AIPMT 2005 Chemistry Question Paper with Answer and Solution

(A) The intensity $I$ of sound from a point source in a non-absorbing medium is inversely proportional to the square of the distance $r$ from the source,given by $I \propto \frac{1}{r^2}$.
Let $I_P$ and $I_Q$ be the intensities at points $P$ and $Q$ respectively.
Given distances are $r_P = 2 \ m$ and $r_Q = 3 \ m$.
Therefore,the ratio of intensities is $\frac{I_P}{I_Q} = \frac{r_Q^2}{r_P^2} = \left( \frac{3}{2} \right)^2 = \frac{9}{4}$.
Thus,the ratio is $9:4$.

(B) The stability of a nucleus is determined by its binding energy per nucleon $(B.E./A)$.
According to the binding energy curve,for heavy nuclei (high mass number $A$),the binding energy per nucleon decreases as the mass number increases.
Nuclear fission occurs when a heavy nucleus splits into two lighter nuclei of intermediate mass,which have higher binding energy per nucleon than the original heavy nucleus.
This increase in binding energy per nucleon results in the release of energy,making the process energetically favorable.
Therefore,the correct option is $B$.

(D) The energy of an orbit in a hydrogen atom is given by the formula $E_n = \frac{-1312}{n^2} \ kJ \ mol^{-1}$.
Given,for the second orbit $(n=2)$,$E_2 = -328 \ kJ \ mol^{-1}$.
For the fourth orbit $(n=4)$,the energy $E_4$ is related to $E_2$ by the ratio $\frac{E_4}{E_2} = \frac{n_2^2}{n_4^2} = \frac{2^2}{4^2} = \frac{4}{16} = \frac{1}{4}$.
Therefore,$E_4 = \frac{E_2}{4} = \frac{-328}{4} = -82 \ kJ \ mol^{-1}$.

(C) According to Fajan's rule,the covalent character increases with an increase in the polarising power of the cation.
Polarising power is directly proportional to the charge density of the cation.
The charge-to-size ratio follows the order: $Be^{2+} > Li^{+} > Na^{+}$.
Therefore,the order of increasing covalent character is $NaCl < LiCl < BeCl_2$.

(C) The dipole moment of a molecule depends on its geometry and the polarity of its bonds.
$BF_3$ has a trigonal planar geometry,$SiF_4$ has a tetrahedral geometry,and $XeF_4$ has a square planar geometry. Due to their high symmetry,the bond dipoles in these molecules cancel each other out,resulting in a net dipole moment of zero.
$SF_4$ has a see-saw geometry due to the presence of one lone pair on the sulfur atom. This lack of symmetry prevents the bond dipoles from canceling out,resulting in a permanent dipole moment.

(D) To determine the geometry,we calculate the number of electron pairs around the central atom using the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
For $BF_3$:
Central atom $B$ has $3$ valence electrons.
Number of monovalent atoms $(F)$ = $3$.
$H = \frac{1}{2}(3 + 3) = 3$.
Since there are $3$ bond pairs and $0$ lone pairs,the hybridization is $sp^2$ and the geometry is trigonal planar.
For $IF_3$: $H = \frac{1}{2}(7 + 3) = 5$ ($sp^3d$,$T$-shaped).
For $PCl_3$: $H = \frac{1}{2}(5 + 3) = 4$ ($sp^3$,trigonal pyramidal).
For $NH_3$: $H = \frac{1}{2}(5 + 3) = 4$ ($sp^3$,trigonal pyramidal).
Therefore,the correct option is $D$.

(C) The paramagnetic behavior of a molecule is determined by the presence of unpaired electrons.
$CO_2$ has $16$ valence electrons and all are paired.
$SO_2$ has $18$ valence electrons and all are paired.
$SiO_2$ is a covalent network solid where all electrons are paired.
$ClO_2$ has $19$ valence electrons,meaning it contains an odd number of electrons,which results in at least one unpaired electron.
Therefore,$ClO_2$ exhibits paramagnetic behavior.

(C) The bond lengths are as follows:
$1$. In $O_2$, the bond order is $2$, resulting in a bond length of approximately $121 \text{ pm}$.
$2$. In $O_3$, resonance results in a bond order of $1.5$, leading to a bond length of approximately $128 \text{ pm}$.
$3$. In $H_2O_2$, the $O-O$ bond is a single bond (bond order $1$), resulting in a bond length of approximately $148 \text{ pm}$.
Therefore, the increasing order of $O-O$ bond length is $O_2 < O_3 < H_2O_2$.

(D) For the first equilibrium: $NO_{(g)} + 1/2 O_{2(g)} \rightleftharpoons NO_{2(g)}$,the equilibrium constant is $K_1 = \frac{[NO_2]}{[NO][O_2]^{1/2}}$.
For the second equilibrium: $2NO_{2(g)} \rightleftharpoons 2NO_{(g)} + O_{2(g)}$,the equilibrium constant is $K_2 = \frac{[NO]^2[O_2]}{[NO_2]^2}$.
Observe that the second reaction is the reverse of the first reaction multiplied by $2$.
If a reaction is reversed,the equilibrium constant becomes $1/K$. If a reaction is multiplied by a factor $n$,the equilibrium constant becomes $K^n$.
Here,the first reaction is reversed and multiplied by $2$,so $K_2 = (1/K_1)^2 = 1/K_1^2$.

