AP EAMCET 2024 Chemistry Question Paper with Answer and Solution

(A) $GeO_2$ is acidic in nature.
$CO$ is a neutral oxide.
$PbO_2$ is amphoteric in nature.
$SnO$ is amphoteric in nature.

(C) In group $14$,most $MX_4$ halides are covalent. However,$SnF_4$ and $PbF_4$ are ionic due to the large electronegativity difference between the metal and fluorine. Thus,Statement $I$ is correct.
For group $14$ elements,the stability of the $+2$ oxidation state increases down the group due to the inert pair effect. $Ge$ is at the top of the group,so the $+4$ oxidation state is more stable than the $+2$ oxidation state. Therefore,$GeCl_4$ is more stable than $GeCl_2$. Thus,Statement $II$ is incorrect.

(B) Most of the $MX_4$ $(X = F, Cl, Br, I)$ compounds of group $14$ elements are covalent in nature.
However,$SnF_4$ and $PbF_4$ are exceptions as they exhibit ionic character due to the high electronegativity of fluorine and the size of the central metal atoms.

(A) $i$. The reaction $H_{2(g)} + F_{2(g)} \longrightarrow 2HF_{(g)}$ is highly spontaneous and occurs even in the dark due to the high reactivity of $F_2$.
$ii$. The Haber process,$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,is an exothermic reaction $(\Delta H = -92.4 \ kJ/mol)$,not endothermic.
$iii$. $HF$ is an electron-rich hydride because the fluorine atom has three lone pairs of electrons.
Therefore,statements $i$ and $iii$ are correct.

(B) Ammonium nitrate $(NH_4NO_3)$ on thermal decomposition produces nitrous oxide $(N_2O)$ and water $(H_2O)$.
$NH_4NO_3 \longrightarrow N_2O + 2H_2O$
$N_2O$ is a colourless,neutral gas (laughing gas).
The structure of $N_2O$ is $N \equiv N^+ - O^-$,which is linear in shape.

(A) $Ortho-phosphorous$ acid $(H_3PO_3)$ on heating undergoes disproportionation to give $ortho-phosphoric$ acid $(H_3PO_4)$ and phosphine $(PH_3)$.
The chemical equation is:
$4H_3PO_3 \xrightarrow{\Delta} 3H_3PO_4 + PH_3$

(B) The balanced stoichiometric equation is:
$P_4 + 3 OH^{-} + 3 H_2 O \rightarrow PH_3 + 3 H_2 PO_2^{-}$
Comparing this to the given equation,$X^{-}$ is $H_2 PO_2^{-}$.
The conjugate acid is formed by adding a proton $(H^{+})$ to the base.
Conjugate acid of $H_2 PO_2^{-}$ is $H_3 PO_2$.
$H_3 PO_2$ is known as hypophosphorous acid.

(B) In acidic medium,$KMnO_4$ acts as a strong oxidizing agent.
When $H_2S$ reacts with $KMnO_4$ in an acidic medium,$H_2S$ is oxidized to elemental sulphur,which precipitates out.
The reaction is:
$5H_2S + 2MnO_4^- + 6H^+ \longrightarrow 2Mn^{2+} + 8H_2O + 5S$
Thus,the precipitation of sulphur from $H_2S$ is a characteristic reaction of $KMnO_4$ in acidic medium.

(D) In neutral or faintly alkaline solutions,$KMnO_4$ acts as an oxidizing agent where $Mn$ is reduced from $+7$ to $+4$ (forming $MnO_2$).
One of the characteristic reactions in this medium is the oxidation of iodide $(I^-)$ to iodate $(IO_3^-)$.
The balanced chemical equation is:
$2MnO_4^- + H_2O + I^- \rightarrow 2MnO_2 + 2OH^- + IO_3^-$

(A) polygon of $n$ sides has the number of diagonals given by the formula: $\frac{n(n-3)}{2} = 275$.
Multiplying by $2$,we get $n(n-3) = 550$.
$n^2 - 3n - 550 = 0$.
Factoring the quadratic equation: $n^2 - 25n + 22n - 550 = 0$.
$n(n - 25) + 22(n - 25) = 0$.
$(n - 25)(n + 22) = 0$.
Since the number of sides $n$ must be positive,$n = 25$ (as $n = -22$ is not possible).

(B) Let $S = \sin 1^{\circ} + \sin 2^{\circ} + \ldots + \sin 89^{\circ}$.
We can pair terms as $(\sin 1^{\circ} + \sin 89^{\circ}) + (\sin 2^{\circ} + \sin 88^{\circ}) + \ldots + (\sin 44^{\circ} + \sin 46^{\circ}) + \sin 45^{\circ}$.
Using the identity $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$,we have $\sin k^{\circ} + \sin(90-k)^{\circ} = 2 \sin 45^{\circ} \cos(\frac{90-2k}{2}) = 2 \cdot \frac{1}{\sqrt{2}} \cos(45-k)^{\circ} = \sqrt{2} \cos(45-k)^{\circ}$.
Alternatively,$\sin k^{\circ} + \sin(90-k)^{\circ} = \sin k^{\circ} + \cos k^{\circ}$.
Actually,$\sin k^{\circ} + \sin(90-k)^{\circ} = 2 \sin 45^{\circ} \cos(k-45)^{\circ} = \sqrt{2} \cos(k-45)^{\circ}$.
Using $\sin A + \sin B = 2 \sin 45^{\circ} \cos(\frac{A-B}{2})$,the numerator becomes $\sqrt{2} \cos 44^{\circ} + \sqrt{2} \cos 43^{\circ} + \ldots + \sqrt{2} \cos 1^{\circ} + \sin 45^{\circ}$.
Numerator $= \sqrt{2} (\cos 1^{\circ} + \cos 2^{\circ} + \ldots + \cos 44^{\circ}) + \frac{1}{\sqrt{2}}$.
Numerator $= \frac{1}{\sqrt{2}} [2(\cos 1^{\circ} + \cos 2^{\circ} + \ldots + \cos 44^{\circ}) + 1]$.
Thus,the expression simplifies to $\frac{1}{\sqrt{2}}$.

(B) Let the points be $A(1, -2)$ and $B(\alpha, \beta)$. The midpoint of $AB$ is $M = \left(\frac{1+\alpha}{2}, \frac{-2+\beta}{2}\right)$.
Since $M$ lies on the line $2x - 3y + 5 = 0$,we have $2\left(\frac{1+\alpha}{2}\right) - 3\left(\frac{-2+\beta}{2}\right) + 5 = 0$,which simplifies to $1 + \alpha + 3 - 1.5\beta + 5 = 0$,or $2\alpha - 3\beta = -18$.
The slope of $AB$ is $m_{AB} = \frac{\beta - (-2)}{\alpha - 1} = \frac{\beta + 2}{\alpha - 1}$.
The slope of the given line $2x - 3y + 5 = 0$ is $m_L = \frac{2}{3}$.
Since the line is the perpendicular bisector,$m_{AB} \times m_L = -1$,so $\left(\frac{\beta + 2}{\alpha - 1}\right) \times \left(\frac{2}{3}\right) = -1$.
This gives $2\beta + 4 = -3\alpha + 3$,or $3\alpha + 2\beta = -1$.
Solving the system of equations $2\alpha - 3\beta = -18$ and $3\alpha + 2\beta = -1$:
Multiply the first by $2$ and the second by $3$: $4\alpha - 6\beta = -36$ and $9\alpha + 6\beta = -3$.
Adding them gives $13\alpha = -39$,so $\alpha = -3$.
Substituting $\alpha = -3$ into $3\alpha + 2\beta = -1$ gives $3(-3) + 2\beta = -1$,so $2\beta = 8$,$\beta = 4$.
Thus,$\alpha + \beta = -3 + 4 = 1$.

(D) Given the equation of the pair of lines is $2x^2 + hxy + 6y^2 = 0$.
Comparing this with the general form $ax^2 + 2h'xy + by^2 = 0$,we have $a = 2$,$2h' = h$,and $b = 6$.
Let the slopes of the two lines be $m_1$ and $m_2$.
We know that $m_1 + m_2 = -\frac{2h'}{b} = -\frac{h}{6}$ and $m_1 m_2 = \frac{a}{b} = \frac{2}{6} = \frac{1}{3}$.
Given that one slope is thrice the other,let $m_1 = 3m_2$.
Substituting this into the product equation: $(3m_2) \cdot m_2 = \frac{1}{3}$ $\Rightarrow 3m_2^2 = \frac{1}{3}$ $\Rightarrow m_2^2 = \frac{1}{9}$ $\Rightarrow m_2 = \pm \frac{1}{3}$.
If $m_2 = \frac{1}{3}$,then $m_1 = 1$.
If $m_2 = -\frac{1}{3}$,then $m_1 = -1$.
Using the sum of slopes: $m_1 + m_2 = 4m_2 = -\frac{h}{6}$.
For $m_2 = \frac{1}{3}$,$4(\frac{1}{3}) = -\frac{h}{6}$ $\Rightarrow \frac{4}{3} = -\frac{h}{6}$ $\Rightarrow h = -8$.
For $m_2 = -\frac{1}{3}$,$4(-\frac{1}{3}) = -\frac{h}{6}$ $\Rightarrow -\frac{4}{3} = -\frac{h}{6}$ $\Rightarrow h = 8$.
Thus,$h = \pm 8$.

