AIPMT 2012 Chemistry Question Paper with Answer and Solution

(D) Let us analyze each statement:
$(1)$ $Equisetum$ is a pteridophyte. In pteridophytes,the female gametophyte is retained on the parent sporophyte for variable periods. This statement is correct.
$(2)$ $Ginkgo$ is a gymnosperm. In gymnosperms,the male gametophyte is highly reduced and is not independent. This statement is correct.
$(3)$ $Polytrichum$ (a moss) has a more complex and differentiated sporophyte (foot,seta,capsule) compared to the simple,undifferentiated sporophyte of $Riccia$ (a liverwort). This statement is correct.
$(4)$ Sexual reproduction in $Volvox$ is oogamous,not isogamous. This statement is incorrect.
$(5)$ Slime mold spores possess true cell walls (made of cellulose). This statement is incorrect.
Therefore,statements $(1), (2),$ and $(3)$ are correct. The total number of correct statements is $3$.

(A) According to the theorem of parallel axes,the moment of inertia $I$ about an axis parallel to the axis passing through the center of mass is given by $I = I_{CM} + Ma^2$,where $M$ is the mass of the disc,$I_{CM}$ is the moment of inertia about the center of mass,and $a$ is the perpendicular distance between the two axes.
Since the axis is perpendicular to the disc and passes through the given points,the distance $a$ is the distance of the point from the center of the disc $(A)$.
For the moment of inertia $I$ to be maximum,the distance $a$ must be maximum.
Looking at the figure,point $B$ is on the circumference of the disc,making its distance from the center $A$ the radius of the disc,which is the maximum distance among the given points.
Therefore,the moment of inertia is maximum about an axis passing through point $B$.

(B) The fundamental frequency of a string of length $L$ is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the mass per unit length.
Since $T$ and $\mu$ are constant for all segments,we have $L = \frac{1}{2n} \sqrt{\frac{T}{\mu}} = \frac{k}{n}$,where $k = \frac{1}{2} \sqrt{\frac{T}{\mu}}$ is a constant.
The string is divided into three segments of lengths $L_1, L_2$,and $L_3$ such that $L = L_1 + L_2 + L_3$.
Substituting the expression for length in terms of frequency: $\frac{k}{n} = \frac{k}{n_1} + \frac{k}{n_2} + \frac{k}{n_3}$.
Dividing both sides by $k$,we get $\frac{1}{n} = \frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3}$.

(C) The correct answer is $(C)$.
Tubectomy is a surgical sterilization method for females.
In this procedure,a small part of the Fallopian tubes is removed or tied up through a small incision in the abdomen or through the vagina.
This blocks the passage of the ovum,thereby preventing fertilization by preventing the sperm from reaching the egg.
It is a highly effective and permanent method of contraception.

(B) : Streptokinase (Tissue Plasminogen Activator or $TPA$) is an enzyme obtained from the culture of some haemolytic bacterium $Streptococcus$ which is modified genetically to function as a clot buster.
It helps in clearing blood clots inside the blood vessels through the dissolution of intravascular fibrin during myocardial infarction.

(A) $n$ represents the main energy level and $l$ represents the subshell.
If $n = 4$ and $l = 3,$ the subshell is $4f.$
In an $f$ subshell,there are $2l + 1 = 2(3) + 1 = 7$ orbitals.
Since each orbital can accommodate a maximum of $2$ electrons,the maximum number of electrons in the $4f$ subshell is $7 \times 2 = 14$.

(C) The electronic configuration of rubidium atom $(Z = 37)$ is $[Kr] 5s^1$.
The valence electron is in the $5s$ orbital.
For the $5s$ orbital,the principal quantum number $(n)$ is $5$.
For an $s$-orbital,the azimuthal quantum number $(l)$ is $0$.
Since $l = 0$,the magnetic quantum number $(m_l)$ is $0$.
The spin quantum number $(m_s)$ can be either $+1/2$ or $-1/2$.
Thus,the set of quantum numbers is $(n = 5, l = 0, m_l = 0, m_s = +1/2)$.
Therefore,the correct option is $C$.

(A) The formula for orbital angular momentum is given by $\sqrt{l(l+1)} \times \frac{h}{2 \pi}$.
For a $p-$ electron,the azimuthal quantum number $l = 1$.
Substituting the value of $l$ into the formula:
Orbital angular momentum $= \sqrt{1(1+1)} \times \frac{h}{2 \pi}$
$= \sqrt{2} \times \frac{h}{2 \pi}$
$= \frac{\sqrt{2} h}{2 \pi} = \frac{h}{\sqrt{2} \pi}$.

(A) Atomic radius of the elements decreases across a period from left to right due to an increase in effective nuclear charge.
On moving down a group,the number of shells increases,so the atomic radius increases.
Amongst isoelectronic species,the ionic radius increases with an increase in negative charge or a decrease in positive charge.
Therefore,the statement that 'smaller the positive charge on the cation,smaller is the ionic radius' is incorrect,because a smaller positive charge means a larger ionic radius.

(A) The bond order is calculated using the formula: $Bond \ order = \frac{N_b - N_a}{2}$,where $N_b$ is the number of electrons in bonding molecular orbitals and $N_a$ is the number of electrons in antibonding molecular orbitals.
For $CO$ ($6+8=14$ electrons): Bond order $= \frac{10-4}{2} = 3$.
For $NO^{+}$ ($7+8-1=14$ electrons): Bond order $= \frac{10-4}{2} = 3$.
Since both $CO$ and $NO^{+}$ have $14$ electrons,they have the same bond order of $3$.
For other options:
$NO^{-}$ ($16$ electrons,$BO=2$),$CN^{-}$ ($14$ electrons,$BO=3$).
$O_2$ ($16$ electrons,$BO=2$),$N_2$ ($14$ electrons,$BO=3$).
$O_2$ ($16$ electrons,$BO=2$),$B_2$ ($10$ electrons,$BO=1$).

(D) In $BF_4^-$,the central atom $B$ has $4$ bond pairs and $0$ lone pairs,resulting in $sp^3$ hybridization and a tetrahedral shape.
In $NH_4^+$,the central atom $N$ has $4$ bond pairs and $0$ lone pairs,resulting in $sp^3$ hybridization and a tetrahedral shape.
Since both species have the same hybridization and shape,they are isostructural.

(B) The bond order is calculated using the formula: $\text{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
For $O_{2}^{+}$ (total $15$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Bond order $= \frac{10 - 5}{2} = 2.5$.
For $O_{2}^{-}$ (total $17$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. Bond order $= \frac{10 - 7}{2} = 1.5$.
For $O_{2}^{2-}$ (total $18$ electrons): Bond order $= \frac{10 - 8}{2} = 1.0$.
For $O_{2}$ (total $16$ electrons): Bond order $= \frac{10 - 6}{2} = 2.0$.
Thus,$O_{2}^{-}$ has a bond order of $1.5$.

(D) To determine the number of bond pairs and lone pairs,we analyze the Lewis structure of each species:
$1$. $H_2O$: The central oxygen atom has $2$ bond pairs (with $H$) and $2$ lone pairs.
$2$. $BF_3$: The central boron atom has $3$ bond pairs (with $F$) and $0$ lone pairs.
$3$. $NH_2^-$: The central nitrogen atom has $2$ bond pairs (with $H$) and $2$ lone pairs.
$4$. $PCl_3$: The central phosphorus atom has $3$ bond pairs (with $Cl$) and $1$ lone pair.
Thus,$PCl_3$ is the species that contains three bond pairs and one lone pair around the central atom.

