AP EAMCET 2016 Chemistry Question Paper with Answer and Solution

(A) For a concave mirror,the mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Using sign convention,$u$ is replaced by $-u$ and $f$ is replaced by $-f$.
Thus,$\frac{1}{v} - \frac{1}{u} = -\frac{1}{f}$,which gives $\frac{1}{v} = \frac{1}{u} - \frac{1}{f} = \frac{f - u}{uf}$.
So,$v = \frac{uf}{f - u} = -\frac{uf}{u - f}$.
The far end of the rod is at infinity $(u = \infty)$,so its image is formed at the focus $(v = f)$.
The near end of the rod is at distance $u$,so its image is at distance $v = \frac{uf}{u - f}$ from the pole.
The length of the image is the difference between the positions of the two images: $L = |v - f|$.
$L = |\frac{uf}{u - f} - f| = |\frac{uf - f(u - f)}{u - f}| = |\frac{uf - uf + f^2}{u - f}| = \frac{f^2}{u - f}$.

(D) For the near end of the rod,the mirror formula is $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$.
Using the sign convention for a concave mirror,$u$ and $f$ are negative.
Thus,$\frac{1}{-f} = \frac{1}{-u} + \frac{1}{v}$,which gives $\frac{1}{v} = \frac{1}{u} - \frac{1}{f} = \frac{f - u}{uf}$.
So,$v = \frac{uf}{f - u} = -\frac{uf}{u - f}$. The distance of the image from the mirror is $|v| = \frac{uf}{u - f}$.
The far end of the rod is at infinity $(u = \infty)$. According to the mirror formula,the image of the far end is formed at the focus $(v = f)$.
The length of the image is the difference between the image positions of the two ends: $L = |v| - f$.
$L = \frac{uf}{u - f} - f = \frac{uf - f(u - f)}{u - f} = \frac{uf - uf + f^2}{u - f} = \frac{f^2}{u - f}$.

(D) To determine the number of lone pairs on the central atom,we use the formula: $\text{Lone pairs} = \frac{1}{2} \times (V - N)$,where $V$ is the number of valence electrons of the central atom and $N$ is the number of electrons involved in bonding (monovalent atoms).
$1$. For $NH_3$: Central atom $N$ has $5$ valence electrons. It forms $3$ bonds. $\text{Lone pairs} = \frac{1}{2} \times (5 - 3) = 1$.
$2$. For $H_2O$: Central atom $O$ has $6$ valence electrons. It forms $2$ bonds. $\text{Lone pairs} = \frac{1}{2} \times (6 - 2) = 2$.
$3$. For $ClF_3$: Central atom $Cl$ has $7$ valence electrons. It forms $3$ bonds. $\text{Lone pairs} = \frac{1}{2} \times (7 - 3) = 2$.
$4$. For $XeF_2$: Central atom $Xe$ has $8$ valence electrons. It forms $2$ bonds. $\text{Lone pairs} = \frac{1}{2} \times (8 - 2) = 3$.
Thus,$XeF_2$ has the maximum number of lone pairs $(3)$ on the central atom.

(B) In $PCl_5$ (phosphorus pentachloride),the geometry is trigonal bipyramidal.
Due to the presence of three equatorial bonds and two axial bonds,the axial $P-Cl$ bonds are longer than the equatorial $P-Cl$ bonds because of greater repulsion from the equatorial bond pairs.
In $SF_6$,$BCl_3$,and $CCl_4$,all bond lengths are equivalent due to the high symmetry of the molecules.

(B) The reaction is $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$.
Let the initial moles of $N_2O_4$ be $1$.
At equilibrium,$20 \%$ of $N_2O_4$ is dissociated,so the moles of $N_2O_4$ remaining $= 1 - 0.2 = 0.8$.
The moles of $NO_2$ formed $= 2 \times 0.2 = 0.4$.
Total moles at equilibrium $= 0.8 + 0.4 = 1.2$.
Partial pressure of $N_2O_4$ $(P_{N_2O_4})$ $= \frac{0.8}{1.2} \times 600 \ mm \ Hg = 400 \ mm \ Hg$.
Partial pressure of $NO_2$ $(P_{NO_2})$ $= \frac{0.4}{1.2} \times 600 \ mm \ Hg = 200 \ mm \ Hg$.
$K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{(200)^2}{400} = \frac{40000}{400} = 100$.

