KCET 2023 Chemistry Question Paper with Answer and Solution

(D) Given $y = a \sin x + b \cos x$.
Differentiating with respect to $x$,we get:
$\frac{dy}{dx} = a \cos x - b \sin x$.
Now,calculating the square of the derivative:
$\left( \frac{dy}{dx} \right)^2 = (a \cos x - b \sin x)^2 = a^2 \cos^2 x + b^2 \sin^2 x - 2ab \sin x \cos x$.
Calculating the square of $y$:
$y^2 = (a \sin x + b \cos x)^2 = a^2 \sin^2 x + b^2 \cos^2 x + 2ab \sin x \cos x$.
Adding both expressions:
$y^2 + \left( \frac{dy}{dx} \right)^2 = (a^2 \sin^2 x + b^2 \cos^2 x + 2ab \sin x \cos x) + (a^2 \cos^2 x + b^2 \sin^2 x - 2ab \sin x \cos x)$.
Grouping the terms:
$y^2 + \left( \frac{dy}{dx} \right)^2 = a^2(\sin^2 x + \cos^2 x) + b^2(\cos^2 x + \sin^2 x)$.
Since $\sin^2 x + \cos^2 x = 1$,we get:
$y^2 + \left( \frac{dy}{dx} \right)^2 = a^2(1) + b^2(1) = a^2 + b^2$.
Since $a$ and $b$ are constants,$a^2 + b^2$ is a constant.
Therefore,the correct option is $D$.

(A) $1$. The reaction sequence is as follows:
$CH_3CHO \xrightarrow[(ii) H_3O^+]{(i) CH_3MgBr} CH_3-CH(OH)-CH_3 (P)$
$CH_3-CH(OH)-CH_3 \xrightarrow[\Delta]{conc. H_2SO_4} CH_3-CH=CH_2 (Q)$
$CH_3-CH=CH_2 \xrightarrow[(ii) H_2O_2/OH^-]{(i) B_2H_6} CH_3-CH_2-CH_2OH (R)$
$2$. Compound $P$ is propan$-2-$ol $(CH_3-CH(OH)-CH_3)$ and compound $R$ is propan$-1-$ol $(CH_3-CH_2-CH_2OH)$.
$3$. Both compounds have the same molecular formula $(C_3H_8O)$ and the same functional group $(-OH)$,but the position of the hydroxyl group is different (carbon-$2$ in $P$ and carbon-$1$ in $R$).
$4$. Therefore,$P$ and $R$ are position isomers.

(C) In the solid state,$PCl_5$ exists as an ionic solid.
It consists of a tetrahedral $[PCl_4]^{+}$ cation and an octahedral $[PCl_6]^{-}$ anion.
This structure is formed due to the transfer of a chloride ion from one $PCl_5$ molecule to another.

(B) The bond order values for the given species are as follows:
$O_2^{2+} = 3.0$
$O_2^+ = 2.5$
$O_2 = 2.0$
$O_2^- = 1.5$
$O_2^{2-} = 1.0$
Bond order is inversely proportional to bond length (Bond length $\propto \frac{1}{\text{Bond order}}$).
Evaluating the options:
$(A)$ Bond length of $O_2 > O_2^{2+}$: Correct (since $2.0 < 3.0$ in bond order).
$(B)$ Bond order of $O_2^+ < O_2^{2-}$: Incorrect ($2.5 < 1.0$ is false).
$(C)$ Bond length of $O_2 > O_2^{2-}$: Correct (since $2.0 < 1.0$ in bond order).
$(D)$ Bond order of $O_2 > O_2^{2-}$: Correct ($2.0 > 1.0$ is true).
Thus,the incorrect statement is option $(B)$.

(D) The value of the equilibrium constant $K_C$ depends only on the temperature of the reaction.
Catalysts increase the rate of both the forward and backward reactions equally,allowing the system to reach equilibrium faster,but they do not change the position of the equilibrium or the value of $K_C$.
Since the temperature remains constant at $500 \ K$,the value of $K_C$ remains unchanged.
Therefore,the equilibrium constant $K_C$ in the presence of a catalyst is $2 \times 10^{-5}$.

(A) The first ionisation enthalpy generally increases as we move from left to right across a period due to an increase in effective nuclear charge.
However,$Be$ $(1s^2 2s^2)$ has a fully-filled $2s$ subshell,which is more stable than the $2p^1$ configuration of $B$ $(1s^2 2s^2 2p^1)$.
Therefore,$Be$ has a higher first ionisation enthalpy than $B$.
The correct order is $Li < B < Be < C$.