(A) The given pair of lines is $xy-x-y+1=0$.
Factorizing the equation: $x(y-1)-1(y-1)=0$,which gives $(x-1)(y-1)=0$.
Thus,the lines are $x=1$ (a vertical line) and $y=1$ (a horizontal line).
Let the required line be $y=mx+c$.
The angle between the line $y=mx+c$ and $y=1$ (slope $m_1=0$) is $45^{\circ}$.
Using $\tan \theta = |\frac{m-m_1}{1+mm_1}|$,we get $\tan 45^{\circ} = |\frac{m-0}{1+0}| = 1$,so $m=1$ or $m=-1$.
If $m=1$,the line is $y=x+c$,or $x-y+c'=0$.
Checking the options,$x-y=5$ (or $x-y-5=0$) is a line with slope $m=1$.
This line makes an angle of $45^{\circ}$ with both $x=1$ and $y=1$.

(A) Given the equation of the pair of lines is $8x^2 + axy + y^2 = 0$.
Let the slopes of the two lines be $m_1$ and $m_2$.
For a homogeneous equation $Ax^2 + Bxy + Cy^2 = 0$,the sum of slopes is $m_1 + m_2 = -\frac{B}{C}$ and the product of slopes is $m_1 m_2 = \frac{A}{C}$.
Here,$A = 8$,$B = a$,and $C = 1$.
So,$m_1 + m_2 = -a$ and $m_1 m_2 = 8$.
Given that one slope is thrice the other,let $m_1 = 3m_2$.
Substituting this into the product equation: $(3m_2) \times m_2 = 8$ $\Rightarrow 3m_2^2 = 8$ $\Rightarrow m_2^2 = \frac{8}{3}$ $\Rightarrow m_2 = \pm \sqrt{\frac{8}{3}}$.
Substituting into the sum equation: $3m_2 + m_2 = -a$ $\Rightarrow 4m_2 = -a$ $\Rightarrow a = -4m_2$.
Thus,$a = -4 \left( \pm \sqrt{\frac{8}{3}} \right) = \pm 4 \times 2 \sqrt{\frac{2}{3}} = \pm 8 \sqrt{\frac{2}{3}}$.
Considering the positive value,$a = 8 \sqrt{\frac{2}{3}}$.

(A) The given equation is $2x^2 + 3xy + Ky^2 = 0$. Dividing by $Ky^2$ (assuming $K \neq 0$),we get the quadratic in terms of slope $m = \frac{y}{x}$ as $Km^2 + 3m + 2 = 0$.
Since one slope $m_1 = 2$,it must satisfy the equation: $K(2)^2 + 3(2) + 2 = 0$ $\Rightarrow 4K + 8 = 0$ $\Rightarrow K = -2$.
Substituting $K = -2$ into the equation,we get $2x^2 + 3xy - 2y^2 = 0$.
Factoring this,we get $(2x - y)(x + 2y) = 0$.
The slopes of the two lines are $m_1 = 2$ and $m_2 = -\frac{1}{2}$.
Since $m_1 \times m_2 = 2 \times (-\frac{1}{2}) = -1$,the lines are perpendicular to each other.
Therefore,the angle between the pair of lines is $\theta = \frac{\pi}{2}$.

(C) The given equation is $2x^2 + xy - 6y^2 - 2x + 17y - 12 = 0$.
Comparing this with the general equation of a pair of lines $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $a = 2$,$g = -1$,and $c = -12$.
The length of the $x$-intercept made by the pair of lines is given by the formula $\frac{2\sqrt{g^2 - ac}}{|a|}$.
Substituting the values,we get $\text{Length} = \frac{2\sqrt{(-1)^2 - 2(-12)}}{2} = \frac{2\sqrt{1 + 24}}{2} = \sqrt{25} = 5$.

(A) Given equation: $x^2+y^2+xy+x+3y+1=0$ ...$(i)$
And $x+y+2=0 \Rightarrow \frac{x+y}{-2}=1$ ...(ii)
Homogenizing $(i)$ with (ii):
$x^2+xy+y^2+x(1)+3y(1)+1(1)^2=0$
Substituting (ii) into the equation:
$x^2+xy+y^2+x(\frac{x+y}{-2})+3y(\frac{x+y}{-2})+(\frac{x+y}{-2})^2=0$
Multiplying by $4$ to clear the denominator:
$4x^2+4xy+4y^2-2x(x+y)-6y(x+y)+(x+y)^2=0$
$4x^2+4xy+4y^2-2x^2-2xy-6xy-6y^2+x^2+2xy+y^2=0$
$3x^2-2xy-y^2=0$
Comparing with $ax^2+2hxy+by^2=0$,we get $a=3, 2h=-2, b=-1$.
The equation of the angle bisectors is given by $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$.
$\frac{x^2-y^2}{3-(-1)} = \frac{xy}{-1}$
$\frac{x^2-y^2}{4} = -xy$
$x^2-y^2 = -4xy$
$x^2+4xy-y^2=0$.

(A) The given equation is $x^2+y^2+2x+2y-5=0$.
Let the axes be rotated by an angle $\alpha$. The transformation equations are $x = X \cos \alpha - Y \sin \alpha$ and $y = X \sin \alpha + Y \cos \alpha$.
Substituting these into the equation:
$(X \cos \alpha - Y \sin \alpha)^2 + (X \sin \alpha + Y \cos \alpha)^2 + 2(X \cos \alpha - Y \sin \alpha) + 2(X \sin \alpha + Y \cos \alpha) - 5 = 0$.
Simplifying,we get $X^2 + Y^2 + 2X(\cos \alpha + \sin \alpha) + 2Y(\cos \alpha - \sin \alpha) - 5 = 0$.
For the equation to contain no linear terms,the coefficients of $X$ and $Y$ must be zero:
$1) \cos \alpha + \sin \alpha = 0$ $\Rightarrow \tan \alpha = -1$ $\Rightarrow \alpha = n\pi - \frac{\pi}{4}$.
$2) \cos \alpha - \sin \alpha = 0$ $\Rightarrow \tan \alpha = 1$ $\Rightarrow \alpha = n\pi + \frac{\pi}{4}$.
Since there is no value of $\alpha$ that satisfies both conditions simultaneously,the number of such values is $0$.

(A) The equation of line $AB$ passing through $A(1, 0, 2)$ and $B(2, 1, 0)$ is given by $\frac{x-1}{2-1} = \frac{y-0}{1-0} = \frac{z-2}{0-2} = \lambda$ $\Rightarrow \frac{x-1}{1} = \frac{y}{1} = \frac{z-2}{-2} = \lambda$.
Thus,any point on $AB$ is $(\lambda+1, \lambda, -2\lambda+2)$.
The equation of line $CD$ passing through $C(2, -5, 3)$ and $D(0, 3, 2)$ is given by $\frac{x-2}{0-2} = \frac{y+5}{3-(-5)} = \frac{z-3}{2-3} = \mu$ $\Rightarrow \frac{x-2}{-2} = \frac{y+5}{8} = \frac{z-3}{-1} = \mu$.
Thus,any point on $CD$ is $(-2\mu+2, 8\mu-5, -\mu+3)$.
For the intersection point $P$,we equate the coordinates:
$1) \lambda+1 = -2\mu+2 \Rightarrow \lambda + 2\mu = 1$
$2) \lambda = 8\mu-5 \Rightarrow \lambda - 8\mu = -5$
Subtracting $(2)$ from $(1)$: $10\mu = 6 \Rightarrow \mu = \frac{3}{5}$.
Substituting $\mu$ in $(1)$: $\lambda + 2(\frac{3}{5}) = 1 \Rightarrow \lambda = 1 - \frac{6}{5} = -\frac{1}{5}$.
Wait,checking the intersection:
$x = -\frac{1}{5} + 1 = \frac{4}{5}$,$y = -\frac{1}{5}$,$z = -2(-\frac{1}{5}) + 2 = \frac{12}{5}$.
Using $\mu = \frac{3}{5}$: $x = -2(\frac{3}{5}) + 2 = \frac{4}{5}$,$y = 8(\frac{3}{5}) - 5 = -\frac{1}{5}$,$z = -(\frac{3}{5}) + 3 = \frac{12}{5}$.
So $P = (\frac{4}{5}, -\frac{1}{5}, \frac{12}{5})$.
Then $a+b+c = \frac{4-1+12}{5} = \frac{15}{5} = 3$.

