TS EAMCET 2017 Chemistry Question Paper with Answer and Solution

(B) The $p^{th}$ term in the expansion of $(1 + x)^n$ is $T_p = {}^nC_{p-1} x^{p-1}$. The coefficient is $p = {}^nC_{p-1}$.
The $(p + 1)^{th}$ term is $T_{p+1} = {}^nC_p x^p$. The coefficient is $q = {}^nC_p$.
We know that $\frac{q}{p} = \frac{{}^nC_p}{{}^nC_{p-1}} = \frac{n - p + 1}{p}$.
Thus,$pq = p(n - p + 1)$ is not the correct path; rather,we use the property of binomial coefficients:
$\frac{q}{p} = \frac{n - p + 1}{p} \implies qp = p(n - p + 1)$.
Actually,the question implies the coefficients are $p$ and $q$ as variables. Given $p = {}^nC_{p-1}$ and $q = {}^nC_p$,the relation $\frac{q}{p} = \frac{n - p + 1}{p}$ leads to $q = \frac{n - p + 1}{p} \cdot p = n - p + 1$.
Therefore,$p + q = n + 1$.

(B) Given that $\alpha$ and $\beta$ are the roots of $ax^2+bx+c=0$.
Then,$\alpha+\beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
The roots of the equation $px^2+qx+r=0$ are $x_1 = \frac{1-\alpha}{\alpha} = \frac{1}{\alpha}-1$ and $x_2 = \frac{1-\beta}{\beta} = \frac{1}{\beta}-1$.
The equation with roots $x_1$ and $x_2$ is given by $x^2 - (x_1+x_2)x + x_1x_2 = 0$.
Sum of roots: $x_1+x_2 = (\frac{1}{\alpha}-1) + (\frac{1}{\beta}-1) = \frac{\alpha+\beta}{\alpha\beta} - 2 = \frac{-b/a}{c/a} - 2 = -\frac{b}{c} - 2 = -\frac{b+2c}{c}$.
Product of roots: $x_1x_2 = (\frac{1}{\alpha}-1)(\frac{1}{\beta}-1) = \frac{1}{\alpha\beta} - (\frac{1}{\alpha}+\frac{1}{\beta}) + 1 = \frac{1}{\alpha\beta} - \frac{\alpha+\beta}{\alpha\beta} + 1 = \frac{a}{c} - (\frac{-b/a}{c/a}) + 1 = \frac{a}{c} + \frac{b}{c} + 1 = \frac{a+b+c}{c}$.
Substituting these into the quadratic equation: $x^2 - (-\frac{b+2c}{c})x + \frac{a+b+c}{c} = 0$.
Multiplying by $c$: $cx^2 + (b+2c)x + (a+b+c) = 0$.
Comparing this with $px^2+qx+r=0$,we get $r = a+b+c$.

(B) The $p^{th}$ term in the expansion of $(1+x)^n$ is $T_p = { }^n C_{p-1} x^{p-1}$,so its coefficient is $p = { }^n C_{p-1}$.
The $(p+1)^{th}$ term is $T_{p+1} = { }^n C_p x^p$,so its coefficient is $q = { }^n C_p$.
We know the property of binomial coefficients: $\frac{{ }^n C_r}{{ }^n C_{r-1}} = \frac{n-r+1}{r}$.
Here,$\frac{q}{p} = \frac{{ }^n C_p}{{ }^n C_{p-1}} = \frac{n-p+1}{p}$.
Thus,$q = \frac{p(n-p+1)}{p} = n-p+1$.
Rearranging the terms,we get $p+q = n+1$.

(B) The oxidation of $cyclohexene$ with a strong oxidizing agent like acidic $KMnO_4$ results in the cleavage of the double bond and the formation of a dicarboxylic acid.
Specifically,$cyclohexene$ $(C_6H_{10})$ undergoes oxidative cleavage to form $adipic$ $acid$ $(HOOC-(CH_2)_4-COOH)$,which is a $6$-carbon dicarboxylic acid.
The chemical reaction is as follows:
$3C_6H_{10} + 8KMnO_4 + 4H_2O \rightarrow 3HOOC(CH_2)_4COOH + 8KOH + 8MnO_2$
Thus,the product formed is $adipic$ $acid$.

