JEE Main 2022 Chemistry Question Paper with Answer and Solution

(D) Given mass of organic compound $= 120 \ g$.
Mass of $CO_{2} = 330 \ g$.
Mass of $H_{2}O = 270 \ g$.
Mass of carbon $= (\text{moles of } CO_{2}) \times 12 = (330 / 44) \times 12 = 7.5 \times 12 = 90 \ g$.
Percentage of carbon $= (90 / 120) \times 100 = 75 \ \%$.
Mass of hydrogen $= (\text{moles of } H_{2}O) \times 2 = (270 / 18) \times 2 = 15 \times 2 = 30 \ g$.
Percentage of hydrogen $= (30 / 120) \times 100 = 25 \ \%$.
Therefore,the percentages are $75 \ \%$ and $25 \ \%$ respectively.

(C) The energy of one mole of photons is given by the formula: $E = \frac{hcN_A}{\lambda}$
Substituting the given values:
$E = \frac{6.63 \times 10^{-34} \ J \ s \times 3 \times 10^8 \ m \ s^{-1} \times 6.02 \times 10^{23} \ mol^{-1}}{300 \times 10^{-9} \ m}$
$E = \frac{11.976 \times 10^{-2} \ J \ mol^{-1}}{3 \times 10^{-7}}$
$E = 3.9913 \times 10^5 \ J \ mol^{-1}$
$E \approx 399 \ kJ \ mol^{-1}$

(B) To determine the bond order,we use the Molecular Orbital Theory $(MOT)$ configuration:
$1$. For $C_{2}^{2-}$ ($14$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. Bond order = $(10-4)/2 = 3$.
$2$. For $N_{2}^{2-}$ ($16$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. Bond order = $(10-6)/2 = 2$.
$3$. For $O_{2}^{2-}$ ($18$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. Bond order = $(10-8)/2 = 1$.
Thus,the order of bond orders is $O_{2}^{2-} (1) < N_{2}^{2-} (2) < C_{2}^{2-} (3)$.

(C) The formation reaction of ethane is: $2C_{\text{(graphite)}} + 3H_{2(g)} \rightarrow C_{2}H_{6(g)}$
The enthalpy of formation is given by: $\Delta_{f}H^{\Theta}(C_{2}H_{6}) = [2 \times \Delta_{c}H^{\Theta}(C) + 3 \times \Delta_{c}H^{\Theta}(H_{2})] - \Delta_{c}H^{\Theta}(C_{2}H_{6})$
Substituting the given values:
$\Delta_{f}H^{\Theta}(C_{2}H_{6}) = [2 \times (-394.0) + 3 \times (-286.0)] - (-1560.0)$
$\Delta_{f}H^{\Theta}(C_{2}H_{6}) = [-788.0 - 858.0] + 1560.0$
$\Delta_{f}H^{\Theta}(C_{2}H_{6}) = -1646.0 + 1560.0 = -86.0 \ kJ \ mol^{-1}$

(B) The melting points of the given metals are as follows:
$Hg$ (mercury) = $-38.8 \ ^\circ C$
$Ga$ (gallium) = $29.76 \ ^\circ C$
$Cs$ (cesium) = $28.44 \ ^\circ C$
$Ag$ (silver) = $961.78 \ ^\circ C$
Among the given options,$Ag$ has the highest melting point.

(A) Sodium hydrogencarbonate $(NaHCO_{3})$,commonly known as Baking soda,is used in certain types of fire extinguishers because it releases $CO_{2}$ gas upon heating or reacting with an acid,which helps in extinguishing the fire.

(D) $CH_4$,$O_3$,and $H_2O$ are greenhouse gases that contribute to global warming.
$N_2$ is the most abundant gas in the atmosphere and does not absorb infrared radiation,therefore it does not contribute to the greenhouse effect.

(D) The stability of carbocations is determined by resonance and inductive effects.
In structure $B$,the positive charge is on the carbon atom adjacent to the oxygen atom. The lone pair on the oxygen atom can donate electron density through resonance,which significantly stabilizes the carbocation.
In structure $C$,the positive charge is on the carbon atom at the $\beta$-position relative to the oxygen atom. The oxygen atom exerts an electron-withdrawing inductive effect ($-I$ effect),which destabilizes the carbocation.
In structure $A$,the carbocation is a simple secondary carbocation without any adjacent heteroatom effects.
Therefore,the stability order is $B > A > C$.

(D) Statement $I$ is incorrect. Alkenes are generally considered more reactive than alkanes due to the presence of the $\pi$-bond,but the stability of a molecule depends on its total bond energy. Alkenes are not inherently 'less stable' than alkanes in a way that makes them unstable compounds; they are simply more reactive towards electrophilic addition.
Statement $II$ is correct. $A$ carbon-carbon double bond $(C=C)$ consists of one $\sigma$-bond and one $\pi$-bond. The total bond energy of a $C=C$ bond (approx. $610 \ kJ/mol$) is significantly higher than that of a $C-C$ single bond (approx. $348 \ kJ/mol$). Therefore,the strength of the double bond is greater than that of the single bond.

(D) Given: Ideal gas $A$ and $H_2$ gas are at the same pressure $(P)$ and volume $(V)$.
From the ideal gas equation,$PV = nRT$,we have $n = \frac{PV}{RT}$.
Since $P$ and $V$ are the same for both gases,$n_A T_A = n_{H_2} T_{H_2}$.
Here,$n_A = \frac{3.0}{M_A}$ and $n_{H_2} = \frac{0.2}{2.0} = 0.1 \ mol$.
Substituting the values: $\frac{3.0}{M_A} \times 300 = 0.1 \times 200$.
$\frac{900}{M_A} = 20$.
$M_A = \frac{900}{20} = 45 \ g \ mol^{-1}$.