(C) For the reaction $2 \ N_2O_{5(g)} \longrightarrow 4 \ NO_{2(g)} + O_{2(g)}$,the rate expression is given by:
$-\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$
Given that the rate of disappearance of $N_2O_5$ is $-\frac{d[N_2O_5]}{dt} = 1.2 \times 10^{-5} \ mol \ L^{-1} \ s^{-1}$.
From the rate expression,we have:
$\frac{d[NO_2]}{dt} = 2 \times \left(-\frac{d[N_2O_5]}{dt}\right)$
$\frac{d[NO_2]}{dt} = 2 \times (1.2 \times 10^{-5}) = 2.4 \times 10^{-5} \ mol \ L^{-1} \ s^{-1}$.

(A) The electronic configuration of the element in its $+2$ oxidation state is $[Ar] 3d^1$.
To find the neutral state configuration,we add the two electrons back to the $4s$ orbital,which is filled before the $3d$ orbital.
Thus,the neutral state configuration is $[Ar] 3d^1 4s^2$.
The principal quantum number of the outermost shell is $n=4$,which indicates the element belongs to period-$4$.
The total number of valence electrons is $1 (d) + 2 (s) = 3$,which indicates the element belongs to group-$3$ (Scandium).
Therefore,the correct position is period-$4$,group-$3$.

(C) The "blue baby syndrome" (also known as methemoglobinemia) is caused by the presence of high concentrations of nitrates in drinking water.
When nitrates are ingested,they are converted into nitrites in the body,which then react with hemoglobin to form methemoglobin.
Methemoglobin is unable to transport oxygen effectively,leading to a bluish discoloration of the skin and potential health risks.

(A) $1$. Identify the longest carbon chain containing the double bonds. The structure is $(CH_3)_2CH-CH=CH-CH=CH-CH(C_2H_5)-CH_3$.
$2$. The longest chain has $9$ carbon atoms,so the parent alkane is nonane.
$3$. Numbering the chain from the left gives the double bonds at positions $3$ and $5$.
$4$. There are methyl groups at positions $2$ and $7$.
$5$. The $IUPAC$ name is $2, 7$-dimethyl-$3, 5$-nonadiene.

(D) homologous series is a group of organic compounds having the same functional group and similar chemical properties,where each successive member differs by a $CH_2$ unit.
For the given formula $C_5H_8O_2N$,the next member in the series is obtained by adding a $CH_2$ group.
$C_5H_8O_2N + CH_2 = C_6H_{10}O_2N$.
Therefore,$C_6H_{10}O_2N$ belongs to the same homologous series.

(B) In the Dumas method,the percentage of nitrogen is calculated using the formula:
$\% \text{ of nitrogen} = \frac{28 \times V \times 100}{22400 \times W}$
Where $V$ is the volume of $N_2$ at $STP$ in $mL$ and $W$ is the mass of the organic compound in $g$.
Substituting the given values:
$\% \text{ of nitrogen} = \frac{28 \times 45 \times 100}{22400 \times 0.3} = \frac{126000}{6720} = 18.75 \%$
Thus,the percentage of nitrogen is approximately $18.7 \%$.

(C) The reaction sequence is as follows:
$1$. $2-$methyl$-2-$bromopropane reacts with $Mg$ in the presence of dry ether to form a Grignard reagent,$X$,which is $2-$methyl$-2-$propylmagnesium bromide: $(CH_3)_3CBr + Mg \xrightarrow{\text{Dry ether}} (CH_3)_3CMgBr (X)$.
$2$. The Grignard reagent $X$ then reacts with water $(H_2O)$ to undergo hydrolysis,yielding $2-$methyl propane $(Z)$ and magnesium hydroxybromide: $(CH_3)_3CMgBr + H_2O \rightarrow (CH_3)_3CH (Z) + Mg(OH)Br$.
$3$. Therefore,$Z$ is $2-$methyl propane.

