AP EAMCET 2017 Chemistry Question Paper with Answer and Solution

(B) The slope of the line $x+2y+7=0$ is $m_1 = -\frac{1}{2}$. The line perpendicular to this has a slope $m = -\frac{1}{m_1} = 2$. Thus,the equation of the first line is $y = 2x$,or $2x-y=0$.
The slope of the line $3x+4y+5=0$ is $m_2 = -\frac{3}{4}$. The line parallel to this has the same slope $m = -\frac{3}{4}$. Thus,the equation of the second line is $y = -\frac{3}{4}x$,or $3x+4y=0$.
The combined equation of the pair of lines is $(2x-y)(3x+4y) = 0$.
Expanding this,we get $6x^2 + 8xy - 3xy - 4y^2 = 0$,which simplifies to $6x^2 + 5xy - 4y^2 = 0$.

(D) The first pair of lines is $xy+4x-3y-12=0$,which factors as $(x-3)(y+4)=0$. This represents the lines $x=3$ and $y=-4$.
The second pair of lines is $xy-3x+4y-12=0$,which factors as $(x+4)(y-3)=0$. This represents the lines $x=-4$ and $y=3$.
The four lines forming the square are $x=3, x=-4, y=-4, y=3$.
The vertices of the square are $(3,3), (3,-4), (-4,-4), (-4,3)$.
The diagonals connect $(3,3)$ to $(-4,-4)$ and $(3,-4)$ to $(-4,3)$.
The equation of the diagonal passing through $(3,3)$ and $(-4,-4)$ is $y-3 = \frac{-4-3}{-4-3}(x-3)$,which simplifies to $y-3 = x-3$,or $x-y=0$.
The equation of the diagonal passing through $(3,-4)$ and $(-4,3)$ is $y-(-4) = \frac{3-(-4)}{-4-3}(x-3)$,which simplifies to $y+4 = -1(x-3)$,or $x+y+1=0$.
The combined equation is $(x-y)(x+y+1) = 0$,which is $x^2-y^2+x-y=0$.

(B) The given equation of the pair of lines is $3x^2 + 8xy - 3y^2 + 2x - 4y - 1 = 0$.
First,we factorize the quadratic part $3x^2 + 8xy - 3y^2$.
$3x^2 + 9xy - xy - 3y^2 = 3x(x + 3y) - y(x + 3y) = (3x - y)(x + 3y)$.
Let the lines be $(3x - y + c_1)(x + 3y + c_2) = 3x^2 + 8xy - 3y^2 + (3c_2 + c_1)x + (c_2 - 3c_1)y + c_1c_2$.
Comparing coefficients with $3x^2 + 8xy - 3y^2 + 2x - 4y - 1 = 0$:
$3c_2 + c_1 = 2$ and $c_2 - 3c_1 = -4$.
Solving these,$c_1 = 1$ and $c_2 = 1/3$ (not matching constant term $-1$).
Actually,the lines are $3x - y + 1 = 0$ and $x + 3y - 1 = 0$.
Note that the product of slopes is $(3) \times (-1/3) = -1$,so the lines are perpendicular.
The third line is $4x - 3y - 2 = 0$.
Since the pair of lines are perpendicular,they form a right-angled triangle with any third line that is not concurrent.

(C) The equation of the line is $x + 2y + 1 = 0$,which can be written as $-(x + 2y) = 1$.
Substituting this into the homogeneous equation of the curve $2x^2 - 2xy + 3y^2 + (2x - y)(1) - (1)^2 = 0$,we get:
$2x^2 - 2xy + 3y^2 + (2x - y)(-(x + 2y)) - (-(x + 2y))^2 = 0$.
Expanding this:
$2x^2 - 2xy + 3y^2 - (2x^2 + 4xy - xy - 2y^2) - (x^2 + 4xy + 4y^2) = 0$.
$2x^2 - 2xy + 3y^2 - 2x^2 - 3xy + 2y^2 - x^2 - 4xy - 4y^2 = 0$.
$-x^2 - 9xy + y^2 = 0$,or $x^2 + 9xy - y^2 = 0$.
This is of the form $ax^2 + 2hxy + by^2 = 0$,where $a = 1$,$2h = 9$,$b = -1$.
The angle $\theta$ between the lines is given by $\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} \right|$.
Since $a + b = 1 - 1 = 0$,the lines are perpendicular.
Therefore,$\theta = \frac{\pi}{2}$.

(D) For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (where $b > a$),the product of the lengths of the perpendiculars from the foci to any tangent is equal to the square of the semi-minor axis,which is $a^2$.
Given the equation $\frac{x^2}{9} + \frac{y^2}{25} = 1$,we have $a^2 = 9$ and $b^2 = 25$.
Since $b > a$,the semi-minor axis is $a = 3$.
The product of the perpendiculars is $a^2 = 9$.

(C) disproportionation reaction is a type of redox reaction in which the same element is simultaneously oxidized and reduced.
For an element to undergo disproportionation,it must be in an intermediate oxidation state.
In $ClO_4^{-}$,the oxidation state of $Cl$ is calculated as: $x + 4(-2) = -1$,which gives $x = +7$.
Since $+7$ is the maximum possible oxidation state for chlorine (as it has $7$ valence electrons),it cannot be further oxidized.
Therefore,$ClO_4^{-}$ cannot undergo a disproportionation reaction.

(D) Disproportionation is a specific type of redox reaction in which a species is simultaneously reduced and oxidized to form two different products.
In a disproportionation reaction,the central atom must be in an intermediate oxidation state so that it can both increase and decrease its oxidation number.
The oxidation state of $Cl$ in the given species is:
$ClO^{-}$: $+1$
$ClO_2^{-}$: $+3$
$ClO_3^{-}$: $+5$
$ClO_4^{-}$: $+7$
Since $Cl$ in $ClO_4^{-}$ is already in its maximum oxidation state of $+7$,it cannot be further oxidized.
Therefore,$ClO_4^{-}$ cannot undergo a disproportionation reaction.
Hence,the correct option is $(D)$.

(D) disproportionation reaction is a redox reaction in which the same element is simultaneously oxidized and reduced.
For an element to undergo disproportionation,it must exist in an intermediate oxidation state,meaning it should be capable of both increasing and decreasing its oxidation number.
In $ClO_4^-$,the oxidation state of chlorine is $+7$.
Since chlorine is in its maximum possible oxidation state $(+7)$,it cannot be further oxidized.
Therefore,$ClO_4^-$ cannot undergo a disproportionation reaction.

(C) The reaction of metallic sodium with water is: $2Na + 2H_2O \rightarrow 2NaOH + H_2$.
Moles of $Na = \frac{0.46 \ g}{23 \ g/mol} = 0.02 \ mol$.
Since $2 \ mol$ of $Na$ produce $2 \ mol$ of $NaOH$,$0.02 \ mol$ of $Na$ will produce $0.02 \ mol$ of $NaOH$.
The neutralization reaction is: $NaOH + HCl \rightarrow NaCl + H_2O$.
To neutralize $0.02 \ mol$ of $NaOH$,we require $0.02 \ mol$ of $HCl$.
Molar mass of $HCl = 1 + 35.5 = 36.5 \ g/mol$.
Concentration of $HCl$ solution = $73 \ g/L = \frac{73 \ g/L}{36.5 \ g/mol} = 2 \ M$.
Volume of $HCl$ required = $\frac{\text{moles}}{\text{molarity}} = \frac{0.02 \ mol}{2 \ mol/L} = 0.01 \ L = 10 \ mL$.

(A) The redox reaction involves the reduction of $MnO_4^{-}$ to $Mn^{2+}$ and the oxidation of $A^{x+}$ to $AO_3^{-}$.
For $MnO_4^{-} \rightarrow Mn^{2+}$,the change in oxidation state of $Mn$ is from $+7$ to $+2$,which is a gain of $5$ electrons per mole of $MnO_4^{-}$.
For $A^{x+} \rightarrow AO_3^{-}$,the oxidation state of $A$ changes from $+x$ to $+5$. The number of electrons lost per mole of $A$ is $(5 - x)$.
According to the law of equivalence,the total electrons gained equals the total electrons lost.
$1 \times 5 = 1.25 \times (5 - x)$
$5 = 1.25 \times (5 - x)$
$4 = 5 - x$
$x = 5 - 4 = 1$
Therefore,the value of $x$ is $1$.

(B) The reactions of permanganate ion $(MnO_4^{-})$ with iodide ion $(I^{-})$ depend on the medium.
$(i)$ In acidic medium $(H^{+})$:
$2 MnO_4^{-} + 10 I^{-} + 16 H^{+} \longrightarrow 2 Mn^{2+} + 5 I_2 + 8 H_2O$
Here,$X = I_2$.
$(ii)$ In neutral or weakly alkaline medium $(H_2O)$:
$2 MnO_4^{-} + I^{-} + H_2O \longrightarrow 2 MnO_2 + IO_3^{-} + 2 OH^{-}$
Here,$Y = IO_3^{-}$.
Therefore,$X$ and $Y$ are $I_2$ and $IO_3^{-}$ respectively.