(A) Carbon monoxide $(CO)$ converts haemoglobin into carboxyhaemoglobin.
When $CO$ is inhaled,it binds to haemoglobin in red blood cells with an affinity approximately $200$ to $300$ times higher than that of oxygen.
This binding forms carboxyhaemoglobin,which is a stable complex that prevents the normal binding of oxygen to haemoglobin.
Consequently,this leads to a significantly decreased oxygen-carrying capacity of the blood,which can be fatal.

(A) The given structure is a hydrocarbon with a double bond between two carbon atoms.
To name it according to $IUPAC$ rules:
$1$. Identify the longest carbon chain containing the double bond. The longest chain has $4$ carbon atoms,so the parent alkane is butane,and the alkene is but$-2-$ene.
$2$. Number the chain from the end that gives the double bond the lowest locant. In this case,numbering from either side gives the double bond the position $2$.
$3$. Identify the substituents. There are two methyl groups attached at positions $2$ and $3$.
$4$. Combining these,the $IUPAC$ name is $2,3-$dimethylbut$-2-$ene.

(D) species is aromatic if it follows $H$ückel's rule,which states it must be planar,cyclic,fully conjugated,and contain $(4n+2)$ $\pi$-electrons,where $n = 0, 1, 2, \dots$
$I$ (Benzene): Planar,cyclic,fully conjugated,$6$ $\pi$-electrons $(n=1)$. It is aromatic.
$II$ (Naphthalene): Planar,cyclic,fully conjugated,$10$ $\pi$-electrons $(n=2)$. It is aromatic.
$III$ (Cyclopentadiene): Not fully conjugated because of the $sp^3$ hybridized carbon. It is non-aromatic.
$IV$ (Cyclopentadienyl anion): Planar,cyclic,fully conjugated,$6$ $\pi$-electrons $(n=1)$. It is aromatic.
$V$ (Cyclooctatetraene): It has $8$ $\pi$-electrons ($4n$ system,$n=2$) and is non-planar (tub-shaped). It is non-aromatic.
Thus,the aromatic species are $I, II$ and $IV$.

(D) Enantiomers,such as $(+)$ butan-$2$-ol and $(-)$ butan-$2$-ol,are non-superimposable mirror images of each other.
They possess identical physical properties,including boiling points,melting points,and refractive indices,in an achiral environment.
They differ only in their interaction with plane-polarized light and their reactivity with other chiral substances.
Therefore,$(+)$ butan-$2$-ol and $(-)$ butan-$2$-ol have the same boiling point.

(D) Water gas is primarily composed of carbon monoxide and hydrogen gas,i.e.,$CO_{(g)} + H_{2(g)}$.
It is a valuable industrial feedstock and is used in the production of chemicals,fuels,and as a reducing agent in metallurgical applications.

(D) $30 \% (w/v)$ solution of $H_2O_2$ means that $30 \ g$ of $H_2O_2$ is present in $100 \ mL$ of the solution.
The molarity of this solution is calculated as: $M = \frac{30 \ g / 34 \ g/mol}{0.1 \ L} \approx 8.82 \ M$.
The volume strength of $H_2O_2$ is given by the formula: $\text{Volume strength} = 5.6 \times \text{Molarity}$.
Therefore,$\text{Volume strength} = 5.6 \times 8.82 \approx 49.4 \approx 50 \ V$ (often approximated as $100 \ V$ in some contexts,but strictly $30 \% (w/v)$ corresponds to approximately $100 \ V$ based on standard industrial definitions where $1 \ mL$ of $30 \% H_2O_2$ yields $100 \ mL$ of $O_2$ gas).

(D) The $pH$ of a salt solution formed by a weak acid and a weak base is given by the formula:
$pH = 7 + \frac{1}{2}(pK_a - pK_b)$
Given: $pK_a = 5.9$,$pK_b = 5.8$
Substituting the values:
$pH = 7 + \frac{1}{2}(5.9 - 5.8)$
$pH = 7 + \frac{1}{2}(0.1)$
$pH = 7 + 0.05 = 7.05$

(D) $BeO$,$Al_2O_3$,and $ZnO$ are amphoteric in nature.
$MgO$ and $Li_2O$ are basic in nature,whereas $B_2O_3$ is acidic.
Thus,the pair of amphoteric oxides is $BeO$ and $ZnO$.

