MHT CET 2021 Chemistry Question Paper with Answer and Solution

(A) The given equation is ${y^2} - 9xy + 18{x^2} = 0$.
Factoring the quadratic expression: $(y - 6x)(y - 3x) = 0$.
So,the two lines are $y = 6x$ and $y = 3x$.
The third line is $y = 9$.
To find the vertices of the triangle,we find the intersection points:
$1$. Intersection of $y = 6x$ and $y = 3x$ is $(0, 0)$.
$2$. Intersection of $y = 6x$ and $y = 9$ is $x = \frac{9}{6} = \frac{3}{2}$,so point is $(\frac{3}{2}, 9)$.
$3$. Intersection of $y = 3x$ and $y = 9$ is $x = \frac{9}{3} = 3$,so point is $(3, 9)$.
The area of the triangle with vertices $(0, 0)$,$(\frac{3}{2}, 9)$,and $(3, 9)$ is given by:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
$\text{Area} = \frac{1}{2} |0(9 - 9) + \frac{3}{2}(9 - 0) + 3(0 - 9)|$
$\text{Area} = \frac{1}{2} |0 + \frac{27}{2} - 27| = \frac{1}{2} |-\frac{27}{2}| = \frac{27}{4} \text{ sq. units}$.

(C) We know the Lagrange's identity for vectors: $(a \times b)^2 + (a \cdot b)^2 = |a|^2 |b|^2$.
Given that $(a \times b)^2 + (a \cdot b)^2 = 144$ and $|a| = 4$.
Substituting these values into the identity:
$144 = (4)^2 |b|^2$
$144 = 16 |b|^2$
$|b|^2 = \frac{144}{16} = 9$
$|b| = \sqrt{9} = 3$.
Thus,$|b| = 3$.

(B) Given $x = \frac{1 - t^2}{1 + t^2}$ and $y = \frac{2at}{1 + t^2}$.
Differentiating $x$ with respect to $t$ using the quotient rule:
$\frac{dx}{dt} = \frac{(1 + t^2)(-2t) - (1 - t^2)(2t)}{(1 + t^2)^2} = \frac{-2t - 2t^3 - 2t + 2t^3}{(1 + t^2)^2} = \frac{-4t}{(1 + t^2)^2}$.
Differentiating $y$ with respect to $t$ using the quotient rule:
$\frac{dy}{dt} = \frac{(1 + t^2)(2a) - (2at)(2t)}{(1 + t^2)^2} = \frac{2a + 2at^2 - 4at^2}{(1 + t^2)^2} = \frac{2a(1 - t^2)}{(1 + t^2)^2}$.
Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{2a(1 - t^2)}{(1 + t^2)^2}}{\frac{-4t}{(1 + t^2)^2}} = \frac{2a(1 - t^2)}{-4t} = \frac{a(1 - t^2)}{-2t} = \frac{a(t^2 - 1)}{2t}$.

(D) Let $I = \int [\sin (\log x) + \cos (\log x)] \;dx$.
We can split the integral as $I = \int \sin (\log x) \;dx + \int \cos (\log x) \;dx$.
Consider the first integral $\int \sin (\log x) \;dx$. Let $u = \log x$,then $x = e^u$ and $dx = e^u \;du$.
So,$\int \sin (\log x) \;dx = \int e^u \sin u \;du$.
Using the formula $\int e^u \sin u \;du = \frac{e^u}{2} (\sin u - \cos u) + c_1$,we get $\int \sin (\log x) \;dx = \frac{x}{2} [\sin (\log x) - \cos (\log x)] + c_1$.
Similarly,$\int \cos (\log x) \;dx = \int e^u \cos u \;du = \frac{e^u}{2} (\sin u + \cos u) + c_2 = \frac{x}{2} [\sin (\log x) + \cos (\log x)] + c_2$.
Adding these,$I = \frac{x}{2} [\sin (\log x) - \cos (\log x)] + \frac{x}{2} [\sin (\log x) + \cos (\log x)] + c$.
$I = \frac{x}{2} [2 \sin (\log x)] + c = x \sin (\log x) + c$.

(B) Let $I = \int_0^1 {{{\tan }^{ - 1}}} \left( {\frac{{2x - 1}}{{1 + x - {x^2}}}} \right)\,dx$.
We can rewrite the argument of the $\tan^{-1}$ function as:
$\frac{2x - 1}{1 + x - x^2} = \frac{x + (x - 1)}{1 - x(x - 1)}$.
Using the identity $\tan^{-1}(A) + \tan^{-1}(B) = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$,we get:
$I = \int_0^1 (\tan^{-1}x + \tan^{-1}(x - 1)) \,dx$.
Now,split the integral:
$I = \int_0^1 \tan^{-1}x \,dx + \int_0^1 \tan^{-1}(x - 1) \,dx$.
Apply the property $\int_0^a f(x) \,dx = \int_0^a f(a - x) \,dx$ to the second integral:
Let $J = \int_0^1 \tan^{-1}(x - 1) \,dx$. Substituting $x = 1 - t$,$dx = -dt$:
$J = \int_1^0 \tan^{-1}(1 - t - 1) \,(-dt) = \int_0^1 \tan^{-1}(-t) \,dt = -\int_0^1 \tan^{-1}t \,dt$.
Thus,$I = \int_0^1 \tan^{-1}x \,dx - \int_0^1 \tan^{-1}x \,dx = 0$.

(B) The equation of a family of parabolas with focus at $(0, 0)$ and the $x$-axis as the axis is given by $y^2 = 4a(x + a)$,where $a$ is a parameter.
Differentiating with respect to $x$:
$2y \frac{dy}{dx} = 4a$
$Rightarrow a = \frac{y}{2} \frac{dy}{dx}$.
Substituting the value of $a$ back into the original equation:
$y^2 = 4 \left( \frac{y}{2} \frac{dy}{dx} \right) \left( x + \frac{y}{2} \frac{dy}{dx} \right)$
$y^2 = 2y \frac{dy}{dx} \left( x + \frac{y}{2} \frac{dy}{dx} \right)$
Dividing by $y$ (assuming $y \neq 0$):
$y = 2x \frac{dy}{dx} + y \left( \frac{dy}{dx} \right)^2$
Rearranging the terms:
$-y \left( \frac{dy}{dx} \right)^2 = 2x \frac{dy}{dx} - y$.
Thus,the correct option is $B$.

(C) Let the two unknown observations be $x$ and $y$.
Given the mean of $5$ observations is $4$,we have:
$\frac{1 + 2 + 6 + x + y}{5} = 4$
$9 + x + y = 20$
$x + y = 11$ .....$(i)$
Given the variance is $5.2$,we use the formula $\text{Variance} = \frac{\sum x_i^2}{n} - (\text{mean})^2$:
$5.2 = \frac{1^2 + 2^2 + 6^2 + x^2 + y^2}{5} - (4)^2$
$5.2 = \frac{1 + 4 + 36 + x^2 + y^2}{5} - 16$
$21.2 = \frac{41 + x^2 + y^2}{5}$
$106 = 41 + x^2 + y^2$
$x^2 + y^2 = 65$ .....$(ii)$
From $(i)$,$y = 11 - x$. Substituting into $(ii)$:
$x^2 + (11 - x)^2 = 65$
$x^2 + 121 - 22x + x^2 = 65$
$2x^2 - 22x + 56 = 0$
$x^2 - 11x + 28 = 0$
$(x - 4)(x - 7) = 0$
Thus,$x = 4$ or $x = 7$. If $x = 4$,then $y = 7$. If $x = 7$,then $y = 4$.
Therefore,the other two observations are $4$ and $7$.

