AIPMT 1989 Chemistry Question Paper with Answer and Solution

(A) The ratio of heat capacities $\frac{C_p}{C_v} = 1.4$ indicates that the gas $X$ is diatomic.
At $NTP$,$22.4 \ L$ of any gas contains $1 \ mol$ of molecules.
Therefore,$11.2 \ L$ of gas $X$ corresponds to $\frac{11.2}{22.4} = 0.5 \ mol$ of molecules.
Number of molecules $= 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}$ molecules.
Since the gas is diatomic,each molecule contains $2$ atoms.
Number of atoms $= 2 \times 3.011 \times 10^{23} = 6.022 \times 10^{23}$ atoms.

(B) The balanced chemical equation for the combustion of ethylene $(C_2H_4)$ is:
$C_2H_4 + 3O_2 \to 2CO_2 + 2H_2O$
From the stoichiometry of the reaction:
$1 \, \text{mole}$ of $C_2H_4$ $(28 \, g)$ requires $3 \, \text{moles}$ of $O_2$ $(3 \times 32 = 96 \, g)$.
Therefore,$28 \, g$ of $C_2H_4$ requires $96 \, g$ of $O_2$.
For $2.8 \, kg$ $(2800 \, g)$ of $C_2H_4$,the mass of $O_2$ required is:
$\text{Mass of } O_2 = \frac{96 \, g}{28 \, g} \times 2800 \, g = 9600 \, g = 9.6 \, kg$.

(D) The experimental evidence for the existence of the atomic nucleus was provided by the Rutherford $\alpha$-particle scattering experiment.
In this experiment,a beam of $\alpha$-particles was directed at a thin gold foil.
The observation that some $\alpha$-particles were deflected at large angles and some even bounced back led to the conclusion that the positive charge and most of the mass of the atom are concentrated in a very small central region called the nucleus.

(D) The correct answer is $(D)$.
Bohr's model assumes that electrons move in well-defined circular orbits with definite radii and velocities.
Statement $(D)$ describes Heisenberg's uncertainty principle,which states that it is impossible to determine simultaneously the exact position and exact momentum of an electron.
This principle contradicts the Bohr model's assumption of defined orbits.

(D) The $37^{th}$ electron will enter the $5s$ sub-level.
According to the Aufbau principle,electrons fill orbitals in the order of increasing energy.
For the $4p$ orbital,$n=4$ and $l=1$,so $(n+l) = 4+1 = 5$.
For the $5s$ orbital,$n=5$ and $l=0$,so $(n+l) = 5+0 = 5$.
When $(n+l)$ values are equal,the orbital with the lower value of $n$ is filled first. Since $4p$ has $n=4$ and $5s$ has $n=5$,the $4p$ orbital is filled before the $5s$ orbital.
Therefore,after the $4p^6$ configuration is completed in Krypton,the next electron $(37^{th})$ enters the $5s$ orbital.

(D) The magnetic quantum number $m$ depends on the azimuthal quantum number $l$ such that $m$ ranges from $-l$ to $+l$.
For an $s$-orbital,$l = 0$,which implies $m = 0$.
Since the given magnetic quantum number is $m = - 1$,it is impossible for an electron to be in an $s$-orbital.

(A) For a neutral atom,the number of electrons is equal to the number of protons.
Since the atom has $20$ protons,it must also have $20$ electrons to maintain electrical neutrality.
Therefore,the correct option is $A$.

(B) The correct answer is $(B)$.
According to Pauli's exclusion principle,no two electrons in an atom can have the same set of all four quantum numbers $(n, l, m_l, m_s)$.
Since an orbital is defined by the first three quantum numbers $(n, l, m_l)$,the only way to distinguish two electrons in the same orbital is by their spin quantum number $(m_s)$,which can only have values of $+1/2$ or $-1/2$.
Therefore,an orbital can accommodate a maximum of two electrons with opposite spins.

(D) In $sp^2$ hybridization,the central atom forms three hybrid orbitals that are directed towards the corners of an equilateral triangle.
This arrangement results in a bond angle of $120^{\circ}$ between the orbitals.
Therefore,the geometry of a molecule with $sp^2$ hybridization and no lone pairs on the central atom is Trigonal planar.

(C) In the compound $CH_2 = C = CH_2$ (allene),the central carbon atom is bonded to two other carbon atoms by double bonds.
Since the central carbon atom forms two $\pi$ bonds and two $\sigma$ bonds,it undergoes $sp$ hybridization.
The other options contain only $sp^2$ hybridized carbon atoms.

(A) The hybridization of a carbon atom is determined by the number of $\pi$ bonds attached to it.
$1$. The first carbon $(CH_3-)$ is bonded to $3$ hydrogens and $1$ carbon via single bonds ($4$ $\sigma$ bonds),so it is $sp^3$ hybridized.
$2$. The second carbon $(-CH=)$ is bonded to $1$ hydrogen,$1$ single bond,and $1$ double bond ($1$ $\pi$ bond),so it is $sp^2$ hybridized.
$3$. The third carbon $(=C=)$ is bonded to two double bonds ($2$ $\pi$ bonds),so it is $sp$ hybridized.
$4$. The fourth carbon $(=CH_2)$ is bonded to $2$ hydrogens and $1$ double bond ($1$ $\pi$ bond),so it is $sp^2$ hybridized.
Therefore,the hybridization order from left to right is $sp^3, sp^2, sp, sp^2$.