(A) In the second reaction,$Zn(s)$ is oxidized to $[Zn(CN)_{4}]^{2-}(aq)$ as its oxidation state increases from $0$ to $+2$.
Since $Zn$ undergoes oxidation,it acts as a reducing agent by reducing $Ag^{+}$ ions to $Ag(s)$.

(D) The structure of $H_2S_2O_8$ (peroxodisulphuric acid) contains a peroxide linkage $(-O-O-)$.
In this structure,there are two sulphur atoms,two hydroxyl groups $(-OH)$,two peroxide oxygen atoms (oxidation state $-1$ each),and six terminal oxygen atoms (oxidation state $-2$ each).
Let the oxidation state of sulphur be $x$.
The sum of oxidation states in a neutral molecule is zero:
$2(+1) + 2(x) + 2(-1) + 6(-2) = 0$
$2 + 2x - 2 - 12 = 0$
$2x - 12 = 0$
$2x = 12$
$x = +6$
Thus,the oxidation state of $S$ in $H_2S_2O_8$ is $+6$.

(B) To balance the redox reaction in basic medium:
$1$. Oxidation half-reaction: $S_2 O_3^{2-} \longrightarrow 2 SO_4^{2-} + 8 e^-$
$2$. Reduction half-reaction: $MnO_4^{-} + 2 H_2 O + 3 e^- \longrightarrow MnO_2 + 4 OH^-$
$3$. To balance electrons,multiply the oxidation half-reaction by $3$ and the reduction half-reaction by $8$:
$3 S_2 O_3^{2-} \longrightarrow 6 SO_4^{2-} + 24 e^-$
$8 MnO_4^{-} + 16 H_2 O + 24 e^- \longrightarrow 8 MnO_2 + 32 OH^-$
$4$. Adding both equations:
$8 MnO_4^{-} + 3 S_2 O_3^{2-} + 16 H_2 O \longrightarrow 8 MnO_2 + 6 SO_4^{2-} + 32 OH^-$
Simplifying $H_2 O$ and $OH^-$ on both sides:
$8 MnO_4^{-} + 3 S_2 O_3^{2-} + H_2 O \longrightarrow 8 MnO_2 + 6 SO_4^{2-} + 2 OH^-$
Comparing with the given equation,$a = 8$ and $y = 6$.

(A) The formula to convert Celsius to Fahrenheit is: $F = (\frac{9}{5} \times ^{\circ} C) + 32$
Substituting $25^{\circ} C$: $F = (\frac{9}{5} \times 25) + 32 = 45 + 32 = 77^{\circ} F$
The formula to convert Celsius to Kelvin is: $K = ^{\circ} C + 273.15$
Substituting $25^{\circ} C$: $K = 25 + 273.15 = 298.15 \ K$

(B) Given: $p_1 = 2 \ atm$,$T_1 = 25 + 273 = 298 \ K$,$T_2 = 323 + 273 = 596 \ K$.
Let initial volume $V_1 = V$. Then final volume $V_2 = \frac{2}{3} V$.
Using the combined gas law: $\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$.
Substituting the values: $\frac{2 \times V}{298} = \frac{p_2 \times (2/3) V}{596}$.
$p_2 = \frac{2 \times 596}{298 \times (2/3)} = \frac{2 \times 2}{2/3} = 2 \times 3 = 6 \ atm$.

(B) For the ion ${ }_{16}^{32} S^{2-}$:
Number of protons $= Z = 16$.
Number of neutrons $= A - Z = 32 - 16 = 16$.
Number of electrons $= Z + (\text{charge magnitude}) = 16 + 2 = 18$.
Therefore,the number of protons,neutrons,and electrons are $16, 16, 18$ respectively.

(B) The enthalpy of solution $(\Delta H_{\text{sol}})$ is given by the sum of the lattice enthalpy and the enthalpy of hydration:
$\Delta H_{\text{sol}} = \Delta H_{\text{lattice}} + \Delta H_{\text{hyd}}$
Given: $\Delta H_{\text{lattice}} = +788 \ kJ \ mol^{-1}$ and $\Delta H_{\text{hyd}} = -784 \ kJ \ mol^{-1}$
$\Delta H_{\text{sol}} = 788 \ kJ \ mol^{-1} + (-784 \ kJ \ mol^{-1}) = +4 \ kJ \ mol^{-1}$

(D) The reaction of phenyl methyl ether (anisole) with $HI$ involves the cleavage of the $C-O$ bond,yielding phenol $(A)$ and methyl iodide $(CH_3I)$.
Phenol $(A)$ reacts with concentrated $HNO_3$ (nitration) to form $2,4,6-$trinitrophenol,commonly known as picric acid $(B)$.