(D) The second equation is $6 x^2 - x y - 5 y^2 = 0$.
Factoring this,we get $(6 x + 5 y)(x - y) = 0$,which gives two lines: $y = x$ and $y = -\frac{6 x}{5}$.
Case $1$: If $y = x$ is a common line,substitute $y = x$ into $3 x^2 - 5 x y + P y^2 = 0$:
$3 x^2 - 5 x^2 + P x^2 = 0 \Rightarrow (P - 2) x^2 = 0$.
For this to hold for all $x$,$P = 2$.
Case $2$: If $y = -\frac{6 x}{5}$ is a common line,substitute $y = -\frac{6 x}{5}$ into $3 x^2 - 5 x y + P y^2 = 0$:
$3 x^2 - 5 x(-\frac{6 x}{5}) + P(-\frac{6 x}{5})^2 = 0$
$3 x^2 + 6 x^2 + P(\frac{36 x^2}{25}) = 0$
$9 x^2 + \frac{36 P x^2}{25} = 0 \Rightarrow 9 + \frac{36 P}{25} = 0$
$36 P = -225 \Rightarrow P = -\frac{225}{36} = -\frac{25}{4}$.
The sum of all possible values of $P$ is $2 + (-\frac{25}{4}) = \frac{8 - 25}{4} = -\frac{17}{4}$.

(A) Decomposition of potassium chlorate: $2 KClO_{3(s)} \xrightarrow{\Delta} 2 KCl_{(s)} + 3 O_{2(g)}$.
In $KClO_3$,the oxidation state of $Cl$ is $+5$,and in $KCl$,it is $-1$. Since the oxidation state decreases,$Cl$ is reduced. Thus,Statement-$I$ is correct.
Reaction of $Na$ with $O_2$: $4 Na_{(s)} + O_{2(g)} \rightarrow 2 Na_2O_{(s)}$.
Here,$Na$ is oxidized from $0$ to $+1$ and $O$ is reduced from $0$ to $-2$. Since both oxidation and reduction occur,it is a redox reaction. Thus,Statement-$II$ is correct.

(B) The balanced chemical equation for the reaction is:
$S + 2H_2SO_4 \rightarrow 3SO_2 + 2H_2O$
In this reaction,the oxidation state of sulphur $(S)$ changes from $0$ to $+4$ in $SO_2$,which is an oxidation process.
The oxidation state of sulphur in $H_2SO_4$ changes from $+6$ to $+4$ in $SO_2$,which is a reduction process.
Therefore,both the oxidised product and the reduced product are $SO_2$.

(A) In reaction $(i)$,the oxidation state of $Cl$ changes from $+5$ to $-1$ and $O$ changes from $-2$ to $0$. It is a redox reaction.
In reaction $(ii)$,the oxidation state of $O$ in $H_2O_2$ changes from $-1$ to $-2$ (in $H_2O$) and $0$ (in $O_2$). It is a redox reaction.
In reaction $(iii)$,there is no change in the oxidation states of any elements. It is a precipitation reaction,not a redox reaction.
In reaction $(iv)$,the oxidation state of $Na$ changes from $0$ to $+1$ and $O$ changes from $0$ to $-2$. It is a redox reaction.
Therefore,reactions $(i)$,$(ii)$,and $(iv)$ are redox reactions. The total number of redox reactions is $3$.

(D) The balanced redox reaction is: $Cr_2O_7^{2-} + 14 H^{+} + 6 Fe^{2+} \longrightarrow 2 Cr^{3+} + 6 Fe^{3+} + 7 H_2O$
At equivalence point,the number of equivalents of $Cr_2O_7^{2-}$ equals the number of equivalents of $Fe^{2+}$.
Equivalents = $Molarity \times n\text{-factor} \times Volume (L)$.
For $Fe^{2+}$,$n\text{-factor} = 1$ (as $Fe^{2+} \rightarrow Fe^{3+} + e^-$).
For $Cr_2O_7^{2-}$,$n\text{-factor} = 6$ (as $Cr_2O_7^{2-} + 14 H^+ + 6e^- \rightarrow 2 Cr^{3+} + 7 H_2O$).
Let $V$ be the volume in $L$ of $Cr_2O_7^{2-}$ solution.
$\frac{1}{60} \times 6 \times V = 0.1 \times 1 \times 0.1$
$0.1 \times V = 0.01$
$V = 0.1 \ L$.

(A) The alloy of lithium and magnesium $(Li-Mg)$ is used to make armour plates.
This alloy is known for being extremely light and strong,possessing one of the lowest densities among metallic materials.

(A) The reaction of $Na_2SO_3$ with $HCl$ is: $Na_2SO_3(s) + 2HCl(aq) \longrightarrow 2NaCl(aq) + H_2O(l) + SO_2(g)$.
However,the question implies a reaction where two gaseous products are formed.
Consider $Na_2S_2O_3 + 2HCl \longrightarrow 2NaCl + H_2O + SO_2(g) + S(s)$ (not two gases).
Consider $Na_2SO_3 + 2HCl \longrightarrow 2NaCl + H_2O + SO_2(g)$.
Actually,for $Na_2SO_3$,only $SO_2$ is a gas.
Let us re-evaluate: $Na_2S + 2HCl \longrightarrow 2NaCl + H_2S(g)$. Only one gas.
Let us check $NaNO_2 + HCl \longrightarrow NaCl + HNO_2$. $HNO_2$ decomposes to $NO(g)$ and $NO_2(g)$ and $H_2O$. Thus,$NO$ and $NO_2$ are two gases.
Therefore,'$X$' is $NaNO_2$.

(A) The thermal stability of alkaline earth metal carbonates increases as the size of the cation increases down the group.
As the size of the cation increases,the polarizing power of the cation decreases,making the $CO_3^{2-}$ ion more stable.
The ionic radii order is $Mg^{2+} < Ca^{2+} < Ba^{2+}$.
Therefore,the thermal stability order is $MgCO_3 < CaCO_3 < BaCO_3$,which corresponds to $X < Z < Y$ or $Y > Z > X$.

(A) The densities of the alkaline earth metals are as follows:
$Be: 1.84 \ g \ cm^{-3}$
$Mg: 1.74 \ g \ cm^{-3}$
$Ca: 1.55 \ g \ cm^{-3}$
$Sr: 2.63 \ g \ cm^{-3}$
Comparing these values,the order of density is $Sr (2.63) > Be (1.84) > Mg (1.74) > Ca (1.55)$.
Therefore,the correct order is $Sr > Be > Mg > Ca$.

(B) $Be(OH)_2$ is amphoteric,but the hydroxides of $Mg$,$Ca$,$Sr$ and $Ba$ are basic. The strength increases from $Mg$ to $Ba$.
Carbonates of group $2$ elements become more thermally stable down the group. The larger compounds require more heat than the lighter compounds in order to decompose.

$BeCO_3$	$< 100 \ ^\circ C$
$MgCO_3$	$540 \ ^\circ C$
$CaCO_3$	$900 \ ^\circ C$
$SrCO_3$	$1290 \ ^\circ C$
$BaCO_3$	$1360 \ ^\circ C$

The solubility of the sulphates in water decreases down the group. $Be > Mg \gg Ca > Sr > Ba$. The higher solubilities of $BeSO_4$ and $MgSO_4$ are due to the high enthalpy of solvation of the smaller $Be^{2+}$ and $Mg^{2+}$ ions.
On heating,the alkaline earth metal nitrates decompose to give metal oxide,nitrogen dioxide and oxygen.
$2 M(NO_3)_2 \xrightarrow{\Delta} 2 MO + 4 NO_2 + O_2$ $(M = Be, Mg, Ca, Sr, Ba)$

(A) Among the alkaline earth metals,$Be$ (Beryllium) is the only element that does not react directly with hydrogen to form its hydride $(BeH_2)$.
This is due to the high ionization energy and small size of the $Be^{2+}$ ion,which makes the formation of the ionic hydride energetically unfavorable.
Other alkaline earth metals like $Mg$,$Ca$,$Sr$,and $Ba$ react directly with hydrogen upon heating to form their respective ionic hydrides $(MH_2)$.

(A) $2 NaNO_3 \longrightarrow 2 NaNO_2 + O_2$ ($Na_2O$ is not formed).
$4 LiNO_3 \longrightarrow 2 Li_2O \text{ (basic)} + 4 NO_2 \text{ (acidic)} + O_2$.
$2 Be(NO_3)_2 \longrightarrow 2 BeO \text{ (amphoteric)} + 4 NO_2 \text{ (acidic)} + O_2$.
$2 Ca(NO_3)_2 \longrightarrow 2 CaO \text{ (basic)} + 4 NO_2 \text{ (acidic)} + O_2$.
$NO_2$ is an acidic oxide. $Li_2O$ and $CaO$ are basic oxides. $BeO$ is amphoteric. Thus,$Ca(NO_3)_2$ and $LiNO_3$ produce both acidic and basic oxides along with $O_2$.