(C) Initial moles: $n(PCl_{5}) = 5$,$n(N_{2}) = 2$. Total moles at equilibrium: $n_{total} = (5-x) + x + x + 2 = 7+x$.
Using ideal gas law $PV = nRT$ at equilibrium:
$2.46 \times 200 = (7+x) \times 0.082 \times 600$
$492 = (7+x) \times 49.2$
$7+x = 10 \implies x = 3$.
At equilibrium: $n(PCl_{5}) = 2$,$n(PCl_{3}) = 3$,$n(Cl_{2}) = 3$,$n(N_{2}) = 2$. Total moles $= 10$.
Partial pressures: $P(PCl_{5}) = (2/10) \times 2.46 = 0.492 \, atm$,$P(PCl_{3}) = (3/10) \times 2.46 = 0.738 \, atm$,$P(Cl_{2}) = (3/10) \times 2.46 = 0.738 \, atm$.
$K_{p} = \frac{P(PCl_{3}) \times P(Cl_{2})}{P(PCl_{5})} = \frac{0.738 \times 0.738}{0.492} = 1.107$.
$K_{p} = 1107 \times 10^{-3}$.

(A) The disproportionation reaction of manganate $(VI)$ ion $(MnO_{4}^{2-})$ in an acidic medium is given by:
$3MnO_{4}^{2-} + 4H^{+} \longrightarrow 2MnO_{4}^{-} + MnO_{2} + 2H_{2}O$
In $MnO_{4}^{-}$,the oxidation state of $Mn$ is $+7$.
In $MnO_{2}$,the oxidation state of $Mn$ is $+4$.
The difference in oxidation states is $|7 - 4| = 3$.

(A) Weight of organic compound $= 0.2 \ g$.
Volume of $N_2$ at $STP$ $= 22.400 \ mL = 22.4 \times 10^{-3} \ L$.
Moles of $N_2 = \frac{22.4 \times 10^{-3} \ L}{22.4 \ L \ mol^{-1}} = 10^{-3} \ mol$.
Mass of $N_2$ evolved $= 10^{-3} \ mol \times 28 \ g \ mol^{-1} = 28 \times 10^{-3} \ g$.
Percentage of nitrogen $= \frac{\text{Mass of } N_2}{\text{Mass of compound}} \times 100$.
Percentage of nitrogen $= \frac{28 \times 10^{-3} \ g}{0.2 \ g} \times 100 = \frac{0.028}{0.2} \times 100 = 14 \%$.

(C) The balanced chemical equation for the combustion of the fuel is:
$C_{15}H_{30} + \frac{45}{2} O_{2} \rightarrow 15 CO_{2} + 15 H_{2}O$
First,calculate the mass of $1 \ L$ $(1000 \ mL)$ of fuel:
$\text{Mass} = \text{density} \times \text{volume} = 0.756 \ g/mL \times 1000 \ mL = 756 \ g$
The molar mass of $C_{15}H_{30}$ is $(15 \times 12) + (30 \times 1) = 210 \ g/mol$.
Number of moles of fuel $= \frac{756 \ g}{210 \ g/mol} = 3.6 \ mol$.
According to the stoichiometry,$1 \ mol$ of fuel requires $\frac{45}{2} = 22.5 \ mol$ of $O_{2}$.
So,$3.6 \ mol$ of fuel requires $3.6 \times 22.5 = 81 \ mol$ of $O_{2}$.
Weight of $O_{2} = 81 \ mol \times 32 \ g/mol = 2592 \ g$.
According to the stoichiometry,$1 \ mol$ of fuel produces $15 \ mol$ of $CO_{2}$.
So,$3.6 \ mol$ of fuel produces $3.6 \times 15 = 54 \ mol$ of $CO_{2}$.
Weight of $CO_{2} = 54 \ mol \times 44 \ g/mol = 2376 \ g$.

(B) Degenerate orbitals are orbitals that have the same energy level.
For a hydrogen-like atom,energy depends only on the principal quantum number $n$.
For multi-electron atoms,orbitals are degenerate if they have the same $n$ and $l$ values.
In pair $(A)$: $(a)$ has $n=3, l=1$ ($3p$ orbital) and $(b)$ has $n=3, l=2$ ($3d$ orbital). These have different $l$ values,so they are not degenerate.
In pair $(B)$: $(a)$ has $n=3, l=2$ ($3d$ orbital) and $(b)$ has $n=3, l=2$ ($3d$ orbital). Since both have the same $n=3$ and $l=2$,they belong to the same subshell and are degenerate.
In pair $(C)$: $(a)$ has $n=4, l=2$ ($4d$ orbital) and $(b)$ has $n=3, l=2$ ($3d$ orbital). These have different $n$ values,so they are not degenerate.
Therefore,only pair $(B)$ consists of electrons in degenerate orbitals.

(A) For the reaction: $A_{(g)} \rightleftharpoons B_{(g)} + \frac{1}{2} C_{(g)}$
Initial moles: $1, 0, 0$
At equilibrium: $(1 - \alpha), \alpha, \frac{\alpha}{2}$
Total moles at equilibrium = $(1 - \alpha) + \alpha + \frac{\alpha}{2} = 1 + \frac{\alpha}{2}$
Partial pressures are given by $P_i = x_i \cdot p$,where $x_i$ is the mole fraction and $p$ is the total pressure.
$P_A = \frac{1 - \alpha}{1 + \frac{\alpha}{2}} p$
$P_B = \frac{\alpha}{1 + \frac{\alpha}{2}} p$
$P_C = \frac{\alpha/2}{1 + \frac{\alpha}{2}} p$
$K_p = \frac{P_B \cdot P_C^{1/2}}{P_A} = \frac{(\frac{\alpha}{1 + \alpha/2} p) \cdot (\frac{\alpha/2}{1 + \alpha/2} p)^{1/2}}{\frac{1 - \alpha}{1 + \alpha/2} p}$
Simplifying the expression:
$K_p = \frac{\alpha \cdot (\alpha/2)^{1/2} \cdot p^{1/2}}{1 - \alpha} \cdot \frac{(1 + \alpha/2)^{1/2}}{(1 + \alpha/2)^{1/2}} = \frac{\alpha \cdot \alpha^{1/2} \cdot p^{1/2}}{\sqrt{2} \cdot (1 - \alpha) \cdot (\frac{2 + \alpha}{2})^{1/2}} = \frac{\alpha^{3/2} p^{1/2}}{(2 + \alpha)^{1/2}(1 - \alpha)}$

(B) The nature of the given oxides is as follows:
$Na_{2}O$: Basic oxide (alkali metal oxide).
$As_{2}O_{3}$: Amphoteric oxide.
$N_{2}O$: Neutral oxide.
$NO$: Neutral oxide.
$Cl_{2}O_{7}$: Acidic oxide (non-metal oxide).
Thus,only $As_{2}O_{3}$ is an amphoteric oxide.
The total number of amphoteric oxides is $1$.