(B) $1$. The reaction of phenol with $NaOH$ gives sodium phenoxide $(A)$.
$2$. The reaction of sodium phenoxide with $CO_2$ followed by acidification gives salicylic acid $(B)$ (Kolbe's reaction).
$3$. The acetylation of salicylic acid $(B)$ with acetic anhydride $(CH_3CO)_2O$ in the presence of an acid catalyst gives acetylsalicylic acid,which is commonly known as aspirin $(Y)$.

(B) In the reactant $CH_3-CH=CH-CH_2OH$,the carbon atom at position $C-1$ is bonded to two hydrogen atoms,one oxygen atom,and one carbon atom. It is $sp^3$ hybridised.
In the product $CH_3-CH=CH-CHO$,the carbon atom at position $C-1$ is part of the aldehyde group $(-CHO)$,where it forms a double bond with oxygen. It is $sp^2$ hybridised.
Therefore,the hybridisation of $C-1$ changes from $sp^3$ to $sp^2$.

(A) $PCC$ (pyridinium chlorochromate) is a commonly used reagent for the oxidation of primary alcohols to aldehydes.
It is favored due to its selectivity,mild reaction conditions,compatibility with other functional groups,and its ability to stop the oxidation at the aldehyde stage without further oxidation to carboxylic acids.

(D) The given reaction is: $C_6H_5CN$ $\xrightarrow[(ii) H_3O^+]{(i) SnCl_2 + HCl} C_6H_5CHO (X)$ $\xrightarrow{conc. KOH} C_6H_5COOK (Y) + C_6H_5CH_2OH (Z)$.
$1$. The conversion of nitrile $(C_6H_5CN)$ to aldehyde $(C_6H_5CHO)$ using $SnCl_2 + HCl$ followed by hydrolysis is known as the Stephen reaction.
$2$. Benzaldehyde $(C_6H_5CHO)$ lacks $\alpha$-hydrogen atoms,so it undergoes the Cannizzaro reaction in the presence of concentrated alkali $(KOH)$ to form a mixture of alcohol $(C_6H_5CH_2OH)$ and salt of carboxylic acid $(C_6H_5COOK)$.

(A) The reaction sequence is as follows:
$1$. $CH_3CH_2Br + KCN \rightarrow CH_3CH_2CN (A) + KBr$
$2$. $CH_3CH_2CN + 4[H] \xrightarrow{LiAlH_4} CH_3CH_2CH_2NH_2 (B)$
$3$. $CH_3CH_2CH_2NH_2 + HNO_2$ $\xrightarrow{0^{\circ}C} [CH_3CH_2CH_2N_2^+]$ $\rightarrow CH_3CH_2CH_2^+$ $\xrightarrow{H_2O} CH_3CH_2CH_2OH (C)$
(Note: While rearrangement to the more stable isopropyl carbocation can occur,the primary alcohol $CH_3CH_2CH_2OH$ is the major product expected in this context.)

(C) Aniline does not undergo Friedel-Craft's reaction because the lone pair on the nitrogen atom of the $-NH_2$ group forms a coordinate bond with the Lewis acid (catalyst) like $AlCl_3$.
This results in the formation of a positively charged species on the nitrogen atom,which acts as a strong deactivating group for the aromatic ring,thereby preventing the reaction.

(A) The reaction sequence is as follows:
$1$. The conversion of aniline to benzene diazonium chloride $(C_6H_5N_2^+Cl^-)$ is achieved using $NaNO_2 + \text{dil. } HCl$ at $0-5 \ ^\circ C$. Thus,$P = NaNO_2 + \text{dil. } HCl$.
$2$. The conversion of benzene diazonium chloride to benzene diazonium fluoroborate $(C_6H_5N_2^+BF_4^-)$ is achieved by adding fluoroboric acid $(HBF_4)$. Thus,$Q = HBF_4$.
$3$. The conversion of benzene diazonium fluoroborate to nitrobenzene $(C_6H_5NO_2)$ is achieved by heating with sodium nitrite in the presence of copper powder. Thus,$R = Cu + NaNO_2$.
Therefore,$P, Q$ and $R$ are $NaNO_2 + \text{dil. } HCl, HBF_4, Cu + NaNO_2$ respectively.