(A) The electronic configuration of $O_2$ is $KK \sigma(2s)^2 \sigma^*(2s)^2 \sigma(2p_z)^2 \pi(2p_x)^2 \pi(2p_y)^2 \pi^*(2p_x)^1 \pi^*(2p_y)^1$.
When $O_2$ gains an electron to form $O_2^-$,the incoming electron enters the next available orbital,which is the antibonding $\pi^*$ orbital.
Thus,the configuration of $O_2^-$ becomes $KK \sigma(2s)^2 \sigma^*(2s)^2 \sigma(2p_z)^2 \pi(2p_x)^2 \pi(2p_y)^2 \pi^*(2p_x)^2 \pi^*(2p_y)^1$.

(D) The bond order is calculated using the formula: $Bond \ Order = \frac{1}{2} (N_b - N_a)$.
For $He_2^+$ ($3$ electrons): $\sigma 1s^2, \sigma^* 1s^1$,$BO = \frac{2-1}{2} = 0.5$.
For $O_2^-$ ($17$ electrons): $BO = \frac{10-7}{2} = 1.5$.
For $NO$ ($15$ electrons): $BO = \frac{10-5}{2} = 2.5$.
For $C_2^{2-}$ ($14$ electrons): $BO = \frac{10-4}{2} = 3.0$.
Thus,the increasing order of bond order is: $He_2^+ < O_2^- < NO < C_2^{2-}$.

(A) According to Graham's law of effusion,the rate of effusion $r$ is inversely proportional to the square root of the molecular mass $M$: $\frac{r_A}{r_B} = \sqrt{\frac{M_B}{M_A}}$.
Since $r = \frac{V}{t}$ and the volumes $V$ are equal,$\frac{r_A}{r_B} = \frac{t_B}{t_A}$.
Substituting the given values: $\frac{200}{150} = \sqrt{\frac{36}{M_A}}$.
Simplifying the ratio: $\frac{4}{3} = \sqrt{\frac{36}{M_A}}$.
Squaring both sides: $\frac{16}{9} = \frac{36}{M_A}$.
Solving for $M_A$: $M_A = \frac{36 \times 9}{16} = \frac{324}{16} = 20.25$.

(B) According to Graham's Law of Effusion,the rate of effusion $r$ is inversely proportional to the square root of the molar mass $M_W$: $\frac{r_1}{r_2} = \sqrt{\frac{M_{W_2}}{M_{W_1}}}$.
Since rate $r = \frac{V}{t}$,for the same volume $V$,the expression becomes $\frac{t_2}{t_1} = \sqrt{\frac{M_{W_1}}{M_{W_2}}}$.
Given that the time taken by the gas $(t_1)$ is $3$ times the time taken by helium $(t_2)$,we have $\frac{t_1}{t_2} = 3$.
Substituting the values: $3 = \sqrt{\frac{M_{W_1}}{4}}$.
Squaring both sides: $9 = \frac{M_{W_1}}{4}$.
Therefore,$M_{W_1} = 9 \times 4 = 36 \ u$.

(A) The constant $b$ represents the excluded volume,which increases with the size of the molecule. The order of $b$ for set-$I$ is: $H_2 (0.0266) < He (0.0237)$ (Note: $He$ is smaller than $H_2$,so $He < H_2$) $< O_2 (0.0318) < CO_2 (0.0427)$. Thus,$I. He < H_2 < O_2 < CO_2$.
The constant $a$ represents the magnitude of intermolecular forces. Larger molecules with more electrons have stronger van der Waals forces. The order of $a$ for set-$II$ is: $CH_4 (2.25) > O_2 (1.36) > H_2 (0.244)$. Thus,$II. CH_4 > O_2 > H_2$.

(D) For a monoatomic gas,the adiabatic index $\gamma = C_P/C_V = 5/3 \approx 1.67$.
Since both gases $A$ and $B$ are monoatomic and are mixed in equal volumes at the same temperature and pressure,the resulting mixture will also behave as a monoatomic gas.
Therefore,the ratio of specific heats $(C_P/C_V)$ for the mixture remains $1.67$.

(A) For a reaction,$\Delta G^o = \Delta H^o - T \Delta S^o$. If $\Delta S^o > 0$,then as temperature $T$ increases,the term $-T \Delta S^o$ becomes more negative,causing $\Delta G^o$ to decrease sharply.
$\Delta S^o$ is positive when the number of moles of gaseous products is greater than the number of moles of gaseous reactants (i.e.,$\Delta n_g > 0$).
In option $A$: $C_{(graphite)} + \frac{1}{2} O_{2(g)} \to CO_{(g)}$,$\Delta n_g = 1 - 0.5 = 0.5 > 0$.
In option $B$: $CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)}$,$\Delta n_g = 1 - 1.5 = -0.5 < 0$.
In option $C$: $Mg_{(s)} + \frac{1}{2} O_{2(g)} \to MgO_{(s)}$,$\Delta n_g = 0 - 0.5 = -0.5 < 0$.
In option $D$: $\frac{1}{2} C_{(graphite)} + \frac{1}{2} O_{2(g)} \to \frac{1}{2} CO_{2(g)}$,$\Delta n_g = 0.5 - 0.5 = 0$.
Thus,only reaction $A$ has a positive $\Delta S^o$.

(C) The process of melting ice at $0 \ ^\circ C$ $(273 \ K)$ is a reversible phase transition.
For a reversible process,the entropy change is given by $\Delta S = \frac{\Delta H_{fus}}{T}$.
Given $\Delta H_{fus} = 1.435 \ kcal/mol = 1435 \ cal/mol$.
Temperature $T = 0 \ ^\circ C = 273 \ K$.
Therefore,$\Delta S = \frac{1435 \ cal/mol}{273 \ K} = 5.256 \ cal/(mol \ K) \approx 5.26 \ cal/(mol \ K)$.

(A) The process is $H_2O_{(l)} \longrightarrow H_2O_{(g)}$.
The relation between enthalpy change and internal energy change is $\Delta H = \Delta U + \Delta n_g RT$.
Here,$\Delta H = 40.66\ kJ\ mol^{-1}$,$T = 100 + 273 = 373\ K$,and $\Delta n_g = 1 - 0 = 1$.
Using $R = 8.314 \times 10^{-3}\ kJ\ K^{-1}\ mol^{-1}$,we have:
$40.66 = \Delta U + (1 \times 8.314 \times 10^{-3} \times 373)$.
$40.66 = \Delta U + 3.101$.
$\Delta U = 40.66 - 3.101 = 37.559\ kJ\ mol^{-1} \approx 37.56\ kJ\ mol^{-1}$.