(C) According to $VSEPR$ theory,the shape of a molecule depends on the number of bonding pairs $(b.p.)$ and lone pairs $(l.p.)$ around the central atom.
$1.$ $SO_3$: Central atom $S$ has $3$ $b.p.$ and $0$ $l.p.$,resulting in a trigonal planar shape.
$2.$ $BrF_3$: Central atom $Br$ has $3$ $b.p.$ and $2$ $l.p.$,resulting in a $T$-shaped geometry.
$3.$ $SO_3^{2-}$: Central atom $S$ has $3$ $b.p.$ and $1$ $l.p.$,which gives it a pyramidal shape.
$4.$ $XeF_2$: Central atom $Xe$ has $2$ $b.p.$ and $3$ $l.p.$,resulting in a linear shape.
Therefore,the correct species with a pyramidal shape is $SO_3^{2-}$.

(A) The bond lengths for the given molecules are as follows:
$H_2 = 74 \ pm$
$F_2 = 144 \ pm$
$Cl_2 = 199 \ pm$
$I_2 = 267 \ pm$
Therefore, the order for $F_2, H_2, Cl_2, I_2$ is $144, 74, 199, 267 \ pm$.
Thus, the correct option is $A$.

(D) Ionic radius depends on the number of shells and the effective nuclear charge.
$Na^{+}$ $(1s^2 2s^2 2p^6)$ has $3$ shells,while $Li^{+}$,$Mg^{2+}$,and $Be^{2+}$ have $2$ shells.
Among ions with $2$ shells ($Li^{+}$,$Mg^{2+}$,$Be^{2+}$),the ionic radius decreases as the positive charge increases.
$Li^{+}$ $(Z=3)$ has the lowest charge,followed by $Mg^{2+}$ ($Z=12$,but smaller due to higher charge),and $Be^{2+}$ $(Z=4)$ is the smallest.
Thus,the correct order is $Na^{+} > Li^{+} > Mg^{2+} > Be^{2+}$.

(B) The melting point order of group $14$ elements is $C > Si > Ge > Sn > Pb$.
Among the given options,$Pb$ $(Lead)$ has the lowest melting point.

(B) The atomic radii of the given elements are approximately: $Al = 143 \ pm$, $Si = 117 \ pm$, $N = 74 \ pm$, and $F = 64 \ pm$.
$Al$ and $Si$ belong to the $3^{rd}$ period, while $N$ and $F$ belong to the $2^{nd}$ period.
Within a period, the atomic radius decreases from left > right due to an increase in effective nuclear charge.
Thus, the order of atomic radii is $Al (143 \ pm) > Si (117 \ pm) > N (74 \ pm) > F (64 \ pm)$.

(B) The cyclopentadienyl anion is a cyclic,planar species with $6 \pi \ e^-$ (where $n=1$ in the $4n+2$ $H$ückel rule).
Since it contains $6 \pi \ e^-$,it follows $H$ückel's rule and is aromatic.
Because it does not contain a benzene ring,it is classified as non-benzenoid.
Therefore,it is non-benzenoid and aromatic.

(C) Given that $\alpha$ and $\beta$ are the roots of $ax^2+bx+c=0$.
Then,$\alpha+\beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
The roots of the new equation are $\gamma = \frac{1-\alpha}{\alpha} = \frac{1}{\alpha}-1$ and $\delta = \frac{1-\beta}{\beta} = \frac{1}{\beta}-1$.
The new equation is $x^2 - (\gamma+\delta)x + \gamma\delta = 0$.
Sum of roots: $\gamma+\delta = (\frac{1}{\alpha}-1) + (\frac{1}{\beta}-1) = \frac{\alpha+\beta}{\alpha\beta} - 2 = \frac{-b/a}{c/a} - 2 = -\frac{b}{c} - 2 = -\frac{b+2c}{c}$.
Product of roots: $\gamma\delta = (\frac{1}{\alpha}-1)(\frac{1}{\beta}-1) = \frac{1}{\alpha\beta} - (\frac{1}{\alpha}+\frac{1}{\beta}) + 1 = \frac{1}{\alpha\beta} - \frac{\alpha+\beta}{\alpha\beta} + 1 = \frac{a}{c} - \frac{-b/a}{c/a} + 1 = \frac{a}{c} + \frac{b}{c} + 1 = \frac{a+b+c}{c}$.
The equation is $x^2 - (-\frac{b+2c}{c})x + \frac{a+b+c}{c} = 0$.
Multiplying by $c$,we get $cx^2 + (b+2c)x + (a+b+c) = 0$.
Comparing with $px^2+qx+r=0$,we find $r = a+b+c$.