(D) The moment of inertia $(I)$ of the system about the axis $YY'$ is the sum of the moments of inertia of the three individual rings.
$I = I_1 + I_2 + I_3$
For ring $1$,the axis $YY'$ passes through its diameter. The moment of inertia of a ring about its diameter is $I_1 = \frac{1}{2}MR^2$.
For rings $2$ and $3$,the axis $YY'$ is tangent to the ring in its own plane. Using the parallel axis theorem,$I = I_{cm} + Md^2$,where $I_{cm} = \frac{1}{2}MR^2$ and $d = R$. Thus,$I_2 = I_3 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.
Adding these together: $I = \frac{1}{2}MR^2 + \frac{3}{2}MR^2 + \frac{3}{2}MR^2 = \frac{7}{2}MR^2$.

(D) The system consists of three rings. Let the top ring be $1$,and the two bottom rings be $2$ and $3$. The axis $YY'$ passes through the center of ring $1$ (diameter) and is tangent to rings $2$ and $3$ (parallel to their diameters).
$1$. For ring $1$,the axis $YY'$ is its diameter. The moment of inertia is $I_1 = \frac{1}{2}MR^2$.
$2$. For rings $2$ and $3$,the axis $YY'$ is parallel to their diameters at a distance $R$. Using the parallel axis theorem,$I = I_{cm} + Md^2$,where $I_{cm} = \frac{1}{2}MR^2$ and $d = R$. Thus,$I_2 = I_3 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.
$3$. The total moment of inertia of the system is $I = I_1 + I_2 + I_3 = \frac{1}{2}MR^2 + \frac{3}{2}MR^2 + \frac{3}{2}MR^2 = \frac{7}{2}MR^2$.

(B) According to the Pauli Exclusion Principle,no two electrons in an atom can have the same set of all four quantum numbers.
For two electrons in the same orbital,the principal quantum number $(n)$,azimuthal quantum number $(l)$,and magnetic quantum number $(m_l)$ are identical.
Therefore,they must differ in their spin quantum number $(m_s)$,which can be either $+1/2$ or $-1/2$.

(A) The force per unit length between two parallel conductors carrying currents $I_1$ and $I_2$ separated by distance $d$ is given by $F = \frac{\mu_0 I_1 I_2}{2\pi d}$.
According to the problem,the initial force is $F = \frac{\mu_0 I_1 I_2}{2\pi d}$.
In the new condition,the current in one conductor becomes $I_1' = -2I_1$ (negative sign indicates reversed direction) and the distance becomes $d' = 3d$.
The new force $F'$ is given by:
$F' = \frac{\mu_0 (-2I_1) I_2}{2\pi (3d)}$
$F' = -\frac{2}{3} \left( \frac{\mu_0 I_1 I_2}{2\pi d} \right)$
Substituting the initial force $F$ into the equation:
$F' = -\frac{2}{3} F$.

(D) Let $I = \int_{0}^{1} \tan^{-1}\left(\frac{2x-1}{1+x-x^2}\right) dx$.
We can rewrite the argument of the $\tan^{-1}$ function as:
$\frac{2x-1}{1+x-x^2} = \frac{x - (1-x)}{1 + x(1-x)}$.
Using the identity $\tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a-b}{1+ab}\right)$,we get:
$I = \int_{0}^{1} [\tan^{-1}(x) - \tan^{-1}(1-x)] dx$ ...... $(1)$
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we have:
$I = \int_{0}^{1} [\tan^{-1}(1-x) - \tan^{-1}(1-(1-x))] dx$
$I = \int_{0}^{1} [\tan^{-1}(1-x) - \tan^{-1}(x)] dx$ ...... $(2)$
Adding $(1)$ and $(2)$:
$2I = \int_{0}^{1} [\tan^{-1}(x) - \tan^{-1}(1-x) + \tan^{-1}(1-x) - \tan^{-1}(x)] dx$
$2I = \int_{0}^{1} 0 dx = 0$.
Therefore,$I = 0$.
Hence,the correct answer is $D$.

(B) Let $X$ be the random variable representing the number obtained on the die.
The total number of faces on the die is $6$.
The probabilities of obtaining each number are:
$P(X=1) = \frac{3}{6} = \frac{1}{2}$
$P(X=2) = \frac{2}{6} = \frac{1}{3}$
$P(X=5) = \frac{1}{6}$
The mean (expected value) $E(X)$ is calculated as $\sum x_i P(x_i)$:
$E(X) = (1 \times \frac{1}{2}) + (2 \times \frac{1}{3}) + (5 \times \frac{1}{6})$
$E(X) = \frac{1}{2} + \frac{2}{3} + \frac{5}{6}$
To add these fractions,find the common denominator,which is $6$:
$E(X) = \frac{3}{6} + \frac{4}{6} + \frac{5}{6}$
$E(X) = \frac{3+4+5}{6} = \frac{12}{6} = 2$
Thus,the mean is $2$. The correct answer is $B$.

(B) Ethers contain an oxygen atom with two lone pairs of electrons,which makes them basic in nature. When ethers are dissolved in cold concentrated $H_2SO_4$,the oxygen atom accepts a proton $(H^+)$ from the acid to form a stable oxonium salt. The reaction is represented as: $R-O-R + H_2SO_4 \rightarrow [R-O^+(H)-R]HSO_4^-$. This product is known as an oxonium salt.

(B) The Cannizzaro reaction is a classic example of a disproportionation reaction (also known as an auto-oxidation-reduction reaction).
In this reaction,an aldehyde that lacks an $\alpha$-hydrogen atom undergoes self-oxidation and reduction in the presence of a concentrated base.
One molecule of the aldehyde is oxidized to form a salt of a carboxylic acid,while another molecule is reduced to form a primary alcohol.

(D) Ozonolysis of propene $(CH_3-CH=CH_2)$ involves the reaction with ozone $(O_3)$ followed by reductive cleavage with zinc and water $(Zn/H_2O)$.
The reaction breaks the double bond and forms carbonyl compounds:
$CH_3-CH=CH_2 \xrightarrow[(ii) Zn/H_2O]{(i) O_3} CH_3CHO + HCHO$
The products obtained are acetaldehyde $(CH_3CHO)$ and formaldehyde $(HCHO)$.

(A) The boiling point of aldehydes depends on the molecular weight of the compound.
As the molecular weight increases,the magnitude of van der Waals forces increases,leading to a higher boiling point.
Comparing the molecular weights:
$Ethanal (C_2H_4O) < Propanal (C_3H_6O) < Pentanal (C_5H_{10}O) < Hexanal (C_6H_{12}O)$.
Therefore,$Hexanal$ has the highest molecular weight and consequently the highest boiling point.