(B) Given,$pH = 12$.
Since $pH + pOH = 14$,we have $pOH = 14 - 12 = 2$.
The concentration of hydroxide ions is $[OH^-] = 10^{-pOH} = 10^{-2} \ M$.
The dissociation of $Ba(OH)_2$ is represented as: $Ba(OH)_2 \rightleftharpoons Ba^{2+} + 2OH^-$.
If the solubility of $Ba(OH)_2$ is $s$,then $[Ba^{2+}] = s$ and $[OH^-] = 2s$.
From the given concentration,$2s = 10^{-2} \ M$,which implies $s = 0.5 \times 10^{-2} = 5 \times 10^{-3} \ M$.
The solubility product expression is $K_{sp} = [Ba^{2+}][OH^-]^2$.
Substituting the values,$K_{sp} = (s)(2s)^2 = 4s^3$.
$K_{sp} = 4 \times (5 \times 10^{-3})^3 = 4 \times 125 \times 10^{-9} = 500 \times 10^{-9} = 5.0 \times 10^{-7}$.

(A) $BaCl_2$ is a salt of a strong acid $(HCl)$ and a strong base $(Ba(OH)_2)$. Therefore,its aqueous solution is neutral with a $pH$ of $7$.
$AlCl_3$,$LiCl$,and $BeCl_2$ undergo cationic hydrolysis in water to produce acidic solutions because they involve weak bases or small,highly charged cations.
Consequently,their $pH$ values are less than $7$.
Thus,the $pH$ value is highest for the solution of $BaCl_2$.

(A) If a small amount of an acid or alkali is added to a buffer solution,it converts them into unionised acid or base.
Thus,the $pH$ remains unaffected or in other words,its acidity/alkalinity remains constant.
For example:
$H_3O^{+} + A^{-} \rightleftharpoons H_2O + HA$
$OH^{-} + HA \rightarrow H_2O + A^{-}$
If an acid is added,it reacts with the conjugate base $A^{-}$ to form undissociated $HA$.
Similarly,if a base/alkali is added,$OH^{-}$ combines with $HA$ to give $H_2O$ and $A^{-}$,and thus,maintains the acidity/alkalinity of the buffer solution.

(C) For the reaction $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$,the equilibrium constant is $K = 278$.
Reversing the reaction gives $2SO_{3(g)} \rightleftharpoons 2SO_{2(g)} + O_{2(g)}$,for which the equilibrium constant is $K' = \frac{1}{K} = \frac{1}{278}$.
Dividing the stoichiometric coefficients of the reversed reaction by $2$ gives the target reaction: $SO_{3(g)} \rightleftharpoons SO_{2(g)} + \frac{1}{2} O_{2(g)}$.
The equilibrium constant for this reaction is $K'' = \sqrt{K'} = \sqrt{\frac{1}{278}} \approx 0.0599 \approx 6.0 \times 10^{-2}$.

(C) The equilibrium reaction is: $A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)}$
The expression for the equilibrium constant $K_c$ is:
$K_c = \frac{[AB]^2}{[A_2][B_2]}$
Substituting the given equilibrium concentrations:
$K_c = \frac{(2.8 \times 10^{-3})^2}{(3.0 \times 10^{-3})(4.2 \times 10^{-3})}$
$K_c = \frac{2.8 \times 2.8}{3.0 \times 4.2} = \frac{7.84}{12.6} \approx 0.622$
Rounding to two decimal places,the value of $K_c$ is $0.62$.

(B) The reaction of chlorine gas with hot and concentrated sodium hydroxide solution is:
$3Cl_2 + 6NaOH \rightarrow NaClO_3 + 5NaCl + 3H_2O$
In $Cl_2$,the oxidation number of $Cl$ is $0$.
In $NaCl$,the oxidation number of $Cl$ is $-1$.
In $NaClO_3$,the oxidation number of $Cl$ is $x + (-2 \times 3) = -1 \Rightarrow x = +5$.
Thus,the oxidation number changes from $0$ to $-1$ and $0$ to $+5$.

(C) The balanced chemical reaction is:
$2KClO_3 + 3H_2C_2O_4 + H_2SO_4 \rightarrow K_2SO_4 + 2KCl + 6CO_2 + 4H_2O$
Calculation of oxidation states:
$1$. For $Cl$ in $KClO_3$,the oxidation state changes from $+5$ to $-1$ in $KCl$. The change is $|-1 - 5| = 6$.
$2$. For $C$ in $H_2C_2O_4$,the oxidation state changes from $+3$ to $+4$ in $CO_2$. The change is $|4 - 3| = 1$.
$3$. The oxidation states of $S$,$H$,$K$,and $O$ remain unchanged in this reaction.
Thus,$Cl$ undergoes the maximum change in oxidation number.

(C) When alkali metals are heated in an atmosphere of oxygen,they ignite and form oxides.
$Li$ reacts with oxygen to form the normal oxide,$Li_2O$.
$Na$ reacts with oxygen to form the peroxide,$Na_2O_2$.
$K$ and $Rb$ react with oxygen to form superoxides,$MO_2$ (where $M = K, Rb$).

(B) The ease of adsorption of hydrated alkali metal ions on an ion-exchange resin depends on the size of the hydrated ion.
Smaller alkali metal ions have a higher charge density,leading to greater hydration. Thus,the size of the hydrated ion follows the order: $Li^{+} > Na^{+} > K^{+} > Rb^{+}$.
Since the ease of adsorption is inversely proportional to the size of the hydrated ion,the order of adsorption is: $Rb^{+} < K^{+} < Na^{+} < Li^{+}$.

(C) The presence of an electron-releasing group like $-CH_3$,$-OH$,etc.,increases the electron density at the $o/p$ positions and thus makes the benzene ring more reactive towards electrophiles.
On the other hand,electron-withdrawing groups like $-COOH$,$-NO_2$,etc.,reduce the electron density of the benzene ring,thereby decreasing its reactivity towards electrophilic substitution.
Comparing the given compounds:
$1$. $-NO_2$ (in nitrobenzene) is a strong electron-withdrawing group.
$2$. $-COOH$ (in benzoic acid) is an electron-withdrawing group.
$3$. $-H$ (in benzene) is the reference.
$4$. $-CH_3$ (in toluene) is an electron-releasing group due to the $+I$ effect and hyperconjugation.
Therefore,the order of reactivity towards electrophilic nitration is: $\text{nitrobenzene} < \text{benzoic acid} < \text{benzene} < \text{toluene}$.
Thus,toluene is the most reactive.

(A) In $Br-CH_2-CH=CH_2$,the double bond is given priority over the halogen substituent for numbering.
The numbering of the carbon chain should start from the end containing the double bond:
$CH_2(1)=CH(2)-CH_2(3)-Br$
Therefore,the correct $IUPAC$ name is $3-$Bromoprop$-1-$ene.
Thus,option $(A)$ is not according to the $IUPAC$ system.

(A) The acid-catalyzed hydration of $3,3-\text{dimethyl}-1-\text{butene}$ proceeds via a carbocation intermediate.
$1$. Protonation of the alkene follows Markovnikov's rule to form a secondary carbocation: $CH_3-C(CH_3)_2-CH^{+}-CH_3$.
$2$. $A$ $1,2-\text{methyl shift}$ occurs to convert the secondary carbocation into a more stable tertiary carbocation: $CH_3-C^{+}(CH_3)-CH(CH_3)-CH_3$.
$3$. Nucleophilic attack by $H_2O$ on the tertiary carbocation followed by deprotonation yields the major product,$2,3-\text{dimethyl}-2-\text{butanol}$: $CH_3-C(OH)(CH_3)-CH(CH_3)-CH_3$.