(A) The enamel on teeth is composed of hydroxyapatite,which is represented as $3 Ca_3(PO_4)_2 \cdot Ca(OH)_2$.
When fluoride ions $(F^-)$ are present in water or toothpaste,they react with the hydroxyapatite to form fluorapatite,which is much harder and more resistant to acid decay.
The chemical reaction is: $3 Ca_3(PO_4)_2 \cdot Ca(OH)_2 + 2 F^- \rightarrow 3 Ca_3(PO_4)_2 \cdot CaF_2 + 2 OH^-$.
Comparing this to the given formula,we identify $X = Ca(OH)_2$ and $Y = CaF_2$.

(B) Since the vectors are coplanar,their scalar triple product is zero:
$\begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 0$
Applying row operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$:
$\begin{vmatrix} a & 1 & 1 \\ 1-a & b-1 & 0 \\ 1-a & 0 & c-1 \end{vmatrix} = 0$
Expanding along the first row:
$a((b-1)(c-1) - 0) - 1((1-a)(c-1) - 0) + 1(0 - (1-a)(b-1)) = 0$
$a(b-1)(c-1) - (1-a)(c-1) - (1-a)(b-1) = 0$
Divide the entire equation by $(1-a)(1-b)(1-c)$ (noting $a, b, c \neq 1$):
$\frac{a(b-1)(c-1)}{(1-a)(1-b)(1-c)} - \frac{(1-a)(c-1)}{(1-a)(1-b)(1-c)} - \frac{(1-a)(b-1)}{(1-a)(1-b)(1-c)} = 0$
$\frac{-a}{(1-a)} - \frac{1}{(1-b)} - \frac{1}{(1-c)} = 0$
Since $\frac{-a}{1-a} = \frac{1-a-1}{1-a} = 1 - \frac{1}{1-a}$,we substitute:
$1 - \frac{1}{1-a} - \frac{1}{1-b} - \frac{1}{1-c} = 0$
Therefore,$\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1$.

(A) The direction ratios of the line segment $OQ$ are $(4-0, -2-0, 1-0)$,which are $(4, -2, 1)$.
The direction ratios of the line segment $OP$ are $(0-0, 1-0, 2-0)$,which are $(0, 1, 2)$.
Let $\theta$ be the angle between $OP$ and $OQ$. The formula for the cosine of the angle between two vectors with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by:
$\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$
Substituting the values:
$\cos \theta = \frac{|(4)(0) + (-2)(1) + (1)(2)|}{\sqrt{4^2 + (-2)^2 + 1^2} \sqrt{0^2 + 1^2 + 2^2}}$
$\cos \theta = \frac{|0 - 2 + 2|}{\sqrt{16 + 4 + 1} \sqrt{0 + 1 + 4}}$
$\cos \theta = \frac{0}{\sqrt{21} \sqrt{5}} = 0$
Since $\cos \theta = 0$,we have $\theta = \frac{\pi}{2}$.

(A) The boiling point is inversely proportional to the vapour pressure at a given temperature.
Given boiling points: $C (308 \ K) < A (350 \ K) < B (373 \ K)$.
Therefore,the order of vapour pressure at any temperature is $C > A > B$.
$I.$ Since $B$ has the highest boiling point,it has the lowest vapour pressure,meaning it remains primarily in the liquid phase in vessel $(I)$. Thus,vessel $(I)$ is rich in liquid $B$. This statement is correct.
$II.$ Since $C$ has the lowest boiling point,it has the highest vapour pressure,meaning it vaporizes most easily. Thus,vessel $(II)$ will be rich in the vapour of $C$. This statement is correct.
$III.$ As established,the order of vapour pressure is $C > A > B$. This statement is correct.
All statements $I, II,$ and $III$ are correct.

(A) The thermal decomposition of potassium permanganate is: $2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2 \uparrow$.
Molar mass of $KMnO_4 = 39 + 55 + (16 \times 4) = 158 \text{ g/mol}$.
Mass of $0.1 \text{ mole}$ of $KMnO_4 = 0.1 \times 158 = 15.8 \text{ g}$.
From the reaction stoichiometry,$2 \text{ moles}$ of $KMnO_4$ produce $1 \text{ mole}$ of $O_2$ gas.
So,$0.1 \text{ mole}$ of $KMnO_4$ produces $0.05 \text{ mole}$ of $O_2$.
Mass of $O_2$ gas $= 0.05 \times 32 = 1.6 \text{ g}$.
Weight of residue = Initial weight - Weight of $O_2$ gas $= 15.8 - 1.6 = 14.2 \text{ g}$.

(B) Initial weight of $H_2SO_4$ $(W)$ = $98 \ mg = 98 \times 10^{-3} \ g$.
Initial number of moles $(n_1)$ = $\frac{98 \times 10^{-3} \ g}{98 \ g/mol} = 10^{-3} \ mol$.
Number of moles removed $(n_2)$ = $\frac{\text{Number of molecules}}{N_A} = \frac{3.01 \times 10^{20}}{6.02 \times 10^{23}} = 0.5 \times 10^{-3} \ mol$.
Remaining moles = $n_1 - n_2 = 10^{-3} - 0.5 \times 10^{-3} = 0.5 \times 10^{-3} \ mol = 5 \times 10^{-4} \ mol$.

(A) $1$. Determine the empirical formula:
$C = 92.3 \%, H = 7.7 \%$.
Moles of $C = \frac{92.3}{12} = 7.69$,Moles of $H = \frac{7.7}{1} = 7.7$.
Ratio $C:H = 1:1$. Empirical formula is $CH$.
$2$. Calculate moles of hydrocarbon:
Given mass $= 39 \ g$. Molar mass of $CH = 13 \ g/mol$.
Moles of $CH = \frac{39}{13} = 3 \ mol$.
$3$. Determine moles of water $(x)$:
Reaction: $2H_2O + 2Na \rightarrow 2NaOH + H_2$.
$2 \ mol$ of $H_2O$ produces $1 \ mol$ of $H_2$.
To produce $0.75 \ mol$ of $H_2$,we need $x = 0.75 \times 2 = 1.5 \ mol$ of $H_2O$.
$4$. Combustion reaction:
For $CH$ (or $C_2H_2$ unit),the combustion is $C_2H_2 + 2.5 O_2 \rightarrow 2 CO_2 + H_2O$.
Since we have $3 \ mol$ of $CH$ (which is $1.5 \ mol$ of $C_2H_2$),the reaction is:
$1.5 C_2H_2 + 3.75 O_2 \rightarrow 3 CO_2 + 1.5 H_2O$.
$5$. Calculate mass of $O_2$:
Moles of $O_2 = 3.75 \ mol$.
Mass of $O_2 = 3.75 \times 32 = 120 \ g$.

(B) The reaction between $CO_2$ and $MOH$ is: $CO_2 + MOH \rightarrow MHCO_3$.
Since $0.2 \ mol$ of $MOH$ reacts completely,$0.2 \ mol$ of $CO_2$ is produced.
The thermal decomposition of metal hydrogen carbonate is: $2MHCO_3 \xrightarrow{\Delta} M_2CO_3 + CO_2 + H_2O$.
From the stoichiometry,$1 \ mol$ of $CO_2$ is produced from $2 \ mol$ of $MHCO_3$.
Therefore,$0.2 \ mol$ of $CO_2$ is produced from $2 \times 0.2 = 0.4 \ mol$ of $MHCO_3$.
The mass '$x$' is calculated as: $x = \text{moles} \times \text{molar mass} = 0.4 \ mol \times 84 \ g \ mol^{-1} = 33.6 \ g$.

(D) Mass of metal $= 10 \text{ g}$.
Mass of oxide $= 11.6 \text{ g}$.
Mass of oxygen $= 11.6 \text{ g} - 10 \text{ g} = 1.6 \text{ g}$.
Equivalent weight of metal $= \frac{\text{Mass of metal}}{\text{Mass of oxygen}} \times 8$.
Equivalent weight of metal $= \frac{10}{1.6} \times 8 = 50$.

(D) The ideal gas equation is $PV = nRT$.
For $n = 1 \ mol$,the equation becomes $P = (RT) \times (V^{-1})$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = P$ and $x = V^{-1}$,we get the slope $m = RT$.
Given that the slope $m = 32.8 \ L \ atm \ mol^{-1}$ and $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$.
Therefore,$RT = 32.8 \ L \ atm \ mol^{-1}$.
$T = \frac{32.8 \ L \ atm \ mol^{-1}}{0.082 \ L \ atm \ K^{-1} \ mol^{-1}} = 400 \ K$.