(A) Grignard reagents $(RMgX)$ are highly reactive towards compounds containing active hydrogen atoms (like $H$ attached to $O, N, S$).
In the given reaction,$C_2H_5MgBr$ reacts with ammonia $(NH_3)$ to form ethane $(C_2H_6)$ and magnesium bromamide $(Mg(NH_2)Br)$.
The reaction is: $C_2H_5MgBr + NH_3 \rightarrow C_2H_6 + Mg(NH_2)Br$.
Thus,the reagent $(A)$ is $NH_3$.

(C) Lactose is a disaccharide composed of $\beta-D-galactose$ and $\beta-D-glucose$ units.
These two monosaccharide units are joined together by a $\beta-1,4-glycosidic$ linkage.
In this linkage,the $C-1$ of $\beta-D-galactose$ is connected to the $C-4$ of $\beta-D-glucose$ through an oxygen atom.

(D) Adipic acid is a dicarboxylic acid with the formula $HOOC-(CH_2)_4-COOH$.
Oxidation of cyclic alkenes with strong oxidizing agents like $KMnO_4$ in acidic medium leads to the cleavage of the double bond and the ring,resulting in the formation of dicarboxylic acids.
Cyclohexene,when treated with $KMnO_4$ in dil. $H_2SO_4$,undergoes oxidative cleavage to form adipic acid (hexanedioic acid).
The reaction is: $\text{Cyclohexene} + [O] \xrightarrow{KMnO_4/H^+} HOOC-(CH_2)_4-COOH$.

(A) Homolytic fission is a type of bond cleavage where the shared pair of electrons in a covalent bond is split equally between the two bonded atoms.
Each atom retains one electron from the shared pair,resulting in the formation of free radicals.
Therefore,the correct statement is that a single electron of the shared pair moves to each of the bonded atoms.

(D) The $SO_2$ molecule has a bent geometry with $sp^2$ hybridization at the sulfur atom.
It contains one lone pair of electrons on the sulfur atom.
According to $VSEPR$ theory,the lone pair-bond pair repulsion is greater than the bond pair-bond pair repulsion.
This causes the bond angle to decrease from the ideal trigonal planar angle of $120^{\circ}$ to approximately $119.5^{\circ}$.

(D) The central atom $Sb$ has $5$ valence electrons and forms $5$ bonds with $F$ atoms,resulting in $5$ bond pairs and $0$ lone pairs.
According to $VSEPR$ theory,a molecule with $5$ bond pairs and $0$ lone pairs has $sp^3d$ hybridization.
The geometry of $SbF_5$ is Trigonal bipyramidal.

(B) To determine the geometry,we calculate the hybridization of the central atom using the formula: $H = \frac{1}{2} (V + M - C + A)$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
$1$. For $SeCl_2$: $H = \frac{1}{2} (6 + 2) = 4$ ($sp^3$ hybridization). With $2$ lone pairs,it has a bent shape.
$2$. For $SeF_4$: $H = \frac{1}{2} (6 + 4) = 5$ ($sp^3d$ hybridization). With $1$ lone pair,it has a seesaw shape,which is derived from a trigonal bipyramidal electron geometry.
$3$. For $SF_6$: $H = \frac{1}{2} (6 + 6) = 6$ ($sp^3d^2$ hybridization). It has an octahedral geometry.
$4$. For $TeF_6$: $H = \frac{1}{2} (6 + 6) = 6$ ($sp^3d^2$ hybridization). It has an octahedral geometry.
Among the given options,$SeF_4$ is the only one associated with the trigonal bipyramidal electron geometry.

(B) The ammonium ion $(NH_4^+)$ is formed by the reaction of ammonia $(NH_3)$ with a hydrogen ion $(H^+)$.
In ammonia $(NH_3)$,the nitrogen atom is bonded to $3$ hydrogen atoms.
When $NH_3$ reacts with $H^+$,the lone pair on the nitrogen atom forms a coordinate covalent bond with the $H^+$ ion.
As a result,the nitrogen atom in the ammonium ion $(NH_4^+)$ is bonded to a total of $4$ hydrogen atoms.

(D) The bond enthalpy is directly related to the bond order or the number of bonds between atoms.
$N_2$ contains a triple bond $(N \equiv N)$,which is significantly stronger than the single bonds in $NH_3$ $(N-H)$ and $CH_4$ $(C-H)$,or the double bond in $O_2$ $(O=O)$.
Therefore,$N_2$ has the highest bond enthalpy.

(D) The concept of the combination of atomic orbitals to form molecular orbitals belongs to the Molecular Orbital Theory $(MOT)$,not the Valence Bond Theory $(VBT)$.

(D) The formal charge $(FC)$ on an atom in a Lewis structure is calculated using the formula:
$FC = V - L - \frac{1}{2}S$
Where:
$V$ = Number of valence electrons of the free atom
$L$ = Number of non-bonding (lone pair) electrons
$S$ = Number of bonding (shared) electrons
For the carbon atom in $CO_3^{2-}$:
$V = 4$ (Carbon belongs to group $14$)
$L = 0$ (Carbon has no lone pairs in the carbonate ion)
$S = 8$ (Carbon forms $4$ bonds: one double bond and two single bonds)
$FC = 4 - 0 - \frac{1}{2}(8) = 4 - 4 = 0$.

(B) The Lewis structure of carbon monoxide is $:C \equiv O:$.
Formal charge $= [\text{Total no. of valence } e^- \text{ in free state}] - [\text{Total no. of lone pair } e^-] - \frac{1}{2}[\text{Total no. of bonding } e^-]$.
For oxygen atom:
Valence electrons $= 6$.
Lone pair electrons $= 2$.
Bonding electrons $= 6$.
Formal charge $= 6 - 2 - \frac{1}{2} \times 6 = 6 - 2 - 3 = +1$.

(A) The formula for calculating the formal charge is:
$\text{Formal charge} = [\text{Total number of valence electrons in free state}] - [\text{Total number of lone pair electrons}] - \frac{1}{2} [\text{Number of bonding electrons on that particular atom}]$
For the $N$ atom in $NH_4^{+}$:
- The number of valence electrons in a free $N$ atom is $5$.
- The number of lone pair electrons on the $N$ atom is $0$.
- The number of bonding electrons around the $N$ atom is $8$ (since there are $4$ covalent bonds).
$\text{Formal charge on } N = 5 - 0 - \frac{1}{2} \times 8 = 5 - 4 = +1$.

(D) The formula for formal charge is: $\text{Formal charge} = [ \text{Total no. of valence } e^- \text{ in free state} ] - [ \text{Total no. of lone pair electrons} ] - \frac{1}{2} [ \text{Total no. of bonding electrons} ]$.
For the carbon atom $(C)$ in the structure $S \equiv C - \ddot{N} :$,the number of valence electrons is $4$.
The carbon atom has $0$ lone pair electrons and forms $4$ bonds (a triple bond with $S$ and a single bond with $N$),which corresponds to $8$ bonding electrons.
Substituting these values: $\text{Formal charge} = 4 - 0 - \frac{1}{2} \times 8 = 4 - 4 = 0$.