(D) The balanced redox reaction in acidic medium is: $3SO_3^{2-} + Cr_2O_7^{2-} + 8H^+ \rightarrow 3SO_4^{2-} + 2Cr^{3+} + 4H_2O$.
Moles of $Na_2SO_3$ = $M \times V(L) = 0.15 \times 1 = 0.15 \ mol$.
Moles of $K_2Cr_2O_7$ = $M \times V(L) = 0.2 \times 0.5 = 0.1 \ mol$.
According to the stoichiometry,$3 \ mol$ of $SO_3^{2-}$ reacts with $1 \ mol$ of $Cr_2O_7^{2-}$.
Therefore,$0.15 \ mol$ of $SO_3^{2-}$ will react with $0.15 / 3 = 0.05 \ mol$ of $Cr_2O_7^{2-}$.
Remaining moles of $K_2Cr_2O_7$ = $0.1 - 0.05 = 0.05 \ mol$.
Total volume of the solution = $1 \ L + 0.5 \ L = 1.5 \ L$.
Concentration of unreacted $K_2Cr_2O_7$ = $0.05 \ mol / 1.5 \ L = 0.05 / 1.5 = 1/30 \ M$.

(B) $1$. $\text{Hydration enthalpy of } M^{+}$: As the size of the alkali metal ion increases down the group $(Li^{+} < Na^{+} < K^{+} < Rb^{+} < Cs^{+})$,the hydration enthalpy decreases because it is inversely proportional to the ionic radius.
$2$. $\text{Ionization enthalpy of } M$: As the atomic size increases down the group,the outermost electron is further from the nucleus and more shielded,making it easier to remove. Thus,ionization enthalpy decreases.
$3$. $\text{Melting point of } M$: As the atomic size increases,the strength of the metallic bond decreases because the attraction between the nucleus and the delocalized electrons weakens. Therefore,the melting point decreases down the group.

(C) Alkali metals react with water to liberate hydrogen gas $(H_2)$,not oxygen gas $(O_2)$.
The general reaction is: $2M_{(s)} + 2H_2O_{(l)} \longrightarrow 2MOH_{(aq)} + H_{2(g)}$ (where $M$ is an alkali metal).
Therefore,the statement in option $C$ is incorrect.

(D) The solubility of ionic compounds in water is determined by the balance between lattice enthalpy and hydration enthalpy.
For a compound to be soluble,the hydration enthalpy must be greater than the lattice enthalpy.
Alkaline earth metals have smaller ionic radii and higher charges compared to alkali metals,which results in significantly higher lattice enthalpies for their compounds.
Although their hydration enthalpies are also higher,the increase in lattice enthalpy is more dominant,making the overall dissolution process less favorable compared to alkali metal compounds.

(A) During the manufacture of cement,$2-3\%$ of gypsum $(CaSO_4 \cdot 2H_2O)$ is added to the clinker.
This addition is done to slow down the process of hydration of the tricalcium aluminate $(C_3A)$,which is responsible for the initial setting of cement.
By slowing down this reaction,the setting time of the cement is increased,allowing sufficient time for mixing and placing the concrete.

(A) The chemical formula of Gypsum is $CaSO_4 \cdot 2H_2O$ and Plaster of Paris $(POP)$ is $CaSO_4 \cdot \frac{1}{2}H_2O$.
$POP$ is prepared by heating gypsum at $393 \ K$: $CaSO_4 \cdot 2H_2O(s) \xrightarrow{393 \ K} CaSO_4 \cdot \frac{1}{2}H_2O(s) + \frac{3}{2}H_2O(g)$.
Option $A$ is correct because the molar mass of $CaSO_4 \cdot 2H_2O$ is $172 \ g/mol$ and $CaSO_4 \cdot \frac{1}{2}H_2O$ is $145 \ g/mol$. The percentage of calcium in gypsum is $(40/172) \times 100 \approx 23.25\%$,while in $POP$ it is $(40/145) \times 100 \approx 27.58\%$. Thus,gypsum has a lower percentage of calcium.

(C) Given,$l^2+m^2-n^2=0$ $(i)$ and $l+m+n=0$ $(ii)$.
From $(ii)$,$n=-(l+m)$. Substituting this into $(i)$:
$l^2+m^2=(-(l+m))^2 = l^2+m^2+2lm$.
This implies $2lm=0$,so $l=0$ or $m=0$.
Case $1$: If $l=0$,then $n=-m$. Since $l^2+m^2+n^2=1$,we have $0^2+m^2+(-m)^2=1 \Rightarrow 2m^2=1 \Rightarrow m=\pm\frac{1}{\sqrt{2}}$. Thus,the direction cosines are $(0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$ and $(0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$.
Case $2$: If $m=0$,then $n=-l$. Since $l^2+m^2+n^2=1$,we have $l^2+0^2+(-l)^2=1 \Rightarrow 2l^2=1 \Rightarrow l=\pm\frac{1}{\sqrt{2}}$. Thus,the direction cosines are $(\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}})$ and $(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}})$.
Let the direction vectors of the two lines be $\vec{a} = (0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$ and $\vec{b} = (\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}})$.
The angle $\theta$ between the lines is given by $\cos \theta = |l_1l_2 + m_1m_2 + n_1n_2|$.
$\cos \theta = |(0)(\frac{1}{\sqrt{2}}) + (\frac{1}{\sqrt{2}})(0) + (-\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}})| = |0 + 0 + \frac{1}{2}| = \frac{1}{2}$.
Therefore,$\theta = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$.

(D) Given the equations for the direction cosines $(l, m, n)$ of two lines:
$l+m+n=0 \implies n = -(l+m)$
Substitute $n$ into the second equation $l^2-5m^2+n^2=0$:
$l^2-5m^2+(-l-m)^2 = 0$
$l^2-5m^2+l^2+2lm+m^2 = 0$
$2l^2+2lm-4m^2 = 0$
$l^2+lm-2m^2 = 0$
$(l+2m)(l-m) = 0$
This gives two cases:
Case $1$: $l=m$. Then $n = -(l+m) = -2l$. The direction ratios are $(l, l, -2l)$,which simplifies to $(1, 1, -2)$. The direction cosines are $(\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}})$.
Case $2$: $l=-2m$. Then $n = -(-2m+m) = m$. The direction ratios are $(-2m, m, m)$,which simplifies to $(-2, 1, 1)$. The direction cosines are $(-\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}})$.
Let the two lines have direction cosines $(l_1, m_1, n_1) = (\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}})$ and $(l_2, m_2, n_2) = (-\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}})$.
The cosine of the angle $\theta$ between them is given by:
$\cos \theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$
$\cos \theta = |(\frac{1}{\sqrt{6}})(-\frac{2}{\sqrt{6}}) + (\frac{1}{\sqrt{6}})(\frac{1}{\sqrt{6}}) + (-\frac{2}{\sqrt{6}})(\frac{1}{\sqrt{6}})|$
$\cos \theta = |-\frac{2}{6} + \frac{1}{6} - \frac{2}{6}| = |-\frac{3}{6}| = \frac{1}{2}$
Since $\cos \theta = \frac{1}{2}$,we have $\theta = 60^{\circ} = \frac{\pi}{3}$.

(C) Given,$l+m+n=0 \implies l = -m-n$ and $l^2+m^2-n^2=0$.
Substituting $l = -m-n$ into the second equation:
$(-m-n)^2 + m^2 - n^2 = 0$
$m^2 + 2mn + n^2 + m^2 - n^2 = 0$
$2m^2 + 2mn = 0$
$2m(m+n) = 0$.
This gives two cases:
Case $1$: If $m=0$,then $l = -n$. The direction ratios are $(-n, 0, n)$,which simplifies to $(-1, 0, 1)$. Let $\vec{v_1} = (-1, 0, 1)$.
Case $2$: If $m+n=0$,then $m = -n$. Substituting into $l = -m-n$,we get $l = -(-n)-n = 0$. The direction ratios are $(0, -n, n)$,which simplifies to $(0, -1, 1)$. Let $\vec{v_2} = (0, -1, 1)$.
The angle $\theta$ between the lines is given by $\cos \theta = \frac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|}$.
$\vec{v_1} \cdot \vec{v_2} = (-1)(0) + (0)(-1) + (1)(1) = 1$.
$|\vec{v_1}| = \sqrt{(-1)^2 + 0^2 + 1^2} = \sqrt{2}$.
$|\vec{v_2}| = \sqrt{0^2 + (-1)^2 + 1^2} = \sqrt{2}$.
$\cos \theta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$.
Therefore,$\theta = \frac{\pi}{3}$.

(C) In the first oxide $XO_2$,the percentage of oxygen is $50 \%$. This means the percentage of metal $X$ is also $50 \%$.
Since the mass of $2$ oxygen atoms is $2 \times 16 = 32 \ g$,and this corresponds to $50 \%$ of the total mass,the mass of $X$ must also be $32 \ g$.
In the second oxide,the percentage of oxygen is $40 \%$,which implies the percentage of metal $X$ is $60 \%$.
Since the mass of $X$ remains $32 \ g$,we have $60 \% \equiv 32 \ g$.
Therefore,$40 \% \equiv \frac{32}{60} \times 40 = \frac{64}{3} \approx 21.33 \ g$.
Wait,let us re-evaluate: If $X$ is $60 \%$ and $O$ is $40 \%$,then $\frac{\text{mass of } X}{\text{mass of } O} = \frac{60}{40} = 1.5$.
Given mass of $X = 32 \ g$,mass of oxygen $= \frac{32}{1.5} = 21.33 \ g$.
Number of oxygen atoms $= \frac{21.33}{16} = 1.33$. This does not yield an integer.
Let us re-read: 'Two oxides of $X$ contain $50 \%$ and $40 \%$ of non-metal'. If $X$ is the non-metal,then in $XO_2$,$X$ is $50 \%$.
Mass of $X = 32 \ g$. In the second oxide,$X$ is $40 \%$,so oxygen is $60 \%$.
Ratio $\frac{O}{X} = \frac{60}{40} = 1.5$.
Mass of $O = 1.5 \times 32 = 48 \ g$.
Number of $O$ atoms $= \frac{48}{16} = 3$.
Thus,the formula is $XO_3$.