(B) The hybridization of the central atom can be determined using the formula: $\text{Steric Number} = (\text{Number of bonded atoms}) + (\text{Number of lone pairs})$.
For $BeF_2$: $\text{Steric Number} = 2 + 0 = 2$ ($sp$ hybridization).
For $BCl_3$: $\text{Steric Number} = 3 + 0 = 3$ ($sp^2$ hybridization).
For $C_2H_2$: The carbon atoms have a steric number of $2$ ($sp$ hybridization).
For $NH_3$: $\text{Steric Number} = 3 + 1 = 4$ ($sp^3$ hybridization).
Thus,the molecule with $sp^2$ hybridization is $BCl_3$.

(A) In $H_2S$,the central sulfur atom is $sp^3$ hybridized and possesses two lone pairs of electrons,resulting in a bent (angular) molecular geometry.
In contrast,$C_2H_2$ ($sp$ hybridization),$BeH_2$ ($sp$ hybridization),and $CO_2$ ($sp$ hybridization) all possess a linear arrangement of atoms due to their $sp$ hybridization and lack of lone pairs on the central atom.

(B) The overall value of the dipole moment of a polar molecule depends on its geometry and shape,i.e.,the vectorial addition of the dipole moments of the constituent bonds.
Water $(H_2O)$ has an angular structure with a bond angle of $105^{\circ}$,which results in a net dipole moment.
However,$BeF_2$ is a linear molecule where the dipole moments of the two $Be-F$ bonds are equal in magnitude and opposite in direction,thus canceling each other out,resulting in a net dipole moment of zero.

(D) Hydrogen bonding occurs when hydrogen is covalently bonded to a highly electronegative atom like $N$,$O$,or $F$.
In $NH_3$,the nitrogen atom is highly electronegative,allowing it to form hydrogen bonds.
Therefore,the correct option is $(d)$.

(D) In a metallic bond,metal cations are surrounded by a sea of mobile valence electrons.
Electrostatic attraction acts equally from all directions,meaning the bond is non-directional.
Therefore,metallic bonds do not possess highly directed characteristics,unlike covalent bonds.

(A) According to Charles's Law,the relationship between volume and temperature at constant pressure is given by: $V_t = V_0(1 + \alpha_v t)$.
Here,$V_t$ is the volume at $t\,^{\circ}C$,$V_0$ is the volume at $0\,^{\circ}C$,and $\alpha_v$ is the coefficient of volume expansion,which is equal to $\frac{1}{273.15}\,^{\circ}C^{-1}$.
For every $1\,^{\circ}C$ rise in temperature,the change in volume $\Delta V$ is given by $\Delta V = V_0 \times \alpha_v \times 1 = \frac{V_0}{273.15}$.
Thus,the volume increases by a definite fraction $\frac{1}{273.15}$ of its volume at $0\,^{\circ}C$.

(B) According to the combined gas law,for a fixed amount of gas,the relationship between pressure $(P)$,volume $(V)$,and temperature $(T)$ is given by $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.
Rearranging this equation,we get $\frac{P_1 V_1}{P_2 V_2} = \frac{T_1}{T_2}$.
Therefore,the correct option is $B$.

(D) The ideal gas equation is given by $PV = nRT$.
Since the number of moles $n = \frac{m}{M}$,where $m$ is the mass and $M$ is the molar mass,we can write $PV = \frac{m}{M}RT$.
Rearranging the terms to solve for density $(d = \frac{m}{V})$,we get $d = \frac{m}{V} = \frac{PM}{RT}$.
Thus,the correct option is $D$.

(A) Given that the ratio $\frac{C_p}{C_v} = 1.4$,the gas '$X$' is diatomic.
At $N.T.P.$,the volume of $1 \text{ mole}$ of any ideal gas is $22.4 \ L$.
Number of moles of gas '$X$' in $11.2 \ L$ = $\frac{11.2 \ L}{22.4 \ L/mol} = 0.5 \ mol$.
Number of molecules = $\text{moles} \times N_A = 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}$.
Since the gas is diatomic,each molecule contains $2$ atoms.
Number of atoms = $2 \times (3.011 \times 10^{23}) = 6.022 \times 10^{23}$.