(D) Sucrose is a disaccharide that is dextrorotatory $(+66.5^{\circ})$.
Upon hydrolysis,it yields an equimolar mixture of $D-(+)$-glucose and $D-(-)$-fructose.
The specific rotation of $D-(+)$-glucose is $+52.5^{\circ}$,while the specific rotation of $D-(-)$-fructose is $-92.4^{\circ}$.
Since the magnitude of the laevorotation of fructose $(-92.4^{\circ})$ is greater than the magnitude of the dextrorotation of glucose $(+52.5^{\circ})$,the resulting mixture is laevorotatory.
Therefore,the correct reason is that the laevorotation of fructose is more than the dextrorotation of glucose.

(D) The correct matches are as follows:
$I$. Vitamin $A$ is associated with $(iii)$ Night-blindness.
$II$. Vitamin $D$ is associated with $(iv)$ Osteomalacia.
$III$. Vitamin $E$ is associated with $(i)$ Muscular weakness.
$IV$. Vitamin $K$ is associated with $(ii)$ Increased blood clotting time.
Therefore,the correct order is $I-iii, II-iv, III-i, IV-ii$.

(C) Thyroxine,also known as tetraiodothyronine or $T_4$,is an iodinated derivative of the amino acid tyrosine.
It is a hormone produced by the thyroid gland and plays a crucial role in regulating the metabolism and growth in the body.

(A) Dimerisation of solute molecules in low dielectric constant solvents is primarily due to the formation of intermolecular $hydrogen \ bonds$.
In solvents with low dielectric constants,the electrostatic screening is weak,which facilitates the association of solute molecules through $hydrogen \ bonding$ to form stable dimers,such as the dimerization of carboxylic acids.

(D) The unit of the rate constant $k$ is given by $(mol \ L^{-1})^{1-n} \ s^{-1}$,where $n$ is the order of the reaction.
For a zero order reaction $(n = 0)$,the unit is $(mol \ L^{-1})^{1-0} \ s^{-1} = mol \ L^{-1} \ s^{-1}$.
Among the given options,the decomposition of $HI$ on the surface of gold is a zero order reaction because it is a heterogeneous catalytic reaction where the rate is independent of the concentration of the reactant.

(D) The Arrhenius equation is given by $k = A e^{-E_a / (RT)}$.
At infinitely high temperature,i.e.,$T \to \infty$,the term $\frac{E_a}{RT} \to 0$.
Therefore,the equation becomes $k = A e^0 = A \times 1 = A$.
Since the Arrhenius factor $A$ is a constant independent of temperature,its value remains the same as the rate constant $k$ at $T \to \infty$.
Given that $k = 6.0 \times 10^5 \ s^{-1}$ at $300 \ K$,and $A$ is independent of temperature,the value of $A$ is $6.0 \times 10^5 \ s^{-1}$.

(C) Given $k_1 = 10^{16} \times e^{-2000/T}$ and $k_2 = 10^{15} \times e^{-1000/T}$.
For $k_1 = k_2$:
$10^{16} \times e^{-2000/T} = 10^{15} \times e^{-1000/T}$
$10 \times e^{-2000/T} = e^{-1000/T}$
$10 = \frac{e^{-1000/T}}{e^{-2000/T}}$
$10 = e^{1000/T}$
Taking the natural logarithm on both sides:
$\ln(10) = \frac{1000}{T}$
$2.303 \times \log_{10}(10) = \frac{1000}{T}$
$2.303 = \frac{1000}{T}$
$T = \frac{1000}{2.303} \text{ K}$

(C) $Aspirin$ is classified as a non-narcotic analgesic because it does not produce the typical effects associated with narcotic or opioid analgesics.
In addition to its analgesic effects,it also possesses anti-inflammatory and antipyretic (fever reducing) properties.
It is commonly used for relieving mild to moderate pain,reducing inflammation,headaches,etc.

(A) Receptors are specialized proteins that are crucial for communication in the body.
They are typically embedded in the $ \text{cell membrane} $ so that their active sites are exposed to the outer surface of the cell,or they may be located intracellularly within the cell.

(C) The formula and name combination given in option $(C)$ is incorrect.
In the complex $[CoCl_2(en)_2]Cl$,the oxidation state of $Co$ is calculated as: $x + 2(-1) + 2(0) = +1$,which gives $x = +3$.
Also,the ligand $en$ (ethylenediamine) is a bidentate ligand,so it is named as 'bis(ethylenediamine)'.
Therefore,the correct $IUPAC$ name is Dichloridobis(ethylenediamine)cobalt$(III)$ chloride.