(B) The highest industrial consumption of molecular hydrogen is in the Haber process for the production of ammonia $(NH_3)$.
Around $55 \, \%$ of the global hydrogen production is utilized for the synthesis of ammonia by reacting it with nitrogen.

(A) $(A)$ $LiCl$ and $MgCl_2$ are covalent in nature due to polarization (Fajans' rule), hence they are soluble in ethanol.
$(B)$ $Li$ and $Mg$ do not form superoxides; they form oxides and peroxides respectively.
$(C)$ $LiF$ has a very high lattice energy due to the small size of both $Li^+$ and $F^-$ ions, making it less soluble in water compared to other alkali metal fluorides.
$(D)$ $Li_2O$ is less soluble in water compared to other alkali metal oxides due to high lattice energy.
Therefore, statements $(A)$ and $(C)$ are correct.

(C) Analysis of statements:
$(A)$ Incorrect: $B_2H_6$ has two types of $B-H$ bonds: terminal $B-H$ bonds (covalent) and bridging $B-H-B$ bonds ($3$-centre-$2$-electron).
$(B)$ Incorrect: $B_2H_6$ contains two $3$-centre-$2$-electron bonds,not four.
$(C)$ Correct: $B_2H_6$ is an electron-deficient molecule and acts as a Lewis acid.
$(D)$ Correct: $B_2H_6$ can be synthesized by the reaction: $3NaBH_4 + 4BF_3 \rightarrow 2B_2H_6 + 3NaBF_4$.
$(E)$ Incorrect: $B_2H_6$ is a non-planar molecule.
Therefore,statements $(C)$ and $(D)$ are correct.

(B) The enamel of human teeth is primarily composed of hydroxyapatite,which is a crystalline form of calcium phosphate,$Ca_{10}(PO_4)_6(OH)_2$.
In this structure,calcium exists in the $Ca^{2+}$ oxidation state.
Phosphorus exists in the phosphate ion $(PO_4)^{3-}$,where the oxidation state of phosphorus is $+5$ $(P^{5+})$.
Fluoride ions $(F^{-})$ are also often incorporated into the enamel structure to form fluorapatite,which strengthens the teeth.
Therefore,$P^{3+}$ is not present in the enamel of the teeth.

(C) For the reaction $2O_{3(g)} \rightleftharpoons 3O_{2(g)}$,let the initial moles of $O_3$ be $1$.
At equilibrium,moles of $O_3 = 1 - 0.5 = 0.5$ and moles of $O_2 = \frac{3}{2} \times 0.5 = 0.75$.
Total moles = $0.5 + 0.75 = 1.25$.
Mole fraction of $O_3 = \frac{0.5}{1.25} = 0.4$ and $O_2 = \frac{0.75}{1.25} = 0.6$.
Partial pressures at $P = 1 \ atm$: $P_{O_3} = 0.4 \ atm$,$P_{O_2} = 0.6 \ atm$.
$K_p = \frac{(P_{O_2})^3}{(P_{O_3})^2} = \frac{(0.6)^3}{(0.4)^2} = \frac{0.216}{0.16} = 1.35$.
$\Delta G^{\circ} = -RT \ln K_p$.
$\Delta G^{\circ} = -8.3 \times 300 \times \ln 1.35$.
$\Delta G^{\circ} = -8.3 \times 300 \times 0.3 = -747 \ J \ mol^{-1}$.

(B) In chromate ion $(CrO_4^{2-})$,let the oxidation state of $Cr$ be $x$.
$x + 4(-2) = -2 \implies x - 8 = -2 \implies x = +6$.
In dichromate ion $(Cr_2O_7^{2-})$,let the oxidation state of $Cr$ be $y$.
$2y + 7(-2) = -2 \implies 2y - 14 = -2 \implies 2y = +12 \implies y = +6$.
The difference in oxidation state is $|+6 - (+6)| = 0$.

(D) An electrophilic center is an atom that is electron-deficient and can accept an electron pair. In the given molecule,we identify the following electrophilic centers:
$1$. The carbonyl carbon $(C=O)$ is electrophilic due to the high electronegativity of oxygen,which creates a partial positive charge on the carbon.
$2$. The $\beta$-carbon of the $\alpha,\beta$-unsaturated carbonyl system is electrophilic due to resonance,which delocalizes the positive charge from the carbonyl carbon to the $\beta$-carbon.
$3$. The carbon atom of the nitrile group $(-CN)$ is electrophilic because the nitrogen atom is more electronegative than carbon,creating a partial positive charge on the carbon.
Thus,there are $3$ electrophilic centers in the molecule.

(D) The reaction involves the acid-catalyzed cyclization of $2,7-\text{dimethyl}-2,6-\text{octadiene}$.
First,the protonation of one of the double bonds forms a tertiary carbocation.
Then,an internal nucleophilic attack by the other double bond occurs,leading to the formation of a five-membered ring with a tertiary carbocation.
Finally,the loss of a proton $(-H^{+})$ from the carbocation results in the formation of the major product,which is $1,1,2-\text{trimethyl}-2-\text{isopropylidenecyclopentane}$ (or a similar isomer depending on the specific mechanism path).
Looking at the final structure shown in the mechanism: the product is $1,1-\text{dimethyl}-2-\text{isopropylidenecyclopentane}$.
In this structure,the double bond is between the ring carbon and the isopropylidene carbon.
This double bond involves $2$ carbon atoms,both of which are $sp^2$ hybridized.
Therefore,the major product has $2$ $sp^2$ hybridized carbon atoms.

(C) The minimum energy required for the photoelectric effect is known as the work function $(W)$.
The formula for the work function is $W = h \nu_0$,where $h$ is Planck's constant and $\nu_0$ is the threshold frequency.
Given:
$h = 6.6 \times 10^{-34} \, J \, s$
$\nu_0 = 1.3 \times 10^{15} \, s^{-1}$
Calculation:
$W = (6.6 \times 10^{-34} \, J \, s) \times (1.3 \times 10^{15} \, s^{-1})$
$W = 8.58 \times 10^{-19} \, J$
Thus,the minimum energy is $8.58 \times 10^{-19} \, J$.