(B) The formula for average speed is $V_{av} = \sqrt{\frac{8RT}{\pi M}}$.
For the average speeds of $CH_4$ and $O_2$ to be equal: $\sqrt{\frac{8RT_{CH_4}}{\pi M_{CH_4}}} = \sqrt{\frac{8RT_{O_2}}{\pi M_{O_2}}}$.
This simplifies to $\frac{T_{CH_4}}{M_{CH_4}} = \frac{T_{O_2}}{M_{O_2}}$.
Given $T_{O_2} = 300 \ K$,$M_{O_2} = 32 \ g/mol$,and $M_{CH_4} = 16 \ g/mol$.
Substituting the values: $\frac{T_{CH_4}}{16} = \frac{300}{32}$.
$T_{CH_4} = \frac{300 \times 16}{32} = 150 \ K$.

(C) The chemical equation is $A + B \rightleftharpoons C + D$.
Initial moles: $A = 1, B = 1, C = 0, D = 0$.
Moles at equilibrium: $A = 1 - 0.6 = 0.4, B = 1 - 0.6 = 0.4, C = 0.6, D = 0.6$.
The equilibrium constant $K_c$ is given by the expression: $K_c = \frac{[C][D]}{[A][B]}$.
Substituting the equilibrium concentrations: $K_c = \frac{0.6 \times 0.6}{0.4 \times 0.4} = \frac{0.36}{0.16} = 2.25$.

(C) For the first equilibrium: $SO_{2(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons SO_{3(g)}$,the equilibrium constant is $K_1 = \frac{[SO_3]}{[SO_2][O_2]^{1/2}}$.
For the second equilibrium: $2SO_{3(g)} \rightleftharpoons 2SO_{2(g)} + O_{2(g)}$,the equilibrium constant is $K_2 = \frac{[SO_2]^2 [O_2]}{[SO_3]^2}$.
Comparing the two expressions,we see that $K_2 = \left( \frac{1}{K_1} \right)^2 = \frac{1}{K_1^2}$.
Therefore,the correct relationship is $K_2 = \frac{1}{K_1^2}$.

(A) For a weak acid $HA$ dissociating as $HA \rightleftharpoons H^{+} + A^{-}$,the dissociation constant is given by $K_a = \frac{[H^{+}][A^{-}]}{[HA]}$.
Let the initial concentration be $c$ and the degree of dissociation be $\alpha$.
Then $[H^{+}] = c\alpha$,$[A^{-}] = c\alpha$,and $[HA] = c(1-\alpha)$.
$K_a = \frac{(c\alpha)(c\alpha)}{c(1-\alpha)} = \frac{c\alpha^2}{1-\alpha}$.
Since the acid is weak,$\alpha <<< 1$,so $1-\alpha \approx 1$.
Thus,$K_a \approx c\alpha^2$,which gives $\alpha = \sqrt{K_a/c}$.
Therefore,$[H^{+}] = c\alpha = c \times \sqrt{K_a/c} = \sqrt{K_a c}$.

(A) According to Ostwald's dilution law,for a weak electrolyte,the degree of dissociation $(\alpha)$ is related to the dilution $(V)$ by the expression $\alpha = \sqrt{K_a \times V}$.
As the dilution $(V)$ increases,the degree of dissociation $(\alpha)$ increases.

(B) The correct option is $(b)$.
When $HCl$ gas is passed through a saturated solution of $NaCl$,it dissociates into $H^+$ and $Cl^-$ ions.
This increases the concentration of $Cl^-$ ions in the solution.
According to the common ion effect,the increased concentration of $Cl^-$ ions shifts the equilibrium $NaCl(s) \rightleftharpoons Na^+(aq) + Cl^-(aq)$ to the left.
Consequently,the solubility of $NaCl$ decreases,leading to the precipitation of $NaCl$.

(B) $NH_4Cl$ is a salt of a weak base $(NH_4OH)$ and a strong acid $(HCl)$.
When dissolved in water,it undergoes hydrolysis to form an acidic solution.
$NH_4Cl + H_2O \rightleftharpoons NH_4OH + HCl$
$NH_4^+ + H_2O \rightleftharpoons NH_4OH + H^+$
$NaCl$,$KCl$,and $Na_2SO_4$ are salts of strong acids and strong bases,so they do not undergo hydrolysis.

(B) $NH_4CN$ is a salt formed from a weak acid $(HCN)$ and a weak base $(NH_4OH)$.
For a salt of a weak acid and a weak base,the hydrolysis constant $(K_h)$ is given by the formula:
$K_h = \frac{K_w}{K_a \times K_b}$
Where $K_w$ is the ionic product of water,$K_a$ is the dissociation constant of the weak acid,and $K_b$ is the dissociation constant of the weak base.
Therefore,the correct option is $(B)$.

(B) At $25\,^oC$,pure water is neutral,meaning the concentration of hydrogen ions is equal to the concentration of hydroxide ions: $[H^{+}] = [OH^{-}]$.
The ionic product of water $(K_w)$ is $1.0 \times 10^{-14}$.
Thus,$[H^{+}] \times [OH^{-}] = 10^{-14}$,which implies $[H^{+}]^2 = 10^{-14}$,so $[H^{+}] = 10^{-7} \, M$.
The $pH$ is calculated as $pH = -\log[H^{+}] = -\log(10^{-7}) = 7$.