(C) Calgon is a commercial name for sodium hexametaphosphate,which has the chemical formula $Na_6P_6O_{18}$.
It is used in water softening to sequester calcium and magnesium ions.

(C) Cationic hydrolysis occurs in salts of a weak base and a strong acid.
In $NH_4Cl$,the salt dissociates as: $NH_4Cl \rightarrow NH_4^{+} + Cl^{-}$.
$Cl^{-}$ is the conjugate base of a strong acid $(HCl)$ and does not undergo hydrolysis.
$NH_4^{+}$ is the conjugate acid of a weak base $(NH_3)$ and undergoes hydrolysis: $NH_4^{+} + H_2O \rightleftharpoons NH_3 + H_3O^{+}$.
Thus,only cationic hydrolysis is involved.

(C) The equivalent current '$I$' produced by the rotating ring is given by the charge per unit time:
$I = \frac{q}{T} = \frac{q \omega}{2 \pi}$
where '$T = \frac{2 \pi}{\omega}$' is the time period of rotation.
The magnetic dipole moment '$M$' of a current loop is given by the product of current '$I$' and the area '$A$' of the loop:
$M = I \times A$
Since the area of the ring is '$A = \pi R^2$',we have:
$M = \left( \frac{q \omega}{2 \pi} \right) \times (\pi R^2)$
$M = \frac{q \omega R^2}{2}$

(A) $AlCl_3$ is an electron-deficient compound with only $6$ electrons in the valence shell of $Al$.
To complete its octet,it forms a dimer $Al_2Cl_6$ where two chlorine atoms act as bridges.
Each bridging chlorine atom donates a lone pair of electrons to the vacant $p$-orbital of the $Al$ atom of the other $AlCl_3$ unit,forming a coordinate bond.
Thus,both the assertion and the reason are true,and the reason correctly explains the stability of the dimer.

(D) Given the circle $x^2+y^2=9$ and the line $x+y=3$.
To find the equation of the pair of lines joining the origin to the points of intersection,we homogenize the equation of the circle using the equation of the line.
Since $x+y=3$,we have $\frac{x+y}{3}=1$.
Substituting this into the circle equation:
$x^2+y^2=9(1)^2$
$x^2+y^2=9\left(\frac{x+y}{3}\right)^2$
$x^2+y^2=9\left(\frac{x^2+y^2+2xy}{9}\right)$
$x^2+y^2=x^2+y^2+2xy$
$2xy=0$
$xy=0$
Thus,the equation of the pair of lines is $xy=0$.

(D) The given circles are $C_1: x^2+y^2-2x-4y+c=0$ and $C_2: x^2+y^2-4x-2y+4=0$.
For $C_1$,the center is $O_1(1, 2)$ and radius $r_1 = \sqrt{1^2+2^2-c} = \sqrt{5-c}$.
For $C_2$,the center is $O_2(2, 1)$ and radius $r_2 = \sqrt{2^2+1^2-4} = 1$.
The distance between centers $d^2 = (2-1)^2 + (1-2)^2 = 1+1 = 2$.
The angle $\theta$ between two circles is given by $\cos \theta = \frac{r_1^2+r_2^2-d^2}{2r_1r_2}$.
Given $\theta = 60^{\circ}$,so $\cos 60^{\circ} = \frac{1}{2}$.
Substituting the values: $\frac{1}{2} = \frac{(5-c) + 1 - 2}{2 \sqrt{5-c} \cdot 1} = \frac{4-c}{2 \sqrt{5-c}}$.
Thus,$\sqrt{5-c} = 4-c$.
Squaring both sides: $5-c = (4-c)^2 = 16 - 8c + c^2$.
Rearranging gives $c^2 - 7c + 11 = 0$.
Using the quadratic formula $c = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get $c = \frac{7 \pm \sqrt{49-44}}{2} = \frac{7 \pm \sqrt{5}}{2}$.