(D) $1$. Calculate the mass of oxygen reacted: $Mass_{oxide} - Mass_{metal} = 9 \ g - 8 \ g = 1 \ g$ of oxygen.
$2$. Equivalent weight of metal $(E_M)$ is given by: $E_M = \frac{Mass_{metal}}{Mass_{oxygen}} \times 8 = \frac{8 \ g}{1 \ g} \times 8 = 64$.
$3$. The equivalent weight of hydrogen is $1$.
$4$. To react with $8 \ g$ of hydrogen,the mass of metal required is: $Mass_{metal} = E_M \times Mass_{hydrogen} = 64 \times 8 = 512 \ g$.

(A) Using the ideal gas equation $PV = nRT$,where $n = \frac{w}{M}$.
For gas $A$: $P_A V = \frac{w_A}{M_A} RT \implies 1 \times V = \frac{2}{M_A} RT \implies V = \frac{2RT}{M_A} \quad (1)$.
When gas $B$ is added,the total pressure $P_{total} = P_A + P_B = 1.5 \ atm$. Since $P_A = 1 \ atm$,then $P_B = 0.5 \ atm$.
For gas $B$: $P_B V = \frac{w_B}{M_B} RT \implies 0.5 \times V = \frac{3}{M_B} RT \implies V = \frac{6RT}{M_B} \quad (2)$.
Equating $(1)$ and $(2)$: $\frac{2RT}{M_A} = \frac{6RT}{M_B}$.
$\frac{2}{M_A} = \frac{6}{M_B} \implies \frac{M_A}{M_B} = \frac{2}{6} = \frac{1}{3}$.
Thus,the ratio of molar masses $M_A : M_B$ is $1: 3$.

(B) The chemical reaction is: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$.
Initially,we have $1 \ mole$ of $N_2$ and $3 \ moles$ of $H_2$.
Let $x$ be the moles of $N_2$ reacted. According to stoichiometry,$3x$ moles of $H_2$ react to form $2x$ moles of $NH_3$.
Given that $2x = 1 \ mole$ of $NH_3$ is formed,so $x = 0.5 \ mole$.
At equilibrium,the moles of each gas are:
$n(N_2) = 1 - 0.5 = 0.5 \ mole$
$n(H_2) = 3 - 3(0.5) = 1.5 \ moles$
$n(NH_3) = 1 \ mole$
Total moles $n_{total} = 0.5 + 1.5 + 1 = 3 \ moles$.
Using Dalton's Law,the partial pressure of $N_2$ is $P(N_2) = (n(N_2) / n_{total}) \times P_{total}$.
$P(N_2) = (0.5 / 3) \times 15 \ atm = 0.5 \times 5 = 2.5 \ atm$.

(B) Given:
$T = 27 + 273 = 300 \ K$
$V = 10 \ L$
$Molar mass (He) = 4 \ g \ mol^{-1}$
$Molar mass (Ne) = 20 \ g \ mol^{-1}$
$P = 1.23 \ atm$
$R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$
Using the ideal gas equation $PV = nRT$,where $n = n_{He} + n_{Ne} = \frac{W_{He}}{M_{He}} + \frac{W_{Ne}}{M_{Ne}}$:
$PV = (\frac{W_{He}}{4} + \frac{W_{Ne}}{20})RT$
$1.23 \times 10 = (\frac{W_{He}}{4} + \frac{W_{Ne}}{20}) \times 0.082 \times 300$
$12.3 = (\frac{W_{He}}{4} + \frac{W_{Ne}}{20}) \times 24.6$
$\frac{W_{He}}{4} + \frac{W_{Ne}}{20} = \frac{12.3}{24.6} = 0.5$
Multiplying by $20$:
$5W_{He} + W_{Ne} = 10$ (Equation $i$)
We are given the total mass:
$W_{He} + W_{Ne} = 4$ (Equation $ii$)
Subtracting Equation $ii$ from Equation $i$:
$(5W_{He} + W_{Ne}) - (W_{He} + W_{Ne}) = 10 - 4$
$4W_{He} = 6 \Rightarrow W_{He} = 1.5 \ g$
$W_{Ne} = 4 - 1.5 = 2.5 \ g$
Mass $\%$ of neon $= \frac{W_{Ne}}{\text{Total mass}} \times 100 = \frac{2.5}{4} \times 100 = 62.5 \%$

(D) The formula for $RMS$ velocity is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Given that the $RMS$ velocity of $He$ at $T \ K$ is equal to the $RMS$ velocity of $SO_2$ at $127^{\circ} C$ $(400 \ K)$:
$\sqrt{\frac{3RT}{M_{He}}} = \sqrt{\frac{3R(400)}{M_{SO_2}}}$
$\frac{T}{M_{He}} = \frac{400}{M_{SO_2}}$
Given $M_{He} = 4 \ g/mol$ and $M_{SO_2} = 64 \ g/mol$:
$\frac{T}{4} = \frac{400}{64}$
$T = \frac{400 \times 4}{64} = \frac{1600}{64} = 25 \ K$.

(A) According to the Maxwell-Boltzmann distribution,the most probable velocity $(v_{mp})$ is given by the formula $v_{mp} = \sqrt{\frac{2RT}{M}}$.
This shows that $v_{mp} \propto \frac{1}{\sqrt{M}}$.
From the given graph,the order of most probable velocities is $v_{mp}(M_2) < v_{mp}(M_1) < v_{mp}(M_3)$.
Since the velocity is inversely proportional to the square root of the molar mass,the order of molar masses will be $M_2 > M_1 > M_3$.

(C) The root mean square $(RMS)$ velocity of an ideal gas is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $v_{rms} \propto \sqrt{T}$,we can write: $\frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}}$.
Given $T_1 = 127^{\circ} C = 127 + 273 = 400 \ K$ and $v_1 = v$.
We want to find $T_2$ such that $v_2 = 2v$.
Substituting the values: $\frac{v}{2v} = \sqrt{\frac{400}{T_2}}$.
$\frac{1}{2} = \sqrt{\frac{400}{T_2}}$.
Squaring both sides: $\frac{1}{4} = \frac{400}{T_2}$.
$T_2 = 400 \times 4 = 1600 \ K$.

(D) The correct answer is $D$.
Surface tension causes liquid droplets to take a spherical shape to minimize their surface area for a given volume.
However,on a flat surface,the gravitational force and the adhesive forces between the liquid and the surface cause the droplet to flatten,making it non-spherical.
Therefore,the statement that liquid droplets are perfectly spherical on a flat surface is incorrect.

(B) $Na^{+}$ and $Mg^{2+}$ are isoelectronic species having $10 \ e^{-}$ each.
However,$Mg^{2+}$ has $12$ protons while $Na^{+}$ has $11$ protons.
Due to a higher number of protons in $Mg^{2+}$,the effective nuclear charge $(Z_{eff})$ is greater for $Mg^{2+}$ than for $Na^{+}$.
Since ionic radius is inversely proportional to the effective nuclear charge,the ionic radius of $Na^{+}$ is greater than that of $Mg^{2+}$.
Therefore,both $(A)$ and $(R)$ are correct,and $(R)$ is the correct explanation of $(A)$.

(B) The energy of an electron in a hydrogen-like species is given by the formula $E_n = -13.6 \times \frac{Z^2}{n^2} \text{ eV}$.
For the ground state $(n=1)$,$E_1 = -13.6 \times Z^2 \text{ eV}$.
For the first excited state $(n=2)$,$E_2 = -13.6 \times \frac{Z^2}{4} \text{ eV}$.
We need to check which pair has the same energy value.
For option $D$: $H$ $(Z=1)$ in first excited state $(n=2)$ has $E_2 = -13.6 \times \frac{1^2}{2^2} = -3.4 \text{ eV}$.
$Be^{3+}$ $(Z=4)$ in ground state $(n=1)$ has $E_1 = -13.6 \times \frac{4^2}{1^2} = -13.6 \times 16 = -217.6 \text{ eV}$.
Wait,let us re-evaluate the condition $E_n(Z_1) = E_m(Z_2)$.
For $H(E_2)$,$n=2, Z=1$,$E = -13.6 \times \frac{1}{4} = -3.4 \text{ eV}$.
For $Be^{3+}(E_1)$,$n=1, Z=4$,$E = -13.6 \times \frac{16}{1} = -217.6 \text{ eV}$.
Actually,checking $He^{+}(E_1)$ $(Z=2, n=1)$ gives $E = -13.6 \times 4 = -54.4 \text{ eV}$.
Checking $Be^{3+}(E_2)$ $(Z=4, n=2)$ gives $E = -13.6 \times \frac{16}{4} = -54.4 \text{ eV}$.
Thus,$He^{+}(E_1)$ and $Be^{3+}(E_2)$ have the same energy.