(B) Let $z = x + iy$. Then $\frac{2z+1}{iz+1} = \frac{2(x+iy)+1}{i(x+iy)+1} = \frac{(2x+1) + 2yi}{(1-y) + xi}$.
Multiplying the numerator and denominator by the conjugate of the denominator $(1-y) - xi$:
$\frac{((2x+1) + 2yi)((1-y) - xi)}{(1-y)^2 + x^2} = \frac{(2x+1)(1-y) + 2xy + i(2y(1-y) - x(2x+1))}{(1-y)^2 + x^2}$.
The imaginary part is given as $-2$:
$\frac{2y - 2y^2 - 2x^2 - x}{(1-y)^2 + x^2} = -2$.
$2y - 2y^2 - 2x^2 - x = -2((1-y)^2 + x^2) = -2(1 - 2y + y^2 + x^2) = -2 + 4y - 2y^2 - 2x^2$.
Simplifying the equation:
$-x - 2y + 2 = 0$,which is $x + 2y - 2 = 0$.
This represents a straight line.

(D) In the expansion of $(1+x)^n$,the general term,i.e.,$(r+1)$-th term is given by $T_{r+1} = { }^n C_r x^r$.
The coefficient of the $p$-th term is ${ }^n C_{p-1}$. Given that this coefficient is $p$,we have $p = { }^n C_{p-1}$.
The coefficient of the $(p+1)$-th term is ${ }^n C_p$. Given that this coefficient is $q$,we have $q = { }^n C_p$.
Now,consider the ratio $\frac{q}{p} = \frac{{ }^n C_p}{{ }^n C_{p-1}}$.
Using the formula $\frac{{ }^n C_r}{{ }^n C_{r-1}} = \frac{n-r+1}{r}$,we get:
$\frac{q}{p} = \frac{n-p+1}{p}$.
Cross-multiplying,we get $qp = p(n-p+1)$.
Wait,let us re-evaluate the given condition: coefficients are $p$ and $q$.
Actually,the problem states the coefficients are $p$ and $q$. Let's use the property $\frac{{ }^n C_p}{{ }^n C_{p-1}} = \frac{n-p+1}{p}$.
Since $q = { }^n C_p$ and $p = { }^n C_{p-1}$,we have $\frac{q}{p} = \frac{n-p+1}{p}$.
This implies $q = n-p+1$,which simplifies to $p+q = n+1$.

(A) Calgon is the trade name for sodium hexametaphosphate,$\left(NaPO_3\right)_6$.
It ionizes in water to provide a complex anion:
$Na_2\left(Na_4P_6O_{18}\right) \longrightarrow 2Na^{+} + \left[Na_4P_6O_{18}\right]^{2-}$
When added to hard water,the $Ca^{2+}$ ions displace the $Na^{+}$ ions from the complex anion to form a soluble complex:
$Ca^{2+} + \left[Na_4P_6O_{18}\right]^{2-} \longrightarrow \left[CaNa_2P_6O_{18}\right]^{2-} + 2Na^{+}$

(D) Given,$|a|=3, |b|=4$ and $|a+b|=5$.
Since $|a+b|=5$,squaring both sides gives $|a+b|^2=25$.
Expanding this,we get $(a+b) \cdot (a+b) = 25$,which simplifies to $|a|^2 + 2(a \cdot b) + |b|^2 = 25$.
Substituting the given values,$3^2 + 2(a \cdot b) + 4^2 = 25$,so $9 + 2(a \cdot b) + 16 = 25$,which implies $2(a \cdot b) = 0$,hence $a \cdot b = 0$.
Now,we need to find $|a-b|$.
We know that $|a-b|^2 = (a-b) \cdot (a-b) = |a|^2 - 2(a \cdot b) + |b|^2$.
Substituting the values,$|a-b|^2 = 3^2 - 2(0) + 4^2 = 9 + 16 = 25$.
Therefore,$|a-b| = \sqrt{25} = 5$.