(B) An acidic buffer solution is prepared by mixing a weak acid and its salt with a strong base.
$CH_3COOH$ is a weak acid and $CH_3COONa$ is its salt with a strong base $(NaOH)$.
Therefore,the mixture of $CH_3COOH + CH_3COONa$ forms an acidic buffer.

(A) For a basic buffer,the formula is $pOH = pK_b + \log \frac{[salt]}{[base]}$.
Given that the volumes are equal,the concentration ratio is equal to the mole ratio.
$pOH = 5 + \log \frac{0.02}{0.2} = 5 + \log(0.1) = 5 - 1 = 4$.
Since $pH + pOH = 14$,we have $pH = 14 - 4 = 10$.

(D) For a weak acid $HA$,the dissociation constant $K_a = 10^{-9}$ and concentration $C = 0.1 \ M$.
The concentration of hydrogen ions $[H^+]$ is given by the formula $[H^+] = \sqrt{K_a \times C}$.
$[H^+] = \sqrt{10^{-9} \times 0.1} = \sqrt{10^{-10}} = 10^{-5} \ M$.
$pH = -\log[H^+] = -\log(10^{-5}) = 5$.
Since $pH + pOH = 14$ at $25^{\circ}C$,we have $pOH = 14 - pH$.
$pOH = 14 - 5 = 9$.

(B) The goal is to find $\Delta H$ for the transition: $C(\text{graphite}) \to C(\text{diamond})$.
Using Hess's Law,we reverse the first equation and add it to the second:
$CO_{2(g)} \to C(\text{diamond}) + O_{2(g)}; \Delta H = +395 \text{ kJ}$
$C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta H = -393.5 \text{ kJ}$
Adding these two equations gives:
$C(\text{graphite}) \to C(\text{diamond}); \Delta H = 395 - 393.5 = +1.5 \text{ kJ}$.

(B) Alkaline pyrogallol is a well-known absorbent for $O_{2}$ gas in gas analysis.
Oil of cinnamon (which contains cinnamaldehyde) is used to absorb $O_{3}$ gas.
Therefore,the correct pair is $O_{2}$ and $O_{3}$.

(A) The correct answer is $A$.
$Cl_2 + 2KBr \to 2KCl + Br_2$
$A$ more electronegative halogen can displace a less electronegative halogen from its salt solution. Since chlorine is more electronegative than bromine,it displaces bromine from $KBr$.

(A) In the titration of $Na_2CO_3$ with $HCl$ using phenolphthalein,the reaction proceeds as: $Na_2CO_3 + HCl \rightarrow NaHCO_3 + NaCl$.
Phenolphthalein changes color in the $pH$ range of $8.2$ to $10.0$.
At the end point of this reaction,the solution contains $NaHCO_3$,which is weakly basic,but the phenolphthalein indicator becomes colourless as the $pH$ drops below $8.2$.
Therefore,the correct observation is that the solution becomes colourless.

(B) Cis-trans isomerism (geometrical isomerism) occurs in compounds containing a carbon-carbon double bond where each carbon atom is attached to two different groups.
In $ClCH=CHCl$ ($1$,$2$-dichloroethene),each carbon atom of the double bond is attached to one $H$ atom and one $Cl$ atom.
Since the groups attached to each carbon are different,it can exist in two forms: cis$-1,2-$dichloroethene and trans$-1,2-$dichloroethene.
Therefore,option $(B)$ is correct.

(D) The molecule $A$ is benzene $(C_6H_6)$.
In benzene,all carbon atoms are $sp^2$ hybridized.
$sp^2$ hybridized carbon atoms have a trigonal planar geometry with a bond angle of $120^{\circ}$.
The carbon-carbon bond length in benzene is $1.39 \ \mathring{A}$,which is intermediate between a single bond $(1.54 \ \mathring{A})$ and a double bond $(1.34 \ \mathring{A})$.
Therefore,the $C-C$ bond angle is $120^{\circ}$.

(A) Pyrolysis (or cracking) is the process of breaking down higher molecular weight hydrocarbons into lower molecular weight hydrocarbons by heating in the presence or absence of a catalyst.
For example:
$C_6H_{14} \xrightarrow[\Delta]{\text{Pyrolysis}} C_2H_4 + C_4H_{10}$
Thus,option $A$ is the correct method for obtaining hydrocarbons with a lower carbon number.

(A) The reaction of an alkene with hypochlorous acid $(HOCl)$ follows electrophilic addition.
$CH_2=CH_2 + HOCl \to CH_2(OH)-CH_2Cl$
This product is $1-$chloro$-2-$hydroxyethane.
Therefore,the hydrocarbon is ethylene $(CH_2=CH_2)$.

(A) The cyclic hydrocarbon $A$ described is benzene $(C_6H_6)$.
In benzene,all carbon atoms are $sp^2$ hybridized and lie in the same plane.
The carbon-carbon bond length in benzene is $1.39 \ \mathring{A}$,which satisfies the condition $1.34 \ \mathring{A} < 1.39 \ \mathring{A} < 1.54 \ \mathring{A}$.
For $sp^2$ hybridized carbon atoms in a planar ring,the $C-C-C$ bond angle is $120^{\circ}$.