(D) Initial moles of $He \ (n_1) = 1 \ mol$.
Initial moles of $CH_4 \ (n_2) = 1 \ mol$.
Moles of $He$ escaped $(n_1') = \frac{2 \ L}{22.4 \ L/mol} \approx 0.089 \ mol$.
Moles of $CH_4$ escaped $(n_2') = \frac{1 \ L}{22.4 \ L/mol} \approx 0.044 \ mol$.
Remaining moles of $He \ (n_1'') = 1 - 0.089 = 0.911 \ mol$.
Remaining moles of $CH_4 \ (n_2'') = 1 - 0.044 = 0.956 \ mol$.
Total remaining moles $= 0.911 + 0.956 = 1.867 \ mol$.
Mole fraction of $He \ (\chi_{He}) = \frac{0.911}{1.867} \approx 0.488$.
Mole fraction of $CH_4 \ (\chi_{CH_4}) = \frac{0.956}{1.867} \approx 0.512$.

(B) According to Graham's Law of diffusion,the rate of diffusion $r$ is given by $r = \frac{V}{t} \propto \frac{1}{\sqrt{M}}$.
Therefore,$\frac{r_1}{r_2} = \frac{V_1/t_1}{V_2/t_2} = \sqrt{\frac{M_2}{M_1}}$.
Given: $V_1 = 60 \ cm^3, M_1 = 64 \ g \ mol^{-1}, t_1 = x$.
$V_2 = 360 \ cm^3, M_2 = 4 \ g \ mol^{-1}, t_2 = y$.
Substituting the values: $\frac{60/x}{360/y} = \sqrt{\frac{4}{64}}$.
$\frac{60}{x} \times \frac{y}{360} = \frac{2}{8} = \frac{1}{4}$.
$\frac{y}{6x} = \frac{1}{4}$.
$\frac{y}{x} = \frac{6}{4} = \frac{3}{2}$.
Therefore,the ratio $\frac{x}{y} = \frac{2}{3}$.

(D) The formula for $RMS$ velocity is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Squaring both sides,we get $v_{rms}^2 = \frac{3RT}{M}$.
Comparing this with the equation of a straight line $y = mx + C$,where $y = v_{rms}^2$ and $x = T$:
The slope $m = \frac{3R}{M}$.
Given slope $m = 249 \ m^2 \ s^{-2} \ K^{-1}$ and $R = 8.3 \ J \ mol^{-1} \ K^{-1}$.
Substituting the values: $249 = \frac{3 \times 8.3}{M}$.
$M = \frac{3 \times 8.3}{249} = \frac{24.9}{249} = 0.1 \ kg \ mol^{-1}$.
Thus,$M = 1 \times 10^{-1} \ kg \ mol^{-1}$.

(A) According to the Maxwell-Boltzmann distribution of molecular speeds,as the temperature increases,the peak of the curve shifts towards higher velocities and the curve becomes broader and flatter.
In the given graph,the peak for $T_2$ is at the highest velocity,followed by $T_1$,and then $T_3$.
Therefore,the correct order of temperature is $T_2 > T_1 > T_3$.

(A) Given: $T = 133.33 \ K$,$M = 0.083 \ kg \ mol^{-1}$,$R = 8.3 \ J \ mol^{-1} \ K^{-1}$.
The formula for the root mean square $(RMS)$ velocity of an ideal gas is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Substituting the given values into the formula:
$v_{rms} = \sqrt{\frac{3 \times 8.3 \times 133.33}{0.083}}$
$v_{rms} = \sqrt{\frac{3323.917}{0.083}}$
$v_{rms} = \sqrt{40047.19} \approx 200.12 \ m \ s^{-1}$.
Thus,the $RMS$ velocity is approximately $200 \ m \ s^{-1}$.

(D) The kinetic energy $(K.E.)$ for one mole of an ideal gas is given by the formula: $K.E. = \frac{1}{2} M v_{rms}^2$
Given:
Molar mass $(M)$ $= 0.1 \ kg \ mol^{-1}$
$v_{rms} = 4 \times 10^2 \ ms^{-1}$
Substituting the values:
$K.E. = \frac{1}{2} \times (0.1 \ kg \ mol^{-1}) \times (4 \times 10^2 \ ms^{-1})^2$
$K.E. = 0.05 \times (16 \times 10^4)$
$K.E. = 0.8 \times 10^4 = 8 \times 10^3 \ J \ mol^{-1}$

(C) The kinetic energy $(KE)$ of an ideal gas is given by the formula: $KE = \frac{3}{2} nRT$.
Since the temperature $(T)$ is constant,the kinetic energy is directly proportional to the number of moles $(n)$,i.e.,$KE \propto n$.
Therefore,the ratio of kinetic energies is: $\frac{KE_{H_2}}{KE_{O_2}} = \frac{n_{H_2}}{n_{O_2}}$.
Calculate the number of moles for each gas:
$n_{H_2} = \frac{\text{mass}}{\text{molar mass}} = \frac{3 \ g}{2 \ g/mol} = 1.5 \ mol$.
$n_{O_2} = \frac{\text{mass}}{\text{molar mass}} = \frac{4 \ g}{32 \ g/mol} = 0.125 \ mol$.
Now,find the ratio: $\frac{KE_{H_2}}{KE_{O_2}} = \frac{1.5}{0.125} = \frac{12}{1}$.
Thus,the ratio is $12: 1$.

(D) $RMS$ velocity $(u_{rms})$ is given by the formula: $u_{rms} = \sqrt{\frac{3 RT}{M}}$
Squaring both sides,we get: $(u_{rms})^2 = \frac{3 RT}{M}$
Comparing this with the equation of a straight line passing through the origin,$y = mx + C$,where $y = (u_{rms})^2$ and $x = T$:
$(u_{rms})^2 = (\frac{3 R}{M}) \cdot T + 0$
Here,the slope $(m)$ is equal to $\frac{3 R}{M}$.

(A) An allylic halide is a compound in which the halogen atom is bonded to an $sp^3$ hybridized carbon atom that is adjacent to a carbon-carbon double bond $(C=C)$.
Let us analyze the structures:
$A$. $4-$chlorobut$-1-$ene $(CH_2=CH-CH_2-CH_2-Cl)$: The $Cl$ atom is attached to a carbon atom that is not adjacent to the double bond. This is a primary alkyl halide,not an allylic halide.
$B$. $1-$chlorobut$-2-$ene $(CH_3-CH=CH-CH_2-Cl)$: The $Cl$ atom is attached to a carbon atom adjacent to the double bond. This is an allylic halide.
$C$. $3-$chloro$-2-$methylbut$-1-$ene $(CH_2=C(CH_3)-CHCl-CH_3)$: The $Cl$ atom is attached to a carbon atom adjacent to the double bond. This is an allylic halide.
$D$. $4-$chloropent$-2-$ene $(CH_3-CH=CH-CH_2-CH_2-Cl)$: The $Cl$ atom is attached to a carbon atom that is not adjacent to the double bond. This is not an allylic halide.
Note: Both $A$ and $D$ are not allylic halides. However,$4-$chlorobut$-1-$ene is the most common example provided in textbooks to illustrate a non-allylic halide.

(A) Carboxylic acids are stronger acids than phenols because the carboxylate ion formed after the loss of a proton is more stable than the phenoxide ion.
In the carboxylate ion $(RCOO^-)$,the negative charge is delocalized over two highly electronegative oxygen atoms,resulting in two equivalent resonance structures.
In the phenoxide ion $(C_6H_5O^-)$,the negative charge is delocalized over the aromatic ring,placing the negative charge on the less electronegative carbon atoms in some resonance structures.
Since equivalent resonance structures provide greater stability,the carboxylate ion is more stable,making carboxylic acids more acidic.

(B) Deactivating groups are those that withdraw electron density from the benzene ring through inductive or resonance effects, making the ring less reactive towards electrophilic substitution reactions.
Among the given groups:
$1$. $-Cl$: Deactivating (due to strong $-I$ effect).
$2$. $-SO_3H$: Deactivating (due to strong $-I$ and $-M$ effects).
$3$. $-OH$: Activating (due to $+M$ effect).
$4$. $-NHC_2H_5$: Activating (due to $+M$ effect).
$5$. $-COOCH_3$: Deactivating (due to $-I$ and $-M$ effects).
$6$. $-CH_3$: Activating (due to $+I$ and hyperconjugation).
Therefore, the deactivating groups are $-Cl, -SO_3H, \text{ and } -COOCH_3$.
The total number of deactivating groups is $3$.

(B) In column chromatography,the substance with the least degree of adsorption is eluted first because it has the weakest interaction with the stationary phase.
Given the order of adsorption: $D > B > C > A$.
Since $A$ has the lowest degree of adsorption,it will be eluted first.
Conversely,$D$ has the highest degree of adsorption,so it will be eluted last.
The elution order is $A, C, B, D$.

(C) Let us analyze the given options based on the chemical composition of the ores:
$A$. Cuprite $(Cu_2O)$ and haematite $(Fe_2O_3)$ are both oxide ores. This is correctly matched.
$B$. Calamine $(ZnCO_3)$ and siderite $(FeCO_3)$ are both carbonate ores. This is correctly matched.
$C$. Magnetite $(Fe_3O_4)$ is an oxide ore,and malachite $(CuCO_3 \cdot Cu(OH)_2)$ is a carbonate ore. Neither is a silicate ore. This is not correctly matched.
$D$. Sphalerite $(ZnS)$ and fool's gold (pyrite,$FeS_2$) are both sulphide ores. This is correctly matched.
Therefore,the set that is not correctly matched is $C$.