(A) The molecular formula of diborane is $B_2H_6$.
In the structure of diborane,there are $4$ terminal $B-H$ bonds which are $2$-centre-$2$-electron $(2C-2e^-)$ bonds.
Additionally,there are $2$ bridging $B-H-B$ bonds,which are $3$-centre-$2$-electron $(3C-2e^-)$ bonds.
Therefore,the bonding consists of $4$ two-centre-two-electron bonds and $2$ three-centre-two-electron bonds.

(D) Among hydrogen halides,the strongest acid is $HI$.
As the size of the halogen increases,the bond length increases,and consequently,the bond dissociation energy decreases.
Therefore,$HI$ releases $H^{\oplus}$ ions most easily,exhibiting the highest acidic nature.

(C) The Joule-Thomson effect describes the temperature change of a real gas as it is forced through a valve or porous plug while kept insulated so that no heat is exchanged with the environment.
For most gases,expansion results in cooling because work is done to overcome intermolecular forces.
However,for gases like $H_2$ and $He$,the inversion temperature is very low (for $He$,it is approximately $40 \ K$).
If the gas temperature is above its inversion temperature during expansion,it exhibits a heating effect instead of cooling.
Since the inversion temperature of $He$ is very low,it shows a heating effect under standard conditions.

(B) To balance the equation $x KNO_3 + y C_{12}H_{22}O_{11} \longrightarrow p N_2 + q CO_2 + r H_2O + s K_2CO_3$:
$1$. Balance $K$: $x = 2s$
$2$. Balance $N$: $x = 2p$
$3$. Balance $C$: $12y = q + s$
$4$. Balance $H$: $22y = 2r \implies 11y = r$
$5$. Balance $O$: $3x + 11y = 2q + r + 3s$
Substituting the values from option $(b)$: $x=48, y=5, p=24, q=36, r=55, s=24$.
$48 KNO_3 + 5 C_{12}H_{22}O_{11} \longrightarrow 24 N_2 + 36 CO_2 + 55 H_2O + 24 K_2CO_3$
Checking atoms:
$K: 48 = 2(24) = 48$ (Balanced)
$N: 48 = 2(24) = 48$ (Balanced)
$C: 12(5) = 60; 36 + 24 = 60$ (Balanced)
$H: 22(5) = 110; 55(2) = 110$ (Balanced)
$O: 3(48) + 11(5) = 144 + 55 = 199; 2(36) + 55 + 3(24) = 72 + 55 + 72 = 199$ (Balanced)
Thus,option $(b)$ is correct.

(B) According to the kinetic theory of gases,the average speed of gas molecules is proportional to $\sqrt{T}$. Since the frequency of collisions is directly proportional to the average speed of the molecules,it follows that the frequency of collisions increases in proportion to the square root of temperature $(\sqrt{T})$.

(C) The energy levels of a hydrogen atom are given by the formula: $E_n = -13.6 / n^2 \ eV$.
For the ground state,$n = 1$,so $E_1 = -13.6 \ eV$.
An excited state occurs when $n > 1$.
For the first excited state,$n = 2$.
Substituting $n = 2$ into the formula: $E_2 = -13.6 / 2^2 = -13.6 / 4 = -3.4 \ eV$.

(C) When an electron in a hydrogen atom is excited to the $n$th energy level,it can return to lower energy levels by emitting photons.
The total number of possible spectral lines is given by the formula:
$\text{Number of spectral lines} = \frac{n(n-1)}{2}$
where $n$ represents the principal quantum number of the excited state.

(C) The standard enthalpy of formation $(\Delta H_f^{\circ})$ is defined as the enthalpy change when $1 \ mol$ of a substance is formed from its elements in their standard states.
For the formation of liquid water: $H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(l)}$.
The standard enthalpy of formation for liquid water at $298 \ K$ $(25^{\circ} C)$ is $-286 \ kJ/mol$.