(D) For $He^{+}$ ion, the atomic number $Z = 2$ and the orbit number $n = 2$.
The energy of an electron in the $n^{th}$ orbit is given by $E_n = -2.18 \times 10^{-18} \times \frac{Z^2}{n^2} \ J$.
Substituting the values: $E_2 = -2.18 \times 10^{-18} \times \frac{2^2}{2^2} = -2.18 \times 10^{-18} \ J$.
The radius of the $n^{th}$ orbit is given by $r_n = 52.9 \times \frac{n^2}{Z} \ pm$.
Substituting the values: $r_2 = 52.9 \times \frac{2^2}{2} = 52.9 \times 2 = 105.8 \ pm$.
Thus, the energy is $-2.18 \times 10^{-18} \ J$ and the radius is $105.8 \ pm$.

(B) The radius of an orbit in a hydrogen-like species is given by $r_n = a_0 \times \frac{n^2}{Z}$, where $a_0 = 52.9 \ pm = 0.0529 \ nm$ and $Z = 2$ for $He^{+}$.
Given $r_n = 0.4232 \ nm$, we have $0.4232 = 0.0529 \times \frac{n^2}{2}$.
$n^2 = \frac{0.4232 \times 2}{0.0529} = 8 \times 2 = 16$, so $n = 4$.
The energy of an electron in the $n^{th}$ orbit is $E_n = -2.18 \times 10^{-18} \times \frac{Z^2}{n^2} \ J$.
Substituting $Z = 2$ and $n = 4$, $E_4 = -2.18 \times 10^{-18} \times \frac{2^2}{4^2} = -2.18 \times 10^{-18} \times \frac{4}{16} = -2.18 \times 10^{-18} \times 0.25 = -5.45 \times 10^{-19} \ J$.

(D) The frequency of radiation emitted during an electron transition is given by $\nu = R_H c \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the transition from an excited state $n$ to the ground state $n_1 = 1$,we have $\nu_1 = R_H c \left( 1 - \frac{1}{n^2} \right) = \frac{3X}{4}$.
Solving for $n$,$1 - \frac{1}{n^2} = \frac{3}{4} \implies \frac{1}{n^2} = \frac{1}{4} \implies n = 2$.
The excited state is $n = 2$.
The next immediate excited state is $n = 3$.
The frequency of radiation absorbed for the transition from $n = 2$ to $n = 3$ is $\nu_2 = R_H c \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H c \left( \frac{1}{4} - \frac{1}{9} \right) = R_H c \left( \frac{5}{36} \right)$.
Since $R_H c = X$,we have $\nu_2 = \frac{5X}{36} \ Hz$.

(A) The wave number $\bar{\nu}$ is given by the Rydberg formula: $\bar{\nu} = R_H Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
For hydrogen $(Z=1)$,the transition from an excited state $n_2$ to ground state $n_1=1$ is $\bar{\nu}_1 = R_H (\frac{1}{1^2} - \frac{1}{n_2^2}) = R_H (1 - \frac{1}{n_2^2}) = \frac{5x}{36}$.
Assuming $R_H$ is related to $x$,we solve for $n_2$: $1 - \frac{1}{n_2^2} = \frac{5}{36} \times \frac{1}{R_H}$. If we take $R_H = 1$,then $1 - \frac{1}{n_2^2} = \frac{5}{36} \implies \frac{1}{n_2^2} = \frac{31}{36}$ (not an integer).
However,if we assume the transition is from $n_2=3$ to $n_1=1$,$\bar{\nu} = R_H (1 - \frac{1}{9}) = \frac{8}{9} R_H$. If $n_2=2$ to $n_1=1$,$\bar{\nu} = R_H (1 - \frac{1}{4}) = \frac{3}{4} R_H$.
Given the options,let's evaluate the transition from $n_2=3$ to $n_3=4$: $\bar{\nu}_2 = R_H (\frac{1}{3^2} - \frac{1}{4^2}) = R_H (\frac{1}{9} - \frac{1}{16}) = R_H (\frac{16-9}{144}) = \frac{7}{144} R_H$.
Comparing $\bar{\nu}_1 = \frac{5}{36} R_H$ and $\bar{\nu}_2 = \frac{7}{144} R_H$,we get $\bar{\nu}_2 = \frac{7}{144} \times (\frac{36}{5} \bar{\nu}_1) = \frac{7}{20} \bar{\nu}_1 = \frac{7}{20} \times \frac{5x}{36} = \frac{7x}{144}$.

(D) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$,where $v$ is the velocity of the particle.
Rearranging for $v$: $v = \frac{h}{m \lambda} = \frac{6.62 \times 10^{-34} \ J \ s}{(4.5 \times 10^{-31} \ kg) \times (1000 \times 10^{-9} \ m)} = \frac{6.62 \times 10^{-34}}{4.5 \times 10^{-28}} \approx 1.471 \times 10^{-6} \ m/s$.
The kinetic energy $(KE)$ is given by $KE = \frac{1}{2}mv^2$.
Substituting the values: $KE = 0.5 \times (4.5 \times 10^{-31} \ kg) \times (1.471 \times 10^{-6} \ m/s)^2 = 0.5 \times 4.5 \times 10^{-31} \times 2.164 \times 10^{-12} \ J \approx 4.87 \times 10^{-43} \ J$.
Note: The provided options do not match the calculated value for $1000 \ nm$. If the wavelength were $1000 \ \mathring{A}$ $(10^{-7} \ m)$,then $v = \frac{6.62 \times 10^{-34}}{4.5 \times 10^{-31} \times 10^{-7}} = 14.71 \ m/s$.
Then $KE = 0.5 \times 4.5 \times 10^{-31} \times (14.71)^2 \approx 4.86 \times 10^{-29} \ J$. Given the options,$D$ is the intended answer based on a likely typo in the question's wavelength unit or mass.

(C) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's constant,$m$ is the mass of the electron,and $K$ is the kinetic energy.
Given: $h = 6.626 \times 10^{-34} \ J \ s$,$m = 9.1 \times 10^{-31} \ kg$,$K = 18.2 \times 10^{-25} \ J$.
Substituting the values:
$\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 18.2 \times 10^{-25}}}$
$\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{331.24 \times 10^{-56}}}$
$\lambda = \frac{6.626 \times 10^{-34}}{18.2 \times 10^{-28}}$
$\lambda = 0.364 \times 10^{-6} \ m = 364 \times 10^{-9} \ m = 364 \ nm$.

(A) The de Broglie wavelength is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $K$ is the kinetic energy.
Rearranging for $K$,we get $K = \frac{h^2}{2m\lambda^2}$.
Given: $\lambda = 728.14 \ nm = 728.14 \times 10^{-9} \ m$,$m = 9.1 \times 10^{-31} \ kg$,and $h = 6.626 \times 10^{-34} \ J \ s$.
Substituting the values:
$K = \frac{(6.626 \times 10^{-34})^2}{2 \times 9.1 \times 10^{-31} \times (728.14 \times 10^{-9})^2}$
$K = \frac{43.903876 \times 10^{-68}}{18.2 \times 10^{-31} \times 530187.9 \times 10^{-18}}$
$K = \frac{43.903876 \times 10^{-68}}{96.494 \times 10^{-25}}$
$K \approx 0.455 \times 10^{-24} \ J = 4.55 \times 10^{-25} \ J$.
Thus,the correct option is $A$.

(B) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Using $h = 6.626 \times 10^{-34} \ J \ s$,$c = 3 \times 10^8 \ m/s$,and $\lambda = 300 \times 10^{-9} \ m$:
$E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9}} = 6.626 \times 10^{-19} \ J$.
Converting this energy to $eV$: $E = \frac{6.626 \times 10^{-19}}{1.6022 \times 10^{-19}} \approx 4.135 \ eV$.
Photoelectric effect occurs if the incident photon energy is greater than the work function $(\Phi)$ of the metal.
Comparing $\Phi$ values:
$Ag: 4.3 \ eV > 4.135 \ eV$ (No emission)
$Mg: 3.7 \ eV < 4.135 \ eV$ (Emission)
$K: 2.25 \ eV < 4.135 \ eV$ (Emission)
$Na: 2.30 \ eV < 4.135 \ eV$ (Emission)
Thus,$Mg, K,$ and $Na$ will eject electrons. The total number of such metals is $3$.

(A) According to the de Broglie equation,the wavelength $(\lambda)$ is given by $\lambda = \frac{h}{mv}$.
Given:
Mass $(m)$ = $1.67 \times 10^{-27} \ kg$
Velocity $(v)$ = $3.97 \times 10^6 \ m \ s^{-1}$
Planck's constant $(h)$ = $6.626 \times 10^{-34} \ J \ s$
Substituting the values:
$\lambda = \frac{6.626 \times 10^{-34}}{1.67 \times 10^{-27} \times 3.97 \times 10^6}$
$\lambda = \frac{6.626 \times 10^{-34}}{6.63 \times 10^{-21}}$
$\lambda \approx 1 \times 10^{-13} \ m$.