(A) The acidic strength of oxoacids of chlorine depends on the oxidation state of the central chlorine atom.
As the oxidation number of the central atom increases,the electron density on the oxygen atom decreases,which facilitates the release of the $H^+$ ion.
The oxidation states are: $HClO (+1)$,$HClO_2 (+3)$,$HClO_3 (+5)$,and $HClO_4 (+7)$.
Therefore,the correct order of acid strength is $HClO < HClO_2 < HClO_3 < HClO_4$.

(D) For a weak base $BOH$,the dissociation is $BOH \rightleftharpoons B^+ + OH^-$.
$K_b = \frac{[B^+][OH^-]}{[BOH]} = \frac{C\alpha \cdot C\alpha}{C(1-\alpha)} \approx C\alpha^2$.
Given $K_b = 1.0 \times 10^{-12}$ and $C = 0.01 \, M = 10^{-2} \, M$.
$1.0 \times 10^{-12} = 10^{-2} \cdot \alpha^2$.
$\alpha^2 = 10^{-10} \implies \alpha = 10^{-5}$.
$[OH^-] = C\alpha = 10^{-2} \times 10^{-5} = 1.0 \times 10^{-7} \, mol \, L^{-1}$.

(D) The salts $Na_2O$,$Na_2S$,$Na_2Se$,and $Na_2Te$ undergo anionic hydrolysis in water to form $NaOH$ and the corresponding weak acid ($H_2O$,$H_2S$,$H_2Se$,$H_2Te$ respectively).
The $pH$ of these solutions depends on the strength of the conjugate acid formed. The weaker the conjugate acid,the stronger the conjugate base,and the higher the $pH$ of the solution.
The acidic strength order of the hydrides is $H_2O < H_2S < H_2Se < H_2Te$.
Therefore,the basic strength of the conjugate bases follows the order $OH^- > HS^- > HSe^- > HTe^-$.
Consequently,the $pH$ values of their isomolar solutions follow the order $pH_1 > pH_2 > pH_3 > pH_4$.

(C) The spontaneity of a reaction is determined by the Gibbs free energy equation: $\Delta G = \Delta H - T\Delta S$.
For a process to be spontaneous,$\Delta G$ must be negative.
In option $(C)$,the reaction is exothermic $(\Delta H < 0)$ and there is an increase in disorder $(\Delta S > 0)$.
Substituting these into the equation: $\Delta G = (-ve) - T(+ve) = -ve$.
Since $\Delta G$ is always negative regardless of the temperature $T$,the reaction is certain to be spontaneous.

(A) The standard enthalpy of neutralisation for a strong acid and a strong base is $-57.33 \ kJ \ mol^{-1}$.
$MgO$ is a metal oxide that acts as a base,but it is not a strong soluble base like $NaOH$ or $KOH$.
When $MgO_{(s)}$ reacts with $HCl_{(aq)}$,energy is consumed to break the crystal lattice of $MgO$ and to convert it into ions.
Therefore,the net heat released is less than the standard value of $-57.33 \ kJ \ mol^{-1}$ (i.e.,it is less exothermic).
Thus,the enthalpy of neutralisation will be less than $-57.33 \ kJ \ mol^{-1}$.

(D) In alkaline medium,the balanced chemical equation for the reaction between $KMnO_4$ and $KI$ is:
$2KMnO_4 + KI + H_2O \to 2MnO_2 + KIO_3 + 2KOH$
From the stoichiometry of the balanced equation,$2$ moles of $KMnO_4$ react with $1$ mole of $KI$.
Therefore,the number of moles of $KMnO_4$ reduced by $1$ mole of $KI$ is $2$.

(C) According to Fajan's rule,the covalent character of an ionic bond increases with the polarising power of the cation.
Polarising power is directly proportional to the charge density of the cation.
The charge and size of the cations are $Na^{+}$ ($+1$,large size),$Li^{+}$ ($+1$,smaller than $Na^{+}$),and $Be^{2+}$ ($+2$,smallest size).
The order of polarising power is $Be^{2+} > Li^{+} > Na^{+}$.
Therefore,the order of increasing covalent character is $NaCl < LiCl < BeCl_2$.

(B) The electron gain enthalpy (with negative sign) represents the energy released when an electron is added to a neutral gaseous atom.
Generally,electron gain enthalpy becomes more negative across a period.
However,for $F$ and $Cl$,the electron gain enthalpy of $Cl$ is more negative than that of $F$ due to the small size of the $F$ atom,which leads to strong inter-electronic repulsions.
The values are: $Cl$ $(349 \ kJ/mol)$,$F$ $(333 \ kJ/mol)$,$S$ $(200 \ kJ/mol)$,and $O$ $(142 \ kJ/mol)$.
Thus,the correct order of increasing magnitude is $O < S < F < Cl$.

