\, B$
At the anode (left): $A \rightarrow A^{+} + e^-$
At the cathode (right): $B^{+} + e^- \rightarrow B$
Adding these two half-reactions,the overall cell reaction is: $A + B^{+} \rightarrow A^{+} + B$
Since the measured $emf$ is positive $(+ 0.20 \ V)$,the reaction is spontaneous in the direction written.

(D) The given half-cell reactions are:
$Fe^{2+} + 2e^{-} \rightarrow Fe, E^{\circ} = -0.441 \ V$ (Anode reaction: $Fe \rightarrow Fe^{2+} + 2e^{-}, E^{\circ} = +0.441 \ V$)
$Fe^{3+} + e^{-} \rightarrow Fe^{2+}, E^{\circ} = +0.771 \ V$ (Cathode reaction: $2Fe^{3+} + 2e^{-} \rightarrow 2Fe^{2+}, E^{\circ} = +0.771 \ V$)
For the overall reaction $Fe + 2Fe^{3+} \rightarrow 3Fe^{2+}$,the standard cell potential is calculated as:
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$
$E^{\circ}_{cell} = E^{\circ}_{Fe^{3+}/Fe^{2+}} - E^{\circ}_{Fe^{2+}/Fe}$
$E^{\circ}_{cell} = 0.771 \ V - (-0.441 \ V)$
$E^{\circ}_{cell} = 1.212 \ V$

(C) For the reaction $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$,the rate of reaction is expressed as:
Rate $= - \frac{d[N_2]}{dt} = - \frac{1}{3} \frac{d[H_2]}{dt} = + \frac{1}{2} \frac{d[NH_3]}{dt}$.
Equating the terms for $NH_3$ and $H_2$:
$+ \frac{1}{2} \frac{d[NH_3]}{dt} = - \frac{1}{3} \frac{d[H_2]}{dt}$.
Multiplying both sides by $2$,we get:
$+ \frac{d[NH_3]}{dt} = - \frac{2}{3} \frac{d[H_2]}{dt}$.

(B) For the reaction $2A + B \rightarrow 3C + D,$ the rate of reaction is expressed as:
Rate $= -\frac{1}{2} \frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{1}{3} \frac{d[C]}{dt} = \frac{d[D]}{dt}$
Comparing this with the given options:
Option $A$ is $-\frac{1}{2} \frac{d[A]}{dt}$ (Correct expression).
Option $B$ is $-\frac{d[C]}{3dt}$,but the correct expression for $C$ is $+\frac{1}{3} \frac{d[C]}{dt}$. Thus,this is incorrect.
Option $C$ is $-\frac{d[B]}{dt}$ (Correct expression).
Option $D$ is $\frac{d[D]}{dt}$ (Correct expression).
Therefore,the expression that does not represent the reaction rate is $-\frac{d[C]}{3dt}$.

(D) According to the Freundlich adsorption isotherm,the relation is given by $x/m = K p^{1/n}$.
Taking logarithm on both sides,we get $\log(x/m) = \log K + \frac{1}{n} \log p$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log(x/m)$,$x = \log p$,slope $m = 1/n$,and intercept $c = \log K$.
Therefore,the slope of the plot is $1/n$.

(A) The order given in option $A$ is incorrect.
The correct order of bond dissociation energy is $Cl_2 > Br_2 > F_2 > I_2$.
This is because the $F-F$ bond is weaker than the $Cl-Cl$ and $Br-Br$ bonds due to the small size of the fluorine atom,which leads to significant interelectronic repulsion between the lone pairs of the two fluorine atoms.

(B) Ions are coloured if they possess unpaired electrons in their $d$-orbitals due to $d-d$ transitions.
$Sc^{3+} \rightarrow [Ar] 3d^{0}$ (No unpaired electrons,colourless)
$Ti^{3+} \rightarrow [Ar] 3d^{1}$ (One unpaired electron,coloured)
$Ni^{2+} \rightarrow [Ar] 3d^{8}$ (Two unpaired electrons,coloured)
$Cu^{+} \rightarrow [Ar] 3d^{10}$ (No unpaired electrons,colourless)
$Co^{2+} \rightarrow [Ar] 3d^{7}$ (Three unpaired electrons,coloured)
Comparing the options:
$A$: $Ni^{2+}$ (coloured),$Cu^{+}$ (colourless)
$B$: $Ni^{2+}$ (coloured),$Ti^{3+}$ (coloured)
$C$: $Sc^{3+}$ (colourless),$Ti^{3+}$ (coloured)
$D$: $Sc^{3+}$ (colourless),$Co^{2+}$ (coloured)
Thus,both ions are coloured in option $B$.