(C) The enthalpy of reaction is calculated using the enthalpies of combustion of reactants and products:
$\Delta H = \sum \Delta H_{\text{combustion}} \text{ (Reactants)} - \sum \Delta H_{\text{combustion}} \text{ (Products)}$
For the reaction $3 C_2H_{2(g)} \rightarrow C_6H_{6(l)}$:
$\Delta H = [3 \times \Delta H_{\text{combustion}}(C_2H_{2(g)})] - [1 \times \Delta H_{\text{combustion}}(C_6H_{6(l)})]$
Substituting the given values:
$\Delta H = [3 \times (-1300 \, kJ \, mol^{-1})] - [-3268 \, kJ \, mol^{-1}]$
$\Delta H = -3900 + 3268 = -632 \, kJ \, mol^{-1}$

(A) The dissociation of bismuth sulphide is given by: $Bi_{2}S_{3}(s) \rightleftharpoons 2Bi^{3+}(aq) + 3S^{2-}(aq)$.
Let the solubility be $s \ mol \ L^{-1}$. Then the concentrations are $[Bi^{3+}] = 2s$ and $[S^{2-}] = 3s$.
The solubility product expression is $K_{sp} = [Bi^{3+}]^{2} [S^{2-}]^{3}$.
Substituting the values: $K_{sp} = (2s)^{2} (3s)^{3} = 4s^{2} \times 27s^{3} = 108s^{5}$.
Given $K_{sp} = 1.08 \times 10^{-73}$,we have $108s^{5} = 1.08 \times 10^{-73}$.
$s^{5} = \frac{1.08 \times 10^{-73}}{108} = 0.01 \times 10^{-73} = 10^{-75}$.
Taking the fifth root: $s = (10^{-75})^{1/5} = 10^{-15} \ mol \ L^{-1}$.

(D) The electron gain enthalpy $(\Delta_{eg}H)$ becomes less negative as we move down a group.
For Group $16$ ($Te$ and $Po$): The order is $Po < Te$ (in terms of magnitude of negative value,$Te$ is more negative than $Po$).
For Group $17$ ($F$ and $Cl$): Due to the small size of $F$,the electron-electron repulsion is high,making $Cl$ have a more negative electron gain enthalpy than $F$.
The values are:
$Cl: -349 \ kJ/mol$
$F: -328 \ kJ/mol$
$Te: -190 \ kJ/mol$
$Po: -174 \ kJ/mol$
Comparing the magnitude of negative values (more negative means more energy released,i.e.,lower enthalpy value):
$Po (-174) < Te (-190) < F (-328) < Cl (-349)$.
Thus,the correct order of electron gain enthalpies is $Po < Te < F < Cl$.

(D) The amphoteric nature of water is explained by the Brønsted-Lowry acid/base concept,not the Lewis concept.
In the reaction $H_{2}O + NH_{3} \rightleftharpoons NH_{4}^{+} + OH^{-}$,water acts as a Brønsted-Lowry acid (proton donor).
In the reaction $H_{2}S + H_{2}O \rightleftharpoons H_{3}O^{+} + HS^{-}$,water acts as a Brønsted-Lowry base (proton acceptor).
Since Assertion $A$ claims the Lewis concept explains this,$A$ is false.
Reason $R$ correctly describes the behavior of water in these reactions,so $R$ is true.
Therefore,$A$ is false but $R$ is true.

(C) Clean water has a $BOD$ value of less than $5 \, ppm$,whereas highly polluted water has a $BOD$ value of $17 \, ppm$ or more. Therefore,Assertion $A$ is correct.
$BOD$ (Biochemical Oxygen Demand) is a measure of the amount of dissolved oxygen required by bacteria to decompose only the biodegradable organic matter present in water. It does not include non-biodegradable organic material. Therefore,Reason $R$ is incorrect.

(D) Benzoic acid is soluble in hot water,while naphthalene is insoluble in water. Therefore,they can be separated by crystallization using hot water,not benzene.
Assertion $A$ is false because benzene would dissolve both benzoic acid and naphthalene,making separation difficult.
Reason $R$ is true as benzoic acid is indeed soluble in hot water.
Thus,$A$ is false but $R$ is true.

(B) The sodium fusion extract is boiled with concentrated $HNO_3$ to decompose sodium cyanide $(NaCN)$ and sodium sulphide $(Na_2S)$ present in the extract.
This is necessary because these ions interfere with the silver nitrate $(AgNO_3)$ test for halogens by forming precipitates like $AgCN$ or $Ag_2S$.

(A) The reaction of $3,4-dihydro-2H-pyran$ with $Br_2$ in the presence of $CH_3OH$ proceeds via the formation of a cyclic bromonium ion intermediate.
$1.$ $Br_2$ adds to the double bond to form a cyclic bromonium ion.
$2.$ The nucleophile $CH_3OH$ attacks the more substituted carbon atom of the bromonium ion due to electronic effects and the stability of the transition state.
$3.$ This results in an anti-addition product where the $-OCH_3$ group and the $-Br$ atom are in a trans-configuration relative to each other.
Thus,the major product is the one where $-Br$ and $-OCH_3$ are trans to each other.

(D) The reaction involves the acid-catalyzed addition of an alcohol to a vinyl ether,specifically $3,4-$dihydro-$2$$H$-pyran.
$1$. The acid catalyst $(H^+)$ protonates the double bond of the vinyl ether to form a stable resonance-stabilized oxocarbenium ion.
$2$. The alcohol (isopentyl alcohol) acts as a nucleophile and attacks the electrophilic carbon of the oxocarbenium ion.
$3$. Subsequent deprotonation yields the final acetal product.
This reaction is commonly used for the protection of alcohols as tetrahydropyranyl $(THP)$ ethers.
The structure of the product is an isopentyl tetrahydropyranyl ether.

(A) The minimum molar mass of a protein is calculated assuming that at least one molecule of the constituent (glycine) is present in the protein molecule.
Given that $0.30\, \%$ of the protein is glycine,we have:
$0.30\, g$ of glycine in $100\, g$ of protein.
Since the molecular weight of glycine is $75\, g\, mol^{-1}$,the minimum molar mass of the protein is calculated as:
$\text{Minimum molar mass} = \frac{\text{Molecular weight of glycine} \times 100}{\text{Percentage of glycine}}$
$\text{Minimum molar mass} = \frac{75 \times 100}{0.30} = \frac{7500}{0.30} = 25000\, g\, mol^{-1}$
Converting this to the required format: $25000\, g\, mol^{-1} = 25 \times 10^{3} \, g\, mol^{-1}$.
Thus,the value is $25$.