(B) The electron gain enthalpy (with negative sign) represents the energy released when an electron is added to a neutral gaseous atom.
Generally,electron gain enthalpy becomes more negative across a period.
However,for halogens,the electron gain enthalpy of $F$ is less negative than that of $Cl$ due to the small size of the $F$ atom,which leads to strong inter-electronic repulsions.
Similarly,for oxygen and sulfur,the electron gain enthalpy of $O$ is less negative than that of $S$ due to the small size of the $O$ atom.
The values are: $Cl$ $(349 \ kJ/mol)$,$F$ $(333 \ kJ/mol)$,$S$ $(200 \ kJ/mol)$,and $O$ $(142 \ kJ/mol)$.
Thus,the correct order of increasing electron gain enthalpy (magnitude) is $O < S < F < Cl$.

(A) An electron-deficient molecule is one in which the central atom has fewer than $8$ electrons in its valence shell,or it lacks sufficient electrons to form conventional covalent bonds.
In $B_2H_6$ (diborane),each boron atom is bonded to four hydrogen atoms,but there are only $12$ valence electrons available for the entire molecule,which is insufficient to form standard two-center-two-electron $(2c-2e)$ bonds for all linkages.
It contains two $3$-center-$2$-electron $(3c-2e)$ bonds,also known as banana bonds,making it an electron-deficient molecule.
In contrast,$C_2H_6$,$PH_3$,and $SiH_4$ are electron-precise molecules where the octet rule is satisfied for the central atoms.

(D) The $pH$ of an aqueous solution of a salt of a strong base and a weak acid depends on the strength of the acid.
Stronger the acid,weaker is its conjugate base,and lower is the $pH$ of its salt solution.
The order of acidic strength of the corresponding hydrides is $H_2O < H_2S < H_2Se < H_2Te$.
Since the acidic strength increases down the group,the basic strength of the conjugate bases $(O^{2-}, S^{2-}, Se^{2-}, Te^{2-})$ decreases in the order $O^{2-} > S^{2-} > Se^{2-} > Te^{2-}$.
Therefore,the $pH$ of the isomolar solutions follows the order $pH_1 > pH_2 > pH_3 > pH_4$.

(A) $H_2S$ is a weak electrolyte and undergoes dissociation as: $H_2S \rightleftharpoons 2H^+ + S^{2-}$.
According to the common ion effect,the presence of $HCl$ (a strong electrolyte) provides a high concentration of $H^+$ ions.
This suppresses the ionization of $H_2S$,thereby significantly decreasing the concentration of $S^{2-}$ ions.
The solubility product $(K_{sp})$ of group $II$ sulphides is very low,so even a low concentration of $S^{2-}$ is sufficient to exceed the ionic product and cause precipitation.
However,the $K_{sp}$ of group $IV$ sulphides is relatively higher,and the suppressed $S^{2-}$ concentration is insufficient to cause their precipitation.

(A) In qualitative analysis,$H_2S$ is a weak acid that dissociates as: $H_2S \rightleftharpoons 2H^+ + S^{2-}$.
When $HCl$ is present,it provides a high concentration of $H^+$ ions due to the common ion effect.
This suppresses the dissociation of $H_2S$,significantly decreasing the concentration of $S^{2-}$ ions.
Group $II$ cations have very low solubility products $(K_{sp})$,so they precipitate even at low $S^{2-}$ concentrations.
Group $IV$ cations have higher $K_{sp}$ values and require a higher concentration of $S^{2-}$ to precipitate,which is not achieved in the presence of $HCl$.

(B) The correct answer is $(B)$.
Benzoic acid is soluble in hot water,whereas naphthalene is insoluble in water.
Therefore,when the mixture is dissolved in hot water and filtered,benzoic acid passes through the filter paper as a filtrate,while naphthalene remains as a residue.
Upon cooling the filtrate,benzoic acid crystallizes out.

(D) In $IUPAC$ nomenclature,substituents are listed in alphabetical order.
For the compound $CH_3-CH_2-CH_2-CH(C_2H_5)-CH(CH_3)-CH_2-CH_3$,the parent chain is heptane.
The substituents are ethyl at position $4$ and methyl at position $3$.
Alphabetically,'ethyl' comes before 'methyl'.
Therefore,the correct name should be $4-\text{Ethyl-}3-\text{methylheptane}$.
Thus,the name $3-\text{Methyl-}4-\text{ethylheptane}$ is incorrect.

(B) The correct answer is $(B)$ $CH_3-C^+(CH_3)_2$.
Carbocation stability is determined by the inductive effect $(+I)$ and hyperconjugation.
$CH_3-C^+(CH_3)_2$ is a $3^\circ$ carbocation,which is stabilized by the electron-donating effect of three methyl groups.
$CH_3-CH^+(CH_3)$ is a $2^\circ$ carbocation,$CH_3-CH_2^+$ is a $1^\circ$ carbocation,and $CH_3^+$ is a methyl carbocation.
Stability order: $3^\circ > 2^\circ > 1^\circ > \text{methyl}$.