(D) In the extraction of $Fe$ (iron),silica $(SiO_2)$ is added as a flux to remove $FeO$ impurity as iron silicate slag $(FeSiO_3)$.
In the extraction of $Cu$ (copper),$FeS$ present as an impurity is removed as iron silicate slag $(FeSiO_3)$ by adding $SiO_2$ flux.
In the extraction of $Zn$ (zinc),$FeO$ impurity is removed as iron silicate slag $(FeSiO_3)$ by adding $SiO_2$ flux.
Therefore,$Fe$,$Cu$,and $Zn$ involve the removal of impurities as slag.

(B) During the electrolytic refining of copper,a block of impure copper is used as the anode,and a thin strip of pure copper is used as the cathode.

(B) In the process of manufacturing wrought iron from cast iron,a reverberatory furnace is lined with limestone $(CaCO_3)$.
$CaCO_3$ acts as a flux to remove impurities like silicon,which is oxidized to $SiO_2$.
$CaO + SiO_2 \longrightarrow CaSiO_3$ (slag).
Thus,$CaCO_3$ is the flux used.

(D) In the blast furnace,the temperature range of $500-800 \ K$ is the lower temperature zone.
In this zone,the following reduction reactions occur:
$(i)$ $3 Fe_2O_3 + CO \rightarrow 2 Fe_3O_4 + CO_2$
$(iii)$ $Fe_3O_4 + 4 CO \rightarrow 3 Fe + 4 CO_2$
$(iv)$ $Fe_2O_3 + CO \rightarrow 2 FeO + CO_2$
Reaction $(ii)$ is not a primary reduction step in this specific temperature range.
Therefore,the correct reactions are $(i)$,$(iii)$,and $(iv)$.

(B) The iron obtained from a blast furnace contains about $4 \%$ carbon and many impurities in smaller amounts (e.g.,$S, P, Si, Mn$).
This form of iron is known as pig iron.

(B) In the blast furnace,the reduction of iron oxides occurs at different temperature ranges.
At the temperature range of $900-1500 \ K$ (lower part of the furnace),the following reactions take place:
$FeO + CO \rightarrow Fe + CO_2$
$C + CO_2 \rightarrow 2CO$
$CaO + SiO_2 \rightarrow CaSiO_3$ (slag formation).
Thus,the reaction $FeO + CO \rightarrow Fe + CO_2$ occurs in this zone.

(B) The reaction of alkyl halide $C_3H_7Cl$ with $KCN$ in the presence of $C_2H_5OH$ (an alcoholic solvent) leads to the formation of an alkyl cyanide (nitrile) as the major product via an $S_N2$ mechanism.
$CH_3CH_2CH_2Cl + KCN \xrightarrow{C_2H_5OH} CH_3CH_2CH_2CN + KCl$
Here,$X = KCN/C_2H_5OH$ and $Y = CH_3CH_2CH_2CN$.
Hydrolysis of alkyl cyanide $(Y)$ yields a carboxylic acid and ammonia gas $(NH_3)$:
$CH_3CH_2CH_2CN + 2H_2O + H^+ \rightarrow CH_3CH_2CH_2COOH + NH_3 \uparrow$
The ammonia gas $(NH_3)$ released is basic in nature and turns red litmus paper blue.

(D) The Swarts reaction is a method used for the synthesis of alkyl fluorides.
It involves the heating of an alkyl chloride or alkyl bromide in the presence of a metallic fluoride such as $AgF$,$Hg_2F_2$,$CoF_2$,or $SbF_3$.
In the given options,$2 R-CH_2-Br + CoF_2 \longrightarrow 2 R-CH_2-F + CoBr_2$ represents the Swarts reaction as it involves the exchange of a halogen atom with fluorine using a metallic fluoride.

(B) The reaction of allylbenzene with $HBr$ in the presence of benzoyl peroxide $((C_6H_5CO)_2O_2)$ proceeds via the peroxide effect (Kharasch effect),which follows Anti-Markovnikov addition.
In this reaction,the $Br$ atom attaches to the less substituted carbon atom of the double bond.
The product formed is $C_6H_5-CH_2-CH_2-CH_2Br$.
Since the bromine atom is attached to a primary carbon atom,the product is a primary alkyl halide.

(C) The $S_N 1$ mechanism proceeds via the formation of a stable carbocation intermediate. The stability of the carbocation determines the ease of the $S_N 1$ reaction. Benzylic carbocations are highly stable due to resonance.
$(A)$ The product is $C_6H_5CH_2CH_2CH_2Cl$ (a primary alkyl halide),which undergoes hydrolysis via $S_N 2$.
$(B)$ The product is $C_6H_5CH(Br)CH_2CH_3$,which forms a stable benzylic carbocation $(C_6H_5CH^+CH_2CH_3)$ upon ionization,thus favoring $S_N 1$.
$(C)$ The product is $C_6H_5C(Br)(CH_3)_2$,which forms a stable benzylic carbocation $(C_6H_5C^+(CH_3)_2)$ upon ionization,thus favoring $S_N 1$.
$(D)$ The product is $C_6H_5CH_2CH_2Br$ (a primary alkyl halide),which undergoes hydrolysis via $S_N 2$.
Therefore,only $(B)$ and $(C)$ form stable benzylic carbocations and undergo hydrolysis by the $S_N 1$ mechanism.

(B) The reactions are analyzed as follows:
$A$: Propene reacts with $HI$ to form $2-$iodopropane via Markovnikov addition.
$B$: Propan-$1-$ol reacts with $NaI$ and $H_3PO_4$ to form $1-$iodopropane.
$C$: $1-$chloropropane reacts with $NaI$ in dry acetone (Finkelstein reaction) to form $1-$iodopropane.
$D$: Propane reacts with $I_2$ in the presence of $hv$ to form $2-$iodopropane as the major product due to the stability of the secondary free radical.
Thus,$B$ and $C$ are the correct methods for the preparation of $1-$iodopropane.

(D) In the first reaction,$(CH_3)_3CONa$ is a strong nucleophile and $CH_3CH_2Br$ is a primary alkyl halide. This undergoes an $S_N2$ reaction to form the ether $(CH_3)_3COCH_2CH_3$ $(X)$.
In the second reaction,$(CH_3)_3CBr$ is a tertiary alkyl halide and $CH_3CH_2ONa$ is a strong base. Tertiary alkyl halides with strong bases undergo an $E2$ elimination reaction to form an alkene,which is $(CH_3)_2C=CH_2$ $(Y)$.

(D) $1$. The isomer of $C_5H_{12}$ that gives only one monobrominated product $C_5H_{11}Br$ upon reaction with $Br_2 / \text{light}$ is neopentane $(2,2-\text{dimethylpropane})$. This is because all $12$ hydrogen atoms in neopentane are equivalent.
$2$. Neopentane reacts with $Br_2 / \text{light}$ to form $1-\text{bromo}-2,2-\text{dimethylpropane}$ $(X)$.
$3$. The reaction of a primary alkyl halide $(X)$ with $AgNO_2$ proceeds via an $S_N2$ mechanism,where the nucleophile is the nitrite ion $(NO_2^-)$. Since $AgNO_2$ is largely covalent,the nitrogen atom acts as the nucleophile,leading to the formation of a nitroalkane $(R-NO_2)$ as the major product.
$4$. Therefore,the reaction of $1-\text{bromo}-2,2-\text{dimethylpropane}$ with $AgNO_2$ yields $2,2-\text{dimethyl}-1-\text{nitropropane}$ $(Y)$ as the major product.

(A) $1$. For the formation of $X$: The reaction of $CH_3CH_2CH_2Br$ with $Mg$ in dry ether forms a Grignard reagent,$CH_3CH_2CH_2MgBr$. Subsequent hydrolysis with $H_2O$ yields propane,$CH_3CH_2CH_3$,as product $X$.
$2$. For the formation of $Y$: The reaction of $CH_3CH_2CH_2Br$ with $NaOEt$ (a strong nucleophile/base) in ethanol follows an $S_N2$ mechanism to produce an ether,$CH_3CH_2CH_2OCH_2CH_3$,as the major product $Y$.

(D) The hydrolysis of an alkyl bromide $C_5H_{11}Br$ following first-order kinetics indicates an $S_N1$ mechanism,which proceeds via the formation of a stable carbocation. For a $C_5$ alkyl bromide to form a stable carbocation,it is likely a tertiary alkyl bromide,specifically $2$-bromo-$2$-methylbutane.
$1$. Formation of Grignard reagent: $CH_3CH_2C(CH_3)_2Br + Mg \xrightarrow{\text{dry ether}} CH_3CH_2C(CH_3)_2MgBr$.
$2$. Reaction with $D_2O$: The Grignard reagent reacts with $D_2O$ to replace the $MgBr$ group with a deuterium atom $(D)$: $CH_3CH_2C(CH_3)_2MgBr + D_2O \rightarrow CH_3CH_2C(CH_3)_2D + Mg(OD)Br$.
Thus,the product $Y$ is $2$-deuterio-$2$-methylbutane,which matches the structure shown in option $D$.