(A) Since the tank is rigid, the volume $V$ remains constant. According to Gay-Lussac's Law, for a fixed amount of gas at constant volume, the pressure is directly proportional to the absolute temperature: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.
Given:
$P_1 = 30 \ atm$
$T_1 = 27^{\circ} C = 27 + 273 = 300 \ K$
$T_2 = 45^{\circ} C = 45 + 273 = 318 \ K$
Substituting the values into the equation:
$\frac{30}{300} = \frac{P_2}{318}$
$0.1 = \frac{P_2}{318}$
$P_2 = 0.1 \times 318 = 31.8 \ atm$.
Rounding to the nearest integer, we get $P_2 = 32 \ atm$.

(A) The net dipole moment $(\mu_{net})$ depends on the molecular geometry and the electronegativity difference between atoms.
$BeF_{2}$ is linear,$BF_{3}$ is trigonal planar,and $CCl_{4}$ is tetrahedral. These molecules have symmetric structures where individual bond dipoles cancel each other out,resulting in $\mu_{net} = 0$.
$H_{2}O$ (bent),$NH_{3}$ (trigonal pyramidal),and $HCl$ (heteronuclear diatomic) have asymmetric charge distributions,resulting in $\mu_{net} \neq 0$.
Therefore,the molecules with non-zero net dipole moment are $H_{2}O$,$NH_{3}$,and $HCl$.
The total count is $3$.

(B) The hydrolysis of $PCl_{3}$ proceeds as follows:
$PCl_{3} + 3H_{2}O \longrightarrow H_{3}PO_{3} (A) + 3HCl$
However,the question implies a stepwise hydrolysis where $A$ is an intermediate like $PCl_{2}(OH)$ or $PCl(OH)_{2}$.
Upon further hydrolysis,$A$ converts to $B$,which is phosphorous acid $(H_{3}PO_{3})$.
The structure of $H_{3}PO_{3}$ is $H-P(=O)(OH)_{2}$.
In $H_{3}PO_{3}$,only the two hydrogen atoms attached to oxygen atoms are ionisable as protons $(H^{+})$.
The hydrogen atom directly attached to the phosphorus atom is not ionisable.
Therefore,the number of ionisable protons in $B$ is $2$.

(C) At the equivalence point,the number of equivalents of acid equals the number of equivalents of base.
The formula for equivalents is $N \times V = M \times n_{factor} \times V$.
For the acid $A$: $M_1 = 0.1 \, M$,$V_1 = 10 \, mL$,$n_{factor} = x$ (basicity).
For the base $M(OH)_2$: $M_2 = 0.05 \, M$,$V_2 = 30 \, mL$,$n_{factor} = 2$ (acidity).
Equating the equivalents: $0.1 \times 10 \times x = 0.05 \times 30 \times 2$.
$1 \times x = 3$.
Therefore,$x = 3$.

(A) According to $M.O.$ Theory,bond strength is directly proportional to bond order $(B.O.)$.
If an electron is removed from an antibonding molecular orbital $(M.O.)$,the bond order increases,making the bond stronger.
Let us analyze the molecules:
$1$. $NO$ ($15$ $e^-$): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Removing an electron from $\pi^* 2p_x$ increases $B.O.$ from $2.5$ to $3.0$.
$2$. $N_2$ ($14$ $e^-$): Removing an electron from a bonding orbital $(\sigma 2p_z)$ decreases $B.O.$ from $3.0$ to $2.5$.
$3$. $O_2$ ($16$ $e^-$): Configuration has electrons in $\pi^* 2p_x$ and $\pi^* 2p_y$. Removing an electron from $\pi^*$ increases $B.O.$ from $2.0$ to $2.5$.
$4$. $C_2$ ($12$ $e^-$): Removing an electron from a bonding orbital $(\pi 2p)$ decreases $B.O.$ from $2.0$ to $1.5$.
$5$. $B_2$ ($10$ $e^-$): Removing an electron from a bonding orbital $(\pi 2p)$ decreases $B.O.$ from $1.0$ to $0.5$.
Thus,the bond becomes stronger for $NO$ and $O_2$.

(D) Isoelectronic species are those that have the same number of electrons.
$Al^{3+}$ has $13 - 3 = 10$ electrons.
$O^{2-}$ has $8 + 2 = 10$ electrons.
$Mg^{2+}$ has $12 - 2 = 10$ electrons.
Since $O^{2-}$ and $Mg^{2+}$ both have $10$ electrons,they are isoelectronic with $Al^{3+}$.

(D) Electron-deficient species are those in which the central atom has an incomplete octet,meaning it has fewer than $8$ electrons in its valence shell.
$1$. $PH_{3}$: $P$ has $8$ electrons (octet complete).
$2$. $B_{2}H_{6}$: Boron has $6$ electrons (electron-deficient).
$3$. $CCl_{4}$: $C$ has $8$ electrons (octet complete).
$4$. $NH_{3}$: $N$ has $8$ electrons (octet complete).
$5$. $LiH$: $Li$ has $2$ electrons (electron-deficient).
$6$. $BCl_{3}$: $B$ has $6$ electrons (electron-deficient).
Therefore,the electron-deficient molecules are $B_{2}H_{6}$,$LiH$,and $BCl_{3}$.
The total number of electron-deficient molecules is $3$.

(D) In aqueous solutions,the ionic mobility is inversely proportional to the degree of hydration.
Smaller ions have a higher charge density,leading to a larger hydration shell,which decreases their mobility.
Among the given ions,$Be^{2+}$ is the smallest and has the highest hydration energy,resulting in the largest hydrated radius and lowest mobility.
Conversely,$Sr^{2+}$ is the largest among the options,has the smallest hydration shell,and therefore exhibits the highest ionic mobility.

(A) Eutrophication is the process in which nutrient enrichment (such as nitrates and phosphates) leads to excessive growth of algae and aquatic plants.
This excessive growth depletes dissolved oxygen in the water,which leads to the death of aquatic organisms and ultimately results in the loss of biodiversity.