(C) When $CuSO_4$ reacts with $KCN$,it first forms a precipitate of cupric cyanide $(Cu(CN)_2)$,which is unstable and decomposes to form cuprous cyanide $(Cu_2(CN)_2)$ and cyanogen gas $((CN)_2)$.
$2CuSO_4 + 4KCN \rightarrow 2Cu(CN)_2 + 2K_2SO_4$
$2Cu(CN)_2 \rightarrow Cu_2(CN)_2 + (CN)_2$
The cuprous cyanide $(Cu_2(CN)_2)$ then dissolves in excess $KCN$ to form the stable soluble complex potassium tetracyanocuprate$(I)$,$K_3[Cu(CN)_4]$.
$Cu_2(CN)_2 + 6KCN \rightarrow 2K_3[Cu(CN)_4]$
Overall reaction:
$2CuSO_4 + 10KCN \rightarrow 2K_3[Cu(CN)_4] + 2K_2SO_4 + (CN)_2$

(C) The actinoids exhibit a greater number of oxidation states compared to the lanthanoids.
This is primarily because the energy difference between the $5f$ and $6d$ orbitals in actinoids is smaller than the energy difference between the $4f$ and $5d$ orbitals in lanthanoids.
Consequently,the $5f$ electrons can participate in bonding more easily than the $4f$ electrons.

(D) Magnetic moment $(\mu) = \sqrt{n(n+2)} \text{ } B.M.$
Given $\mu = 3.83 \text{ } B.M.$
$3.83 = \sqrt{n(n+2)} \implies n \approx 3$
This indicates there are $3$ unpaired electrons.
For the complex $[Cr(H_2O)_6]Cl_3$,the oxidation state of $Cr$ is $+3$.
The electronic configuration of $Cr$ $(Z=24)$ is $[Ar] 3d^5 4s^1$.
Thus,$Cr^{3+}$ is $[Ar] 3d^3$.
According to Hund's rule,the $3$ electrons in the $3d$ orbitals occupy the $t_{2g}$ set $(d_{xy}, d_{yz}, d_{xz})$ singly.
Therefore,the distribution is $3d_{xy}^1, 3d_{yz}^1, 3d_{xz}^1$.

(C) The complex $[Co(NH_3)_4(NO_2)_2]Cl$ exhibits the following types of isomerism:
$1.$ Linkage isomerism: Due to the presence of the ambidentate ligand $NO_2^-$ (it can coordinate through $N$ or $O$).
$2.$ Ionization isomerism: It can exchange ions between the coordination sphere and the ionization sphere to form $[Co(NH_3)_4(NO_2)Cl]NO_2$.
$3.$ Geometrical isomerism: It is an $[MA_4B_2]$ type octahedral complex,which exists in $\text{cis}$ and $\text{trans}$ forms.
It does not show optical isomerism because both $\text{cis}$ and $\text{trans}$ forms possess a plane of symmetry.

(A) Alkanols (alcohols) are derivatives of alkanes where one hydrogen atom is replaced by a hydroxyl $(-OH)$ group.
The general formula for an alkane is $C_nH_{2n+2}$.
Replacing one $H$ with $OH$ gives $C_nH_{2n+1}OH$,which simplifies to $C_nH_{2n+2}O$.
Thus,the correct general molecular formula is $C_nH_{2n+2}O$.

(A) The reaction of ethylene oxide $(CH_2-CH_2-O)$ with a Grignard reagent $(RMgX)$ involves the nucleophilic attack of the alkyl group $(R^-)$ on one of the carbon atoms of the epoxide ring,leading to ring opening.
This forms an intermediate alkoxide $(R-CH_2-CH_2-OMgX)$.
Subsequent acid hydrolysis of this intermediate yields a primary alcohol $(R-CH_2-CH_2-OH)$ as the final product.

(A) In the reaction of an unsymmetrical ether with $HI$,if the alkyl groups are primary or secondary,the reaction follows the $S_N2$ mechanism.
The iodide ion $(I^{-})$ attacks the smaller (less sterically hindered) alkyl group.
In $CH_3-O-CH(CH_3)_2$,the methyl group $(-CH_3)$ is smaller than the isopropyl group $(-CH(CH_3)_2)$.
Therefore,the $I^{-}$ ion attacks the methyl group,leading to the formation of $CH_3I$ and $(CH_3)_2CHOH$ as the major products.