(A) In acidic medium,the balanced redox reaction is:
$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$
Using the equivalence principle,the number of equivalents of $KMnO_4$ equals the number of equivalents of oxalic acid:
$n_1 \times M_1 \times V_1 = n_2 \times M_2 \times V_2$
For $KMnO_4$,the change in oxidation state of $Mn$ is from $+7$ to $+2$,so $n_1 = 5$.
For oxalic acid $(H_2C_2O_4)$,the change in oxidation state of $C$ is from $+3$ to $+4$ for two carbon atoms,so $n_2 = 2$.
Substituting the given values:
$5 \times x \times 40 = 2 \times 0.02 \times 200$
$200x = 8$
$x = \frac{8}{200} = 0.04 \text{ M}$

(B) $I$. Incorrect: $Li^{+}$ has the highest charge density and is most highly hydrated,not $Cs^{+}$.
$II$. Correct: $Li$ is the only alkali metal that reacts directly with $N_2$ to form $Li_3N$.
$III$. Incorrect: $Li$ has the highest melting point among these alkali metals due to its small size and strong metallic bonding.
$IV$. Incorrect: $Li$ forms an oxide $(Li_2O)$,$Na$ forms a peroxide $(Na_2O_2)$,and $K, Rb, Cs$ form superoxides $(MO_2)$.

(B) The number of moles of hydrogen is calculated as $n_{H_2} = \frac{3 \ g}{2 \ g \ mol^{-1}} = 1.5 \ mol$.
The number of moles of oxygen is calculated as $n_{O_2} = \frac{80 \ g}{32 \ g \ mol^{-1}} = 2.5 \ mol$.
The total number of moles in the gaseous mixture is $n = 1.5 + 2.5 = 4 \ mol$.
The total kinetic energy of an ideal gas mixture is given by the formula $KE = \frac{3}{2} n R T$.
Substituting the values,we get $KE = \frac{3}{2} \times 4 \times R T = 6RT$.

(D) The ease of liquefaction of a gas is directly proportional to its critical temperature $(T_c)$.
$C$ has the highest critical temperature,which is $405.5 \ K$.
Therefore,gas $C$ will be the easiest to liquefy and will get liquefied first upon cooling.

(C) According to Heisenberg's uncertainty principle,the product of uncertainty in position $(\Delta x)$ and uncertainty in momentum $(\Delta p)$ is given by:
$\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}$
Since momentum $\Delta p = m \cdot \Delta v$,where $m$ is the mass and $\Delta v$ is the uncertainty in velocity,we substitute this into the equation:
$\Delta x \cdot (m \cdot \Delta v) \geq \frac{h}{4 \pi}$
Rearranging to find the product of uncertainty in position and velocity:
$\Delta x \cdot \Delta v \geq \frac{h}{4 \pi \cdot m}$
Thus,the product is not less than $\frac{h}{4 \pi \cdot m}$.

(B) The enthalpy of formation of $CO$ corresponds to the reaction: $C_{(s)} + \frac{1}{2}O_{2(g)} \longrightarrow CO_{(g)}$.
Given equations:
$(i) \ C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta H^{\circ} = -x \ kJ \ mol^{-1}$
$(ii) \ 2CO_{(g)} + O_{2(g)} \longrightarrow 2CO_{2(g)} ; \Delta H^{\circ} = -y \ kJ \ mol^{-1}$
Divide equation $(ii)$ by $2$:
$CO_{(g)} + \frac{1}{2}O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta H^{\circ} = -\frac{y}{2} \ kJ \ mol^{-1} \ (iii)$
Subtract equation $(iii)$ from equation $(i)$:
$(C_{(s)} + O_{2(g)}) - (CO_{(g)} + \frac{1}{2}O_{2(g)}) \longrightarrow CO_{2(g)} - CO_{2(g)}$
$C_{(s)} + \frac{1}{2}O_{2(g)} \longrightarrow CO_{(g)}$
$\Delta H^{\circ}_{f} = -x - (-\frac{y}{2}) = \frac{y}{2} - x = \frac{y-2x}{2} \ kJ \ mol^{-1}$.