(A) The reaction is an electrophilic addition of acetyl chloride to the alkene catalyzed by a Lewis acid $(ZnCl_2)$.
The acetyl cation $(CH_3CO^+)$ attacks the terminal carbon of $2-$methylpropene to form a stable tertiary carbocation intermediate,$(CH_3)_2C^+-CH_2-COCH_3$.
This intermediate then reacts with the chloride ion $(Cl^-)$ to give $4-$chloro$-4-$methylpentan$-2-$one,$CH_3-C(Cl)(CH_3)-CH_2-COCH_3$.

(A) In the presence of peroxide,the addition of $HBr$ to an unsymmetrical alkene like propene follows the Anti-Markownikoff's rule.
According to this rule,the negative part of the reagent $(Br^-)$ attaches to the carbon atom with the greater number of hydrogen atoms.
Reaction: $CH_3-CH=CH_2 + HBr \xrightarrow{\text{peroxide}} CH_3-CH_2-CH_2-Br$ ($n-$propyl bromide).
Therefore,the major product is $n-$propyl bromide.

(A) The formation of $C-H$ bond in acetylene involves $sp-$hybridised carbon atom.
Since $s-$electrons are closer to the nucleus than $p$ electrons,$sp$ hybrid orbitals are more electronegative due to their smaller (in length) and thicker lobes,which increases the time spent by electrons near the nucleus compared to $sp^2$ and $sp^3$ orbitals.
Thus,the electrons present in a bond having more $s-$character will be closer to the nucleus.
In alkynes $(C \equiv C-H)$ bond,the $s-$character is $50\%$,so the electrons constituting this bond are more strongly attracted by the carbon nucleus.
Thus,the acetylenic carbon atom becomes more electronegative in comparison to $sp^2$ and $sp^3$ hybridized carbons,and hence the hydrogen atom present on the carbon atom $(C \equiv C-H)$ can be easily removed as a proton $(H^+)$.

(D) Terminal alkynes,such as compound $(iii)$,contain an acidic hydrogen atom attached to the $sp$-hybridized carbon.
Ammoniacal silver nitrate (Tollens' reagent) reacts with terminal alkynes to form a white precipitate of silver acetylide.
$CH_3-CH_2-C\equiv CH + [Ag(NH_3)_2]^+ + OH^-$ $\rightarrow CH_3-CH_2-C\equiv C^- Ag^+ \text{ (white precipitate)} + 2NH_3 + H_2O$
Compounds $(i)$,$(ii)$,and $(iv)$ do not possess acidic terminal hydrogen atoms and therefore do not form a precipitate with this reagent.

(C) $(C) \, R-CH_2-CCl_2-R \xrightarrow{KOH \text{ (alc.)}} R-C \equiv C-R + 2HCl$
This reaction is an example of dehydrohalogenation,where two molecules of $HCl$ are removed from a gem-dihalide to form an alkyne.
Alcoholic $KOH$ acts as a strong base to facilitate this elimination reaction.

(C) Ammoniacal cuprous chloride (Tollens' reagent for alkynes) reacts with terminal alkynes to form a red precipitate of copper acetylide.
$2NH_4OH + Cu_2Cl_2 \to 2CuOH + 2NH_4Cl$
$CuOH + 2NH_4OH \to [Cu(NH_3)_2]OH + 2H_2O$
$HC \equiv CH + 2[Cu(NH_3)_2]OH \to CuC \equiv CCu + 4NH_3 + 2H_2O$
Since $C_2H_2$ (acetylene) is a terminal alkyne,it forms the copper derivative (copper acetylide).

(D) double salt is a crystalline compound that contains two different salts combined in a definite molar ratio.
These salts dissociate completely into their constituent ions in an aqueous solution.
Potash alum,with the formula $K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O$,is a classic example of a double salt.
In contrast,$K_4[Fe(CN)_6]$ is a coordination compound (complex salt) that does not dissociate completely into its constituent ions.

(A) $Ortho$-nitrophenol and $para$-nitrophenol can be separated by steam distillation.
$o$-nitrophenol is steam volatile due to the presence of intramolecular hydrogen bonding,which reduces the intermolecular forces of attraction.
In contrast,$p$-nitrophenol exhibits intermolecular hydrogen bonding,leading to the association of molecules,which increases its boiling point and makes it less volatile.
Intramolecular hydrogen bonding occurs in $ortho$-nitrophenol because the $-NO_2$ and $-OH$ groups are in close proximity to each other.

(B) Sleeping sickness, also known as African trypanosomiasis, is a parasitic disease caused by the protozoan $Trypanosoma$ $brucei$.
The disease is transmitted to humans through the bite of an infected $Tsetse$ $fly$ (genus $Glossina$).
Therefore, the $Tsetse$ $fly$ acts as the biological vector for this disease.