(C) Dihedral angle $:$ It is the angle between $2$ specified groups ($-CH_3$ here) in a Newman projection.
Staggered conformation refers to the arrangement where the groups on the front and back carbons are as far apart as possible.
In option $(A)$,the dihedral angle between the two $-CH_3$ groups is $60^{\circ}$ (gauche).
In option $(B)$,the structure is eclipsed.
In option $(C)$,the two $-CH_3$ groups are in the anti-position,meaning the dihedral angle is $180^{\circ}$. This is the staggered conformation with the maximum dihedral angle.
In option $(D)$,the structure is eclipsed.

(B) The reaction involves the acid-catalyzed alkylation of an alkene (isobutylene) with an alkane (isobutane).
$1$. The $H^+$ ion protonates the isobutylene to form a stable tert-butyl carbocation: $(CH_3)_2C=CH_2 + H^+ \rightarrow (CH_3)_3C^+$.
$2$. This carbocation then attacks another molecule of isobutylene to form a larger carbocation: $(CH_3)_3C^+ + (CH_3)_2C=CH_2 \rightarrow (CH_3)_3C-CH_2-C^+(CH_3)_2$.
$3$. This new carbocation then abstracts a hydride ion from a molecule of isobutane: $(CH_3)_3C-CH_2-C^+(CH_3)_2 + (CH_3)_3CH \rightarrow (CH_3)_3C-CH_2-CH(CH_3)_2 + (CH_3)_3C^+$.
The final product is $2,2,4$-trimethylpentane,which is $(CH_3)_3C-CH_2-CH(CH_3)_2$.

(B) The molar mass of $C_{7}H_{5}N_{3}O_{6}$ is $(7 \times 12) + (5 \times 1) + (3 \times 14) + (6 \times 16) = 84 + 5 + 42 + 96 = 227 \ g/mol$.
The number of moles of $C_{7}H_{5}N_{3}O_{6}$ is $n = \frac{681 \ g}{227 \ g/mol} = 3 \ mol$.
Since each molecule of $C_{7}H_{5}N_{3}O_{6}$ contains $3$ atoms of $N$,the total moles of $N$ atoms is $n_{N} = 3 \times 3 = 9 \ mol$.
The number of $N$ atoms is $9 \times N_{A} = 9 \times 6.02 \times 10^{23} = 54.18 \times 10^{23} = 5418 \times 10^{21}$.
Therefore,the value of $x$ is $5418$.

(D) The ionization energy of a hydrogen-like species is given by $E_n = -13.6 \times Z^2 / n^2 \, eV$ or $E_n = -E_H \times Z^2 / n^2$.
For $Li$ $(Z=3)$ in the ground state $(n=1)$,the energy required to remove the electron is $E = E_{\infty} - E_1 = 0 - (-2.2 \times 10^{-18} \times 3^2 / 1^2) = 19.8 \times 10^{-18} \, J$.
The energy of a photon is $E = hc / \lambda$,so $\lambda = hc / E$.
Substituting the values: $\lambda = (6.63 \times 10^{-34} \times 3 \times 10^8) / (19.8 \times 10^{-18}) = 19.89 \times 10^{-26} / 19.8 \times 10^{-18} \approx 1.0045 \times 10^{-8} \, m$.
Comparing this with $x \times 10^{-8} \, m$,we get $x \approx 1$.

(B) At equilibrium,the Gibbs free energy change $\Delta G$ is equal to $0$.
Using the relation $\Delta G = \Delta H - T \Delta S = 0$,we get $T = \frac{\Delta H}{\Delta S}$.
Given $\Delta H = -165 \ kJ \ mol^{-1} = -165000 \ J \ mol^{-1}$ and $\Delta S = -550 \ J K^{-1} mol^{-1}$.
Substituting the values: $T = \frac{-165000 \ J \ mol^{-1}}{-550 \ J K^{-1} mol^{-1}} = 300 \ K$.
Therefore,the temperature at which the reaction attains equilibrium is $300 \ K$.

(C) For a first order reaction,the rate constant $K$ is given by $K = \frac{0.693}{t_{1/2}}$.
The time $t$ required for a reaction to complete $90 \%$ is given by $t_{90\%} = \frac{2.303}{K} \log \left( \frac{100}{100 - 90} \right) = \frac{2.303}{K} \log 10 = \frac{2.303}{K}$.
Substituting $K = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{t_{1/2}}$ into the equation:
$t_{90\%} = \frac{2.303}{0.693} \times t_{1/2}$.
Since $2.303 \approx \ln 10$ and $0.693 \approx \ln 2$,we have $t_{90\%} = \frac{\ln 10}{\ln 2} \times t_{1/2} = \frac{2.303}{2.303 \times 0.3010} \times t_{1/2} = \frac{1}{0.3010} \times t_{1/2} \approx 3.32 \times t_{1/2}$.
Therefore,$x = 3.32$.

(B) The Hall-Heroult process is the major industrial method for the extraction of aluminium from alumina $(Al_2O_3)$.
In this electrolytic process,alumina is dissolved in molten cryolite $(Na_3AlF_6)$ and electrolyzed using carbon electrodes.
The overall chemical reaction is: $2 Al_2O_3 + 3 C \rightarrow 4 Al + 3 CO_2$.

(A) Sodium hydroxide $(NaOH)$ is prepared commercially by the electrolysis of brine ($NaCl$ solution) in a $Castner-Kellner$ cell.
At the cathode,sodium ions are reduced to sodium amalgam $(Na-Hg)$:
$Na^{+} + e^{-} \xrightarrow{Hg} Na-Hg$
At the anode,chloride ions are oxidized to chlorine gas:
$Cl^{-} \longrightarrow \frac{1}{2}Cl_{2} + e^{-}$
The sodium amalgam is then treated with water to produce sodium hydroxide and molecular hydrogen gas as a byproduct:
$2Na(amalgam) + 2H_{2}O \longrightarrow 2NaOH + H_{2} + 2Hg$

(B) $PCl_5$ forms five bonds by using the $d-$orbitals to expand its octet.
$NCl_5$ does not exist because nitrogen belongs to the $2^{nd}$ period and lacks $d-$orbitals in its valence shell.
Therefore,nitrogen cannot expand its octet beyond a covalency of $4$.