(A) The coupling reaction of benzene diazonium chloride with aniline in a mildly acidic medium $(pH \approx 4-5)$ yields $p$-aminoazobenzene,which is a yellow dye.
The reaction is as follows:
$C_6H_5N_2^+Cl^- + C_6H_5NH_2 \xrightarrow{H^+} C_6H_5-N=N-C_6H_4-NH_2 + HCl$
Thus,$X$ is $Aniline$.

(D) The given reaction involves the oxidation of toluene $(C_6H_5CH_3)$ to benzaldehyde $(C_6H_5CHO)$ using chromyl chloride $(CrO_2Cl_2)$ in the presence of carbon disulfide $(CS_2)$ as a solvent,followed by hydrolysis.
This specific chemical transformation is known as the Etard reaction.

(C) In aqueous solution,the basic strength of aliphatic amines depends on the combined effect of inductive effect,solvation effect,and steric hindrance.
The order for methyl-substituted amines is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N$.
Comparing this with $NH_3$ and aniline $(C_6H_5NH_2)$:
$1$. $(CH_3)_2NH$ $(V)$ is the most basic due to the $+I$ effect and favorable solvation.
$2$. $CH_3NH_2$ $(IV)$ is next.
$3$. $(CH_3)_3N$ $(II)$ is less basic than $CH_3NH_2$ and $(CH_3)_2NH$ due to steric hindrance in the aqueous phase.
$4$. $NH_3$ $(III)$ is less basic than the aliphatic amines.
$5$. $C_6H_5NH_2$ $(I)$ is the least basic because the lone pair on nitrogen is delocalized into the benzene ring.
Thus,the correct order is $V > IV > II > III > I$.

(D) $I$. Sucrose is a disaccharide formed by the condensation of $D$-glucose and $D$-fructose,joined by a glycosidic linkage. This statement is correct.
$II$. Cellulose is a structural polysaccharide found exclusively in the cell walls of plants. It is not present in animals. This statement is incorrect.
$III$. Lactose is a disaccharide composed of $D$-galactose and $D$-glucose units linked by a $\beta$-glycosidic bond. This statement is correct.
Therefore,statements $I$ and $III$ are correct.

(C) For the reaction $2N_2O_{5(g)} \longrightarrow 4NO_{2(g)} + O_{2(g)}$,the rate expression is given by:
$-\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$
Given that the rate of disappearance of $N_2O_5$ is $-\frac{d[N_2O_5]}{dt} = 1.2 \times 10^{-5} \ mol \ L^{-1} \ s^{-1}$.
From the rate expression,we have:
$\frac{1}{4} \frac{d[NO_2]}{dt} = \frac{1}{2} \left( -\frac{d[N_2O_5]}{dt} \right)$
$\frac{d[NO_2]}{dt} = 2 \times \left( -\frac{d[N_2O_5]}{dt} \right)$
$\frac{d[NO_2]}{dt} = 2 \times (1.2 \times 10^{-5} \ mol \ L^{-1} \ s^{-1}) = 2.4 \times 10^{-5} \ mol \ L^{-1} \ s^{-1}$.

(D) Antioxidants are food additives that prevent the oxidation of food components,thereby preserving their quality and shelf life.
$BHT$ (Butylated hydroxy toluene) is a widely used synthetic antioxidant in food products to prevent rancidity in fats and oils.
$Aspartame$ is an artificial sweetener.
$Sodium \ benzoate$ is a food preservative.
$Ortho-sulphobenzimide$ (Saccharin) is an artificial sweetener.
Therefore,the correct option is $D$.

(A) $1$. Identify the ligands: $NH_3$ is ammine and $CO_3^{2-}$ is carbonato. Since there are $5$ ammine ligands,we use the prefix 'penta'.
$2$. Determine the oxidation state of the central metal atom $(Co)$: Let the oxidation state be $x$. The charge on $NH_3$ is $0$,$CO_3$ is $-2$,and $Cl$ is $-1$. Thus,$x + 5(0) + (-2) + (-1) = 0$,which gives $x = +3$.
$3$. Assemble the name: The ligands are listed alphabetically (ammine before carbonato). The name is Pentaamminocarbonatocobalt$(III)$ chloride.