(C) In the Hall-$H$éroult process for the extraction of aluminium,pure alumina $(Al_2O_3)$ is dissolved in molten cryolite $(Na_3AlF_6)$.
This mixture is used as the electrolyte because it lowers the melting point of the mixture and increases its electrical conductivity.

(B) Elements like $I_2$,$N_2$,and $O_2$ exist as stable diatomic molecules.
Phosphorus,due to its smaller size and high repulsion between lone pairs,prefers to form a tetrahedral $P_4$ molecule rather than a diatomic $P_2$ molecule.

(A) Ammonia $(NH_3)$ is a basic gas.
To dry a basic gas,we must use a basic drying agent.
If an acidic drying agent is used,it will react with the ammonia to form a salt.
$CaCl_2$ forms an adduct with $NH_3$ $(CaCl_2 \cdot 8NH_3)$.
$P_2O_5$ and $H_2SO_4$ are acidic and will react with $NH_3$ to form ammonium salts.
Therefore,$CaO$ (quicklime) is the only suitable basic drying agent among the given options.

(A) The correct answer is $A$. White phosphorus is soluble in $CS_2$,whereas red phosphorus is insoluble in $CS_2$. Both forms consist of $P_4$ units (same kind of atoms),both can be oxidized by heating in air,and they can be interconverted under specific conditions.

(A) The correct answer is $A$.
$NCl_5$ is not known because nitrogen lacks $d$-orbitals in its valence shell,which prevents it from expanding its covalency beyond $4$.

(D) The anhydride of an acid is formed by the removal of water molecules from the acid.
For nitric acid $(HNO_3)$,the reaction is:
$2HNO_3 \rightarrow N_2O_5 + H_2O$
Thus,$N_2O_5$ is the anhydride of $HNO_3$.

(B) The boiling point of hydrides of Group $15$ elements depends on the molecular mass and the presence of hydrogen bonding.
$NH_3$ has the highest boiling point due to intermolecular hydrogen bonding.
For the remaining hydrides $(PH_3, AsH_3, SbH_3, BiH_3)$,the boiling point increases with an increase in molecular mass due to stronger van der Waals forces.

\text{Hydride}	\text{Boiling point } $(K)$
$NH_3$	$238.5$
$PH_3$	$185.5$
$AsH_3$	$210.6$
$SbH_3$	$254.6$
$BiH_3$	$290.0$

Thus,$PH_3$ has the lowest boiling point.

(B) Polymorphism is the ability of a substance to exist in two or more different crystalline forms or structures.

(C) Fractional distillation of liquid air is based on the difference in the boiling points $(b.p.)$ of its components.
Oxygen has a higher boiling point $(90 \ K)$ compared to Nitrogen $(77 \ K)$.
Therefore,Nitrogen boils off first,leaving behind liquid Oxygen.
Hence,Option $C$ is the correct answer.

(D) Among the given elements,$P$,$Na$,and $S$ react directly with oxygen to form their respective oxides.
$P_4 + 5O_2 \longrightarrow P_4O_{10}$
$S + O_2 \longrightarrow SO_2$
$4Na + O_2 \longrightarrow 2Na_2O$
Chlorine $(Cl)$ does not react directly with oxygen under normal conditions. Therefore,the correct option is $D$.

(A) In $CH_4$ molecules,there exist strong covalent bonds between the atoms within each molecule.
However,in solid $CH_4$,the molecules are held together by weak van der Waals forces.
Since the constituent particles are molecules held by weak intermolecular forces,it is classified as a molecular solid.
Therefore,option $A$ is the correct answer.

(A) The strength of an $o, p-$directing group is determined by its ability to donate electron density to the benzene ring via the resonance effect ($+R$ or $+M$ effect).
Among the given options:
$1$. $-OH$ has a strong $+R$ effect due to the lone pair on oxygen,making it a powerful activating group.
$2$. $-Cl$ and $-Br$ are deactivating groups due to their strong $-I$ effect,although they are $o, p-$directing due to the $+R$ effect.
$3$. $-C_6H_5$ (phenyl group) is an activating group but is weaker than $-OH$ in terms of electron donation to the ring.
Therefore,$-OH$ is the strongest $o, p-$directing group among the choices provided.

(A) The ebullioscopic constant $(K_b)$ is given by the formula: $K_b = \frac{R \times T_b^2 \times M}{1000 \times \Delta H_{vap}}$
Where:
$R = 1.987 \, cal \, K^{-1} \, mol^{-1} \approx 2 \, cal \, K^{-1} \, mol^{-1}$
$T_b = 100 + 273 = 373 \, K$
$\Delta H_{vap} = 9700 \, cal/mol$
$M = 18 \, g/mol$ (Molar mass of water)
Substituting the values:
$K_b = \frac{2 \times (373)^2 \times 18}{1000 \times 9700}$
$K_b = \frac{2 \times 139129 \times 18}{9700000}$
$K_b = \frac{5008644}{9700000} \approx 0.516 \, ^oC \, kg/mol$
Thus,the correct option is $A$.