(B) The formal charge $(FC)$ of an atom in a molecule or ion is calculated using the formula:
$FC = V - L - \frac{B}{2}$
Where:
$V$ = Total number of valence electrons in the free atom.
$L$ = Total number of non-bonding (lone pair) electrons.
$B$ = Total number of bonding (shared) electrons.
For the Carbon $(C)$ atom in the given structure of $SCN^-$ $(S \equiv C - N:)$:
$V = 4$ (Carbon is in group $14$)
$L = 0$ (Carbon has no lone pairs)
$B = 8$ (Carbon forms $3$ bonds with $S$ and $1$ bond with $N$,total $4$ bonds = $8$ electrons)
$FC = 4 - 0 - \frac{8}{2} = 4 - 4 = 0$.
Thus,the formal charge on the $C$ atom is $0$.

(A) The formal charge $(FC)$ is calculated using the formula: $FC = V - L - \frac{B}{2}$.
For the nitrogen $(N)$ atom in the $[S-C \equiv N:]^-$ ion:
Valence electrons $(V)$ of $N = 5$.
Lone pair electrons $(L)$ on $N = 2$.
Bonding electrons $(B)$ for $N = 6$ (due to the triple bond).
$FC = 5 - 2 - \frac{6}{2} = 5 - 2 - 3 = 0$.

(A) According to Fajan's rule,as the size of the anion increases,its polarizability increases,which leads to an increase in covalent character and a decrease in ionic character.
Since the size of the fluoride ion $(F^-)$ is the smallest among the given halide ions $(F^- < Cl^- < Br^- < I^-)$,it has the least polarizability.
Therefore,$MF$ exhibits the highest ionic character.
The order of ionic character is: $MF > MCl > MBr > MI$.

(C) The dipole moment $(\mu)$ of a molecule depends on its geometry and the polarity of its bonds.
$BF_3$ has a trigonal planar geometry,$CCl_4$ and $CH_4$ have tetrahedral geometries. Due to their highly symmetric structures,the individual bond dipoles cancel each other out,resulting in a net dipole moment of $\mu = 0$.
$CHCl_3$ (chloroform) has a tetrahedral geometry,but because the atoms attached to the central carbon atom are not identical ($3$ chlorine atoms and $1$ hydrogen atom),the bond dipoles do not cancel out. Therefore,$CHCl_3$ has a non-zero net dipole moment $(\mu \neq 0)$.

(B) For a weak monobasic acid $HA$,the dissociation equilibrium is $HA \rightleftharpoons H^+ + A^-$.
Let $C$ be the initial molar concentration and $x$ be the degree of dissociation.
$K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(Cx)(Cx)}{C(1-x)} = \frac{Cx^2}{1-x}$.
Since the acid is weak,$x \ll 1$,so $1-x \approx 1$,and $K_a \approx Cx^2$.
Given $K_a = 5 \times 10^{-8}$ and $x = 0.5 \% = 0.005 = 5 \times 10^{-3}$.
$C = \frac{K_a}{x^2} = \frac{5 \times 10^{-8}}{(5 \times 10^{-3})^2}$.
$C = \frac{5 \times 10^{-8}}{25 \times 10^{-6}} = \frac{5}{25} \times 10^{-2} = 0.2 \times 10^{-2} = 0.002 \ M$.

(B) Given: $pH = 2.34$,$c = 0.1 \ M$.
Step $1$: Calculate the concentration of hydrogen ions $[H^+]$.
$[H^+] = 10^{-pH} = 10^{-2.34} = 4.571 \times 10^{-3} \ M$.
Step $2$: Use the relation for a monobasic acid,$[H^+] = c \alpha$.
$\alpha = \frac{[H^+]}{c} = \frac{4.571 \times 10^{-3}}{0.1} = 4.571 \times 10^{-2}$.
Thus,the degree of dissociation is $4.57 \times 10^{-2}$.

(B) For a weak acid,the dissociation constant $K_a$ is related to the degree of dissociation $\alpha$ and concentration $C$ by the formula: $K_a = C\alpha^2$ (assuming $\alpha \ll 1$).
Given $K_a = 1.32 \times 10^{-5}$ and $C = 0.05 \ M$.
$\alpha = \sqrt{\frac{K_a}{C}}$
$\alpha = \sqrt{\frac{1.32 \times 10^{-5}}{0.05}}$
$\alpha = \sqrt{\frac{1.32 \times 10^{-5}}{5 \times 10^{-2}}} = \sqrt{0.264 \times 10^{-3}} = \sqrt{2.64 \times 10^{-4}}$
$\alpha = 1.624 \times 10^{-2} \approx 1.61 \times 10^{-2}$.

(B) For a weak monobasic acid $HA$,the dissociation equilibrium is:
$HA_{(aq)} \rightleftharpoons H^+_{(aq)} + A^-_{(aq)}$
At equilibrium,the concentration of $H^+$ and $A^-$ is $C\alpha$ and the concentration of $HA$ is $C(1-\alpha)$,where $C$ is the initial concentration and $\alpha$ is the degree of dissociation.
The dissociation constant $K_a$ is given by:
$K_a = \frac{[H^+][A^-]}{[HA]} = \frac{C\alpha^2}{1-\alpha}$
Since the acid is weak,$\alpha \ll 1$,so $1-\alpha \approx 1$.
Thus,$K_a = C\alpha^2$.
Rearranging to solve for $C$:
$C = \frac{K_a}{\alpha^2} = \frac{2.7 \times 10^{-5}}{(3 \times 10^{-2})^2} = \frac{2.7 \times 10^{-5}}{9 \times 10^{-4}} = 0.3 \times 10^{-1} = 0.03 \ M$.

(B) For a weak monobasic acid $(HA)$,the dissociation is given by: $HA_{(aq)} \rightleftharpoons H^{+}_{(aq)} + A^{-}_{(aq)}$.
The dissociation constant $K_a$ is given by $K_a = \frac{C \alpha^2}{1-\alpha}$.
Since the acid is weak,$\alpha \ll 1$,so $1-\alpha \approx 1$,and $K_a \approx C \alpha^2$.
For condition $1$: $C_1 = 0.05 \ M$,$\alpha_1 = 0.10$.
$K_a = 0.05 \times (0.1)^2 = 0.05 \times 0.01 = 0.0005$.
For condition $2$: $C_2 = 0.10 \ M$,$\alpha_2 = ?$.
$K_a = C_2 \alpha_2^2 \implies 0.0005 = 0.10 \times \alpha_2^2$.
$\alpha_2^2 = \frac{0.0005}{0.10} = 0.005$.
$\alpha_2 = \sqrt{0.005} \approx 0.0707$.
Percentage dissociation $= 0.0707 \times 100 = 7.07 \%$.
The closest option is $7.17 \%$.