(D) The reaction is: $CO_{2(g)} + H_{2(g)} \longrightarrow CO_{(g)} + H_2 O_{(g)}$
$\Delta H_r^{\circ} = \sum \Delta H_{f, \text{products}}^{\circ} - \sum \Delta H_{f, \text{reactants}}^{\circ}$
$\Delta H_r^{\circ} = [\Delta H_{f, CO}^{\circ} + \Delta H_{f, H_2 O}^{\circ}] - [\Delta H_{f, CO_2}^{\circ} + \Delta H_{f, H_2}^{\circ}]$
Since the standard enthalpy of formation for an element in its standard state is zero,$\Delta H_{f, H_2}^{\circ} = 0$.
$\Delta H_r^{\circ} = [-110.5 + (-241.8)] - [-393.5 + 0]$
$\Delta H_r^{\circ} = -352.3 + 393.5 = 41.2 \ kJ \ mol^{-1}$

(D) Lucas reagent is a mixture of concentrated $HCl$ and anhydrous $ZnCl_2$. The reaction of alcohols with Lucas reagent proceeds via an $S_N1$ mechanism,which involves the formation of a carbocation intermediate. The reactivity of alcohols towards Lucas reagent follows the order: $3^{\circ} > 2^{\circ} > 1^{\circ}$.
$(A)$ $CH_3-CH_2-CH_2-CH_2-OH$ is a $1^{\circ}$ alcohol.
$(B)$ $CH_3-CH_2-CH(OH)-CH_3$ is a $2^{\circ}$ alcohol.
$(C)$ $(CH_3)_2CH-CH_2-OH$ is a $1^{\circ}$ alcohol.
$(D)$ $(CH_3)_3C-OH$ is a $3^{\circ}$ alcohol.
Since $3^{\circ}$ alcohols form the most stable carbocation,they react fastest with Lucas reagent,often at room temperature.

(A) The compound $3-$methylpent$-1-$en$-3-$ol is an allylic alcohol. In the presence of an acid,it undergoes protonation of the $-OH$ group followed by the loss of water to form a resonance-stabilized carbocation: $CH_3-CH_2-C^+(CH_3)-CH=CH_2 \leftrightarrow CH_3-CH_2-C(CH_3)=CH-CH_2^+$.
This carbocation can lose a proton to form a more stable,conjugated diene or a substituted alkene. Specifically,the loss of optical activity occurs because the chiral center at $C-3$ is destroyed upon the formation of the planar carbocation intermediate. The major product formed is $3-$methylpenta$-1,3-$diene or a related stable alkene isomer depending on the specific rearrangement,but among the given options,the most stable alkene formed via rearrangement is $3-$methylpent$-2-$ene $(CH_3-CH_2-C(CH_3)=CH-CH_3)$.

(D) $DIBAL-H$ (Diisobutylaluminium hydride) is a selective reducing agent.
It is specifically used for the partial reduction of esters to aldehydes at low temperatures (typically $195 \ K$ to $203 \ K$),preventing further reduction to primary alcohols.

(D) Aniline is an aromatic primary amine,while cyclohexylamine is an aliphatic primary amine.
The azo dye test is specific to aromatic primary amines.
When aniline reacts with nitrous acid $(HNO_2)$ at $0-5 \ ^{\circ}C$ followed by coupling with $\beta$-naphthol,it forms a bright orange or red azo dye.
Cyclohexylamine,being aliphatic,does not form a stable diazonium salt at low temperatures and consequently does not give the azo dye test.
Therefore,the azo dye test is used to distinguish between them.

(D) In $DNA$,adenine $(A)$ pairs with thymine $(T)$ through $2$ hydrogen bonds $(A=T)$.
In contrast,guanine $(G)$ pairs with cytosine $(C)$ through $3$ hydrogen bonds $(G \equiv C)$.

(B) For a zero order reaction:
$1$. The integrated rate law is $[A] = [A]_0 - kt$. Thus,a plot of $[A]$ versus $t$ is a straight line with a negative slope $(-k)$,which matches graph $(i)$.
$2$. The half-life is $t_{1/2} = \frac{[A]_0}{2k}$. Thus,a plot of $t_{1/2}$ versus $[A]_0$ is a straight line passing through the origin with a slope of $\frac{1}{2k}$,which matches graph $(ii)$.
$3$. The rate law is $\text{Rate} = k[A]^0 = k$. The rate is independent of the reactant concentration. Thus,a plot of $\text{Rate}$ versus $[A]$ is a horizontal line,which matches graph $(iv)$.
Graph $(iii)$ shows $\text{Rate} \propto [A]$,which is characteristic of a first order reaction.
Therefore,the correct representations are $(i), (ii),$ and $(iv)$.