(D) The crystal field splitting energy $(\Delta_{0})$ increases down a group in the periodic table for metal ions with the same charge and same ligand environment.
This is because the $d$-orbitals of the $5d$ series elements are more extended in space compared to the $3d$ and $4d$ series,leading to stronger interaction with the ligands.
Among the given complexes,$Cr^{3+}$ $(3d)$,$Mo^{3+}$ $(4d)$,$Fe^{3+}$ $(3d)$,and $Os^{3+}$ $(5d)$ are present.
Since $Os$ belongs to the $5d$ series,the complex $[Os(H_{2}O)_{6}]^{3+}$ will have the highest $\Delta_{0}$ value.

(C) The conversion of $A$ ($4$-methylmethylenecyclohexane) to $B$ ($4$-methylcyclohexanecarbaldehyde) involves the transformation of an exocyclic double bond into an aldehyde group.
$1$. First,hydroboration-oxidation using $BH_3, H_2O_2 / OH^-$ is performed. This reaction follows anti-Markovnikov addition of water across the double bond to form a primary alcohol,$4$-methylcyclohexylmethanol.
$2$. Second,the primary alcohol is oxidized to an aldehyde using $PCC$ (Pyridinium chlorochromate). $PCC$ is a mild oxidizing agent that stops the oxidation at the aldehyde stage,preventing further oxidation to a carboxylic acid.

(C) $1$. $PCC$ (Pyridinium chlorochromate) oxidizes the secondary alcohol $CH_3-CH=CH-CH_2-CH(OH)-CH_3$ to the ketone $CH_3-CH=CH-CH_2-CO-CH_3$ (Compound $'A'$).
$2$. Compound $'A'$ contains a methyl ketone group,so it undergoes the iodoform reaction with sodium hypoiodite $(NaOI)$ to form the carboxylic acid $CH_3-CH=CH-CH_2-COOH$ (Compound $'B'$) and iodoform $(CHI_3)$.
$3$. Decarboxylation of the carboxylic acid $CH_3-CH=CH-CH_2-COOH$ with soda lime $(NaOH + CaO)$ removes $CO_2$ to yield $CH_3-CH=CH-CH_3$ (Compound $'C'$),which is $but-2-ene$.

(A) The conversion of propan$-1-$ol $(CH_3CH_2CH_2OH)$ to $n-$butylamine $(CH_3CH_2CH_2CH_2NH_2)$ requires increasing the carbon chain length by one carbon atom.
Step $1$: Propan$-1-$ol reacts with $SOCl_2$ to form $1-$chloropropane $(CH_3CH_2CH_2Cl)$.
Step $2$: $1-$chloropropane reacts with $KCN$ (nucleophilic substitution) to form butanenitrile $(CH_3CH_2CH_2CN)$.
Step $3$: Butanenitrile is reduced using $H_2/Ni$ or $Na(Hg)/C_2H_5OH$ (Mendius reduction) to form $n-$butylamine $(CH_3CH_2CH_2CH_2NH_2)$.

(C) Buna$-N$ is an addition copolymer formed by the polymerization of $1,3-$butadiene and acrylonitrile.
Condensation polymers are formed by the repeated condensation reaction between two different bi-functional or tri-functional monomeric units,usually with the elimination of small molecules like $H_2O$,$HCl$,or $NH_3$.
Nylon $6,6$,Dacron,and Silicone are examples of condensation polymers.
Buna$-N$ is an addition polymer,not a condensation polymer.

(C) The structure provided is that of $Cimetidine$.
$Cimetidine$ is a well-known histamine $H_2$-receptor antagonist that inhibits stomach acid production.
It is characterized by an imidazole ring attached to a thioether chain,which terminates in a cyanoguanidine group.

(A) The flame test is a diagnostic tool used to identify the presence of certain metal ions in a sample based on the characteristic color they impart to a non-luminous flame.

Ion	Colour of the flame
$A. Cu^{2+}$	Green flame with blue centre
$B. Sr^{2+}$	Crimson Red
$C. Ba^{2+}$	Apple green
$D. Ca^{2+}$	Brick red

Based on the observation,the presence of a green flame with a blue centre is characteristic of the $Cu^{2+}$ ion.

(D) According to Henry's Law: $P = K_{H} \times \chi_{CO_{2}}$
Given:
$P = 0.835 \ bar$
$K_{H} = 1.67 \ kbar = 1670 \ bar$
Since the amount of $CO_{2}$ is very small,we can approximate the mole fraction $\chi_{CO_{2}} \approx \frac{n_{CO_{2}}}{n_{H_{2}O}}$
$n_{H_{2}O} = \frac{1000 \ g}{18 \ g \ mol^{-1}} = 55.55 \ mol$
$0.835 = 1670 \times \frac{n_{CO_{2}}}{55.55}$
$n_{CO_{2}} = \frac{0.835 \times 55.55}{1670} = 0.02777 \ mol$
Mass of $CO_{2} = n_{CO_{2}} \times \text{Molar mass of } CO_{2} = 0.02777 \times 44 = 1.222 \ g$
$X = 1.222 \times 10^{3} \times 10^{-3} \ g = 1222 \times 10^{-3} \ g$
Thus,$X = 1222$.

(C) The relationship between conductivity $(k)$,resistance $(R)$,and cell constant $(G^*)$ is given by:
$k = \frac{1}{R} \times G^*$
Given:
$R = 1750 \, \Omega$
$k = 0.152 \times 10^{-3} \, S \, cm^{-1}$
Substituting the values:
$0.152 \times 10^{-3} = \frac{1}{1750} \times G^*$
$G^* = 0.152 \times 10^{-3} \times 1750$
$G^* = 266 \times 10^{-3} \, cm^{-1}$

(C) Initial moles of acetic acid $= M_1 \times V_1 = 0.2 \, M \times 0.2 \, L = 0.04 \, mol$.
Final moles of acetic acid $= M_2 \times V_2 = 0.1 \, M \times 0.2 \, L = 0.02 \, mol$.
Moles of acetic acid adsorbed $= 0.04 - 0.02 = 0.02 \, mol$.
Molar mass of acetic acid $(CH_3COOH) = 60 \, g/mol$.
Mass of acetic acid adsorbed $= 0.02 \, mol \times 60 \, g/mol = 1.2 \, g$.
Mass of wood charcoal $= 0.6 \, g$.
Mass of acetic acid adsorbed per gram of carbon $= \frac{1.2 \, g}{0.6 \, g} = 2 \, g/g$.