(C) For $n=4$,total orbitals = $n^2 = 16$. Each orbital can have one electron with $m_s=+\frac{1}{2}$. Thus,$16$ electrons are possible. (Correct)
b) For $n=5$,subshells are $s, p, d, f, g$ $(l=0, 1, 2, 3, 4)$,so there are $5$ subshells. (Incorrect)
c) For $n=2$,$l$ can be $0, 1$. If $l=1$,$m_l$ can be $-1, 0, +1$. Thus,$n=2, l=1, m_l=0, m_s=-\frac{1}{2}$ is valid. (Correct)
d) Radial nodes = $n-l-1$. For $3s$,$n=3, l=0$. Nodes = $3-0-1 = 2$. (Correct)
Therefore,statements $a, c,$ and $d$ are correct.

(D) According to the first law of thermodynamics,$\Delta U = q + w$.
Given: Heat absorbed $q = +100 \ J$.
Work done on the gas $w = -P_{ext} \Delta V$.
$P_{ext} = 1.5 \ atm$,$\Delta V = V_f - V_i = 2.0 \ L - 8.0 \ L = -6.0 \ L$.
$w = -(1.5 \ atm) \times (-6.0 \ L) = +9.0 \ L \cdot atm$.
Convert work to Joules: $w = 9.0 \times 101.32 \ J = 911.88 \ J$.
Therefore,$\Delta U = 100 \ J + 911.88 \ J = 1011.88 \ J \approx 1011.9 \ J$.

(D) The process is the vaporisation of water: $H_2O(l) \rightarrow H_2O(g)$.
Given $\Delta_{vap}H = 41.0 \ kJ \ mol^{-1}$ at $T = 373 \ K$.
The relationship between enthalpy change and internal energy change is $\Delta H = \Delta U + \Delta(PV)$.
Assuming water vapour is an ideal gas and neglecting the volume of liquid water,$\Delta(PV) \approx \Delta(n_gRT) = \Delta n_g RT$.
For $1 \ mol$ of water,$\Delta n_g = 1$.
So,$\Delta U = \Delta H - \Delta n_g RT = 41.0 \ kJ \ mol^{-1} - (1 \ mol \times 8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1} \times 373 \ K) = 41.0 - 3.101 = 37.899 \ kJ \ mol^{-1}$.
For $1.0 \ g$ of water $(1/18 \ mol)$,the internal energy change is $\Delta U = 37.899 \ kJ \ mol^{-1} / 18 \ g \ mol^{-1} \approx 2.106 \ kJ \ g^{-1}$.

(D) The molar mass of $C_2H_6$ is $(2 \times 12) + (6 \times 1) = 30 \ g \ mol^{-1}$.
Number of moles in $150 \ g$ of $C_2H_6$ is $n = \frac{150 \ g}{30 \ g \ mol^{-1}} = 5 \ mol$.
Heat liberated on combustion of $n$ moles is $n \times \Delta H_c$.
Therefore,heat liberated = $5 \times X \ kJ = 5X \ kJ$.
Assertion $(A)$ states the heat liberated is $\frac{X}{5} \ kJ$,which is incorrect.
Enthalpy is indeed an extensive property,meaning it depends on the amount of substance,which is the reason why the heat liberated is proportional to the number of moles. Thus,Reason $(R)$ is correct.

(D) The combustion reaction of methane is: $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$.
The standard enthalpy of combustion $(\Delta_{c} H^{\circ})$ is calculated using the formula: $\Delta_{c} H^{\circ} = [\sum \Delta_{f} H^{\circ}(\text{products}) - \sum \Delta_{f} H^{\circ}(\text{reactants})]$.
Substituting the given values: $\Delta_{c} H^{\circ} = [\Delta_{f} H^{\circ}(CO_2) + 2 \times \Delta_{f} H^{\circ}(H_2O)] - [\Delta_{f} H^{\circ}(CH_4) + 2 \times \Delta_{f} H^{\circ}(O_2)]$.
Since $\Delta_{f} H^{\circ}(O_2) = 0$,we have: $\Delta_{c} H^{\circ} = [-393 + 2 \times (-286)] - [-74.0]$.
$\Delta_{c} H^{\circ} = [-393 - 572] + 74.0$.
$\Delta_{c} H^{\circ} = -965 + 74.0 = -891 \ kJ \ mol^{-1}$.

(D) The chemical equation for the oxidation of methanol is: $CH_3OH(l) + \frac{1}{2} O_2(g) \longrightarrow HCHO(g) + H_2O(l)$
The enthalpy change of the reaction $(\Delta_r H)$ is calculated using the formula: $\Delta_r H = \sum \Delta_f H^{\circ}(\text{products}) - \sum \Delta_f H^{\circ}(\text{reactants})$
Given values:
$\Delta_f H^{\circ}(CH_3OH) = -239 \ kJ \ mol^{-1}$
$\Delta_f H^{\circ}(HCHO) = -116 \ kJ \ mol^{-1}$
$\Delta_f H^{\circ}(H_2O) = -286 \ kJ \ mol^{-1}$
$\Delta_f H^{\circ}(O_2) = 0 \ kJ \ mol^{-1}$ (standard state)
Substituting the values:
$\Delta_r H = [(-116) + (-286)] - [-239 + 0]$
$\Delta_r H = -402 - (-239)$
$\Delta_r H = -402 + 239 = -163 \ kJ \ mol^{-1}$

(D) Entropy $(S)$ is a measure of the randomness or disorder of a system. $A$ decrease in entropy occurs when the system becomes more ordered,such as when gas molecules are consumed to form a solid.
In option $D$,$CaO_{(s)} + CO_{2(g)} \longrightarrow CaCO_{3(s)}$,one mole of gas $(CO_2)$ is consumed to form a solid product $(CaCO_3)$.
Since the number of moles of gaseous species decreases from $1$ to $0$,the randomness of the system decreases,leading to a decrease in entropy $(\Delta S < 0)$.

(D) The standard Gibbs energy change of a reaction is given by $\Delta G^\circ = -RT \ln K$. At equilibrium,the reaction quotient $Q = K$,but the standard Gibbs energy change $\Delta G^\circ$ is not necessarily zero unless the equilibrium constant $K = 1$. Therefore,Assertion $(A)$ is incorrect.
The reason statement is correct because for a spontaneous process at constant temperature and pressure,the Gibbs energy of the system must decrease $(\Delta G < 0)$. Thus,$(A)$ is incorrect but $(R)$ is correct.

(C) $1$. The reaction of $2CH_3CH_2CH_2Br$ with $Na/Ether$ is a Wurtz reaction,which produces $n$-hexane $(X = CH_3CH_2CH_2CH_2CH_2CH_3)$.
$2$. The aromatization of $n$-hexane in the presence of $Mo_2O_3$ at $773K$ and $10-20 \ atm$ yields benzene $(Y = C_6H_6)$.
$3$. The reaction of benzene $(Y)$ with $CH_3Cl$ in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts alkylation,which produces toluene $(Z = C_6H_5CH_3)$.

(B) The acidic strength of oxyacids depends on the oxidation state of the central atom and the stability of the conjugate base.
As the oxidation state of chlorine increases,the electronegativity of the chlorine atom increases,which pulls the electron density away from the $O-H$ bond,making the proton more acidic.
The oxidation states of chlorine in $HOCl$,$HClO_2$,$HClO_3$,and $HClO_4$ are $+1, +3, +5$,and $+7$ respectively.
Therefore,the acidic strength increases in the order: $HOCl < HClO_2 < HClO_3 < HClO_4$.

(B) The hydrolysis of interhalogen compounds like $BrF_5$ involves the reaction with water to form hydrohalic acid and an oxyacid of the halogen with a higher oxidation state.
For $BrF_5$,the balanced chemical equation is: $BrF_5 + 3H_2O \rightarrow HBrO_3 + 5HF$.
Here,$BrF_5$ reacts with $3$ moles of water to produce $1$ mole of bromic acid $(HBrO_3)$ and $5$ moles of hydrogen fluoride $(HF)$.
Therefore,the products are $HF$ and $HBrO_3$.

(A) The hydrolysis of $XeF_2$ occurs according to the following reaction:
$2XeF_2(s) + 2H_2O(l) \rightarrow 2Xe(g) + 4HF(aq) + O_2(g)$
In this reaction,$XeF_2$ reacts with water to produce xenon gas $(Xe)$,hydrogen fluoride $(HF)$,and oxygen gas $(O_2)$.
Therefore,the gaseous products formed are $Xe$ and $O_2$.

(A) The boiling point of noble gases increases down the group due to the increase in the magnitude of van der Waals forces as the atomic size increases.
Helium $(He)$ is the first element in Group $18$ and has the smallest atomic size.
Due to its extremely weak interatomic van der Waals forces,Helium has the lowest boiling point of all known substances,which is approximately $4.2 \ K$.

(B) The reaction of $Xe$ with $F_2$ in a $1:20$ molar ratio at $573 \ K$ and $60-70 \ bar$ pressure yields $XeF_6$ $(A)$.
$Xe_{(g)} + 3F_{2(g)} \xrightarrow{573 \ K, 60-70 \ bar} XeF_{6(s)}$
Complete hydrolysis of $XeF_6$ produces $XeO_3$ $(B)$ and $HF$.
$XeF_{6(s)} + 3H_2O_{(l)} \rightarrow XeO_{3(s)} + 6HF_{(aq)}$
Therefore,$A$ is $XeF_6$ and $B$ is $XeO_3$.