(A) Histamine is a potent vasodilator. It causes the dilation of blood vessels,which leads to a decrease in blood pressure.

(C) In group $16$,the metallic character increases as we move down the group from $O$ to $Po$.
Metallic oxides are generally basic in nature,while non-metallic oxides are acidic.
Since $Po$ is a metal,its oxide $PoO$ is the most basic among the given options.
As we move down the group,the electronegativity of the element decreases,which leads to an increase in the basic character of the corresponding oxides.

(A) In crystal field theory,the distribution of electrons in $d$-orbitals depends on the relative magnitude of crystal field splitting energy $(\Delta_0)$ and pairing energy $(P)$.
When $\Delta_0 > P$,the energy required to pair electrons is less than the energy required to promote an electron to the higher energy $e_g$ orbital.
Therefore,electrons prefer to pair up in the $t_{2g}$ orbitals,resulting in the formation of low-spin complexes.
Conversely,when $\Delta_0 < P$,electrons occupy the higher energy $e_g$ orbitals before pairing,resulting in high-spin complexes.

(C) The half-cell reaction is $2H^{+} + 2e^{-} \rightleftharpoons H_2$.
Using the Nernst equation: $E = E^{\circ} - \frac{0.059}{n} \log \frac{P_{H_2}}{[H^{+}]^2}$.
Given $E = -0.059 \ V$,$E^{\circ} = 0 \ V$ (for $SHE$),$n = 2$,and $P_{H_2} = 1 \ bar$.
Substituting the values: $-0.059 = 0 - \frac{0.059}{2} \log \frac{1}{[H^{+}]^2}$.
$-0.059 = -\frac{0.059}{2} \times (-2 \log [H^{+}])$.
$-0.059 = 0.059 \log [H^{+}]$.
$\log [H^{+}] = -1$.
$[H^{+}] = 10^{-1} = 0.1 \ M$.

(A) The cell reactions are:
Anode (oxidation): $Cl^{-} \rightarrow \frac{1}{2} Cl_2 + e^{-}$
Cathode (reduction): $MCl + e^{-} \rightarrow M + Cl^{-}$
Overall cell reaction: $MCl \rightarrow M + \frac{1}{2} Cl_2$
Using the Nernst equation at equilibrium ($E_{cell} = 0$ is not applicable here as the cell is not at equilibrium,but we use the relation $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log Q$):
Given $E_{cell} = -1.140 \ V$,$E^{\circ}_{cell} = -0.55 \ V$,$n = 1$:
$-1.140 = -0.55 - 0.0591 \log K_c$
$-0.59 = -0.0591 \log K_c$
$\log K_c = \frac{0.59}{0.0591} \approx 10$
$K_c = 10^{10}$
Since the reaction $MCl \rightarrow M + \frac{1}{2} Cl_2$ is the reverse of the solubility product reaction $M^{+} + Cl^{-} \rightarrow MCl$,the equilibrium constant $K_c$ is the reciprocal of $K_{sp}$.
$K_{sp} = \frac{1}{K_c} = \frac{1}{10^{10}} = 10^{-10}$

(D) It is an electrophilic aromatic substitution reaction.
In phenyl benzoate $(Ph-COO-Ph)$,the ester group is attached to two phenyl rings.
The carbonyl carbon is attached to one phenyl ring,making it electron-deficient (deactivated),while the oxygen atom is attached to the other phenyl ring,which acts as an electron-donating group through resonance.
Therefore,the electrophilic nitration occurs on the phenyl ring attached to the oxygen atom.
Due to steric hindrance from the bulky $-COOPh$ group,the para-substitution is favored over ortho-substitution.
Thus,the major product is $4$-nitrophenyl benzoate.

(B) Vapour phase refining is a technique used to purify metals by converting them into a volatile compound and then decomposing it to obtain the pure metal.
$1$. The Van Arkel method is used for refining $Zr$ and $Ti$.
$2$. The Mond process is used for refining $Ni$.
Therefore,$Zr$ is purified by vapour phase refining.