(C) In a face-centred cubic $(FCC)$ unit cell,atoms are present at the corners and at the centre of each face.
Number of atoms at corners = $8 \times \frac{1}{8} = 1$.
Number of atoms at face centres = $6 \times \frac{1}{2} = 3$.
Total number of atoms = $1 + 3 = 4$.

(B) An $\alpha$-particle is a helium nucleus,represented as $_2^4He$.
When a radioactive nucleus emits an $\alpha$-particle,its mass number $(A)$ decreases by $4$ and its atomic number $(Z)$ decreases by $2$.
The nuclear reaction is represented as: $_Z^AX \rightarrow _{Z-2}^{A-4}Y + _2^4He$.

(A) The enzyme $Zymase$ is responsible for the fermentation of glucose into ethanol and carbon dioxide.
The chemical reaction is as follows:
$C_6H_{12}O_6 \xrightarrow{\text{Zymase}} 2C_2H_5OH + 2CO_2$

(C) . $\text{Liquid (dispersed phase)} + \text{Solid (dispersion medium)} = \text{Gel (colloid)}$.
Examples include butter,cheese,and jelly.

(A) Lyophobic sols are more easily coagulated by the addition of suitable electrolytes.
To prevent the coagulation of a lyophobic sol by the addition of an electrolyte,a small amount of a lyophilic colloid is added to it.
This lyophilic colloid acts as a protective colloid,and its protective power is measured in terms of the Gold number.
Thus,the Gold number is a measure of the protective action of a lyophilic colloid on a lyophobic colloid.

(A) When an excess of electrolyte is added to a colloid,the colloidal particles interact with ions carrying a charge opposite to that present on them.
This causes neutralization of the charge on the colloidal particles,which leads to their coagulation or precipitation.

(D) . Iron.
Iron undergoes oxidation in the presence of moisture and oxygen to form hydrated iron$(III)$ oxide,commonly known as rust.
The chemical reaction is: $4Fe(s) + 3O_2(g) + 2xH_2O(l) \to 2Fe_2O_3 \cdot xH_2O(s)$.

(D) . Potash alum is a double salt with the formula $K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O$.
Double salts dissociate completely into their constituent ions in aqueous solution,whereas coordination compounds like $K_4[Fe(CN)_6]$ do not.

(B) The reaction between $CuSO_4$ and $K_4[Fe(CN)_6]$ produces a chocolate brown precipitate of $Cu_2[Fe(CN)_6]$,not a blue colouration.
$2CuSO_4 + K_4[Fe(CN)_6] \to Cu_2[Fe(CN)_6] + 2K_2SO_4$ (Chocolate brown).
In contrast,$4NH_4OH + CuSO_4 \to [Cu(NH_3)_4]SO_4 + 4H_2O$ produces a deep blue solution.
Anhydrous $CuSO_4 + 5H_2O \to CuSO_4 \cdot 5H_2O$ produces a blue colour.
$4FeCl_3 + 3Na_4[Fe(CN)_6] \to Fe_4[Fe(CN)_6]_3 + 12NaCl$ produces a Prussian blue precipitate.

(B) $AgBr$ is most sensitive to light and undergoes photochemical reduction.
$2AgBr \xrightarrow{\text{Light}} 2Ag + Br_2$

(B) The extraction of silver $(Ag)$ from its ore (like $Ag_2S$) involves leaching with a dilute solution of $NaCN$ in the presence of air $(O_2)$:
$Ag_2S + 4NaCN \to 2Na[Ag(CN)_2] + Na_2S$
$2Na_2S + 2O_2 + H_2O \to 2NaOH + Na_2S_2O_3$
Silver is then recovered from the complex by displacement using zinc dust:
$2Na[Ag(CN)_2] + Zn \to Na_2[Zn(CN)_4] + 2Ag$

(D) The extraction of $Ag$ (Silver) from its ore (like $Ag_2S$) involves leaching the finely powdered ore with a dilute solution of $KCN$ in the presence of air $(O_2)$.
The chemical reaction is: $4Ag(s) + 8CN^-(aq) + 2H_2O(aq) + O_2(g) \rightarrow 4[Ag(CN)_2]^-(aq) + 4OH^-(aq)$.
The soluble complex formed is potassium argentocyanide,$K[Ag(CN)_2]$.
Thus,the element being extracted is Silver $(Ag)$.

(C) When phenol reacts with $CCl_4$ in the presence of aqueous $NaOH$,the reaction is a variation of the Riemer-Tiemann reaction.
Instead of forming an aldehyde group (as with $CHCl_3$),the use of $CCl_4$ results in the introduction of a carboxylic acid group $(-COOH)$ at the ortho position of the phenol.
The reaction is:
$C_6H_5OH + CCl_4 + 4NaOH \rightarrow C_6H_4(OH)(COOH) + 4NaCl + 2H_2O$.
The product formed is Salicylic acid.