(B) For a monoacidic weak base $BOH$:
$BOH_{(aq)} \rightleftharpoons B_{(aq)}^{+} + OH_{(aq)}^{-}$
Given $pH = 10.9$,we calculate $pOH$:
$pOH = 14 - pH = 14 - 10.9 = 3.1$
The concentration of hydroxide ions is:
$[OH^{-}] = 10^{-pOH} = 10^{-3.1} = 10^{-4} \times 10^{0.9} \approx 7.94 \times 10^{-4} \ M$
For a weak base,$[OH^{-}] = C \times \alpha$,where $C = 0.02 \ M$:
$\alpha = \frac{[OH^{-}]}{C} = \frac{7.94 \times 10^{-4}}{0.02} = 3.97 \times 10^{-2}$
The percent dissociation is $\alpha \times 100$:
$\% \alpha = 3.97 \times 10^{-2} \times 100 = 3.97 \% \approx 3.95 \%$
Thus,the correct option is $B$.

(D) For the dissociation of acetic acid: $CH_3COOH_{(aq)} \rightleftharpoons CH_3COO^-_{(aq)} + H^+_{(aq)}$
Given concentration $C = 0.01 \ M$ and dissociation constant $K_a = 1.34 \times 10^{-6}$.
The expression for dissociation constant is $K_a = \frac{C \alpha^2}{1-\alpha}$.
Since $K_a$ is very small,$1-\alpha \approx 1$,so $K_a \approx C \alpha^2$.
Thus,$\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.34 \times 10^{-6}}{0.01}} = \sqrt{1.34 \times 10^{-4}} = 1.157 \times 10^{-2}$.
Percent dissociation $\% \alpha = \alpha \times 100 = 1.157 \times 10^{-2} \times 100 = 1.157 \% \approx 1.15 \%$.

(B) For a weak monobasic acid $HA$,the dissociation equilibrium is: $HA_{(aq)} \rightleftharpoons H^+_{(aq)} + A^-_{(aq)}$
Given concentration $C = 0.04 \ M$ and degree of dissociation $\alpha = 3.0 \% = 0.03$.
The dissociation constant $K_a$ is given by the formula: $K_a = \frac{C \alpha^2}{1 - \alpha}$.
Since the acid is weak,$\alpha \ll 1$,so $1 - \alpha \approx 1$.
Thus,$K_a \approx C \alpha^2$.
Substituting the values: $K_a = 0.04 \times (0.03)^2$.
$K_a = 0.04 \times 0.0009 = 3.6 \times 10^{-5}$.

(B) The dissociation of acetic acid is given by: $CH_3COOH_{(aq)} \rightleftharpoons CH_3COO^-_{(aq)} + H^+_{(aq)}$
Given concentration $C = 0.1 \ M$ and degree of ionization $\alpha = 1.34 \ \% = 0.0134$.
The dissociation constant $K_a$ is given by the formula: $K_a = \frac{C\alpha^2}{1-\alpha}$.
Since $\alpha$ is very small,we can approximate $1-\alpha \approx 1$.
Thus,$K_a \approx C\alpha^2$.
Substituting the values: $K_a = 0.1 \times (0.0134)^2$.
$K_a = 0.1 \times 0.00017956 = 1.7956 \times 10^{-5} \approx 1.8 \times 10^{-5}$.

(B) is the correct answer.
Alkali metals are the elements belonging to Group $1$ of the periodic table.
These include lithium $(Li)$,sodium $(Na)$,potassium $(K)$,rubidium $(Rb)$,caesium $(Cs)$,and francium $(Fr)$.
Among the given options,rubidium $(Rb)$ is an alkali metal.

(B) The alkali metals are the elements belonging to Group $1$ of the periodic table.
These include lithium $(Li)$,sodium $(Na)$,potassium $(K)$,rubidium $(Rb)$,caesium $(Cs)$,and francium $(Fr)$.
Among the given options,$Cs$ (caesium) is an alkali metal,while $Ba$,$Ca$,and $Sr$ are alkaline earth metals belonging to Group $2$.

(B) The oxidation states are $X = +2$,$Y = +5$,and $Z = -2$.
For a neutral compound,the sum of oxidation states must be zero.
Checking option $B$: $X_3(YZ_4)_2$.
Total charge $= 3 \times (+2) + 2 \times (+5 + 4 \times (-2)) = 6 + 2 \times (5 - 8) = 6 + 2 \times (-3) = 6 - 6 = 0$.
Since the total charge is zero,the correct formula is $X_3(YZ_4)_2$.

(D) High purity dihydrogen $(> 99.95\%)$ is obtained by the electrolysis of warm aqueous barium hydroxide solution between $Ni$ electrodes.
In this process,the electrolyte is $Ba(OH)_2$ and the electrodes used are $Ni$ electrodes.
This method is preferred for producing high-purity hydrogen gas.

(D) The ozone layer is primarily found in the stratosphere. It has the potential to absorb around $97-99 \%$ of the harmful ultraviolet $(UV)$ radiation coming from the sun,which can damage life on Earth.
Ozone depletion leads to the thinning of this protective layer,allowing a higher amount of harmful $UV$ radiation to reach the Earth's surface.
Therefore,the statement that 'less amount of $UV$ radiation reaches the earth's surface' due to ozone depletion is incorrect.

(B) $1$. Identify the principal functional group: The hydroxyl group $(-OH)$ is the principal functional group,so the parent chain is a cycloheptanol. The carbon attached to the $-OH$ group is assigned position $1$.
$2$. Number the ring: To give the substituents the lowest possible locants,we number the ring starting from the carbon with the $-OH$ group. Moving clockwise,the ethyl group is at position $3$ and the methyl group is at position $6$.
$3$. Alphabetical order: Ethyl comes before methyl. Thus,the name is $3-$ethyl$-6-$methylcycloheptan$-1-$ol.

(A) $LiAlH_4$ is a strong reducing agent that reduces aldehydes to primary alcohols.
It does not reduce isolated carbon-carbon double bonds $(C=C)$.
Therefore,the aldehyde group $(-CHO)$ is reduced to a primary alcohol group $(-CH_2OH)$ while the double bond remains intact.
The reaction is:
$CH_3-CH=CH-CH_2-CHO \xrightarrow[(ii) \ H_3O^{+}]{(i) \ LiAlH_4} CH_3-CH=CH-CH_2-CH_2-OH$
Thus,the correct product is $CH_3-CH=CH-CH_2-CH_2-OH$.

(A) Aldehydes other than formaldehyde $(HCHO)$ react with Grignard reagents $(RMgX)$ to form secondary alcohols after acidic hydrolysis.
$CH_3CHO + RMgX$ $\rightarrow CH_3CH(OMgX)R$ $\xrightarrow{H_3O^+} CH_3CH(OH)R$ (Secondary alcohol).
Formaldehyde $(HCHO)$ forms primary alcohols,while ketones form tertiary alcohols.
Therefore,$CH_3CHO$ is the correct answer.

(A) The melting point of a compound depends on its molecular symmetry and packing in the crystal lattice.
$tert-$Butyl alcohol ($2-$methylpropan$-2-$ol) has a highly symmetrical structure compared to the other isomers.
Due to this high symmetry,it packs more efficiently into the crystal lattice,leading to stronger intermolecular forces and a significantly higher melting point.