(A) Nucleophilic addition reactions are influenced by steric and electronic factors.
Aldehydes are generally more reactive than ketones because they have less steric hindrance and fewer electron-donating alkyl groups ($+I$ effect),which reduces the electrophilicity of the carbonyl carbon.
Among the given options,$CH_3-CHO$ (ethanal) has the smallest alkyl group compared to $CH_3-CH_2-CHO$ (propanal),and it is an aldehyde,whereas the other options are ketones or larger aldehydes.
Therefore,$CH_3-CHO$ is the most reactive toward nucleophilic attack.

(B) . Acetaldehyde $(CH_3CHO)$ reacts with $HCN$ to form acetaldehyde cyanohydrin $(CH_3CH(OH)CN)$.
On hydrolysis,it yields lactic acid $(CH_3CH(OH)COOH)$.
Since the $\alpha$-carbon in lactic acid is chiral (bonded to four different groups: $-H$,$-OH$,$-CH_3$,and $-COOH$),it exists as a racemic mixture when synthesized from an achiral starting material like acetaldehyde.

(A) The reaction sequence is as follows:
$1$. Propionic acid $(CH_3CH_2COOH)$ reacts with $SOCl_2$ to form propionyl chloride $(B = CH_3CH_2COCl)$.
$2$. Propionyl chloride reacts with $NH_3$ to form propionamide $(C = CH_3CH_2CONH_2)$.
$3$. Propionamide reacts with $KOH$ and $Br_2$ (Hofmann bromamide degradation reaction) to form ethylamine $(D = CH_3CH_2NH_2)$.
Thus,the structure of $D$ is $CH_3CH_2NH_2$.

(C) The self-condensation of two moles of ethyl acetate in the presence of a strong base like sodium ethoxide $(NaOC_2H_5)$ is known as the Claisen condensation.
This reaction occurs because ethyl acetate possesses $\alpha$-hydrogen atoms.
The reaction proceeds as follows:
$2CH_3COOC_2H_5 \xrightarrow{NaOC_2H_5} CH_3COCH_2COOC_2H_5 + C_2H_5OH$
The product formed is ethyl acetoacetate,which is commonly referred to as acetoacetic ester.

(A) Benzylamine,$C_{6}H_{5}CH_{2}NH_{2}$,is more basic than aniline because the benzyl group $(C_{6}H_{5}CH_{2}-)$ is an electron-donating group due to the $+I$ effect.
This increases the electron density on the nitrogen atom of the $-NH_{2}$ group,making the lone pair more available for donation.
In contrast,phenyl and nitro groups are electron-withdrawing groups ($-I$ or $-M$ effects),which decrease the electron density on the nitrogen atom.
Therefore,diphenylamine,triphenylamine,and $p-$nitroaniline are less basic than aniline.

(D) In the process of digestion,the proteins present in food material are hydrolysed to amino acids.
In this process,two enzymes,$pepsin$ and $trypsin$,are involved as follows:
$Proteins$ $\xrightarrow{Pepsin \ (A)} Polypeptides$ $\xrightarrow{trypsin \ (B)} Amino \ acids$

(B) Glucagon is a peptide hormone because it consists of a chain of $29$ amino acids linked by peptide bonds.
Adrenaline is an amino acid derivative.
Testosterone is a steroid hormone.
Thyroxine is an iodinated amino acid derivative.

(B) The given structure represents $Nylon-6,6$.
$Nylon-6,6$ is formed by the condensation polymerization of two different monomers: hexamethylenediamine $(H_2N(CH_2)_6NH_2)$ and adipic acid $(HOOC(CH_2)_4COOH)$.
Since it is formed from two different types of monomers,it is classified as a copolymer.

(A) Two solutions are isotonic if they have the same osmotic pressure,which implies they have the same molar concentration at the same temperature.
Molar concentration of urea $= \frac{10 \ g \ L^{-1}}{60 \ g \ mol^{-1}} = \frac{1}{6} \ mol \ L^{-1}$.
$A$ $5 \%$ solution of a non-volatile solute means $5 \ g$ of solute in $100 \ mL$ of solution,which is $50 \ g$ of solute in $1 \ L$ $(dm^3)$ of solution.
Let the molecular mass of the non-volatile solute be $M$.
Molar concentration of the non-volatile solute $= \frac{50 \ g \ L^{-1}}{M \ g \ mol^{-1}} = \frac{50}{M} \ mol \ L^{-1}$.
Since the solutions are isotonic,$\frac{1}{6} = \frac{50}{M}$.
Therefore,$M = 50 \times 6 = 300 \ g \ mol^{-1}$.

(C) Vitamins are a group of biomolecules required for normal metabolic processes,growth,and health in humans and animals.
The human body cannot synthesize most vitamins and must obtain them from external sources such as vegetables,fish,meat,eggs,fruits,and sunlight to prevent deficiency diseases.
Therefore,the human body does not produce vitamins.