(D) The color of a coordination complex is complementary to the color of light absorbed.
For complex $I$,the observed color is pale blue,which corresponds to the absorption of orange light (wavelength range $\approx 600-650 \ nm$).
For complex $III$,the observed color is violet,which corresponds to the absorption of yellow light (wavelength range $\approx 530-580 \ nm$).
Comparing these with the given options,the wavelengths absorbed for $I$ and $III$ are approximately $600 \ nm$ and $535 \ nm$ respectively.

(D) In an octahedral complex,the $d$-orbitals split into $t_{2g}$ and $e_g$ levels.
For $d^9$ (high spin): $t_{2g}^6 e_g^3$,number of unpaired electrons $(n)$ = $1$.
For $d^6$ (low spin): $t_{2g}^6 e_g^0$,$n = 0$.
For $d^4$ (low spin): $t_{2g}^4 e_g^0$,$n = 2$.
For $d^7$ (high spin): $t_{2g}^5 e_g^2$,$n = 3$.
Therefore,the $d^7$ (high spin) system has the maximum number of unpaired electrons.

(B) $CH_3CH_2MgBr$ is an organometallic compound because it contains a direct carbon-metal bond between the carbon atom of the ethyl group and the magnesium atom.
Organometallic compounds are defined as compounds containing at least one direct bond between a carbon atom and a metal atom.
In $CH_3COONa$,$(CH_3COO)_2Ca$,and $CH_3ONa$,the metal atoms are bonded to oxygen atoms,not directly to carbon atoms.

(D) The general electronic configuration of group $12$ elements $(Zn, Cd, Hg, Cn)$ is $(n-1)d^{10}ns^2$.
Copper $(Cu)$ belongs to group $11$ and has an electronic configuration of $3d^{10}4s^1$.
Zinc $(Zn)$ belongs to group $12$ and has an electronic configuration of $3d^{10}4s^2$.
In option $D$,$Cu$ does not have the $(n-1)d^{10}ns^2$ configuration,while $Zn$ does.
Therefore,the pair where both elements do not have this configuration is $Cu, Zn$.

(B) The melting points of transition elements are related to the strength of metallic bonding,which depends on the number of unpaired $d$-electrons available for bonding.
In the $3d$ series,$Cr$ $(3d^5 4s^1)$ has six unpaired electrons,leading to strong metallic bonding and a high melting point.
$Mn$ $(3d^5 4s^2)$ has a stable half-filled $d$-subshell,which makes the $d$-electrons less available for metallic bonding,resulting in a significantly lower melting point compared to $Cr$.
Therefore,the correct order is $Cr > Mn$.

(A) When $FeCl_3$ is added to an excess of hot water,hydrolysis occurs to form a positively charged sol '$X$' of hydrated ferric oxide,which adsorbs $Fe^{3+}$ ions from the solution: $Fe_2O_3 \cdot xH_2O / Fe^{3+}$.
When $FeCl_3$ is added to $NaOH_{(aq)}$ solution,the resulting hydrated ferric oxide sol adsorbs $OH^{-}$ ions from the excess $NaOH$ to form a negatively charged sol '$Y$': $Fe_2O_3 \cdot xH_2O / OH^{-}$.

(D) During the electrolysis of brine ($NaCl$ aqueous solution) using inert electrodes:
At anode: $2 Cl^{-} \longrightarrow Cl_2 + 2 e^{-}$
At cathode: $2 H_2 O + 2 e^{-} \longrightarrow H_2 + 2 OH^{-}$
Overall reaction: $2 NaCl + 2 H_2 O \longrightarrow Cl_2 + H_2 + 2 NaOH$
Thus,$Cl_2$ gas is liberated at the anode and $H_2$ gas is liberated at the cathode.

(A) The Nernst equation for the reduction of $Ag^{+}$ to $Ag_{(s)}$ is given by:
$E_{Ag^{+} / Ag} = E^{\circ}_{Ag^{+} / Ag} + 0.059 \log [Ag^{+}]$
Since $E^{\circ}_{Ag^{+} / Ag}$ is constant $(+0.80 \ V)$,the reduction potential $E_{Ag^{+} / Ag}$ depends directly on the concentration of $Ag^{+}$ ions.
As the concentration of $Ag^{+}$ increases,the value of $\log [Ag^{+}]$ increases,and thus the reduction potential $E_{Ag^{+} / Ag}$ increases.
Comparing the concentrations:
$B: 0.1 \ M$
$C: 0.01 \ M$
$D: 0.001 \ M$
$A: 0.0001 \ M$
Since $0.1 > 0.01 > 0.001 > 0.0001$,the order of reduction potential is $B > C > D > A$.