(D) The preparation of $XeF_6$ involves the reaction of xenon with excess fluorine.
The reaction is given by: $Xe(g) + 3F_2(g) \xrightarrow{573 \ K, 60-70 \ bar} XeF_6(s)$.
To ensure the formation of $XeF_6$,$F_2$ is taken in a large excess,typically in a molar ratio of $1:20$ $(Xe:F_2)$.

(A) Fibre $(X)$ is a polymer that has strong intermolecular forces like hydrogen bonding or dipole-dipole interactions,which lead to close packing and crystalline nature. $Dacron$ (also known as $Terylene$) is a polyester fibre.
Elastomer $(Y)$ is a polymer in which polymer chains are held together by the weakest intermolecular forces,allowing the polymer to be stretched. $Neoprene$ (polychloroprene) is a synthetic rubber,which is an elastomer.
Therefore,$Dacron$ is a fibre and $Neoprene$ is an elastomer.

(D) In ionic polymerisation,the chain initiation involves the formation of ions or ion pairs at the active center.
$NaNH_2$ acts as an initiator for anionic polymerisation,while $SnCl_2$ and $AlCl_3$ are Lewis acids that can act as initiators for cationic polymerisation.
$(C_6H_5CO)_2O_2$ (benzoyl peroxide) is a covalent compound that undergoes homolytic cleavage to produce free radicals.
Therefore,it is used as an initiator in free radical polymerisation,not in ionic polymerisation.

(A) Buna-$N$ is a synthetic rubber formed by the co-polymerization of $1,3-$Butadiene and Acrylonitrile $(CH_2=CH-CN)$.
Therefore,$\underline{X}$ is Acrylonitrile,which corresponds to option $A$.

(C) Ethylene $(CH_2=CH_2)$ reacts with Baeyer's reagent (cold,dilute alkaline $KMnO_4$ solution) to form ethylene glycol $(HO-CH_2-CH_2-OH)$,which is compound $A$.
Ethylene glycol is used as a monomer in the preparation of the co-polymer Glyptal (polyethylene terephthalate is a polyester,but Glyptal is specifically formed from ethylene glycol and phthalic acid).
Therefore,the correct answer is Glyptal.

(D) Addition polymers are formed by the repeated addition of monomer molecules possessing double or triple bonds. $A$ homopolymer is formed from only one type of monomer unit.
$1$. Polythene is formed from ethene $(CH_2=CH_2)$.
$2$. Teflon (Polytetrafluoroethene) is formed from tetrafluoroethene $(CF_2=CF_2)$.
$3$. Orlon (Polyacrylonitrile) is formed from acrylonitrile $(CH_2=CH-CN)$.
All three are addition homopolymers. Therefore,the correct set is Polythene,Teflon,and Orlon.

(C) The monomer that can undergo polymerisation by all three mechanisms (free radical,cationic,and anionic) is $Styrene$ $(C_6H_5CH=CH_2)$.
$1$. Free radical polymerisation: The phenyl group stabilizes the radical intermediate.
$2$. Cationic polymerisation: The phenyl group stabilizes the carbocation intermediate through resonance.
$3$. Anionic polymerisation: The phenyl group stabilizes the carbanion intermediate through resonance.
Therefore,$Styrene$ is the correct answer.

(A) . Natural rubber becomes soft and sticky at high temperature,not hard. So,statement $a$ is incorrect.
$b$. Neoprene is formed by the polymerization of chloroprene ($2-$chloro$-1,3-$butadiene). So,statement $b$ is correct.
$c$. Nylon $6,6$ is formed by the condensation polymerization of hexamethylenediamine and adipic acid,which contains amide linkages. So,statement $c$ is correct.
$d$. Buna$-S$ is a copolymer formed from $1,3-$butadiene and styrene. So,statement $d$ is incorrect.
Therefore,statements $b$ and $c$ are correct.

(A) The standard reduction potentials are: $E^{\circ}_{Li^+/Li} = -3.05 \ V$,$E^{\circ}_{Mg^{2+}/Mg} = -2.36 \ V$,$E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$,and $E^{\circ}_{Ni^{2+}/Ni} = -0.25 \ V$.
$A$ metal with a more negative reduction potential acts as a stronger reducing agent and can displace a metal with a less negative (or more positive) reduction potential from its salt solution.
Comparing the values: $E^{\circ}_{Mg^{2+}/Mg} (-2.36 \ V) < E^{\circ}_{Zn^{2+}/Zn} (-0.76 \ V)$.
Since $Mg$ has a lower reduction potential than $Zn$,$Mg$ can displace $Zn$ from its solution.
Therefore,the statement '$Mg$ displaces $Zn$ from its solution' is correct.

(A, B) In a unit cell,the edge lengths are $a$,$b$,and $c$. The angle between $b$ and $c$ is $\alpha$,between $a$ and $c$ is $\beta$,and between $a$ and $b$ is $\gamma$. Thus,option $A$ is correct.
In a $bcc$ (body-centered cubic) lattice,the number of atoms per unit cell is $1$ (at center) $+ 8 \times (1/8)$ (at corners) $= 2$. Thus,option $B$ is also correct.
$SiC$ (Silicon carbide) is a covalent network solid,not an ionic solid. Thus,option $C$ is incorrect.
For a triclinic lattice,the relationship is $\alpha \neq \beta \neq \gamma \neq 90^{\circ}$. Thus,option $D$ is incorrect.
Note: Both $A$ and $B$ are factually correct statements.

(A) The density $(d)$ of a unit cell is given by the formula: $d = \frac{Z \times M}{N \times a^3}$,where $Z$ is the number of atoms per unit cell,$M$ is the molar mass,$N$ is Avogadro's number,and $a$ is the edge length of the unit cell.
For a $bcc$ (body-centered cubic) lattice,the number of atoms per unit cell $(Z)$ is $2$.
Substituting $Z = 2$ into the density formula: $d = \frac{2M}{N \times a^3}$.
Rearranging for $a^3$: $a^3 = \frac{2M}{Nd}$.
Taking the cube root on both sides: $a = \left( \frac{2M}{Nd} \right)^{\frac{1}{3}}$.

(B) In a cubic close packing $(CCP)$ or face-centered cubic $(FCC)$ unit cell,the atoms touch each other along the face diagonal.
The length of the face diagonal is given by $\sqrt{2} a$,where $a$ is the edge length of the unit cell.
Since the face diagonal consists of the radius of the corner atom,the diameter of the face-centered atom,and the radius of the other corner atom,we have: $4r = \sqrt{2} a$.
Therefore,the radius $r$ is given by $r = \frac{\sqrt{2} a}{4} = \frac{a}{2 \sqrt{2}}$.

(D) For a body-centered cubic $(bcc)$ structure, the number of atoms per unit cell $(Z)$ is $2$.
Given edge length $(a)$ = $400 \ pm = 400 \times 10^{-10} \ cm = 4 \times 10^{-8} \ cm$.
Atomic mass $(M)$ = $24 \ g \ mol^{-1}$.
Avogadro's number $(N_A)$ = $6 \times 10^{23} \ mol^{-1}$.
The formula for density $(d)$ is:
$d = \frac{Z \times M}{a^3 \times N_A}$
Substituting the values:
$d = \frac{2 \times 24}{(4 \times 10^{-8})^3 \times 6 \times 10^{23}}$
$d = \frac{48}{64 \times 10^{-24} \times 6 \times 10^{23}}$
$d = \frac{48}{384 \times 10^{-1}}$
$d = \frac{48}{38.4} = 1.25 \ g \ cm^{-3}$.

(A) The density of a unit cell is given by the formula: $d = \frac{Z \times M}{a^3 \times N_A}$.
Given:
Density $d = 2.7 \times 10^3 \ kg \ m^{-3} = 2.7 \ \text{g} \ cm^{-3}$.
Molar mass $M = 2.7 \times 10^{-2} \ kg \ mol^{-1} = 27 \ \text{g} \ mol^{-1}$.
Edge length $a = 405 \ pm = 405 \times 10^{-10} \ cm = 4.05 \times 10^{-8} \ cm$.
Avogadro's number $N_A = 6.02 \times 10^{23} \ mol^{-1}$.
Rearranging the formula to find $Z$:
$Z = \frac{d \times a^3 \times N_A}{M}$.
Substituting the values:
$Z = \frac{2.7 \times (4.05 \times 10^{-8})^3 \times 6.02 \times 10^{23}}{27}$.
$Z = \frac{2.7 \times 66.43 \times 10^{-24} \times 6.02 \times 10^{23}}{27}$.
$Z = \frac{1079.7 \times 10^{-1}}{27} = \frac{107.97}{27} \approx 4$.
Since the number of atoms per unit cell $Z = 4$, the unit cell is face-centered cubic $(FCC)$.

(C) Given: $P_A^0 = 450 \ mm \ Hg$,$P_B^0 = 700 \ mm \ Hg$,$P_{total} = 600 \ mm \ Hg$.
According to Raoult's Law: $P_{total} = P_A^0 x_A + P_B^0 x_B$.
Since $x_A + x_B = 1$,we have $x_B = 1 - x_A$.
Substituting values: $600 = 450 x_A + 700(1 - x_A)$.
$600 = 450 x_A + 700 - 700 x_A$.
$250 x_A = 100$,so $x_A = 100 / 250 = 0.4$.
Thus,$x_B = 1 - 0.4 = 0.6$.
Partial pressures in vapour phase: $P_A = P_A^0 x_A = 450 \times 0.4 = 180 \ mm \ Hg$.
$P_B = P_B^0 x_B = 700 \times 0.6 = 420 \ mm \ Hg$.
Mole fractions in vapour phase $(y_A, y_B)$:
$y_A = P_A / P_{total} = 180 / 600 = 0.3$.
$y_B = P_B / P_{total} = 420 / 600 = 0.7$.