(D) $1$. The reaction of $4-\text{bromoethylbenzene}$ with $Br_2$ in the presence of heat $(\Delta)$ undergoes free-radical substitution at the benzylic position to form $1-(4-\text{bromophenyl})\text{bromoethane}$ (compound $B$).
$2$. The subsequent reaction of compound $B$ with aqueous $KOH$ is a nucleophilic substitution reaction ($S_N1$ or $S_N2$),where the bromine atom at the benzylic position is replaced by a hydroxyl group $(-OH)$ to form $1-(4-\text{bromophenyl})\text{ethanol}$ (compound $C$).

(D) The correct matches are:
$(A)$ Sandmeyer reaction: Used to prepare aryl halides like $Ar-Cl$ or $Ar-Br$ from diazonium salts. Thus,$(A-III)$.
$(B)$ Finkelstein reaction: $R-X + NaI \longrightarrow R-I + NaX$ (where $X = Cl, Br$). Thus,$(B-I)$.
$(C)$ Swarts reaction: $R-X + AgF \longrightarrow R-F + AgX$. Thus,$(C-II)$.
Therefore,the correct sequence is $A-III, B-I, C-II$.

(A) $1$. Conversion of chlorobenzene to phenol is done by Dow's process,which requires $NaOH$ at $623 \ K$ and $300 \ atm$ pressure,followed by acidification with $H^+$.
$2$. The major product of nitration of chlorobenzene is $p$-nitrochlorobenzene. This compound is more reactive towards nucleophilic substitution due to the electron-withdrawing $-NO_2$ group at the para position. It can be converted to $p$-nitrophenol using $NaOH$ at a milder condition of $443 \ K$,followed by acidification with $H^+$.
$3$. Thus,$A = (i) \ NaOH, 623 \ K, 300 \ atm \ (ii) \ H^+$ and $B = (i) \ NaOH, 443 \ K \ (ii) \ H^+$.

(D) The conversion of benzene diazonium chloride to diphenyl occurs in two steps:
$1$. The conversion of benzene diazonium chloride to chlorobenzene using $CuCl$ is known as the $Sandmeyer$ reaction.
$2$. The coupling of two aryl halide molecules (chlorobenzene) in the presence of sodium $(Na)$ and dry ether to form diphenyl is known as the $Fittig$ reaction.
Therefore,the correct sequence is $Sandmeyer$ and $Fittig$.

(B) $1$. The nitration of chlorobenzene with $conc. HNO_3$ and $conc. H_2SO_4$ yields $1$-chloro-$4$-nitrobenzene as the major product $(X)$ because the $-Cl$ group is ortho/para directing and the para-isomer is sterically favored.
$2$. The conversion of $1$-chloro-$4$-nitrobenzene to $4$-nitrophenol involves nucleophilic aromatic substitution.
$3$. The reaction requires strong nucleophilic conditions,which are provided by $(i) NaOH$ at $443 \ K$ followed by $(ii) H^+$ (acidification) to convert the phenoxide ion to the phenol.
$4$. Thus,$Y$ is $(i) NaOH, 443 \ K, (ii) H^+$ and $Z$ is $4$-nitrophenol.

(D) The first step is the addition of $HBr$ to propene $(C_3H_6)$ in the presence of benzoyl peroxide,which follows the anti-Markovnikov rule. This results in the formation of $1$-bromopropane $(CH_3CH_2CH_2Br)$ as $X$.
In the second step,$1$-bromopropane reacts with bromobenzene in the presence of $Na$ and dry ether. This is a Wurtz-Fittig reaction,which couples the alkyl group to the benzene ring,forming propylbenzene as $Y$.

(B) The Wurtz-Fittig reaction involves the reaction of an aryl halide with an alkyl halide in the presence of sodium metal and dry ether to form an alkyl-substituted aromatic compound.
The general reaction is:
$Ar-X + R-X + 2Na \xrightarrow{\text{dry ether}} Ar-R + 2NaX$
In this reaction,the compound $X$ must be an aryl halide (such as chlorobenzene) to react with the alkyl halide $(R-X)$ to form the substituted benzene product.
Therefore,the correct compound $X$ is chlorobenzene.

(C) The reaction sequence is as follows:
$1$. Benzene reacts with $Cl_2$ in the presence of $FeCl_3$ (a Lewis acid) to form chlorobenzene $(X)$ via electrophilic aromatic substitution.
$2$. Chlorobenzene $(X)$ reacts with an alkyl halide $(R-X)$ in the presence of sodium metal and dry ether to form an alkylbenzene $(Y)$.
This specific reaction between an aryl halide and an alkyl halide in the presence of sodium metal and dry ether is known as the Wurtz-Fittig reaction.

(A) The reaction of $4$-bromotoluene with $Br_2$ in the presence of $UV$ light is a free radical substitution reaction. This reaction occurs at the benzylic position,not on the aromatic ring. Therefore,$X$ is $4$-bromobenzyl bromide $(Br-C_6H_4-CH_2Br)$.
When $4$-bromobenzyl bromide is treated with $OH^-$,it undergoes a nucleophilic substitution reaction $(S_N2)$ where the $Br$ atom on the side chain is replaced by an $OH$ group. Therefore,$Y$ is $4$-bromobenzyl alcohol $(Br-C_6H_4-CH_2OH)$.

(C) The conversion of $3-$hexene to propane involves two main steps:
Step-$I$: Ozonolysis of $3-$hexene $(CH_3CH_2CH=CHCH_2CH_3)$ using $(i) O_3$ followed by $(ii) Zn, H_2O$ yields two molecules of propanal $(CH_3CH_2CHO)$.
Step-$II$: Reduction of propanal $(CH_3CH_2CHO)$ to propane $(CH_3CH_2CH_3)$ is achieved using Clemmensen reduction,which employs $Zn(Hg)$ and $HCl$.

(D) $1$. The ozonolysis of $C_4H_8$ gives acetone $(CH_3COCH_3)$ and formaldehyde $(HCHO)$. This confirms that $C_4H_8$ is isobutylene,$(CH_3)_2C=CH_2$.
$2$. Reaction of isobutylene with $HBr$ in the presence of peroxide follows anti-Markovnikov addition: $(CH_3)_2C=CH_2 + HBr \xrightarrow{\text{peroxide}} (CH_3)_2CH-CH_2Br$ ($X$ is $1-$bromo$-2-$methylpropane).
$3$. $X$ reacts with $Mg$ in dry ether to form the Grignard reagent: $(CH_3)_2CH-CH_2MgBr$.
$4$. The Grignard reagent reacts with $CO_2$ followed by acid hydrolysis to form a carboxylic acid: $(CH_3)_2CH-CH_2MgBr + CO_2$ $\rightarrow (CH_3)_2CH-CH_2COOMgBr$ $\xrightarrow{H_3O^+} (CH_3)_2CH-CH_2COOH$.
$5$. The final product $Y$ is $3-$methylbutanoic acid.

(A) $1$. The alkene $X$ is $2$-methylpropene $(CH_3-C(CH_3)=CH_2)$.
$2$. Reaction of $2$-methylpropene with $HBr$ follows Markovnikov's rule to give $Y$,which is $tert$-butyl bromide $(CH_3-C(Br)(CH_3)-CH_3)$.
$3$. Reaction of $tert$-butyl bromide with benzene in the presence of anhydrous $AlCl_3$ (Friedel-Crafts alkylation) gives $Z$,which is $tert$-butylbenzene $(C_6H_5-C(CH_3)_3)$.
$4$. $tert$-butylbenzene has no benzylic hydrogen atoms,so it is resistant to oxidation by $KMnO_4-KOH$.

(A) $(i)$ Calcium phosphide reacts with water to produce phosphine: $Ca_3P_2 + 6H_2O \rightarrow 3Ca(OH)_2 + 2PH_3$.
$(ii)$ White phosphorus reacts with concentrated $NaOH$ solution in an inert atmosphere to produce phosphine: $P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2$.
$(iii)$ Red phosphorus is much less reactive than white phosphorus and does not react with alkali to give phosphine under these conditions.