(A) To determine the chirality ($R/S$ configuration) of the chiral center,we assign priorities to the groups attached to the central carbon atom based on the Cahn-Ingold-Prelog $(CIP)$ sequence rules based on atomic number:
$1$. $-Br$ (atomic number $35$) is priority $1$.
$2$. $-Cl$ (atomic number $17$) is priority $2$.
$3$. $-CH_3$ (atomic number $6$ for $C$) is priority $3$.
$4$. $-H$ (atomic number $1$) is priority $4$.
Since the lowest priority group $(-H)$ is on a dashed bond (pointing away from the viewer),we look at the sequence $1$ $\rightarrow 2$ $\rightarrow 3$.
The sequence $Br$ $\rightarrow Cl$ $\rightarrow CH_3$ follows a clockwise direction.
Therefore,the configuration is $R$.

(D) Stereoisomerism occurs when isomers have the same structural formula but differ in the relative arrangement of atoms or groups in space within the molecule.
Stereoisomerism is primarily classified into three types:
$(i)$ Geometrical isomerism
$(ii)$ Optical isomerism
$(iii)$ Conformational isomerism.
Therefore,both optical isomerism and geometric isomerism are types of stereoisomerism.

(C) Ozonolysis of an internal alkyne followed by hydrolysis yields diketones as intermediate products,which are further oxidized to carboxylic acids.
The reaction proceeds as follows:
$CH_3-C \equiv C-CH_2-CH_3$ $\xrightarrow{O_3} \text{ozonide intermediate}$ $\xrightarrow{H_2O} CH_3-CO-CO-CH_2-CH_3 + H_2O_2$
The diketone $CH_3-CO-CO-CH_2-CH_3$ is then cleaved to form carboxylic acids:
$CH_3-CO-CO-CH_2-CH_3 \rightarrow CH_3COOH + CH_3CH_2COOH$
Thus,the final products are acetic acid and propanoic acid.

(A) Surface tension is primarily determined by the strength of intermolecular forces,specifically hydrogen bonding.
Water $(H_2O)$ molecules exhibit strong hydrogen bonding due to their high polarity and the presence of two hydrogen atoms capable of forming bonds with oxygen atoms of adjacent molecules.
This extensive network of hydrogen bonds results in a higher cohesive force compared to organic solvents like benzene $(C_6H_6)$,methanol $(CH_3OH)$,or ethanol $(C_2H_5OH)$.
Therefore,water has the maximum surface tension among the given options.

(D) In alkaline medium,the balanced chemical equation for the reaction between $KMnO_4$ and $KI$ is:
$2KMnO_4 + KI + H_2O \to 2MnO_2 + KIO_3 + 2KOH$
From the stoichiometry of the balanced equation,$2$ moles of $KMnO_4$ react with $1$ mole of $KI$.
Therefore,the number of moles of $KMnO_4$ reduced by $1$ mole of $KI$ is $2$.

(C) The reaction of an alkyne with ozone followed by hydrolysis is an oxidative cleavage reaction.
For an internal alkyne $R-C\equiv C-R'$,the products are two carboxylic acids: $RCOOH$ and $R'COOH$.
In the given reaction,$CH_3-C\equiv C-CH_2-CH_3$ undergoes oxidative cleavage at the triple bond.
The $CH_3-C$ part is oxidized to acetic acid $(CH_3COOH)$.
The $-C-CH_2-CH_3$ part is oxidized to propanoic acid $(CH_3CH_2COOH)$.
Thus,the products are $CH_3COOH + CH_3CH_2COOH$.

(C) $Auxospores$ are specialized cells formed by many diatoms to restore their normal size after repeated cell divisions.
$Hormocysts$ are thick-walled,resting structures formed by some cyanobacteria (like $Westiellopsis$) under unfavorable conditions to survive and propagate.
Therefore,$Auxospores$ are formed by diatoms and $Hormocysts$ are formed by cyanobacteria.

(B) In an ectophloic siphonostele,the central pith is surrounded by a ring of xylem,which is further surrounded by a ring of phloem,pericycle,and endodermis.
In this type,the phloem is present only on the outer side of the xylem.
Examples of plants exhibiting this type of stele include $Osmunda$ and $Equisetum$.

(D) Annelids are coelomate animals,meaning they possess a true body cavity (coelom) lined by mesoderm.
In contrast,Platyhelminthes are acoelomate,meaning they lack a body cavity entirely.
Therefore,the absence of a body cavity is a characteristic feature of Platyhelminthes that distinguishes them from annelids.

(C) The correct answer is $C$.
Most members of the phylum $Porifera$ (sponges) are marine,but they are not exclusively marine.
Some sponges,such as $Spongilla$,are found in freshwater environments.
Therefore,the statement that animals belonging to phylum $Porifera$ are exclusively marine is incorrect.

(D) The correct answer is $D$. Thromboplastin is a complex substance released by damaged tissues or platelets that initiates the extrinsic pathway of blood coagulation. When introduced into the bloodstream,it triggers the conversion of prothrombin into thrombin in the presence of calcium ions $(Ca^{2+})$. Thrombin then converts soluble fibrinogen into insoluble fibrin threads,leading to the formation of a blood clot at the site of introduction.