(A) Optical isomerism requires the presence of at least one chiral carbon atom (a carbon atom bonded to four different groups).
Maleic acid $(HOOC-CH=CH-COOH)$ contains $sp^2$ hybridized carbons in a double bond and lacks a chiral carbon atom,so it does not exhibit optical isomerism.
Tartaric acid,lactic acid,and $\alpha$-amino acids all contain at least one chiral carbon atom (indicated by $*$ in the structure),allowing them to exhibit optical isomerism.

(A) $But-1-yne$ $(CH_3CH_2C \equiv CH)$ contains a terminal acidic hydrogen atom attached to an $sp$ hybridized carbon atom.
$But-2-yne$ $(CH_3C \equiv CCH_3)$ does not have any terminal acidic hydrogen atom.
$NaNH_2$ is a strong base that can abstract the acidic proton from $but-1-yne$ to form a sodium acetylide salt,whereas $but-2-yne$ does not react with $NaNH_2$.
Reaction: $CH_3CH_2C \equiv CH + NaNH_2 \rightarrow CH_3CH_2C \equiv C^-Na^+ + NH_3$.
Therefore,$NaNH_2$ is the correct reagent to distinguish between them.

(D) Photochemical smog is formed in warm and sunny climates during the daytime by the action of sunlight on primary pollutants like nitrogen oxides and hydrocarbons.
It contains components such as nitrogen oxides,ozone $(O_3)$,and peroxyacetyl nitrate $(PAN)$,which are oxidising in nature.
Therefore,photochemical smog is an oxidising agent in character.
It is well-known to cause irritation in the eyes and throat,making statement $D$ incorrect.

(C) To find the oxidation state of nitrogen $(N)$ in each compound,we assign hydrogen $(H)$ a value of $+1$ and oxygen $(O)$ a value of $-2$:
$1$. In $N_2H_4$: $2x + 4(+1) = 0 \implies 2x = -4 \implies x = -2$.
$2$. In $NH_3$: $x + 3(+1) = 0 \implies x = -3$.
$3$. In $N_3H$: $3x + 1(+1) = 0 \implies 3x = -1 \implies x = -1/3$.
$4$. In $NH_2OH$: $x + 2(+1) + (-2) + 1(+1) = 0 \implies x + 1 = 0 \implies x = -1$.
Comparing the values $-2, -3, -1/3, -1$,the highest value is $-1/3$.

(D) Condensation polymers are formed by the repeated condensation reaction between bifunctional monomers,often accompanied by the elimination of small molecules like $H_2O$ or $HCl$.
Addition polymers are formed by the polymerization of monomers containing multiple bonds (double or triple bonds) without the loss of any small molecules.
$A$. Melamine is a condensation polymer formed from melamine and formaldehyde.
$B$. Glyptal is a polyester formed by the condensation of ethylene glycol and phthalic acid.
$C$. Dacron (Terylene) is a polyester formed by the condensation of ethylene glycol and terephthalic acid.
$D$. Neoprene is a polymer of chloroprene $(CH_2=C(Cl)-CH=CH_2)$,which is an addition polymer.
Therefore,Neoprene is not a condensation polymer.

(B) Comparing the given equations with the standard forms $i = I_0 \sin(\omega t)$ and $e = E_0 \sin(\omega t + \phi)$,we get:
$I_0 = \frac{1}{\sqrt{2}} \text{ A}$
$E_0 = \frac{1}{\sqrt{2}} \text{ V}$
$\phi = \frac{\pi}{3}$
The $RMS$ values are calculated as:
$I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{1/\sqrt{2}}{\sqrt{2}} = \frac{1}{2} \text{ A}$
$E_{rms} = \frac{E_0}{\sqrt{2}} = \frac{1/\sqrt{2}}{\sqrt{2}} = \frac{1}{2} \text{ V}$
The power factor is $\cos \phi = \cos(\frac{\pi}{3}) = \frac{1}{2}$.
The average power consumed is given by $P_{av} = E_{rms} I_{rms} \cos \phi$.
$P_{av} = (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) = \frac{1}{8} \text{ W}$.

(C) Yeast,specifically $Saccharomyces$ $cerevisiae$ (also known as brewer's yeast),is widely used in the food and beverage industry.
It is used in the production of bread,where it acts as a leavening agent by fermenting sugars to produce $CO_2$ and ethanol,causing the dough to rise.
It is also used in the production of beer and other alcoholic beverages through the process of fermentation.
Therefore,the correct answer is bread and beer.

(A) Manganese $(Mn^{2+})$ plays a crucial role in the process of photosynthesis.
Its most significant and well-defined function is the photolysis of water (splitting of water molecules) during the light-dependent reactions.
This process releases electrons,protons $(H^+)$,and oxygen $(O_2)$,which are essential for the electron transport chain and the subsequent synthesis of $ATP$ and $NADPH$.

(C) $ABA$ (Abscisic Acid) acts as a general plant growth inhibitor. It functions as an antagonist to gibberellins in several physiological processes,such as seed germination and bud dormancy. While gibberellins promote seed germination and break dormancy,$ABA$ inhibits germination and induces dormancy.

(C) Vernalization is the process of inducing flowering by exposure to prolonged periods of low temperature. It is commonly observed in biennial plants such as $carrot$,$cabbage$,and $sugar$ $beet$. These plants typically have a life cycle that spans two years; in the first year,they undergo vegetative growth,and in the second year,they flower and produce seeds. Exposure to cold temperatures during the winter is essential to trigger the transition from vegetative to reproductive growth in these species.

(A) Indigestion is a condition in which the food is not properly digested,leading to a feeling of fullness. The causes of indigestion include inadequate enzyme secretion,anxiety,food poisoning,overeating,and eating spicy food. Therefore,when anxiety and spicy food are combined in an otherwise normal human,it frequently results in indigestion.

(C) The $Proximal \text{ } Convoluted \text{ } Tubule$ $(PCT)$ is lined by simple cuboidal brush border epithelium which increases the surface area for reabsorption.
Nearly all of the essential nutrients and $70-80$ percent of electrolytes and water are reabsorbed in the $PCT$.
Therefore, the correct answer is $C$.

(C) The $GFR$ (Glomerular Filtration Rate) is regulated by the $JGA$ (Juxta-Glomerular Apparatus).
When there is a significant fall in $GFR$,the $JG$ cells are activated.
These $JG$ cells release an enzyme called renin into the bloodstream.
Renin converts angiotensinogen in the blood to angiotensin $I$ and further to angiotensin $II$.
Angiotensin $II$ is a powerful vasoconstrictor that increases the glomerular blood pressure and thereby $GFR$ back to normal levels.

(B) The correct statement is $B$.
$1$. Osteoporosis is an age-related disorder characterized by decreased bone mass and increased chances of fractures. It is often caused by decreased levels of estrogen.
$2$. Muscular dystrophy is a genetic disorder resulting in progressive degeneration of skeletal muscle,not age-related shortening.
$3$. Myasthenia gravis is an autoimmune disorder affecting the neuromuscular junction,leading to fatigue,weakening,and paralysis of skeletal muscle,not specifically the sliding of myosin filaments.
$4$. Gout is the inflammation of joints due to the accumulation of uric acid crystals,not calcium deposition.