(C) The boiling point of isomeric alcohols depends on the extent of branching in the carbon chain.
As the branching increases,the surface area of the molecule decreases,which reduces the magnitude of van der Waals forces of attraction.
Consequently,the boiling point decreases with an increase in branching.
Among the given isomers:
$n$-Butyl alcohol $(CH_3CH_2CH_2CH_2OH)$ has no branching.
Isobutyl alcohol $((CH_3)_2CHCH_2OH)$ has one branch.
sec-Butyl alcohol $(CH_3CH_2CH(OH)CH_3)$ has one branch.
tert-Butyl alcohol $((CH_3)_3COH)$ has the maximum branching (three methyl groups attached to the central carbon).
Therefore,tert-Butyl alcohol has the lowest boiling point.

(B) $p$-Nitrophenol has the highest melting point due to strong intermolecular hydrogen bonding in the solid state,which leads to a more stable crystal lattice structure compared to the other isomers. $o$-Nitrophenol exhibits intramolecular hydrogen bonding,which reduces its melting point.

(D) Racemization occurs during the alkaline hydrolysis of an alkyl halide if the reaction proceeds via an $S_N1$ mechanism,which involves the formation of a planar carbocation intermediate. This requires the starting alkyl halide to be chiral (optically active). Among the given options,$CH_3-CH_2-CH(Cl)-CH_3$ ($2$-chlorobutane) is a chiral molecule because the carbon atom attached to the chlorine is bonded to four different groups $(-H, -Cl, -CH_3, -CH_2CH_3)$. Therefore,it can form a planar carbocation and undergo racemization.

(D) The reaction of acetaldehyde $(CH_3CHO)$ with a Grignard reagent $(RMgBr)$ followed by acid hydrolysis $(H_3O^+)$ yields a secondary alcohol.
To obtain $3-$methylbutan$-2-$ol $(CH_3-CH(OH)-CH(CH_3)_2)$,the Grignard reagent must provide the isopropyl group $(CH(CH_3)_2)$.
The reaction is as follows:
$CH_3-CH=O + (CH_3)_2CHMgBr \rightarrow CH_3-CH(OMgBr)-CH(CH_3)_2$
$CH_3-CH(OMgBr)-CH(CH_3)_2 + H_2O/H^+ \rightarrow CH_3-CH(OH)-CH(CH_3)_2 + Mg(OH)Br$
Thus,the required Grignard reagent is isopropylmagnesium bromide,which is $(CH_3)_2CHMgBr$.

(D) The reaction of a Grignard reagent $(C_2H_5MgBr)$ with a carbonyl compound $(A)$ followed by acid hydrolysis $(H_3O^{+})$ yields an alcohol.
In the final product,$3-\text{methylpentan}-3-\text{ol}$,the carbon atom bearing the hydroxyl group is bonded to two ethyl groups $(C_2H_5-)$ and one methyl group $(CH_3-)$.
Since one ethyl group comes from the Grignard reagent $(C_2H_5MgBr)$,the carbonyl compound $(A)$ must provide the remaining part of the structure,which is a ketone with one methyl group and one ethyl group attached to the carbonyl carbon.
Thus,the structure of $A$ is $CH_3-CO-CH_2-CH_3$,which is butanone.
The reaction is:
$CH_3-CO-CH_2-CH_3 + C_2H_5MgBr$ $\rightarrow CH_3-C(OMgBr)(C_2H_5)-CH_2-CH_3$ $\xrightarrow{H_3O^{+}} CH_3-C(OH)(C_2H_5)-CH_2-CH_3$ (which is $3-\text{methylpentan}-3-\text{ol}$).

(B) $(i)$ Boiling point $\propto$ Molar mass. Since all are isomers of $C_4H_9OH$,molar mass is the same.
$(ii)$ Boiling point $\propto \frac{1}{\text{Branching}}$.
$(iii)$ $tert-Butyl$ alcohol has the maximum branching,which reduces the surface area and decreases the van der Waals forces of attraction.
$(iv)$ Therefore,$tert-Butyl$ alcohol has the lowest boiling point.

(A) In Kolbe's reaction,sodium phenoxide $(C_6H_5ONa)$ is treated with carbon dioxide $(CO_2)$ at a temperature of $398\ K$ and a pressure of $6\ atm$,followed by acidification to yield salicylic acid.
Therefore,the correct reactants and conditions are $C_6H_5ONa, CO_2$ and $398\ K, 6\ atm$.

(B) When phenol reacts with concentrated sulphuric acid at a low temperature of $293 \ K$,the electrophilic substitution reaction occurs primarily at the ortho position.
This results in the formation of $o-$phenol sulphonic acid as the major product.
At higher temperatures (e.g.,$373 \ K$),the para isomer becomes the major product due to thermodynamic stability.

(D) When phenol is treated with concentrated $HNO_3$ in the presence of concentrated $H_2SO_4$,it undergoes electrophilic aromatic substitution (nitration) at all available ortho and para positions.
This reaction results in the formation of $2, 4, 6-$trinitrophenol,which is commonly known as picric acid.
Therefore,the correct product $(A)$ is $2, 4, 6-$trinitrophenol.

(D) When phenol $(C_6H_5OH)$ is hydrogenated in the presence of a nickel catalyst at $433 \ K$,the aromatic ring undergoes reduction to form a saturated cyclic alcohol.
The reaction is as follows:
$C_6H_5OH + 3H_2 \xrightarrow{Ni, 433 \ K} C_6H_{11}OH$
Thus,the product obtained is cyclohexanol.

(B) Anisole $(C_6H_5OCH_3)$ undergoes electrophilic aromatic substitution when treated with $Br_2$ in the presence of acetic acid $(CH_3COOH)$.
The methoxy group $(-OCH_3)$ is an activating group and is ortho/para-directing.
Due to steric hindrance at the ortho position,the para-isomer is formed as the major product.
Thus,the reaction is: $C_6H_5OCH_3 + Br_2 \xrightarrow{CH_3COOH} p\text{-bromoanisole} + HBr$.

(D) Ethers contain an oxygen atom with two lone pairs of electrons,which makes them act as weak Lewis bases.
When ethers are dissolved in cold concentrated sulfuric acid $(H_2SO_4)$,the oxygen atom of the ether gets protonated by the acid to form an oxonium salt.
The reaction is as follows:
$R-O-R' + H_2SO_4 \rightarrow [R-O(H)-R']^+ HSO_4^-$
Here,the product formed is an oxonium salt.

(B) The reaction of anisole with a mixture of concentrated $HNO_3$ and concentrated $H_2SO_4$ is an electrophilic aromatic substitution reaction known as nitration.
In anisole $(C_6H_5OCH_3)$,the methoxy group $(-OCH_3)$ is an ortho/para-directing group due to its $+M$ effect.
Due to steric hindrance at the ortho position,the para-substituted product is formed as the major product.
Therefore,the major product $x$ is $4-$nitro anisole.