(D) The correct answer is $D$.
In eukaryotes,the transcription initiation complex involves the binding of $RNA$ polymerase to the promoter region.
The $TATA$ box (also known as the Goldberg-Hogness box) is a specific $DNA$ sequence found in the promoter region of genes.
When $RNA$ polymerase and transcription factors bind to the $TATA$ box,the $DNA$ double helix undergoes a conformational change,bending into a saddle-like structure to facilitate the initiation of transcription.

(D) Induced mutagenesis is a technique used in plant breeding to create genetic variability.
Gamma rays,emitted from radioactive isotopes like cobalt-$60$,are the most commonly used physical mutagens for inducing mutations in crop plants.
They have high penetrating power,which allows them to reach the meristematic tissues of seeds or plant parts effectively,leading to chromosomal changes and mutations.

(C) The simplest amino acid is $Glycine$ $(NH_2-CH_2-COOH)$.
It is the simplest because its side chain ($R$-group) consists of only a single hydrogen atom $(H)$,making it the only amino acid without a chiral carbon atom.

(C) The correct answer is $C$. Solar energy is a clean and renewable source of energy and is not responsible for the greenhouse effect. In contrast,the greenhouse effect is primarily caused by the accumulation of greenhouse gases like $CO_2$,$CH_4$,and $CFCs$ in the atmosphere,which trap heat. Fossil fuel burning and biomass burning both release significant amounts of $CO_2$,contributing to global warming. Nuclear power generation produces hazardous radioactive waste.

(C) Carbohydrates are primarily synthesized through the process of photosynthesis.
Photosynthesis is performed by organisms that possess chlorophyll or similar pigments capable of capturing light energy.
These organisms include green plants,algae,and certain photosynthetic bacteria (such as cyanobacteria).
Fungi and viruses are heterotrophic or parasitic and do not perform photosynthesis; therefore,they do not produce carbohydrates via this pathway.

(C) The enzyme $Reverse$ $transcriptase$ is an $RNA$-dependent $DNA$ polymerase. It catalyzes the synthesis of a $DNA$ molecule from an $RNA$ template. This process is known as reverse transcription and is commonly observed in retroviruses like $HIV$.

(C) The general structure of an amino acid is $R-CH(NH_2)-COOH$.
In this structure,$R$ represents the side chain or functional group.
For the simplest amino acid,the $R$ group must be the smallest possible substituent,which is a hydrogen atom $(H)$.
When $R = H$,the amino acid is $H-CH(NH_2)-COOH$,which is known as Glycine.
Therefore,Glycine is the simplest amino acid because it has the smallest side chain.

(B) To determine the genotype of an organism or to identify the types of gametes produced by an individual with a dominant phenotype,a test cross is performed.
In a test cross,the individual in question is crossed with a homozygous recessive individual.
For a dihybrid genotype $AaBb$,the homozygous recessive genotype is $aabb$.
By crossing $AaBb$ with $aabb$,the resulting offspring will reflect the types of gametes produced by the $AaBb$ parent,as the $aabb$ parent only contributes $ab$ gametes.

(A) The Biodiversity Act of India was enacted by the Parliament of India in the year $2002$. This act aims to provide for the conservation of biological diversity,sustainable use of its components,and fair and equitable sharing of the benefits arising out of the use of biological resources and knowledge.

(A) More than $70\%$ of the world's freshwater is stored in the form of ice caps and glaciers. Specifically,the vast majority of this is located in Antarctica and Greenland. Among the given options,Antarctica represents the largest single reservoir of freshwater on Earth.

(B) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative.
$\Delta G = \Delta H - T\Delta S$.
Given that both $\Delta H$ and $\Delta S$ are positive $(+ve)$,
$\Delta G = (+ve) - T(+ve)$.
If $T\Delta S > \Delta H$,then the term $T\Delta S$ is larger than $\Delta H$,making $\Delta G$ negative.
Therefore,the reaction is spontaneous when $T\Delta S > \Delta H$.

(A) The rate of catalytic hydrogenation of an alkene is inversely proportional to its stability.
More substituted alkenes are more stable due to hyperconjugation and inductive effects,and therefore,they react more slowly with $H_2$.
The order of stability of alkenes is:
Tetrasubstituted > Trisubstituted > Disubstituted > Monosubstituted > Ethene.
Conversely,the order of reactivity towards catalytic hydrogenation is:
Monosubstituted > Disubstituted > Trisubstituted > Tetrasubstituted.
Among the given options,the alkene with the least number of substituents (monosubstituted or the least substituted one) will react fastest.
Based on the provided structures,the alkene with the fewest $R$ groups (most $H$ atoms) will be the most reactive.

(D) The acidic strength of oxyacids increases with an increase in the oxidation state of the central atom.
The oxidation states of chlorine in the given oxyacids are:
$HOCl$ $(+1)$,
$HOClO$ $(+3)$,
$HOClO_2$ $(+5)$,
$HOClO_3$ $(+7)$.
Since the oxidation state increases from $HOCl$ to $HOClO_3$,the acidic strength increases in the order: $HOCl < HOClO < HOClO_2 < HOClO_3$.