(D) The catalytic activity of transition elements is primarily attributed to their ability to exhibit $Variable \ oxidation \ states$ and their capacity to form $complexes$. These properties allow them to provide a large surface area and form intermediate compounds with reactants,thereby lowering the activation energy of the reaction.

(A) The correct matching is based on the magnetic properties of the given substances:
$(A)$ Ferromagnetic: $CrO_2$ is a well-known ferromagnetic substance.
$(B)$ Antiferromagnetic: $MnO$ exhibits antiferromagnetism due to the alignment of magnetic moments in opposite directions.
$(C)$ Ferrimagnetic: $Fe_3O_4$ (magnetite) is a classic example of a ferrimagnetic substance.
$(D)$ Paramagnetic: $O_2$ is paramagnetic due to the presence of unpaired electrons in its molecular orbitals.
Therefore,the correct sequence is $A-2, B-3, C-4, D-1$.

(B) The aqueous solution of $K_2SO_4$ dissociates as: $K_2SO_4(aq) \rightarrow 2K^+(aq) + SO_4^{2-}(aq)$.
Water also undergoes self-ionization: $H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)$.
At the cathode,the reduction potential of $H^+$ is higher than that of $K^+$,so $H^+$ ions are reduced: $2H^+(aq) + 2e^- \rightarrow H_2(g)$.
At the anode,the oxidation potential of $OH^-$ (or $H_2O$) is higher than that of $SO_4^{2-}$,so $H_2O$ is oxidized: $2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-$.
Therefore,$H_2$ gas is produced at the cathode and $O_2$ gas is produced at the anode.

(B) The chemical formula for Siderite is $FeCO_3$,which is an iron carbonate ore.
Cuprite is $Cu_2O$ (oxide ore).
Zincite is $ZnO$ (oxide ore).
Bauxite is $AlO_x(OH)_{3-2x}$ (oxide/hydroxide ore).
Therefore,the correct option is $B$.

(C) The reaction sequence is as follows:
$1$. $CH_3OH + PCl_3 \rightarrow CH_3Cl (A) + H_3PO_3$
$2$. $CH_3Cl + KCN \rightarrow CH_3CN (B) + KCl$
$3$. $CH_3CN + 2H_2O \xrightarrow{H_3O^+} CH_3COOH (C) + NH_3$
Therefore,$C$ is $CH_3COOH$.

(A) The reaction involves the coupling of an aryl halide $(C_6H_5Cl)$ and an alkyl halide $(CH_3Cl)$ in the presence of sodium metal $(Na)$ and dry ether to form an alkylbenzene (toluene).
This specific type of coupling reaction between an aryl halide and an alkyl halide is known as the $Wurtz-Fittig$ reaction.

(B) Let us analyze each reaction:
$(A)$ $CH_3CHO$ is reduced by $LiAlH_4$ to ethanol $(CH_3CH_2OH)$. This is correct.
$(B)$ $CH_3COCH_3$ undergoes Clemmensen reduction with $Zn(Hg)/HCl$ to form propane $(CH_3CH_2CH_3)$. The product given in the option is incorrect as it shows a diol.
$(C)$ $CH_3CHO$ undergoes the iodoform reaction with $I_2/NaOH$ to form iodoform $(CHI_3)$ and sodium formate $(HCOONa)$. This is correct.
$(D)$ $CH_3CH_2CHO$ is oxidized by $KMnO_4$ to propanoic acid $(CH_3CH_2COOH)$. This is correct.
Therefore,the incorrect product is in option $(B)$.

(B) In molecular oxygen $(O_2)$, the bond order is $2$, resulting in a shorter bond length $(121 \text{ pm})$.
In ozone $(O_3)$, due to resonance, the $O-O$ bonds have partial double bond character with a bond order of $1.5$, resulting in a longer bond length $(128 \text{ pm})$.
Therefore, the statement that the $O-O$ bond length in $O_3$ is identical to that of molecular oxygen is incorrect.