(C) The reaction of alcohols with $ZnCl_2 +$ conc. $HCl$ is known as the $Lucas$ test.
$3^o$ alcohols react immediately to form turbidity.
$2^o$ alcohols react after about $5$ minutes.
$1^o$ alcohols do not react at room temperature.
Among the given options,$(CH_3)_3C-OH$ ($2-$methylpropan$-2-$ol) is a $3^o$ alcohol,which reacts immediately to produce turbidity.

(A) When ethyl alcohol is passed over heated alumina $(Al_2O_3)$ in the presence of ammonia,it undergoes an ammonolysis reaction to form ethylamine.
The chemical reaction is:
$C_2H_5OH + NH_3 \xrightarrow{Al_2O_3, \Delta} C_2H_5NH_2 + H_2O$
Thus,the correct product is ethylamine $(C_2H_5NH_2)$.

(A) When acetone $(CH_3COCH_3)$ is saturated with dry hydrogen chloride gas,it undergoes an aldol condensation followed by dehydration to form Phorone.
The reaction is:
$3 CH_3COCH_3 \xrightarrow{HCl} (CH_3)_2C = CH - CO - CH = C(CH_3)_2 + 2 H_2O$
The product formed is Phorone.

(D) The $Wolf-Kishner$ reduction is a method used to convert carbonyl compounds (aldehydes and ketones) into their corresponding alkanes (hydrocarbons).
In this reaction, the ketone is treated with hydrazine $(NH_2NH_2)$ followed by heating with a strong base like $KOH$ in a high-boiling solvent such as ethylene glycol.
The reaction for acetone is:
$CH_3COCH_3 \xrightarrow{NH_2NH_2 / KOH, \text{glycol}, \Delta} CH_3CH_2CH_3 + N_2 + H_2O$

(A) Paraldehyde is primarily used as a sedative and hypnotic medication.
It was historically used to treat conditions such as insomnia and epilepsy,although its use has decreased over time due to the development of safer and more effective alternatives.
Paraldehyde is known for its sedative effects,which can induce sleep and reduce anxiety.
Therefore,it is classified as a medicine.

(A) The structure shown in the image is a cyclic trimer of formaldehyde $(HCHO)$,which is known as $1,3,5$-trioxane or simply trioxane.
It is formed by the trimerization of formaldehyde in the presence of an acid catalyst.
The reaction is: $3HCHO \rightleftharpoons (CH_2O)_3$ (Trioxane).

(D) The reduction of a carboxylic acid $(RCOOH)$ to a primary alcohol $(RCH_2OH)$ requires a strong reducing agent.
$LiAlH_4$ (Lithium aluminium hydride) is a powerful reducing agent capable of reducing carboxylic acids directly to primary alcohols.
Other reagents like $Zn/HCl$ (Clemmensen reduction) reduce carbonyl groups to methylene groups,and $Na/\text{alcohol}$ (Bouveault-Blanc reduction) is typically used for esters.
Therefore,the correct option is $(D)$.

(C) The Hofmann bromamide degradation reaction is a characteristic reaction of primary amides.
In this reaction,a primary amide $(R-CONH_2)$ reacts with bromine $(Br_2)$ in the presence of a strong base like potassium hydroxide $(KOH)$ to form a primary amine $(R-NH_2)$ with one carbon atom less than the original amide.
The balanced chemical equation is:
$R-CONH_2 + Br_2 + 4KOH \xrightarrow{} R-NH_2 + 2KBr + K_2CO_3 + 2H_2O$
Therefore,the compound that undergoes this reaction is $R-CONH_2$.

(B) To improve physical properties such as water repellency,reduced water absorption,and resistance to oxidising agents,a process called vulcanisation is carried out.
This process consists of heating a mixture of raw rubber with sulphur and an appropriate additive at a temperature range between $373 \ K$ to $415 \ K$.
On vulcanisation,sulphur forms cross-links at the reactive sites of double bonds,which causes the rubber to become stiffened.

(D) The thermal decomposition of orthophosphoric acid $(H_3PO_4)$ at $600\,^{\circ}C$ results in the formation of metaphosphoric acid $(HPO_3)$ and water vapor.
The chemical equation is: $H_3PO_4 \xrightarrow{600\,^{\circ}C} HPO_3 + H_2O$.
Therefore,the correct option is $(D)$.

(A) The iodoform test is given by compounds containing the $CH_3-C(=O)-$ group or the $CH_3-CH(OH)-$ group.
$A$. Pentan-$1$-one (which is actually $Pentanal$) does not contain the $CH_3-C(=O)-$ group,as the carbonyl group is at the terminal position.
$B$. Pentan-$2$-one contains the $CH_3-C(=O)-$ group and gives a positive iodoform test.
$C$. Propan-$2$-one (acetone) contains the $CH_3-C(=O)-$ group and gives a positive iodoform test.
$D$. Ethanol contains the $CH_3-CH(OH)-$ group and gives a positive iodoform test.
Therefore,the correct option is $A$.