(A) The acid strength of oxyacids increases with an increase in the oxidation state of the central atom.
The oxidation states of chlorine in $HClO$,$HClO_2$,$HClO_3$,and $HClO_4$ are $+1$,$+3$,$+5$,and $+7$ respectively.
As the oxidation state increases,the electron-withdrawing power of the chlorine atom increases,which stabilizes the conjugate base $(ClO_n^-)$ by dispersing the negative charge.
Therefore,the correct order of acid strength is $HClO < HClO_2 < HClO_3 < HClO_4$.

(C) Paramagnetism is observed in species that contain one or more unpaired electrons.
In $ClO_2$,the chlorine atom has an odd number of valence electrons,resulting in an unpaired electron in the molecule.
Therefore,$ClO_2$ is paramagnetic.
$CO_2$,$SO_2$,and $SiO_2$ have all their electrons paired,making them diamagnetic.

(B) The acidity of a compound is determined by the stability of its conjugate base.
Electron-withdrawing groups $(EWG)$ stabilize the phenoxide ion by dispersing the negative charge through $-I$ (inductive) and $-R$ (resonance) effects,thereby increasing acidity.
$NO_2$ is a strong electron-withdrawing group.
In $o$-nitrophenol,the $-NO_2$ group is present at the ortho position,which exerts a strong $-I$ and $-R$ effect,making it the most acidic among the given options.
Therefore,$o$-nitrophenol is the most acidic compound.

(C) $1 \ m$ solution means $1 \ mole$ of solute is dissolved in $1000 \ g$ of solvent $(H_2O)$.
The number of moles of water $(N)$ is calculated as: $N = \frac{1000 \ g}{18 \ g/mol} = 55.55 \ mol$.
The mole fraction of the solute $(x_{solute})$ is given by the formula: $x_{solute} = \frac{n}{n + N}$.
Substituting the values: $x_{solute} = \frac{1}{1 + 55.55} = \frac{1}{56.55} \approx 0.0177 \approx 0.018$.

(D) Given mole ratio of pentane $(n_p)$ to hexane $(n_h)$ is $1:4$.
Mole fraction of pentane $(x_p)$ = $\frac{1}{1+4} = 0.2$.
Mole fraction of hexane $(x_h)$ = $\frac{4}{1+4} = 0.8$.
Total vapour pressure $(P_T)$ = $P_p^0 x_p + P_h^0 x_h$.
$P_T = (440 \times 0.2) + (120 \times 0.8) = 88 + 96 = 184 \ mm \ Hg$.
According to Dalton's law,the mole fraction of pentane in the vapour phase $(y_p)$ is given by $y_p = \frac{P_p^0 x_p}{P_T}$.
$y_p = \frac{88}{184} \approx 0.478$.

(D) According to Raoult's law,the total vapour pressure $P_T$ of an ideal solution is given by:
$P_T = P_P^0 X_P + P_Q^0 X_Q$
Where $P_P^0 = 80 \, torr$ and $P_Q^0 = 60 \, torr$.
The mole fractions are:
$X_P = \frac{3}{3+2} = \frac{3}{5} = 0.6$
$X_Q = \frac{2}{3+2} = \frac{2}{5} = 0.4$
Substituting the values:
$P_T = (80 \times 0.6) + (60 \times 0.4)$
$P_T = 48 + 24 = 72 \, torr$.

(D) In a face-centered cubic $(F.C.C)$ lattice,each face of the unit cell is shared by $2$ adjacent unit cells.
Since there are $6$ faces in a cubic unit cell,each face is shared equally by $6$ other unit cells.

(B) Alkaline earth metals belong to $Group$ $2$ of the periodic table.
When a radioactive element emits one $\alpha$-particle $(_{2}He^{4})$,its atomic number decreases by $2$,shifting its position in the periodic table by two groups to the left.
Starting from $Group$ $2$:
After the first $\alpha$-emission: $2 - 2 = 0$ (which corresponds to $Group$ $18$ in the periodic table).
After the second $\alpha$-emission: $18 - 2 = 16$.
Therefore,the resulting daughter element belongs to $Group$ $16$.

(D) For a first-order reaction,the rate law is given by $Rate = K[A]$.
Given: $Rate = 2.0 \times 10^{-5} \ mol \ L^{-1} \ s^{-1}$ and $[A] = 0.01 \ M = 10^{-2} \ M$.
Substituting the values: $2.0 \times 10^{-5} = K \times 10^{-2}$.
Solving for the rate constant $K$: $K = \frac{2.0 \times 10^{-5}}{10^{-2}} = 2.0 \times 10^{-3} \ s^{-1}$.
The half-life period $t_{1/2}$ for a first-order reaction is calculated as $t_{1/2} = \frac{0.693}{K}$.
$t_{1/2} = \frac{0.693}{2.0 \times 10^{-3}} = \frac{693}{2} = 346.5 \ s \approx 347 \ s$.