(B) The statement that the oxidation state of sulphur is never less than $+4$ is incorrect. Sulphur exhibits a wide range of oxidation states in its compounds,ranging from $-2$ (as in $H_2S$) to $+6$ (as in $H_2SO_4$).
At high temperatures (around $1000 \ K$),$S_2$ is the dominant species and it is paramagnetic,similar to $O_2$.
Rhombic sulphur is the most stable allotrope at room temperature and is readily soluble in $CS_2$.

(D) The hydrolysis reactions of xenon fluorides are as follows:
$1$. $2XeF_2 + 2H_2O \rightarrow 2Xe + 4HF + O_2$ (Liberates $O_2$)
$2$. $6XeF_4 + 12H_2O \rightarrow 4Xe + 2XeO_3 + 24HF + 3O_2$ (Liberates $O_2$)
$3$. $XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF$ (Does not liberate $O_2$)
$4$. $XeF_4 + O_2F_2 \rightarrow XeF_6 + O_2$ (Liberates $O_2$)
Therefore,the reaction $XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF$ does not involve the liberation of oxygen.

(B) Biodegradable polymers are those that can be decomposed by microorganisms.
Among the given polymers:
$I$. $PHBV$ (Poly-$\beta$-hydroxybutyrate-co-$\beta$-hydroxyvalerate) is a well-known biodegradable polymer.
$II$. Nylon-$2$-nylon-$6$ is a polyamide copolymer of glycine and amino caproic acid,which is also biodegradable.
$III$. Glyptal is a condensation polymer of ethylene glycol and phthalic acid,which is non-biodegradable.
$IV$. Bakelite is a thermosetting phenol-formaldehyde resin,which is non-biodegradable.
Therefore,$I$ and $II$ are biodegradable polymers.

(D) The ability of a halogen to act as an oxidizing agent decreases down the group $(F_2 > Cl_2 > Br_2 > I_2)$.
$F_2$ is the strongest oxidizing agent and can oxidize $Cl^{-}$,$Br^{-}$,and $I^{-}$ to their respective halogens.
However,$Cl_2$ is a weaker oxidizing agent than $F_2$ and cannot oxidize the fluoride ion $(F^{-})$ to fluorine gas $(F_2)$.
Therefore,the reaction $Cl_2 + 2F^{-} \longrightarrow 2Cl^{-} + F_2$ does not occur.

(D) For an equimolar solution,the mole fractions in the liquid phase are $\chi_b = \chi_t = 0.5$.
The partial vapour pressure of benzene is $p_b = \chi_b \times p_b^0 = 0.5 \times 160 = 80 \ mm \ Hg$.
The partial vapour pressure of toluene is $p_t = \chi_t \times p_t^0 = 0.5 \times 60 = 30 \ mm \ Hg$.
The total vapour pressure is $p_{\text{total}} = p_b + p_t = 80 + 30 = 110 \ mm \ Hg$.
The mole fraction of benzene in the vapour phase $(y_b)$ is given by $y_b = \frac{p_b}{p_{\text{total}}} = \frac{80}{110} \approx 0.727 \approx 0.73$.

(D) The depression in freezing point is given by the formula: $\Delta T_f = K_f \times m$
Where $\Delta T_f = T_f^{\circ} - T_f = 0 - (-0.93) = 0.93 \ K$.
Molality $m = \frac{w_2 \times 1000}{M_2 \times w_1}$,where $w_2 = 6 \ g$,$w_1 = 100 \ g$,and $M_2$ is the molar mass of $X$.
Substituting the values: $0.93 = 1.86 \times \frac{6 \times 1000}{M_2 \times 100}$.
$0.93 = \frac{1.86 \times 60}{M_2}$.
$M_2 = \frac{1.86 \times 60}{0.93} = 2 \times 60 = 120 \ g \ mol^{-1}$.

(C) Chemisorption is characterized by the formation of chemical bonds between the adsorbate and the adsorbent.
It is highly specific in nature and is generally irreversible.
Due to the formation of chemical bonds,it involves a high enthalpy of adsorption (typically $80-240 \ kJ \ mol^{-1}$).
However,chemisorption is a unimolecular process,meaning it forms a monolayer,not a multilayer.
Therefore,the statement that chemisorption is a multilayered adsorption is incorrect.