(C) For an ideal solution,the total vapour pressure $P_T$ is given by Raoult's law: $P_T = P_A^0 x_A + P_B^0 x_B$,where $x_A + x_B = 1$.
Case $1$: Molar ratio $1:1$,so $x_A = 0.5$ and $x_B = 0.5$. $P_T = 400 \ mm$. Thus,$0.5 P_A^0 + 0.5 P_B^0 = 400$,which simplifies to $P_A^0 + P_B^0 = 800$ (Equation $1$).
Case $2$: Molar ratio $1:2$,so $x_A = 1/3$ and $x_B = 2/3$. $P_T = 350 \ mm$. Thus,$(1/3) P_A^0 + (2/3) P_B^0 = 350$,which simplifies to $P_A^0 + 2 P_B^0 = 1050$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(P_A^0 + 2 P_B^0) - (P_A^0 + P_B^0) = 1050 - 800$,so $P_B^0 = 250 \ mm$.
Substituting $P_B^0 = 250$ into Equation $1$: $P_A^0 + 250 = 800$,so $P_A^0 = 550 \ mm$.
Therefore,the vapour pressures of pure liquids $A$ and $B$ are $550 \ mm$ and $250 \ mm$ respectively.

(A) According to Raoult's law for an ideal solution,the total vapour pressure $P_{total}$ is given by the sum of the partial pressures of the components: $P_{total} = P_A + P_B$.
Given:
Mole fraction of $A$ $(x_A)$ = $0.67$
Mole fraction of $B$ $(x_B)$ = $0.33$
Vapour pressure of pure $A$ $(P^{\circ}_A)$ = $300 \ mm$
Vapour pressure of pure $B$ $(P^{\circ}_B)$ = $450 \ mm$
Using the formula $P_{total} = x_A P^{\circ}_A + x_B P^{\circ}_B$:
$P_{total} = (0.67 \times 300) + (0.33 \times 450)$
$P_{total} = 201 + 148.5 = 349.5 \ mm$.

(A) $1$. Calculate the molar masses: Heptane $(C_7H_{16})$ = $7 \times 12 + 16 \times 1 = 100 \ g/mol$. Octane $(C_8H_{18})$ = $8 \times 12 + 18 \times 1 = 114 \ g/mol$.
$2$. Calculate the number of moles: $n_{\text{heptane}} = \frac{25 \ g}{100 \ g/mol} = 0.25 \ mol$. $n_{\text{octane}} = \frac{57 \ g}{114 \ g/mol} = 0.50 \ mol$.
$3$. Calculate the mole fractions: $x_{\text{heptane}} = \frac{0.25}{0.25 + 0.50} = \frac{0.25}{0.75} = \frac{1}{3}$. $x_{\text{octane}} = \frac{0.50}{0.75} = \frac{2}{3}$.
$4$. Apply Raoult's Law: $P_{\text{total}} = P^{\circ}_{\text{heptane}} \times x_{\text{heptane}} + P^{\circ}_{\text{octane}} \times x_{\text{octane}}$.
$5$. $P_{\text{total}} = 106 \times \frac{1}{3} + 47 \times \frac{2}{3} = \frac{106 + 94}{3} = \frac{200}{3} = 66.66 \ kPa$.

(D) According to Henry's law,$P = K_H \times \chi$,where $P$ is the partial pressure of the gas,$K_H$ is Henry's law constant,and $\chi$ is the mole fraction of the gas in the solution.
Given: $P = 1.67 \times 10^2 \ kPa = 1.67 \times 10^5 \ Pa$,$K_H = 1.67 \times 10^8 \ Pa$.
Calculating mole fraction $\chi = P / K_H = (1.67 \times 10^5) / (1.67 \times 10^8) = 10^{-3}$.
Since the solution is dilute,the mole fraction $\chi \approx n_{CO_2} / n_{H_2O}$.
For $1000 \ mL$ of water,mass $= 1000 \ g$,so $n_{H_2O} = 1000 / 18 = 55.55 \ mol$.
Thus,$n_{CO_2} = \chi \times n_{H_2O} = 10^{-3} \times 55.55 = 5.555 \times 10^{-2} \ mol$.
Since the volume is $1 \ L$,the concentration is $5.55 \times 10^{-2} \ mol \ L^{-1}$.

(D) Sugar is a non-electrolyte,so $0.1 \ M$ sugar solution contains $0.1 \ M$ particles.
$KCl$ is a strong electrolyte that dissociates as $KCl \rightarrow K^+ + Cl^-$,so $0.1 \ M$ $KCl$ solution contains $0.1 + 0.1 = 0.2 \ M$ particles.
Lowering of vapour pressure is a colligative property,which depends on the number of solute particles.
Since $KCl$ has more particles,it causes a greater lowering of vapour pressure,meaning the vapour pressure of $KCl$ solution is lower than that of the sugar solution.
Therefore,Assertion $(A)$ is incorrect,while Reason $(R)$ is a correct statement.

(B) The depression in freezing point is given by $\Delta T_f = K_f \times m$,where $m$ is the molality of the solution.
Given $\Delta T_f = 0 - (-0.93) = 0.93 \ K$.
$K_f = 1.86 \ K \ kg \ mol^{-1}$.
For a $7 \%$ by mass solution,$7 \ g$ of solute is present in $93 \ g$ of solvent (water).
Molality $m = \frac{\text{mass of solute}}{\text{molar mass of solute} \times \text{mass of solvent in kg}} = \frac{7}{M \times 0.093}$.
Substituting the values: $0.93 = 1.86 \times \frac{7}{M \times 0.093}$.
$M = \frac{1.86 \times 7}{0.93 \times 0.093} = \frac{2 \times 7}{0.093} = \frac{14}{0.093} \approx 150.53 \ g \ mol^{-1}$.
Thus,the molar mass is approximately $150.5 \ g \ mol^{-1}$.

(A) $1$. Calculate the mass of acetic acid $(CH_3COOH)$: $\text{Mass} = \text{Density} \times \text{Volume} = 1.06 \ g \ mL^{-1} \times 0.8 \ mL = 0.848 \ g$.
$2$. Calculate the moles of acetic acid: $\text{Molar mass of } CH_3COOH = 60 \ g \ mol^{-1}$. $\text{Moles} = \frac{0.848 \ g}{60 \ g \ mol^{-1}} \approx 0.01413 \ mol$.
$3$. Calculate the molality $(m)$: Since the solvent is $1 \ kg$ of water,$m = 0.01413 \ mol \ kg^{-1}$.
$4$. Calculate the theoretical depression in freezing point $(\Delta T_f)$: $\Delta T_f = K_f \times m = 1.86 \ K \ kg \ mol^{-1} \times 0.01413 \ mol \ kg^{-1} \approx 0.02628 \ K$.
$5$. Calculate the Van't Hoff factor $(i)$: $i = \frac{\Delta T_f \text{ (observed)}}{\Delta T_f \text{ (theoretical)}} = \frac{0.0325}{0.02628} \approx 1.236 \approx 1.24$.

(D) According to Raoult's law for non-volatile solutes: $\frac{P^o - P_s}{P^o} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$.
Given: $P^o = 0.85 \ bar$,$P_s = 0.83 \ bar$,$W_2 = 2.0 \ g$,$W_1 = 39 \ g$,$M_1 = 78 \ g \ mol^{-1}$.
$\frac{0.85 - 0.83}{0.85} = \frac{2.0 / M_2}{39 / 78}$.
$\frac{0.02}{0.85} = \frac{2.0 / M_2}{0.5} \implies \frac{0.02}{0.85} = \frac{4}{M_2}$.
$M_2 = \frac{4 \times 0.85}{0.02} = 200 \ g \ mol^{-1}$.
Now,calculate molality $(m)$: $m = \frac{W_2 \times 1000}{M_2 \times W_1(g)} = \frac{2.0 \times 1000}{200 \times 39} = \frac{10}{39} \approx 0.2564 \ mol \ kg^{-1}$.
Elevation in boiling point: $\Delta T_b = K_b \times m = 2.6 \times \frac{10}{39} = \frac{26}{39} = \frac{2}{3} \approx 0.667 \ K$.
Wait,re-calculating: $\frac{0.02}{0.85} = \frac{n_2}{0.5} \implies n_2 = \frac{0.01}{0.85} \approx 0.01176 \ mol$.
$M_2 = \frac{2.0}{0.01176} = 170 \ g \ mol^{-1}$.
$m = \frac{0.01176}{0.039} = 0.3015 \ mol \ kg^{-1}$.
$\Delta T_b = 2.6 \times 0.3015 = 0.784 \ K$. Thus,the correct option is $D$.