(A) The reaction of ethoxybenzene $(C_6H_5-O-CH_2CH_3)$ with hot and concentrated $HI$ involves the cleavage of the $C-O$ bond between the oxygen atom and the ethyl group.
This occurs because the $C-O$ bond between the oxygen and the phenyl ring has partial double bond character due to resonance,making it stronger and harder to break.
Therefore,the $I^-$ ion attacks the less sterically hindered ethyl group via an $S_N2$ mechanism.
The products formed are phenol $(C_6H_5OH)$ and ethyl iodide $(CH_3CH_2I)$.

(C) The reaction of acid chlorides with dialkyl cadmium reagents is a standard method for the preparation of ketones.
The chemical equation is:
$2CH_3COCl + (CH_3)_2Cd \rightarrow 2CH_3COCH_3 + CdCl_2$
Here,$CH_3COCl$ is ethanoyl chloride (also known as acetyl chloride),which reacts with dimethyl cadmium to form propanone (acetone).
Therefore,'$A$' is ethanoyl chloride.

(A) The silver mirror test,also known as Tollen's test,is given by aldehydes.
Ethanal $(CH_3CHO)$ is an aldehyde,so it reacts with ammoniacal silver nitrate $([Ag(NH_3)_2]^+)$ to form a silver mirror (precipitate of silver).
The reaction is: $CH_3CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow CH_3COO^- + 2Ag(s) + 4NH_3 + 2H_2O$.

(B) Schiff's reagent is a specific test used to detect the presence of an aldehydic group $(-CHO)$.
When an aldehyde reacts with Schiff's reagent,it restores the pink colour of the reagent,confirming the presence of the $-CHO$ group.

(A) The side chain chlorination of toluene $(C_6H_5CH_3)$ with $Cl_2$ in the presence of light $(hv)$ gives benzal chloride $(C_6H_5CHCl_2)$.
Subsequent acid hydrolysis of benzal chloride at $373 \ K$ leads to the formation of benzaldehyde $(C_6H_5CHO)$ as the final product.
The reaction sequence is:
$C_6H_5CH_3$ $\xrightarrow{Cl_2/hv} C_6H_5CHCl_2$ $\xrightarrow{H_2O/H^+, 373 \ K} C_6H_5CHO$.

(C) In the Gatterman-Koch reaction,benzene or its derivatives are treated with carbon monoxide $(CO)$ and hydrogen chloride $(HCl)$ in the presence of anhydrous aluminum chloride $(AlCl_3)$ or cuprous chloride $(CuCl)$ under high pressure to form benzaldehyde or substituted benzaldehydes.
Thus,the reagents used are $CO$ and $HCl$.

(C) The given reaction is the reduction of an ester (methyl $4$-nitrobenzoate) to an aldehyde ($4$-nitrobenzaldehyde).
$DIBAL-H$ (Diisobutylaluminium hydride),represented as $AlH(i-Bu)_2$,is a selective reducing agent that reduces esters to aldehydes at low temperatures,followed by acidic workup $(H_3 O^{+})$.
Therefore,the correct reagent is $AlH(i-Bu)_2, H_3 O^{+}$.

(C) The reaction shown is the Rosenmund reduction,which is used to convert acid chlorides to aldehydes.
In this reaction,benzoyl chloride is hydrogenated using $H_2$ 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 correct reagent is $H_2 / Pd-BaSO_4$.

(D) The reaction of benzoyl chloride with hydrogen in the presence of palladium $(Pd)$ catalyst supported on barium sulphate $(BaSO_4)$ is known as the Rosenmund reduction.
In this reaction,the acid chloride is selectively reduced to an aldehyde.
The chemical equation is:
$C_6H_5COCl + H_2 \xrightarrow{Pd/BaSO_4} C_6H_5CHO + HCl$
Thus,the product formed is benzaldehyde.

(C) The Rosenmund reduction is a hydrogenation process where an acid chloride $(R-CO-Cl)$ is reduced to an aldehyde $(R-CHO)$ using hydrogen gas in the presence of a poisoned palladium catalyst,typically $Pd$ supported on $BaSO_4$. The reaction is: $R-CO-Cl + \text{H}_2 \xrightarrow{\text{Pd/BaSO}_4} R-CHO + \text{HCl}$. Therefore,option $C$ represents the Rosenmund reduction.

(A) The conversion of a nitrile $(-CN)$ to an aldehyde $(-CHO)$ while preserving other functional groups like a double bond is achieved using Diisobutylaluminium hydride ($DIBAL-H$ or $AlH(i-Bu)_2$),followed by acid hydrolysis $(H_3O^{+})$.
$DIBAL-H$ is a selective reducing agent that reduces nitriles to imines,which upon hydrolysis yield aldehydes.
Therefore,the correct reagent is $AlH(i-Bu)_2 / H_3O^{+}$.

(D) The boiling point of aldehydes and ketones depends on the molecular mass and the strength of intermolecular forces.
As the carbon chain length increases,the molecular mass increases,which leads to an increase in the magnitude of van der Waals forces of attraction.
Among the given options,$Hexanal$ $(C_6H_{12}O)$ has the longest carbon chain and the highest molecular mass compared to $Propanal$ $(C_3H_6O)$,$Ethanal$ $(C_2H_4O)$,and $Pentanal$ $(C_5H_{10}O)$.
Therefore,$Hexanal$ has the highest boiling point.

(C) The reaction is an Aldol condensation followed by dehydration.
Step $1$: Two molecules of acetaldehyde $(CH_3CHO)$ react in the presence of dilute $NaOH$ to form $3-$hydroxybutanal $(A)$,which is $CH_3CH(OH)CH_2CHO$.
Step $2$: Upon heating with acid or base,$3-$hydroxybutanal undergoes dehydration (loss of $H_2O$) to form an $\alpha,\beta-$unsaturated aldehyde,which is but$-2-$enal $(CH_3CH=CHCHO)$,also known as crotonaldehyde.
Therefore,the product '$B$' is but$-2-$enal.

(A) The reactivity of aldehydes towards nucleophilic addition reactions depends on two main factors: steric hindrance and electronic effects.
$1$. Steric hindrance: As the size of the alkyl group attached to the carbonyl carbon increases,the attack of the nucleophile becomes more difficult.
$2$. Electronic effects: Electron-donating groups ($+I$ or $+M$ effect) decrease the electrophilicity of the carbonyl carbon,thereby reducing reactivity.
In $C_6H_5CHO$ (Benzaldehyde),the phenyl group is bulky (high steric hindrance) and also exerts a $+M$ effect (resonance),which stabilizes the carbonyl carbon and reduces its electrophilicity.
Therefore,the decreasing order of reactivity is: $HCHO > CH_3CHO > CH_3CH_2CH_2CHO > C_6H_5CHO$.
Thus,$C_6H_5CHO$ is the least reactive.

(C) Butyraldehyde $(CH_3CH_2CH_2CHO)$ is known for its characteristic buttery or rancid butter-like odour.