(C) The human brain is divided into three main parts: forebrain,midbrain,and hindbrain.
The hindbrain consists of three specific structures: $Pons$,$Cerebellum$,and $Medulla$ $oblongata$.
$Spinal$ $cord$ is a part of the central nervous system but not the brain.
$Corpus$ $callosum$ is a tract of nerve fibers connecting the two cerebral hemispheres.
$Hypothalamus$ is a part of the forebrain.
Therefore,$Cerebellum$ is the correct part of the hindbrain.

(C) The human ear consists of three parts: the outer,middle,and inner ear.
$1$. The $Organ$ $of$ $Corti$ and $Cochlear$ $duct$ are essential components of the cochlea,which is responsible for the process of hearing.
$2$. The $Eustachian$ $tube$ connects the middle ear to the pharynx and helps in equalizing the pressure on either side of the eardrum.
$3$. The $Vestibular$ $apparatus$ is located in the inner ear and is composed of the semicircular canals and the otolith organ (saccule and utricle). Its primary function is the maintenance of body balance and posture,not hearing.
Therefore,the $Vestibular$ $apparatus$ is the part that plays no role in hearing but is essential for equilibrium.

(A) For a face-centred cubic $(fcc)$ lattice, the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by $4r = \sqrt{2}a$.
The diameter $(d)$ of the atom is $2r$. Therefore, $d = 2r = \frac{\sqrt{2}a}{2} = \frac{a}{\sqrt{2}}$.
Given the edge length $a = 408 \, pm$, we calculate the diameter as:
$d = \frac{408 \, pm}{1.414} \approx 288.5 \, pm$.
Since $288.5 \, pm$ is the correct calculated value, option $A$ is the closest match.

(A) In a cubic close-packed $(CCP)$ or face-centered cubic $(FCC)$ structure,the number of octahedral voids is equal to the number of atoms present in the unit cell.
If the number of atoms per unit cell is $N$,then the number of octahedral voids is $N$.
Therefore,the number of octahedral voids per atom is $\frac{N}{N} = 1$.

(D) In a $ccp$ lattice,the number of oxide ions $(O^{2-})$ per unit cell is $4$.
For a $ccp$ structure,the number of tetrahedral voids is $2 \times 4 = 8$ and the number of octahedral voids is $4$.
Given that $A$ ions occupy $\frac{1}{4}$ of the tetrahedral voids,the number of $A$ ions $= \frac{1}{4} \times 8 = 2$.
Given that $B$ ions occupy all the octahedral voids,the number of $B$ ions $= 4$.
The ratio of $A : B : O$ is $2 : 4 : 4$,which simplifies to $1 : 2 : 2$.
Therefore,the formula of the oxide is $AB_2O_2$.

(D) According to Raoult's Law,the total pressure $P_T$ of an ideal binary solution is given by the sum of the partial pressures of the components:
$P_T = P_A + P_B$
We know that $P_A = p_A x_A$ and $P_B = p_B x_B$,where $p_A$ and $p_B$ are the vapour pressures of pure components.
Substituting these into the equation:
$P_T = p_A x_A + p_B x_B$
Since $x_A + x_B = 1$,we have $x_B = 1 - x_A$.
Substituting $x_B$ in the equation:
$P_T = p_A x_A + p_B(1 - x_A)$
$P_T = p_A x_A + p_B - p_B x_A$
$P_T = p_B + x_A(p_A - p_B)$

(D) $1$. Calculate the moles of each component:
$n_{CHCl_3} = \frac{25.5 \ g}{119.5 \ g/mol} \approx 0.213 \ mol$
$n_{CH_2Cl_2} = \frac{40 \ g}{85 \ g/mol} \approx 0.471 \ mol$
$2$. Calculate the mole fractions:
$n_{total} = 0.213 + 0.471 = 0.684 \ mol$
$x_{CHCl_3} = \frac{0.213}{0.684} \approx 0.311$
$x_{CH_2Cl_2} = \frac{0.471}{0.684} \approx 0.689$
$3$. Calculate the total vapour pressure using Raoult's Law:
$P_{total} = P^0_{CHCl_3} \cdot x_{CHCl_3} + P^0_{CH_2Cl_2} \cdot x_{CH_2Cl_2}$
$P_{total} = (200 \ mm \ Hg \times 0.311) + (41.5 \ mm \ Hg \times 0.689)$
$P_{total} = 62.2 + 28.5935 = 90.7935 \ mm \ Hg$
Since $90.7935 \ mm \ Hg$ is not among the options,the correct choice is $D$.

(D) According to Kohlrausch's law of independent migration of ions,the limiting molar conductivity of a weak electrolyte can be calculated using the limiting molar conductivities of strong electrolytes.
For $NH_4OH$,we can express it as:
$\Lambda ^o_{m(NH_4OH)} = \lambda ^o_{NH_4^+} + \lambda ^o_{OH^-}$
Using strong electrolytes $NH_4Cl$,$NaOH$,and $NaCl$:
$\Lambda ^o_{m(NH_4Cl)} = \lambda ^o_{NH_4^+} + \lambda ^o_{Cl^-}$
$\Lambda ^o_{m(NaOH)} = \lambda ^o_{Na^+} + \lambda ^o_{OH^-}$
$\Lambda ^o_{m(NaCl)} = \lambda ^o_{Na^+} + \lambda ^o_{Cl^-}$
By performing the operation $\Lambda ^o_{m(NH_4Cl)} + \Lambda ^o_{m(NaOH)} - \Lambda ^o_{m(NaCl)}$,we get:
$(\lambda ^o_{NH_4^+} + \lambda ^o_{Cl^-}) + (\lambda ^o_{Na^+} + \lambda ^o_{OH^-}) - (\lambda ^o_{Na^+} + \lambda ^o_{Cl^-}) = \lambda ^o_{NH_4^+} + \lambda ^o_{OH^-} = \Lambda ^o_{m(NH_4OH)}$
Therefore,the correct expression is $\Lambda ^o_{m(NH_4Cl)} + \Lambda ^o_{m(NaOH)} - \Lambda ^o_{m(NaCl)}$.

(A) The strength of an oxidising agent is directly proportional to the standard reduction potential $(E^o)$. $A$ higher positive $E^o$ value indicates a stronger oxidising agent.
The strength of a reducing agent is inversely proportional to the standard reduction potential $(E^o)$. $A$ lower (or more negative) $E^o$ value indicates a stronger reducing agent.
Comparing the given values:
$F_2$ has the highest $E^o$ $(+2.85 \ V)$,so it is the strongest oxidising agent.
$I^-$ has the lowest $E^o$ ($+0.53 \ V$ for the corresponding $I_2/I^-$ couple),making it the strongest reducing agent among the species listed.
Therefore,the strongest oxidising and reducing agents are $F_2$ and $I^-$ respectively.