(B) Given:
$W_B$ (mass of ethylene glycol) = $31 \ g$
$W_A$ (mass of water) = $600 \ g$
$K_f$ (for water) = $1.86 \ K \ kg \ mol^{-1}$
$M_B$ (molar mass of $C_2H_6O_2$) = $(2 \times 12) + (6 \times 1) + (2 \times 16) = 62 \ g \ mol^{-1}$
Formula for freezing point depression: $\Delta T_f = K_f \times \frac{W_B}{M_B} \times \frac{1000}{W_A(g)}$
Substituting the values:
$\Delta T_f = \frac{1.86 \times 31 \times 1000}{62 \times 600}$
$\Delta T_f = \frac{1.86 \times 31000}{37200} = \frac{57660}{37200} = 1.55 \ K$

(D) The osmotic pressure is given by the formula $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is the molar concentration,$R$ is the gas constant $(0.0821 \ L \ atm \ K^{-1} \ mol^{-1})$,and $T$ is the temperature in Kelvin.
$(a)$ For $5.0 \ M$ urea: $\pi = 1 \times 5.0 \times 0.0821 \times (67 + 273) = 1 \times 5.0 \times 0.0821 \times 340 = 139.57 \ atm$.
$(b)$ For $1.5 \ M A_2 B_3$: $\pi = 4.1 \times 1.5 \times 0.0821 \times (27 + 273) = 4.1 \times 1.5 \times 0.0821 \times 300 = 151.47 \ atm$.
$(c)$ For $3.0 \ M AB$: $\pi = 1.6 \times 3.0 \times 0.0821 \times (27 + 273) = 1.6 \times 3.0 \times 0.0821 \times 300 = 118.22 \ atm$.
$(d)$ For $2.5 \ M AB_2$: $\pi = 2.5 \times 2.5 \times 0.0821 \times (57 + 273) = 2.5 \times 2.5 \times 0.0821 \times 330 = 169.33 \ atm$.
Comparing the values,the solution with the highest osmotic pressure is $2.5 \ M AB_2$.

(C) The formula for osmotic pressure $(\pi)$ is $\pi = CRT$,where $C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature in Kelvin.
Given: $\pi = 0.4 \ bar$,$w = 4 \ g$,$V = 1.0 \ L$,$R = 0.083 \ L \ bar \ K^{-1} \ mol^{-1}$,$T = 27 + 273 = 300 \ K$.
The molar concentration $C = \frac{n}{V} = \frac{w}{M \times V}$,where $M$ is the molar mass.
Substituting the values: $0.4 = \frac{4}{M \times 1.0} \times 0.083 \times 300$.
$0.4 = \frac{4 \times 0.083 \times 300}{M}$.
$M = \frac{4 \times 0.083 \times 300}{0.4} = \frac{99.6}{0.4} = 249 \ g \ mol^{-1}$.

(B) Statement $a$ is correct: As the atomic number increases in the lanthanide series,the ionic radius decreases due to lanthanide contraction. This increases the covalent character of the $M-OH$ bond,thereby decreasing the basic strength. Thus,the order $Ce(OH)_3 > Gd(OH)_3 > Lu(OH)_3$ is correct.
Statement $b$ is incorrect: $O^{2-}$,$N^{3-}$,$F^{-}$,and $Na^{+}$ all have $10$ electrons,but the original statement listed $O^{-}$ and $Mg^{+}$,which are not isoelectronic with the others.
Statement $c$ is correct: Due to lanthanide contraction,the atomic radii of $Zr$ $(160 \ pm)$ and $Hf$ $(159 \ pm)$ are approximately the same.
Therefore,statements $a$ and $c$ are correct.

(C) Statement $a$ is incorrect: The protective power of a lyophilic sol is inversely proportional to its gold number. $A$ lower gold number indicates higher protective power.
Statement $b$ is incorrect: According to the Hardy-Schulze rule,the coagulating power of an ion increases with the increase in the magnitude of the charge on the ion. For negative sols,the order should be $Al^{3+} > Ba^{2+} > Na^{+}$.
Statement $c$ is incorrect: $A$ cloud is a liquid in gas type of colloid (aerosol).
Statement $d$ is correct: Physical adsorption is non-specific in nature and can form multiple layers on the surface of the adsorbent at high pressure.

(C) The Freundlich adsorption isotherm equation is given by $\frac{x}{m} = kP^{1/n}$.
For $1/n = 0$,$\frac{x}{m} = k$,which means adsorption is independent of pressure.
For $1/n = 1$,$\frac{x}{m} = kP$,which means adsorption varies directly with pressure.
Physical adsorption is an exothermic process,so according to Le Chatelier's principle,the extent of adsorption decreases with an increase in temperature.
Therefore,the statement that the extent of adsorption increases with an increase in temperature is incorrect.

(B) The colour of colloidal solutions depends on the wavelength of light scattered by the dispersed particles.
In the case of gold sol,the colour changes with the size of the gold particles.
As the particle size increases,the wavelength of the scattered light changes,resulting in different colours like red,purple,blue,and golden.

(B) $1$. $As_2S_3$ is a negatively charged sol,not positively charged. So,option $A$ is incorrect.
$2$. For the Tyndall effect to be observed,the refractive indices of the dispersed phase and the dispersion medium must differ significantly in magnitude. This is a correct statement.
$3$. An ultramicroscope is used to detect the presence of colloidal particles,but it does not provide information about their size and shape. So,option $C$ is incorrect.
$4$. The colour of gold sol depends on the size of the particles. The finest gold sol is red in colour,while coarser sols appear purple or blue. So,option $D$ is incorrect.

(D) . Noble gases can indeed be separated by selective adsorption on coconut charcoal at different temperatures due to differences in their van der Waals forces.
$B$. Animal charcoal is a well-known adsorbent used to remove colored impurities from solutions.
$C$. Heterogeneous catalysis involves the adsorption of reactants on the catalyst surface,which increases the local concentration and lowers the activation energy,thereby increasing the reaction rate.
$D$. Silica gel and alumina gel are used as desiccants to remove moisture from the air,not to increase it. Therefore,this statement is incorrect.

(A) The reduction of a metal oxide is represented by the reaction: $MO(s/l) + C(s) \rightarrow M(s/l) + CO(g)$.
According to the Gibbs free energy equation,$\Delta G = \Delta H - T\Delta S$.
For the reaction to be spontaneous,$\Delta G$ must be negative.
When the metal oxide is in the liquid state,the entropy of the system is higher compared to the solid state.
Consequently,the entropy change $(\Delta S)$ for the reduction process becomes more positive (or less negative) when the reactant is in the liquid state,making the term $-T\Delta S$ more negative,which facilitates the reduction process.

(B) The oxidation of phenol with chromic acid $(Na_2Cr_2O_7 / H_2SO_4)$ yields $p$-benzoquinone as the product.
Therefore,$X$ is phenol and $Y$ is $Na_2Cr_2O_7 / H_2SO_4$.

(B) $1$. Benzoic acid reacts with concentrated $HNO_3$ and concentrated $H_2SO_4$ (nitration) to form $m$-nitrobenzoic acid $(X)$.
$2$. Reduction of $m$-nitrobenzoic acid $(X)$ with $Sn/HCl$ converts the $-NO_2$ group to an $-NH_2$ group,forming $m$-aminobenzoic acid $(Y)$.
$3$. $m$-Aminobenzoic acid $(Y)$ reacts with $NaNO_2/HCl$ at $0-5 \ ^\circ C$ to form the diazonium salt,which then reacts with $Cu_2Cl_2/HCl$ (Sandmeyer reaction) to replace the diazonium group with a chlorine atom,resulting in $m$-chlorobenzoic acid $(Z)$.

(C) The reduction of nitrobenzene depends on the medium used:
$1$. In neutral medium $(Zn/NH_4Cl)$: Nitrobenzene is reduced to phenylhydroxylamine $(C_6H_5NHOH)$. Thus,$X$ is phenylhydroxylamine.
$2$. In alkaline medium $(Zn + KOH/C_2H_5OH)$: Nitrobenzene undergoes reduction to form azoxybenzene,azobenzene,and finally hydrazobenzene $(C_6H_5NH-NHC_6H_5)$. Thus,$Y$ is hydrazobenzene.
Therefore,the correct option is $C$.

(B) The temperature coefficient is defined as the ratio of rate constants at temperatures differing by $10^{\circ} C$: $\text{Temperature coefficient} = \frac{k_{(t+10)}}{k_t}$.
Given,$\text{Temperature coefficient} = 2$,$k_{(27^{\circ} C)} = 10^{-3} \ min^{-1}$,and we need to find $k_{(17^{\circ} C)}$.
Substituting the values: $2 = \frac{k_{(27^{\circ} C)}}{k_{(17^{\circ} C)}}$.
$2 = \frac{10^{-3}}{k_{(17^{\circ} C)}}$.
$k_{(17^{\circ} C)} = \frac{10^{-3}}{2} = 0.5 \times 10^{-3} = 5 \times 10^{-4} \ min^{-1}$.

(A) The chemical compositions of the given ores are as follows:
$A$. Calamine: $ZnCO_3$ (matches with $ii$)
$B$. Chalcopyrite: $CuFeS_2$ (matches with $i$)
$C$. Bauxite: $Al_2O_3 \cdot 2H_2O$ (matches with $iv$)
$D$. Haematite: $Fe_2O_3$ (matches with $iii$)
Therefore,the correct matching is $A-ii, B-i, C-iv, D-iii$.