(C) The reaction of propanone with $\text{Ba(OH)}_2$ is an aldol condensation reaction.
Two molecules of propanone undergo self-aldol condensation to form $4-$hydroxy$-4-$methylpentan$-2-$one (product $A$).
Upon heating $(A)$,it undergoes dehydration (loss of water) to form an $\alpha,\beta-$unsaturated ketone,which is $4-$methylpent$-3-$en$-2-$one (product $B$).

(C) The reaction of dimethylamine $(CH_3)_2NH$ with methyl iodide $(CH_3I)$ proceeds via nucleophilic substitution (Hofmann alkylation).
Step $1$: $(CH_3)_2NH + CH_3I \rightarrow (CH_3)_3N + HI$
Step $2$: $(CH_3)_3N + CH_3I \rightarrow (CH_3)_4N^{+}I^{-}$
Thus,a total of $2$ molecules of methyl iodide are required to convert dimethylamine into tetramethylammonium iodide.

(C) Gabriel phthalimide synthesis is used for the preparation of primary aliphatic amines.
It cannot be used for the preparation of aromatic amines because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide due to the partial double bond character of the $C-X$ bond and the instability of the phenyl cation.

(B) The boiling point of compounds depends on intermolecular forces such as hydrogen bonding and molar mass.
$1$. Primary amines $(n-C_4H_9NH_2)$ and secondary amines $((C_2H_5)_2NH)$ can form intermolecular hydrogen bonds,leading to higher boiling points.
$2$. Tertiary amines $(C_2H_5N(CH_3)_2)$ cannot form intermolecular hydrogen bonds because they lack a hydrogen atom attached to the nitrogen.
$3$. Alkanes $(C_2H_5CH(CH_3)_2)$ have only weak van der Waals forces.
$4$. Comparing $C_2H_5N(CH_3)_2$ (molar mass $\approx 73 \ g/mol$) and $C_2H_5CH(CH_3)_2$ (molar mass $\approx 72 \ g/mol$),the tertiary amine has a higher boiling point due to the dipole-dipole interaction of the $C-N$ bond compared to the non-polar alkane.
$5$. Therefore,$C_2H_5CH(CH_3)_2$ has the lowest boiling point.

(B) The reduction of acetamide $(CH_3CONH_2)$ to ethylamine $(CH_3CH_2NH_2)$ using a strong reducing agent like $LiAlH_4$ or $Na/C_2H_5OH$ is represented by the following chemical equation:
$CH_3CONH_2 + 4[H] \xrightarrow{LiAlH_4/ether} CH_3CH_2NH_2 + H_2O$
From the stoichiometry of the reaction,it is clear that $4$ moles of nascent hydrogen $([H])$ are required to reduce $1$ mole of acetamide to $1$ mole of ethylamine.
Therefore,the correct option is $B$.

(B) trisaccharide is a carbohydrate that yields three monosaccharide units upon hydrolysis.
$Raffinose$ is a trisaccharide composed of galactose,glucose,and fructose units.

(D) In a nucleoside,the nitrogenous base is attached to the $C-1$ position of the ribose sugar via a $\beta$-glycosidic linkage.

(D) Raffinose is a trisaccharide with the molecular formula $C_{18}H_{32}O_{16}$. Upon hydrolysis,it yields one molecule of each of the following monosaccharides: $D$-Glucose,$D$-Fructose,and $D$-Galactose.

(B) An aldohexose is a monosaccharide containing six carbon atoms and an aldehyde group $(-CHO)$.
Glucose has the molecular formula $C_6H_{12}O_6$ and contains an aldehyde group at the $C-1$ position,making it an aldohexose.
Ribose is an aldopentose $(C_5H_{10}O_5)$.
Fructose is a ketohexose $(C_6H_{12}O_6)$ containing a ketone group.
Threose is an aldotetrose $(C_4H_8O_4)$.

(C) Glucose $(CHO-(CHOH)_4-CH_2OH)$ and gluconic acid $(COOH-(CHOH)_4-CH_2OH)$ both undergo oxidation with dilute $HNO_3$ to form saccharic acid $(COOH-(CHOH)_4-COOH)$.
In this reaction,the terminal primary alcoholic group $(-CH_2OH)$ is oxidized to a carboxylic acid group $(-COOH)$.
Since only one $-CH_2OH$ group is present in the glucose molecule and it gets converted to $-COOH$,this reaction confirms the presence of one primary alcoholic group in glucose.

(A) Starch is a polysaccharide that acts as the main storage carbohydrate in plants.
Upon complete hydrolysis,starch breaks down into its monomeric units.
The monomeric unit of starch is $\alpha-D-glucose$.

(A) Glucose $(CHO(CHOH)_4CH_2OH)$ on oxidation with strong oxidizing agents like dilute nitric acid $(HNO_3)$ gets oxidized to a dicarboxylic acid known as saccharic acid (or glucaric acid).
The reaction is:
$CHO(CHOH)_4CH_2OH + [O] \xrightarrow{\text{Dil. } HNO_3} COOH(CHOH)_4COOH$
Thus,dilute $HNO_3$ is the correct reagent.

(B) $Glucose$ $(xylose)$ $isomerase$ catalyzes the reversible isomerization of $glucose$ to $fructose$ and that of $xylose$ to $xylulose$.
It is an important enzyme used in the industrial production of high-$fructose$ corn syrup.

(A) When glucose is heated with concentrated $HI$ (hydroiodic acid) for a long time,it undergoes reduction to form $n$-hexane.
This reaction confirms the presence of a straight chain of six carbon atoms in the glucose molecule.
The reaction is as follows:
$CHO-(CHOH)_4-CH_2OH \xrightarrow{HI, \Delta} CH_3-(CH_2)_4-CH_3$ (n-hexane).

(A) When glucose is heated with $HI$ (hydroiodic acid) and red phosphorus,it undergoes reduction to form $n$-hexane $(CH_3(CH_2)_4CH_3)$.
This reaction confirms that the six carbon atoms in glucose are linked in a straight chain.

(C) When glucose is treated with dilute nitric acid $(HNO_3)$,both the aldehyde group $(-CHO)$ and the primary alcoholic group $(-CH_2OH)$ are oxidized to carboxylic acid groups $(-COOH)$.
This reaction results in the formation of a dicarboxylic acid known as saccharic acid (also called glucaric acid).
The reaction is represented as:
$CHO-(CHOH)_4-CH_2OH + [O] \xrightarrow{Dil. HNO_3} COOH-(CHOH)_4-COOH$
Thus,the correct product is saccharic acid.

(B) The synthesis of adipic acid from renewable resources was developed by $Drath$ and $Frost$ using an engineered strain of $Escherichia \text{ } coli$. The process utilizes $Glucose$ as the primary carbon source to produce adipic acid through a microbial fermentation pathway.

(C) Lactose is a disaccharide composed of $\beta-D-galactose$ and $\beta-D-glucose$.
The glycosidic linkage is formed between the $C-1$ of $\beta-D-galactose$ and the $C-4$ of $\beta-D-glucose$.
Therefore,the linkage is a $\beta-1,4-glycosidic$ linkage.

(C) Lactose is a disaccharide sugar composed of galactose and glucose units. It is commonly known as milk sugar because it is found in the milk of mammals.