(D) Given: Concentration $C = 0.1 \ M$,Molar conductivity at infinite dilution $\Lambda_{m}^{\circ} = 390 \ S \ cm^2 \ mol^{-1}$,Resistance $R = 2 \times 10^3 \ \Omega$,Cell constant $G^* = 0.78 \ cm^{-1}$.
First,calculate the conductivity $K$:
$K = \frac{G^*}{R} = \frac{0.78}{2 \times 10^3} = 3.9 \times 10^{-4} \ S \ cm^{-1}$.
Next,calculate the molar conductivity $\Lambda_{m}$:
$\Lambda_{m} = \frac{K \times 1000}{C} = \frac{3.9 \times 10^{-4} \times 1000}{0.1} = 3.9 \ S \ cm^2 \ mol^{-1}$.
Calculate the degree of dissociation $\alpha$:
$\alpha = \frac{\Lambda_{m}}{\Lambda_{m}^{\circ}} = \frac{3.9}{390} = 0.01 = 10^{-2}$.
Calculate the concentration of $H^{+}$ ions:
$[H^{+}] = C \times \alpha = 0.1 \times 10^{-2} = 10^{-3} \ M$.
Finally,calculate the $pH$:
$pH = -\log[H^{+}] = -\log(10^{-3}) = 3$.

(A) In the Ellingham diagram,the slope of the line for the formation of $CO$ is negative because the reaction $2C(s) + O_2(g) \rightarrow 2CO(g)$ involves an increase in the number of moles of gas $(\Delta n_g = 2 - 1 = 1)$.
According to the equation $\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$,since $\Delta S^{\circ}$ is positive,the term $-T\Delta S^{\circ}$ becomes more negative as temperature $(T)$ increases.
Therefore,the formation of $CO$ becomes more thermodynamically stable and favorable at higher temperatures compared to metal oxides like $FeO$,$ZnO$,or $Cu_2O$,which show positive slopes.

(C) Among the given compounds,only ammonium nitrate $(NH_4NO_3)$ does not produce $N_2$ gas on decomposition; instead,it produces nitrous oxide $(N_2O)$.
$Ba(N_3)_2 \xrightarrow{\Delta} Ba + 3N_2$
$NH_4NO_2 \xrightarrow{\Delta} N_2 + 2H_2O$
$NH_4NO_3 \xrightarrow{\Delta} N_2O + 2H_2O$
$(NH_4)_2Cr_2O_7 \xrightarrow{\Delta} Cr_2O_3 + N_2 + 4H_2O$

(D) The basicity of group-$15$ hydrides depends on the availability of the lone pair of electrons on the central atom for donation.
As we move down the group from $N$ to $Bi$,the size of the central atom increases.
This leads to the dispersion of the lone pair over a larger volume,making it less available for donation.
Therefore,the basicity decreases down the group.
The correct decreasing order is $NH_3 > PH_3 > AsH_3 > SbH_3$.

(A) Hypophosphorus acid $(H_3PO_2)$ acts as a strong reducing agent because it contains two $P-H$ bonds.
It can effectively reduce $Ag^{+}$ ions to metallic $Ag$ and itself gets oxidized to phosphorus acid $(H_3PO_3)$.
The chemical reaction is as follows:
$3AgNO_3 + H_3PO_2 + 2H_2O \longrightarrow 3Ag + H_3PO_3 + 3HNO_3$

(D) $PHBV$ (Poly-$\beta$-hydroxybutyrate-co-$\beta$-hydroxyvalerate) is a biodegradable polymer that belongs to the family of polyhydroxyalkanoates $(PHAs)$.
It is synthesized by the copolymerization of two monomeric units,namely $3-$hydroxybutanoic acid and $3-$hydroxypentanoic acid.

(D) The correct matches are:
$1$. Caprolactam polymerizes to form Nylon-$6$,which has the structure $(-CO-(CH_2)_5-NH-)_n$ $(d)$.
$2$. Vinyl chloride polymerizes to form Polyvinyl chloride $(PVC)$,which has the structure $(-CH_2-CH(Cl)-)_n$ $(c)$.
$3$. Styrene polymerizes to form Polystyrene,which has the structure $(-CH_2-CH(C_6H_5)-)_n$ $(b)$.
$4$. Propene polymerizes to form Polypropylene,which has the structure $(-CH_2-CH(CH_3)-)_n$ $(a)$.
Thus,the correct match is $1-d, 2-c, 3-b, 4-a$.