(D) According to Kohlrausch's Law of independent migration of ions,the molar conductivity of a weak electrolyte can be calculated using the molar conductivities of strong electrolytes.
$\Lambda ^o_m(CH_3COOH) = \Lambda ^o_m(CH_3COONa) + \Lambda ^o_m(HCl) - \Lambda ^o_m(NaCl)$
Substituting the given values:
$\Lambda ^o_m(CH_3COOH) = 91.0 + 425.9 - 126.4$
$\Lambda ^o_m(CH_3COOH) = 516.9 - 126.4 = 390.5 \ S \ cm^2 \ mol^{-1}$

(C) The relationship between Gibbs energy and cell potential is given by $\Delta G = -nFE_{cell}$.
For the reaction $\frac{2}{3} Al_2O_3 \rightarrow \frac{4}{3} Al + O_2$,the number of electrons transferred $(n)$ is calculated as follows:
$Al$ goes from $+3$ to $0$ (change of $3$ per $Al$ atom).
For $\frac{4}{3}$ moles of $Al$,$n = \frac{4}{3} \times 3 = 4$.
Given $\Delta G = +960 \ kJ \ mol^{-1} = 960000 \ J \ mol^{-1}$.
Using $\Delta G = -nFE_{cell}$:
$960000 = -4 \times 96500 \times E_{cell}$.
$E_{cell} = -\frac{960000}{386000} \approx -2.487 \ V$.
The magnitude of the potential difference required for the electrolytic reduction is approximately $2.5 \ V$.

(D) Let the order of reaction with respect to $A$ and $B$ be $x$ and $y$ respectively.
So,the rate law can be given as:
$R = k[A]^{x}[B]^{y} \dots (i)$
When the concentration of only $B$ is doubled,the rate is doubled:
$2R = k[A]^{x}[2B]^{y} \dots (ii)$
Dividing $(ii)$ by $(i)$:
$2 = 2^{y} \Rightarrow y = 1$
When concentrations of both $A$ and $B$ are doubled,the rate increases by a factor of $8$:
$8R = k[2A]^{x}[2B]^{y} \dots (iii)$
Dividing $(iii)$ by $(i)$:
$8 = 2^{x} \times 2^{y}$
Substituting $y = 1$:
$8 = 2^{x} \times 2^{1}$ $\Rightarrow 4 = 2^{x}$ $\Rightarrow x = 2$
Thus,the rate law is $R = k[A]^{2}[B]$.

(B) The number of $10\,^{\circ}C$ intervals $(n)$ is calculated as:
$n = \frac{T_2 - T_1}{10} = \frac{100 - 10}{10} = \frac{90}{10} = 9$
Since the rate doubles for every $10\,^{\circ}C$ rise,the rate increases by a factor of $2^n$.
Rate factor $= 2^9 = 512$
Therefore,the rate of the reaction will become $512$ times.

(D) The Arrhenius equation is given by: $k = A e^{-E_a / RT}$.
For two different temperatures $T_1$ and $T_2$ with corresponding rate constants $k_1$ and $k_2$:
$\ln k_1 = \ln A - \frac{E_a}{RT_1}$
$\ln k_2 = \ln A - \frac{E_a}{RT_2}$
Subtracting the first equation from the second:
$\ln k_2 - \ln k_1 = \left( \ln A - \frac{E_a}{RT_2} \right) - \left( \ln A - \frac{E_a}{RT_1} \right)$
$\ln \frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$
Alternatively,this can be written as:
$\ln \frac{k_2}{k_1} = \frac{-E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)$
Thus,both expressions $(b)$ and $(c)$ are correct.

(A) In Freundlich adsorption isotherm,the equation is given by $\frac{x}{m} = k p^{1/n}$.
Here,$x$ is the mass of the adsorbate,$m$ is the mass of the adsorbent,$p$ is the pressure,and $k$ and $n$ are constants that depend on the nature of the adsorbent and adsorbate at a particular temperature.
The value of $n$ is always greater than $1$,which implies that $n > 1$.
Therefore,the value of the exponent $1/n$ must lie between $0$ and $1$ in all cases.

(D) Most enzymes are proteinous in nature.
They are highly specific in their action.
They get denatured by high temperatures or $UV$ rays.
At the optimum temperature,enzyme activity is at its maximum,not minimum.

(B) Lyophilic sols are stabilized by adding lyophilic colloids,which protect them from precipitation. These are called protecting colloids.
Their protecting power is expressed in terms of $gold \ number$.
$Gold \ number$ is defined as the minimum amount of lyophilic colloid in milligrams that prevents the flocculation of $10 \ mL$ of gold sol by the addition of $1 \ mL$ of $10 \ \% \ NaCl$ solution.
Lesser the $gold \ number$,higher is the protecting power.

(B) In the extraction of copper from its sulphide ore,the ore is subjected to roasting,where some of it is oxidized to $Cu_{2}O$.
This $Cu_{2}O$ then reacts with the remaining $Cu_{2}S$ (cuprous sulphide) to produce copper metal.
The chemical equation for this self-reduction process is:
$2Cu_{2}O + Cu_{2}S \longrightarrow 6Cu + SO_{2} \uparrow$
In this reaction,$Cu_{2}S$ acts as the reducing agent for $Cu_{2}O$.

(C) Alumina,$Al_2O_3$,is a poor conductor of electricity and has a very high melting point.
To facilitate electrolysis,it is mixed with cryolite $(Na_3AlF_6)$ and fluorspar $(CaF_2)$.
These substances lower the melting point of the mixture and increase its electrical conductivity,allowing the electrolysis to occur at a lower temperature.

(D) The chemical compositions of the given minerals are as follows:
$1$. Malachite: $CuCO_{3} \cdot Cu(OH)_{2}$ (Copper ore)
$2$. Cassiterite: $SnO_{2}$ (Tin ore)
$3$. Pyrolusite: $MnO_{2}$ (Manganese ore)
$4$. Magnetite: $Fe_{3}O_{4}$ (Iron ore)
Therefore,$Magnetite$ is a mineral of iron.

(B) Hypophosphorous acid,$H_{3}PO_{2}$,contains only one replaceable $H$-atom (that is attached to $O$,not with $P$ directly),so it is a monoprotic acid. Therefore,the statement that it is a diprotic acid is incorrect.

(B) The thermal decomposition of ferric sulphate is a standard laboratory method for the preparation of sulphur trioxide:
$Fe_2(SO_4)_3 \xrightarrow{\Delta} Fe_2O_3 + 3SO_3$
Analysis of other options:
$(A)$ $CaSO_4 + C \xrightarrow{\Delta} CaO + SO_2 + CO$ (Produces $SO_2$)
$(B)$ $Fe_2(SO_4)_3 \xrightarrow{\Delta} Fe_2O_3 + 3SO_3$ (Produces $SO_3$)
$(C)$ $S + 2H_2SO_4 \xrightarrow{\Delta} 3SO_2 + 2H_2O$ (Produces $SO_2$)
$(D)$ $H_2SO_4 + PCl_5 \xrightarrow{\Delta} SO_3HCl + POCl_3 + HCl$ (Produces chlorosulphonic acid)
Thus,$SO_3$ is obtained by heating $Fe_2(SO_4)_3$.

(B) The acidic strength of hydrides of group $16$ elements increases down the group as the bond dissociation energy decreases: $H_2O < H_2S < H_2Se < H_2Te$.
Since $pK_a = -\log(K_a)$,a higher acidic strength corresponds to a lower $pK_a$ value.
Therefore,the correct order for $pK_a$ values should be $H_2O > H_2S > H_2Se > H_2Te$.
Option $B$ shows an increasing order of $pK_a$ values,which is incorrect.