(B) The rate law for the reaction with respect to reactant $B$ is given by $R = k[B]^n$,where $n$ is the order of reaction with respect to $B$.
According to the problem,if the concentration of $B$ is doubled $([B]' = 2[B])$,the rate becomes one-fourth of the original rate $(R' = \frac{1}{4}R)$.
Substituting these into the rate law:
$\frac{1}{4}R = k(2[B])^n$
Dividing the new rate equation by the original rate equation:
$\frac{\frac{1}{4}R}{R} = \frac{k(2[B])^n}{k[B]^n}$
$\frac{1}{4} = 2^n$
$2^{-2} = 2^n$
Therefore,$n = -2$.

(D) The chemical reaction at the anode is: $C(s) + 2O^{2-} \rightarrow CO_2(g) + 4e^-$.
The reaction for the production of aluminium is: $Al^{3+} + 3e^- \rightarrow Al(s)$.
According to Faraday's laws of electrolysis,the number of equivalents of $Al$ produced must equal the number of equivalents of $C$ consumed.
The equivalent weight of $Al$ is $E_{Al} = \frac{27}{3} = 9$.
The equivalent weight of $C$ is $E_C = \frac{12}{4} = 3$.
Using the relation $\frac{w_{Al}}{E_{Al}} = \frac{w_C}{E_C}$:
$\frac{270}{9} = \frac{w_C}{3}$.
$w_C = \frac{270 \times 3}{9} = 90 \ kg$.

(C) According to Faraday's law of electrolysis,the number of equivalents of substances deposited or produced by the same quantity of charge is equal.
Equivalents of $Al = \text{Equivalents of } H_2$
$\text{Equivalents of } Al = \frac{\text{mass}}{\text{equivalent mass}} = \frac{4.5}{27/3} = \frac{4.5}{9} = 0.5$
Since the number of equivalents of $H_2$ is also $0.5$,we use the relation: $\text{Equivalents of } H_2 = \text{moles of } H_2 \times n\text{-factor}$.
For the reaction $2H^+ + 2e^- \to H_2$,the $n$-factor is $2$.
$0.5 = n_{H_2} \times 2 \implies n_{H_2} = 0.25 \text{ moles}$.
At $STP$,the volume of $1 \text{ mole}$ of gas is $22.4 \ L$.
$V_{H_2} = 0.25 \times 22.4 = 5.6 \ L$.

(B) Micelles are formed by surfactants or detergents in an aqueous solution above a specific concentration known as the Critical Micelle Concentration $(CMC)$.
Among the given options,$Dodecyl trimethyl ammonium chloride$ is a cationic surfactant (detergent).
$Urea$,$Pyridinium chloride$,and $Glucose$ do not form micelles in aqueous solutions.
Therefore,the correct answer is $Dodecyl trimethyl ammonium chloride$.

(A) Unlike lanthanoids,actinoids exhibit a wider range of oxidation states (from $+3$ to $+7$) because the energy gap between the $5f$,$6d$,and $7s$ subshells is very small.
In contrast,the $4f$ orbitals in lanthanoids are more deeply buried and have a larger energy gap relative to the $5d$ and $6s$ orbitals,which limits their oxidation states primarily to $+3$.

(D) The electronic configuration of Manganese $(Z = 25)$ is $[Ar] 3d^5 4s^2$.
After the removal of two electrons,the configuration of $Mn^{2+}$ becomes $[Ar] 3d^5$.
This $3d^5$ configuration is exceptionally stable due to the half-filled $d$-subshell.
Therefore,removing the third electron requires a significantly higher amount of energy,making the third ionization enthalpy of Manganese the highest among the given elements.

(A) An ion is colourless if it does not have any unpaired electrons in its $d$-orbitals.
$1. _{21}Sc^{3+}: [Ar] 3d^0$. It has no unpaired electrons,so it is colourless.
$2. _{26}Fe^{2+}: [Ar] 3d^6$. It has $4$ unpaired electrons,so it is coloured.
$3. _{22}Ti^{3+}: [Ar] 3d^1$. It has $1$ unpaired electron,so it is coloured.
$4. _{25}Mn^{2+}: [Ar] 3d^5$. It has $5$ unpaired electrons,so it is coloured.
Therefore,the correct option is $A$.

(C) Optical isomerism is shown by coordination compounds that lack a plane of symmetry and a center of symmetry.
$A$. $[Cu(NH_3)_4]^{2+}$ is a square planar complex,which is achiral.
$B$. $[ZnCl_4]^{2-}$ is a tetrahedral complex with identical ligands,which is achiral.
$C$. $[Cr(C_2O_4)_3]^{3-}$ is an octahedral complex with three bidentate oxalate ligands. It exists as a pair of non-superimposable mirror images (enantiomers),thus showing optical isomerism.
$D$. $[Co(CN)_6]^{3-}$ is an octahedral complex with identical ligands,which is achiral.
Therefore,the correct option is $C$.

(B) The reaction proceeds in two main steps:
$1$. The ketone,$3$-methylbutan-$2$-one,reacts with methylamine $(CH_3NH_2)$ to form an imine intermediate $(CH_3-CH(CH_3)-C(CH_3)=NCH_3)$ via nucleophilic addition followed by the elimination of water.
$2$. The imine is then reduced by lithium aluminium hydride $(LiAlH_4)$ to the corresponding secondary amine.
$3$. The final product is $N,2,3$-trimethylbutan-$2$-amine.