(D) The correct matches are as follows:
$P$. Molecular solid: $H_2O$ $(iii)$
$Q$. Ionic solid: $MgO$ $(iv)$
$R$. Metallic solid: $Mg$ $(ii)$
$S$. Network solid: $SiC$ $(i)$
Therefore,the correct sequence is $P-iii, Q-iv, R-ii, S-i$.

(B) For a simple cubic unit cell,the relation between radius '$r$' and edge length '$a$' is $r = \frac{a}{2}$.
For a body-centered cubic $(bcc)$ unit cell,the relation is $r = \frac{a \sqrt{3}}{4}$.
For a face-centered cubic $(fcc)$ unit cell,the relation is $r = \frac{a}{2 \sqrt{2}}$.
Thus,the ratio of the radii for simple cubic : $bcc$ : $fcc$ is $\frac{1}{2} \ a : \frac{\sqrt{3}}{4} \ a : \frac{1}{2 \sqrt{2}} \ a$.

(C) Given,metallic radius $r = \sqrt{3} \ \mathring{A} = \sqrt{3} \times 10^{-10} \ m$.
For a body-centred cubic $(BCC)$ lattice,the relationship between radius $r$ and edge length $a$ is $r = \frac{\sqrt{3} a}{4}$.
Rearranging for $a$: $a = \frac{4r}{\sqrt{3}} = \frac{4 \times \sqrt{3} \times 10^{-10} \ m}{\sqrt{3}} = 4 \times 10^{-10} \ m$.
The volume of the unit cell is $V = a^3 = (4 \times 10^{-10} \ m)^3 = 64 \times 10^{-30} \ m^3 = 6.4 \times 10^{-29} \ m^3$.

(B) The swelling in the feet is caused by the accumulation of excess fluid in the tissues,a condition known as edema.
When the feet are soaked in a concentrated salt solution,the concentration of salt outside the skin is higher than that inside the tissues.
Through the process of $osmosis$,water molecules move from the region of higher water concentration (the swollen tissues) to the region of lower water concentration (the salt solution) across the semi-permeable membrane of the skin.
This movement of water out of the tissues helps to reduce the swelling.

(A) $1$. Calculate molality $(m)$ of each solute in $100 \ g$ of solution (assuming $84.65 \ g$ of water solvent):
$m_{AB} = \frac{5.85 \ g / 58.5 \ g \ mol^{-1}}{0.08465 \ kg} = 1.181 \ mol \ kg^{-1}$
$m_{XY_2} = \frac{9.50 \ g / 95 \ g \ mol^{-1}}{0.08465 \ kg} = 1.181 \ mol \ kg^{-1}$
$2$. Calculate van't Hoff factor $(i)$ for each:
For $AB \rightarrow A^+ + B^-$, $i_1 = 1 + (2-1)0.8 = 1.8$
For $XY_2 \rightarrow X^{2+} + 2Y^-$, $i_2 = 1 + (3-1)0.6 = 2.2$
$3$. Calculate total depression in freezing point $(\Delta T_f)$:
$\Delta T_f = K_f \times (i_1 m_1 + i_2 m_2) = 1.86 \times (1.8 \times 1.181 + 2.2 \times 1.181) = 1.86 \times (4.724) \approx 8.786 \ K$
$4$. Calculate final freezing point:
$T_f = 273 \ K - 8.786 \ K = 264.214 \ K \approx 264.25 \ K$.

(C) In the process of decolorization of raw sugar,$\text{Animal charcoal}$ acts as the $\text{adsorbent}$ because it provides the surface for the impurities to stick to.
The $\text{colouring substances}$ (impurities) present in the sugar solution are the $\text{adsorbate}$ as they get accumulated on the surface of the charcoal.
This phenomenon is known as $\text{adsorption}$.
Therefore,the correct set is: $\text{Adsorbent} = \text{Animal charcoal}$,$\text{Adsorbate} = \text{Colouring substance}$,$\text{Process} = \text{Adsorption}$.

(B) The Freundlich adsorption isotherm equation is given by $\frac{x}{m} = k p^{1/n}$.
Taking logarithm on both sides,we get $\log (x / m) = \log k + \frac{1}{n} \log p$.
Comparing this with the equation of a straight line,$y = mx + c$,where $y = \log (x / m)$,$x = \log p$,$m = \frac{1}{n}$,and $c = \log k$.
Thus,the slope of the line is $\frac{1}{n}$ and the $Y$-axis intercept is $\log k$.