(C) $1$. The reaction of $P_2O_3$ with water is: $P_2O_3 + 3H_2O \rightarrow 2H_3PO_3$ ($X = H_3PO_3$,Phosphorous acid).
$2$. The structure of $H_3PO_3$ contains two $P-OH$ bonds.
$3$. The reaction of $Red \ P_4$ with alkali is: $P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2$ ($Y = H_3PO_2$,Hypophosphorous acid).
$4$. The structure of $H_3PO_2$ contains one $P-OH$ bond.
$5$. Therefore,the number of $P-OH$ bonds in $X$ $(H_3PO_3)$ and $Y$ $(H_3PO_2)$ are $2$ and $1$ respectively. Note: The provided options do not match this result exactly. Re-evaluating the question,if $X$ is $H_3PO_3$ ($2$ $P-OH$) and $Y$ is $H_3PO_2$ ($1$ $P-OH$),the answer should be $2, 1$. Given the options,there might be a typo in the question or options. Based on standard chemistry,$H_3PO_3$ has $2$ $P-OH$ and $H_3PO_2$ has $1$ $P-OH$.

(D) The basicity of an oxoacid of phosphorus is determined by the number of $P-OH$ groups present in its structure.
$1$. $H_3PO_2$ (Hypophosphorous acid): It has one $P-OH$ group,so its basicity is $1$.
$2$. $H_3PO_3$ (Orthophosphorous acid): It has two $P-OH$ groups,so its basicity is $2$.
$3$. $H_3PO_4$ (Orthophosphoric acid): It has three $P-OH$ groups,so its basicity is $3$.
Therefore,the basicity of $H_3PO_2, H_3PO_3, H_3PO_4$ is $1, 2, 3$ respectively.

(A) When chlorine reacts with hot and concentrated $NaOH$,it undergoes a disproportionation reaction.
The balanced chemical equation is:
$3 Cl_2 + 6 NaOH \longrightarrow 5 NaCl + NaClO_3 + 3 H_2 O$
Thus,the products formed are sodium chloride $(NaCl)$,sodium chlorate $(NaClO_3)$,and water $(H_2 O)$.

(B) Xenon reacts with fluorine under different conditions to form $XeF_2$,$XeF_4$,and $XeF_6$.
The reaction for the formation of $XeF_4$ is given by:
$Xe_{(g)} + 2 F_{2(g)} \xrightarrow{873 \text{ K}, 7 \text{ bar }} XeF_{4(s)}$
In this reaction,$1 \text{ mole}$ of $Xe$ reacts with $2 \text{ moles}$ of $F_2$ to form $XeF_4$.
However,in the industrial preparation of $XeF_4$,a mixture of $Xe$ and $F_2$ in a $1 : 5$ ratio is used to ensure the complete conversion of $Xe$ to $XeF_4$ and to prevent the formation of $XeF_2$ or $XeF_6$ as impurities.
Therefore,the required ratio is $1 : 5$.

(A) The hydrolysis reaction of $XeF_4$ is given by:
$6XeF_4 + 12H_2O \longrightarrow 4Xe + 2XeO_3 + 24HF + 3O_2$
Here,'$X$' is $XeO_3$.
In $XeO_3$,the central atom $Xe$ has $8$ valence electrons. It forms $3$ double bonds with $O$ atoms and has $1$ lone pair.
According to $VSEPR$ theory,the hybridization is $sp^3$ with one lone pair,resulting in a pyramidal geometry.

(C) The biodegradable polymer $PHBV$ (poly $\beta$-hydroxybutyrate-co-$\beta$-hydroxyvalerate) is obtained by the copolymerization of $3$-hydroxybutanoic acid and $3$-hydroxypentanoic acid.
The monomers are:
$Y = CH_3-CH(OH)-CH_2-COOH$ ($3$-hydroxybutanoic acid)
$Z = CH_3-CH_2-CH(OH)-CH_2-COOH$ ($3$-hydroxypentanoic acid)
Thus,option $C$ is the correct answer.

(B) $PHBV$ (Poly-$\beta$-hydroxybutyrate-co-$\beta$-hydroxyvalerate) is a copolymer formed by the condensation polymerization of two hydroxy acids:
$1$. $3$-Hydroxybutanoic acid $(CH_3-CH(OH)-CH_2-COOH)$
$2$. $3$-Hydroxypentanoic acid $(CH_3-CH_2-CH(OH)-CH_2-COOH)$
Thus,$X$ is $3$-hydroxybutanoic acid and $Y$ is $3$-hydroxypentanoic acid.
Comparing this with the given options,option $B$ matches these structures.

(C) Let us analyze the given polymers based on their type and classification:
$1$. $Polystyrene$ is a homopolymer of styrene and is a thermoplastic.
$2$. $Bakelite$ is a condensation polymer (phenol-formaldehyde resin) and is thermosetting.
$3$. $Nylon$ $6$ is a homopolymer of caprolactam and is classified as a fibre due to strong intermolecular forces like hydrogen bonding.
$4$. $Buna-N$ is a copolymer of $1,3-butadiene$ and $acrylonitrile$ and is an elastomer.
Comparing these with the options:
- Option $A$: $Polystyrene$ is a homopolymer,not a copolymer.
- Option $B$: $Bakelite$ is a condensation polymer,not an addition polymer.
- Option $C$: $Nylon$ $6$ is a homopolymer and is a fibre. This is correct.
- Option $D$: $Buna-N$ is a copolymer,not a homopolymer.
Therefore,the correct set is $Nylon$ $6$ - Homopolymer - fibre.

(A) The correct matches are as follows:
$A$. Buna-$N$-Rubber is an Elastomer $(III)$.
$B$. Terylene is a Fibre $(I)$.
$C$. Polystyrene is a Thermoplastic polymer $(IV)$.
$D$. Urea-Formaldehyde resin is a Thermosetting polymer $(II)$.
Thus,the correct sequence is $A-III, B-I, C-IV, D-II$.

(D) Thermoplastic polymers are linear or slightly branched long-chain molecules that are capable of repeatedly softening on heating and hardening on cooling.
Examples include $Polythene$,$polystyrene$,and $polyvinyls$.

(D) Fibres are polymers that have strong intermolecular forces between the chains,such as hydrogen bonds or dipole-dipole interactions. These forces lead to close packing,which imparts crystalline character and makes them long,thin,and thread-like,suitable for weaving into fabrics. Examples include nylon-$66$,silk,and terylene (also known as dacron). The structure given in option $D$ represents terylene (dacron),which is a polyester fibre formed by the condensation polymerization of ethylene glycol and terephthalic acid.

(C) Neoprene (polychloroprene) is prepared by the free radical polymerisation of chloroprene ($2$-chloro-buta-$1,3$-diene) in the presence of a peroxide initiator like benzoyl peroxide,$(C_6H_5CO)_2O_2$.
The reaction is:
$n CH_2=C(Cl)-CH=CH_2 \xrightarrow{(C_6H_5CO)_2O_2} [-CH_2-C(Cl)=CH-CH_2-]_n$
Thus,option $C$ represents the correct polymerisation reaction.

(A) $Terylene$:
The monomers for $Terylene$ are ethylene glycol and terephthalic acid. $Terylene$ is also known as $Dacron$ or polyester.
$Glyptal$:
The monomers for $Glyptal$ are ethylene glycol and phthalic acid. $Glyptal$ is an alkyd resin and a polyester.
$\therefore$ Ethylene glycol $(HO-CH_2-CH_2-OH)$ is the common monomer for both.

(B) The reaction of phenol with formaldehyde in the presence of a base $(OH^-)$ and heat leads to the formation of $Bakelite$.
$Bakelite$ is a thermosetting polymer formed by the condensation polymerization of phenol and formaldehyde.
It has a highly branched,three-dimensional structure,which is classified as a cross-linked polymer.
Therefore,$X$ is $Bakelite$ and it is a cross-linked polymer.

(B) $Quartz$ is a piezoelectric material.
Piezoelectric materials develop an electric charge due to the displacement of ions or movement of electrons upon the application of mechanical stress.

(C) According to Bragg's law: $n \lambda = 2 d \sin \theta$.
Given: $n = 1$,$\lambda = 1.54 \mathring{A}$,and $2 \theta = 60^{\circ}$,so $\theta = 30^{\circ}$.
Substituting the values into the formula: $1 \times 1.54 \mathring{A} = 2 \times d \times \sin 30^{\circ}$.
Since $\sin 30^{\circ} = 0.5$,we have: $1.54 \mathring{A} = 2 \times d \times 0.5$.
$1.54 \mathring{A} = d$.
Converting $\mathring{A}$ to $cm$ $(1 \mathring{A} = 10^{-8} \ cm)$:
$d = 1.54 \times 10^{-8} \ cm$.

(C) In an $hcp$ lattice,the number of atoms of $O$ is $N = 6$.
Number of octahedral voids $= N = 6$.
Number of tetrahedral voids $= 2N = 12$.
Atoms of $A$ occupy $25\%$ of tetrahedral voids: $A = 0.25 \times 12 = 3$.
Atoms of $B$ occupy $50\%$ of octahedral voids: $B = 0.50 \times 6 = 3$.
Atoms of $O$ in $hcp$ lattice $= 6$.
The ratio of atoms $A:B:O$ is $3:3:6$,which simplifies to $1:1:2$.
Therefore,the molecular formula is $ABO_2$.