(B) Steel is an alloy of iron $(Fe)$ and carbon $(C)$.
Carbon is a non-metal.
Invar $(Fe-Ni)$,Bell metal $(Cu-Sn)$,and Bronze $(Cu-Sn)$ consist only of metals.

(B) . Passing $H_2S$ through acidified $K_2Cr_2O_7$ produces $S$ (sulfur) which causes a milky appearance. This is true.
$B$. $Na_2Cr_2O_7$ is more soluble than $K_2Cr_2O_7$,but it is hygroscopic,making it unsuitable for primary standards in volumetric analysis. Thus,$K_2Cr_2O_7$ is preferred. This statement is false.
$C$. $K_2Cr_2O_7$ exists as dichromate ions in acidic medium,which are orange. This is true.
$D$. In basic medium $(pH > 7)$,dichromate ions $(Cr_2O_7^{2-})$ convert to chromate ions $(CrO_4^{2-})$,which are yellow. This is true.

(C) The catalytic activity of transition metals and their compounds is primarily attributed to their ability to adopt variable oxidation states and form complexes.
For example,in the contact process,$V_{2}O_{5}$ acts as a catalyst:
$2SO_{2} + O_{2} \xrightarrow{V_{2}O_{5}} 2SO_{3}$
This occurs because vanadium can easily switch between oxidation states:
$SO_{2} + V_{2}O_{5} \longrightarrow SO_{3} + 2VO_{2}$
$2VO_{2} + \frac{1}{2}O_{2} \longrightarrow V_{2}O_{5}$

(C) Actinium $(Ac)$ has the atomic number $89$.
The electronic configuration of $Ac$ is $[Rn] \ 6d^1 \ 7s^2$.
Due to the stability of its electronic configuration,it exhibits only a $+3$ oxidation state.
In contrast,$Th$ $(Z=90)$ exhibits $+3$ and $+4$,$Pa$ $(Z=91)$ exhibits $+3, +4, +5$,and $U$ $(Z=92)$ exhibits $+3, +4, +5, +6$ oxidation states.

(C) For $Ti, V, Cr, Mn$,the number of oxidation states are $3, 4, 5, 6$ respectively. Thus,$Ti < V < Cr < Mn$ is correct.
For magnetic moment $(\mu) = \sqrt{n(n+2)} \ B.M$:
$Ti^{3+} (3d^1): n=1, \mu = \sqrt{3} \ B.M$
$V^{3+} (3d^2): n=2, \mu = \sqrt{8} \ B.M$
$Cr^{3+} (3d^3): n=3, \mu = \sqrt{15} \ B.M$
$Mn^{3+} (3d^4): n=4, \mu = \sqrt{24} \ B.M$
Thus,$Ti^{3+} < V^{3+} < Cr^{3+} < Mn^{3+}$ is correct.
The actual order of melting points is $Mn < Ti < Cr < V$. Therefore,$Ti < V < Cr < Mn$ is incorrect.
The $2^{nd}$ ionization enthalpy order is $Ti < V < Mn < Cr$,which is correct.

(C) The standard electrode potential $(E^o_{M^{2+}/M})$ depends on the sum of the enthalpy of atomization,ionization enthalpy,and hydration enthalpy.
For most transition metals,the value of $(E^o_{M^{2+}/M})$ is negative.
However,for $Cu$,the high energy to transform $Cu(s)$ to $Cu^{2+}(aq)$ is not compensated by its hydration enthalpy,resulting in a positive $(E^o_{M^{2+}/M})$ value of $+0.34 \ V$.
Therefore,$Cu$ is the correct answer.

(A) For $[Ni(NH_3)_6]^{2+}$,the oxidation state of $Ni$ is $+2$ ($3d^8$ configuration). It uses $4s, 4p,$ and $4d$ orbitals to form $sp^3d^2$ hybrid orbitals,making it an outer orbital complex. It has two unpaired electrons in the $3d$ orbitals,so it is paramagnetic.
For $[Zn(NH_3)_6]^{2+}$,the oxidation state of $Zn$ is $+2$ ($3d^{10}$ configuration). It is an outer orbital complex $(sp^3d^2)$,but it is diamagnetic due to no unpaired electrons.
For $[Cr(NH_3)_6]^{3+}$,the oxidation state of $Cr$ is $+3$ ($3d^3$ configuration). It uses $3d, 4s,$ and $4p$ orbitals to form $d^2sp^3$ hybrid orbitals,making it an inner orbital complex. It is paramagnetic.
For $[Co(NH_3)_6]^{3+}$,the oxidation state of $Co$ is $+3$ ($3d^6$ configuration). It forms $d^2sp^3$ hybrid orbitals,making it an inner orbital complex. It is diamagnetic.

(C) The reaction of $Ni^{2+}$ with dimethylglyoxime $(DMG)$ in an ammoniacal medium produces a red-coloured complex,$[Ni(DMG)_2]$.
This complex has a square planar geometry,not a tetrahedral one.
Dimethylglyoxime acts as a bidentate ligand,coordinating through the nitrogen atoms.
The complex is highly stable due to the presence of strong intramolecular symmetrical hydrogen bonding between the oxime groups of the two ligands.

(B) For a low spin complex of a $d^6$ cation in an octahedral field,the condition is $\Delta_o > P$.
The electronic configuration is $t_{2g}^6 e_g^0$.
In this configuration,all $6$ electrons occupy the $t_{2g}$ orbitals,resulting in $3$ pairs of electrons.
The Crystal Field Stabilization Energy $(CFSE)$ is calculated as:
$CFSE = (6 \times -0.4 \Delta_o) + 3P$
$CFSE = (6 \times -\frac{2}{5} \Delta_o) + 3P$
$CFSE = -\frac{12}{5} \Delta_o + 3P$

(C) The given reaction is a Swarts reaction,which is used to prepare alkyl fluorides from alkyl chlorides or bromides.
The reaction is driven by the precipitation of the metal halide $(MCl)$ formed as a byproduct.
For the reaction to be most favourable,the metal fluoride $(MF)$ should be soluble in the reaction medium,and the metal chloride $(MCl)$ should be insoluble (precipitate out).
Among the given options,$RbF$ is the most soluble in organic solvents due to its lower lattice energy compared to $LiF$,$NaF$,and $KF$. Consequently,$RbCl$ is more likely to precipitate,driving the reaction forward.

(D) The reaction sequence is as follows:
$1$. $CH_3Br + KCN \rightarrow CH_3CN (A) + KBr$
$2$. $CH_3CN + 2H_2O + H^{+} \rightarrow CH_3COOH (B) + NH_4^{+}$
$3$. $CH_3COOH + LiAlH_4 \xrightarrow{ether} CH_3CH_2OH (C)$
$LiAlH_4$ is a strong reducing agent that reduces carboxylic acids to primary alcohols. Therefore,the final product $(C)$ is $CH_3CH_2OH$,which is ethyl alcohol.

(B) Ethylene glycol (commonly known as $Glycol$) is widely used as an antifreeze in automobile radiators because it lowers the freezing point of water and raises its boiling point.