(C) The reaction between $Cu_2O$ and $Cu_2S$ is a self-reduction process used in the extraction of copper.
The chemical equation is: $2Cu_2O + Cu_2S \longrightarrow 6Cu + SO_2$.
In this process,cuprous sulphide $(Cu_2S)$ acts as a reducing agent for cuprous oxide $(Cu_2O)$.
This reaction occurs in the $Bessemer$ converter during the metallurgy of copper.

(A) In dry acetone,the reaction with iodide ion follows the $S_N2$ mechanism.
In an $S_N2$ reaction,the rate of reaction is inversely proportional to the steric hindrance around the electrophilic carbon atom.
The order of steric hindrance is: $Methyl < Primary < Secondary < Tertiary$.
Therefore,the order of reactivity is: $CH_3Br (IV) > CH_3CH_2Br (I) > (CH_3)_2CHBr (II) > (CH_3)_3CBr (III)$.
Thus,the correct order is $IV > I > II > III$.

(D) The stability of the lower oxidation state increases down the group in Group $13$ due to the inert pair effect.
$Tl$ (Thallium) belongs to Group $13$ and exhibits a stable $+1$ oxidation state,which is lower than the group oxidation state of $+3$.
Therefore,$Tl$ forms stable compounds in the low oxidation state.

(C) Buna-$S$ is a synthetic copolymer formed by the polymerization of two monomers:
$(I)$ $1,3-$butadiene $(CH_2=CH-CH=CH_2)$
$(II)$ Styrene $(C_6H_5-CH=CH_2)$
Thus,the correct option is $C$.

(B) In a close-packed structure ($CCP$ or $HCP$),if there are $n$ spheres (atoms or ions) in the packing,then:
Number of octahedral voids $= n$
Number of tetrahedral voids $= 2n$
In a $CCP$ unit cell,the number of atoms per unit cell is $4$.
Therefore,the number of octahedral voids $= 4$.
The number of tetrahedral voids $= 2 \times 4 = 8$.
Thus,the number of tetrahedral and octahedral voids are $8$ and $4$ respectively.

(A) The given graph shows a minimum in the vapour pressure curve at $x = 0.5$.
This indicates that the solution exhibits a large positive deviation from Raoult's law.
For solutions showing a large positive deviation from Raoult's law,the boiling point of the mixture is lower than the boiling points of the individual components,resulting in a minimum boiling azeotrope.
Therefore,the boiling point $(B.P.)$ vs. mole fraction $(x)$ curve must show a minimum at the same composition $(x = 0.5)$.
Among the given options,option $A$ represents a minimum boiling azeotrope curve.

(D) For real solutions showing negative deviation from Raoult's law,the solute-solvent interactions are stronger than the solute-solute or solvent-solvent interactions.
This leads to a decrease in the total volume of the solution,meaning $\Delta V_{\text{mix}} < 0$.
Additionally,the formation of stronger bonds releases energy,making the process exothermic,which means $\Delta H_{\text{mix}} < 0$.

(A) Given: Weight of $H_2SO_4 = 98 \ g$,Molar mass of $H_2SO_4 = 98 \ g \ mol^{-1}$,Weight of solvent (water) $= 2 \ g$.
Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
Moles of $H_2SO_4 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{98 \ g}{98 \ g \ mol^{-1}} = 1 \ mol$.
Weight of solvent in $kg = \frac{2 \ g}{1000} = 0.002 \ kg$.
Molality $(m) = \frac{\text{Moles of solute}}{\text{Mass of solvent in } kg} = \frac{1 \ mol}{0.002 \ kg} = 500 \ m$.

(C) The atomic number of $Praseodymium$ $(Pr)$ is $59$.
Following the $Aufbau$ principle,we fill the orbitals after the $Xenon$ ($Xe$,atomic number $54$) core.
The remaining $5$ electrons occupy the $4f$ and $6s$ orbitals.
The correct electronic configuration is $[Xe] 4f^3 6s^2$.

(A) Adsorption is a spontaneous process where gas molecules are trapped on a solid surface,leading to a decrease in the randomness of the system. Thus,$\Delta S < 0$.
Since the process is spontaneous,the Gibbs free energy change must be negative $(\Delta G < 0)$.
From the equation $\Delta G = \Delta H - T\Delta S$,for $\Delta G$ to be negative when $\Delta S$ is negative,$\Delta H$ must be negative.
Therefore,the process is exothermic $(\Delta H < 0)$.