(A) The reaction sequence is as follows:
$1$. $CH_3COOH + SOCl_2 \rightarrow CH_3COCl$ ($A$ is acetyl chloride).
$2$. $CH_3COCl + C_6H_6 \xrightarrow{Anhyd. AlCl_3} C_6H_5COCH_3$ ($B$ is acetophenone,a Friedel-Crafts acylation reaction).
$3$. $C_6H_5COCH_3 + HCN \rightarrow C_6H_5C(OH)(CH_3)CN$ ($C$ is acetophenone cyanohydrin,a nucleophilic addition reaction).
$4$. $C_6H_5C(OH)(CH_3)CN + H_3O^+ \rightarrow C_6H_5C(OH)(CH_3)COOH$ ($D$ is $2-$hydroxy$-2-$phenylpropanoic acid,hydrolysis of the nitrile group).
Thus,the final product $D$ is $2-$hydroxy$-2-$phenylpropanoic acid.

(D) The reaction sequence is as follows:
$1$. Aniline reacts with $NaNO_2/HCl$ at $0-5^{\circ}C$ to form benzene diazonium chloride $(A)$: $C_6H_5NH_2 \xrightarrow{NaNO_2/HCl} C_6H_5N_2^+Cl^-$.
$2$. Benzene diazonium chloride reacts with $CuCN$ to form benzonitrile $(B)$: $C_6H_5N_2^+Cl^- \xrightarrow{CuCN} C_6H_5CN$.
$3$. Benzonitrile is reduced by $H_2/Ni$ to form benzylamine $(C)$: $C_6H_5CN \xrightarrow{H_2/Ni} C_6H_5CH_2NH_2$.
$4$. Benzylamine reacts with nitrous acid $(HNO_2)$ to form benzyl alcohol $(D)$: $C_6H_5CH_2NH_2 \xrightarrow{HNO_2} C_6H_5CH_2OH$.
Thus,the final product $D$ is benzyl alcohol $(C_6H_5CH_2OH)$.

(C) The electrolytic reduction of nitrobenzene $(C_6H_5NO_2)$ in a weakly acidic medium leads to the formation of $N$-phenylhydroxylamine $(C_6H_5NHOH)$.
In a strongly acidic medium,it would rearrange to form $p$-hydroxyaniline,but in a weakly acidic medium,the reduction stops at $N$-phenylhydroxylamine.

(A) The given structure represents a segment of polyisobutylene (or polyisobutene).
By observing the repeating unit $-[CH_2-C(CH_3)_2]-$,we can identify the monomer by introducing a double bond between the carbon atoms to satisfy valency.
Thus,the monomer is isobutylene ($2$-methylpropene),which is $CH_2=C(CH_3)_2$.
Therefore,the correct option is $A$.

(B) The correct answer is $B$. Cell membranes are primarily composed of proteins and lipids. The protein content in cell membranes typically varies from $46-76\%$,while the lipid content (including phospholipids) ranges from $20-53\%$,and carbohydrates constitute $1-8\%$. Although phospholipids are a major structural component,proteins are generally the most abundant component by mass in most cell membranes.

(A) . Benzyl chloride undergoes $S_N1$ mechanism most readily because it forms a resonance-stabilized benzyl carbocation $(C_6H_5CH_2^+)$ as an intermediate.
Primary alkyl halides like ethyl chloride and secondary alkyl halides like isopropyl chloride prefer $S_N2$ or $E2$ mechanisms,while chlorobenzene is highly unreactive towards nucleophilic substitution due to partial double bond character.

(C) The correct option is $C$.
$1$. For $[Co(NH_3)_6]^{3+}$,the oxidation state of $Co$ is $+3$. The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
$2$. $NH_3$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.
$3$. The configuration becomes $t_{2g}^6 e_g^0$,utilizing $d^2sp^3$ hybridization,which makes it an inner orbital complex.
$4$. Since all electrons are paired,the complex is diamagnetic.

(B) The thiol $(-SH)$ functional group present in the amino acid cysteine participates in the formation of disulphide bonds $(-S-S-)$ in proteins through an oxidation reaction.

(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.

(B) Given: Boiling point of solution = $100.18 \, ^oC$. Boiling point of pure water = $100 \, ^oC$.
Elevation in boiling point,$\Delta T_b = 100.18 - 100 = 0.18 \, ^oC$.
We know that $\Delta T_b = K_b \cdot m$ and $\Delta T_f = K_f \cdot m$.
Therefore,$\frac{\Delta T_f}{\Delta T_b} = \frac{K_f}{K_b}$.
$\Delta T_f = \Delta T_b \times \frac{K_f}{K_b} = 0.18 \times \frac{1.86}{0.512} \approx 0.654 \, ^oC$.
Since $\Delta T_f = T_f^{\circ} - T_f$,where $T_f^{\circ} = 0 \, ^oC$ for water:
$0.654 = 0 - T_f$.
$T_f = -0.654 \, ^oC$.
Rounding to two decimal places,the freezing point is $-0.65 \, ^oC$.