(A) The given reaction is a Cannizzaro reaction because the reactant,$3$-chlorobenzaldehyde,does not have any $\alpha$-hydrogen atoms. When treated with $50 \% \ KOH$,it undergoes disproportionation (self-oxidation and reduction). The aldehyde group is oxidized to a carboxylate salt $(COOK)$ and reduced to an alcohol group $(-CH_2OH)$. The products are $3$-chlorobenzyl alcohol and potassium $3$-chlorobenzoate.

(A) Acetone reacts with excess ethanol in the presence of dry $HCl$ gas to form a ketal (a type of acetal). The reaction is as follows:
$CH_3COCH_3 + 2C_2H_5OH \xrightarrow{dry \ HCl} CH_3C(OC_2H_5)_2CH_3 + H_2O$
Thus,the product obtained is $CH_3C(OC_2H_5)_2CH_3$.

(B) $CH_3CHO$ and $C_6H_5CH_2CHO$ are both aldehydes and react with Tollen's reagent,Fehling's solution,and Benedict's solution,so these cannot distinguish them.
$CH_3CHO$ contains a $CH_3CO-$ group,which gives a positive Iodoform test with $I_2$ and $NaOH$,forming a yellow precipitate of $CHI_3$.
$CH_3CHO + 3 I_2 + 4 NaOH \longrightarrow CHI_3 (\text{yellow ppt}) + HCOONa + 3 NaI + 3 H_2O$
$C_6H_5CH_2CHO$ does not contain the $CH_3CO-$ group and does not give the Iodoform test.
$\therefore$ The Iodoform test is the correct method to distinguish between them.

(A) The acid strength of carboxylic acids is determined by the stability of the carboxylate anion formed after the loss of a $H^+$ ion.
Electron-withdrawing groups ($-I$ effect) stabilize the carboxylate anion,thereby increasing the acidity.
Electron-donating groups ($+I$ effect) destabilize the carboxylate anion,thereby decreasing the acidity.
Comparing the substituents:
$CF_3-$ (trifluoroacetic acid,$B$) has a stronger $-I$ effect than $CCl_3-$ (trichloroacetic acid,$A$) due to the higher electronegativity of fluorine.
Formic acid ($D$,$HCOOH$) has no alkyl group,while acetic acid ($C$,$CH_3COOH$) has a methyl group which exerts a $+I$ effect.
Therefore,the order of decreasing acid strength is $CF_3COOH (B) > CCl_3COOH (A) > HCOOH (D) > CH_3COOH (C)$.

(A) The given reaction is the Rosenmund reduction.
In this reaction,an acid chloride $(RCOCl)$ is hydrogenated to an aldehyde $(RCHO)$ using hydrogen gas in the presence of a palladium catalyst supported on barium sulfate $(Pd/BaSO_4)$.
The $BaSO_4$ acts as a poison to prevent the further reduction of the aldehyde to an alcohol.
Therefore,the reaction of benzoyl chloride $(C_6H_5COCl)$ with $H_2$ in the presence of $Pd/BaSO_4$ yields benzaldehyde $(C_6H_5CHO)$.

(D) The reaction $RCHO + NH_2NH_2 \rightarrow RCH=NNH_2 + H_2O$ involves the attack of the nucleophilic nitrogen atom of hydrazine $(NH_2NH_2)$ on the electrophilic carbonyl carbon of the aldehyde $(RCHO)$.
This step is a nucleophilic addition to the carbonyl group.
Subsequently,the elimination of a water molecule $(H_2O)$ occurs to form the final product,which is a hydrazone.
Therefore,the overall mechanism is classified as a nucleophilic addition-elimination reaction.

(D) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
Acetophenone $(C_6H_5COCH_3)$ contains the $CH_3CO-$ group and gives a yellow precipitate of iodoform $(CHI_3)$.
$2-$Hydroxypropane $(CH_3CH(OH)CH_3)$ contains the $CH_3CH(OH)-$ group and also gives a yellow precipitate of iodoform $(CHI_3)$.
Methyl acetate $(CH_3COOCH_3)$ does not give this test.
Therefore,both $(a)$ and $(c)$ give the yellow precipitate.

(A) The molecular formula $(C_3H_9N)$ corresponds to a primary amine.
When treated with nitrous acid $(HNO_2)$,primary aliphatic amines evolve $N_2$ gas and form an alcohol.
The carbylamine reaction (warming with $CHCl_3$ and $KOH$) is a characteristic test for primary amines,forming an isocyanide $(C)$.
Reduction of the isocyanide $(CH_3)_2CH-NC$ gives isopropylmethylamine $(CH_3)_2CH-NH-CH_3$.
Therefore,$(A)$ is isopropylamine,$(CH_3)_2CH-NH_2$.

(B) deficiency of vitamin $B_1$ (thiamine) leads to the disease known as beri-beri.
This condition is characterized by damage to the peripheral nerves and muscle wasting.
Symptoms may include weight loss,anorexia,confusion,and short-term memory loss.

(A) Sucrose is a disaccharide composed of $\alpha-D$-glucopyranose and $\beta-D$-fructofuranose units.
These units are linked by an $\alpha, \beta-1,2$-glycosidic linkage between the $C-1$ of the glucose unit and the $C-2$ of the fructose unit.

(B) Elastomers are polymers that possess weak intermolecular forces of attraction,allowing them to be stretched.
Nylon-$6,6$ is a fibre,not an elastomer,because it has strong intermolecular forces like hydrogen bonding.
Artificial silk (rayon) is indeed derived from cellulose.
Natural rubber is a polymer of isoprene ($2$-methyl-$1,3$-butadiene).
Both starch and cellulose are natural polymers of glucose.
Therefore,the statement that Nylon-$6,6$ is an elastomer is false.

(B) Biodegradable polymers are polymers that can be decomposed by microorganisms.
Option $B$ involves $H_2N-CH_2-COOH$ (Glycine) and $H_2N-(CH_2)_5-COOH$ (Aminocaproic acid). These monomers undergo condensation polymerization to form Nylon-$2$-Nylon-$6$,which is a well-known biodegradable polyamide polymer.
Therefore,the correct set is $B$.

(D) . Antifertility drugs are used to control pregnancy. These drugs prevent conception or fertilization,e.g.,$Mifepristone$,$norethindrone$,$mestranol$ etc.
$B$. Antihistamines are used for the relief of allergies,e.g.,$diphenhydramine$,$chlorpheniramine$,$promethazine$ etc.
$C$. Antiseptics are used to reduce the number and growth of microorganisms,e.g.,$Dettol$ soap,whereas disinfectants kill bacteria and are used for sterilization of inanimate objects like instruments,utensils,clothes,floors,etc.,e.g.,$phenol$,$dettol$,$iodoform$ etc.
$D$. Broad spectrum antibiotics are such antibiotics that act against a wide range of disease-causing bacteria. They act against both $Gram$-positive and $Gram$-negative bacteria,e.g.,$ampicillin$,$chloramphenicol$ etc.
Therefore,$Chloramphenicol$ is a broad spectrum antibiotic.

(D) Neoprene is an addition polymer formed by the polymerization of chloroprene $(CH_2=C(Cl)-CH=CH_2)$.
Melamine,Glyptal,and Dacron are condensation polymers.