(A) $(1)$ Baryte: $BaSO_4$ (Sulphate ore)
$(2)$ Galena: $PbS$ (Sulphide ore)
$(3)$ Zinc blende: $ZnS$ (Sulphide ore)
$(4)$ Copper pyrites: $CuFeS_2$ (Sulphide ore)
Thus,$(2)$,$(3)$,and $(4)$ are sulphide $(S^{2-})$ based ores.
Therefore,the total number of sulphide-based minerals is $3$.

(B) The reaction of $3$-chloro-$2$-methylpent-$4$-ene with aqueous $NaOH$ is a nucleophilic substitution reaction ($S_N1$ or $S_N2$ depending on conditions,but here it leads to the substitution of $-Cl$ by $-OH$).
The product '$P$' is $3$-methylpent-$4$-en-$2$-ol.
The structure of the product contains one double bond $(C=C)$.
Since each double bond consists of one $\pi$ bond,and each $\pi$ bond contains $2$ electrons,the total number of $\pi$ electrons in the product is $2$.

(B) The given peptide is a pentapeptide,which consists of $5$ amino acid residues: $\text{alanine}$,$\text{glycine}$,$\text{leucine}$,$\text{alanine}$,and $\text{valine}$.
In a polypeptide chain,the number of peptide linkages is always $(n-1)$,where $n$ is the number of amino acid residues.
Here,$n = 5$,so the number of peptide linkages $= 5 - 1 = 4$.

(B) The hybridization of the given species is as follows:
$1$. $[PtCl_4]^{2-}$: $Pt^{2+}$ has a $d^8$ configuration. It forms a square planar complex with $dsp^2$ hybridization. $(A-III)$
$2$. $BrF_5$: The central atom $Br$ has $7$ valence electrons. It forms $5$ bonds and has $1$ lone pair,resulting in $sp^3d^2$ hybridization. $(B-IV)$
$3$. $PCl_5$: The central atom $P$ has $5$ valence electrons. It forms $5$ bonds with no lone pairs,resulting in $sp^3d$ hybridization. $(C-I)$
$4$. $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ has a $d^6$ configuration. $NH_3$ is a strong field ligand,forcing pairing of electrons,resulting in $d^2sp^3$ hybridization. $(D-II)$
Thus,the correct match is $A-III, B-IV, C-I, D-II$.

(C) Statement $I$ is correct because emulsions are thermodynamically unstable systems,and oil droplets tend to coalesce and separate into two layers over time.
Statement $II$ is incorrect because adding an excess of electrolyte to an emulsion typically causes coagulation or breaking of the emulsion,not stabilization. Emulsions are stabilized by adding emulsifying agents (emulsifiers) like soaps,proteins,or gums,which form a protective film around the dispersed droplets.

(A) The correct matches for the ores are as follows:
- Sphalerite is $ZnS$ $(iv)$.
- Calamine is $ZnCO_3$ $(iii)$.
- Galena is $PbS$ $(ii)$.
- Siderite is $FeCO_3$ $(i)$.
Therefore,the correct sequence is $(a)-iv, (b)-iii, (c)-ii, (d)-i$.

(A) The stability of nitrogen trihalides decreases as the size of the halogen atom increases.
The bond energy of the $N-X$ bond decreases from $F$ to $I$ due to the increasing mismatch in the size of the orbitals of $N$ and $X$ atoms.
Therefore,the order of stability is: $NF_{3} > NCl_{3} > NBr_{3} > NI_{3}$.
Thus,$NF_{3}$ is the most stable trihalide.

(B) The starting material $C_8H_{10}$ is ethylbenzene.
$1$. Nitration of ethylbenzene with $HNO_3/H_2SO_4$ gives $p$-nitroethylbenzene as the major product $(A)$.
$2$. Radical bromination of $p$-nitroethylbenzene with $Br_2/\Delta$ occurs at the benzylic position to form $1-(4-nitrophenyl)ethyl$ bromide $(B)$.
$3$. Dehydrohalogenation of $B$ with alcoholic $KOH$ yields $p$-nitrostyrene $(C)$ as the major product.

(A) $I.$ Monocarboxylic acids with an even number of carbon atoms exhibit better packing efficiency in the crystal lattice,which results in a higher melting point $(M.P.)$ compared to their odd-numbered counterparts.
$II.$ As the molar mass of monocarboxylic acids increases,the size of the hydrophobic alkyl chain increases,which reduces the interaction with water molecules,thereby decreasing solubility.

(C) conjugated diketone is a compound where the two carbonyl groups are separated by a system of alternating single and double bonds,allowing for resonance delocalization between the two carbonyl groups.
In $p-Benzoquinone$,the two carbonyl groups are part of a cyclic system where they are conjugated with the carbon-carbon double bonds in the ring.
Therefore,$p-Benzoquinone$ is a conjugated diketone.

(A) Step $1$: $4$-Methoxybenzyl bromide reacts with $NaCN$ to form $4$-methoxybenzyl cyanide $(CH_3O-C_6H_4-CH_2CN)$.
Step $2$: In the presence of $OH^-$,the alpha-hydrogen of the nitrile is abstracted to form a carbanion,which then performs a nucleophilic attack on the carbonyl carbon of cyclohexanone.
Step $3$: This results in the formation of a cyanohydrin derivative: $CH_3O-C_6H_4-CH(CN)-C(OH)(C_6H_{10})$.
Step $4$: Finally,catalytic hydrogenation using $H_2, Ni$ reduces the nitrile group $(-CN)$ to a primary amine group $(-CH_2NH_2)$.
The final product is $1-(4-methoxyphenyl)-2-(cyclohexyl)-2-hydroxyethanamine$ derivative,which matches the structure in option $A$.

(D) . Polyesters are polymers that contain the ester functional group in their main chain.
$PHBV$ (Poly-$ \beta $-hydroxybutyrate-co-$ \beta $-hydroxy valerate) is a biodegradable aliphatic polyester formed by the copolymerization of $3$-hydroxybutanoic acid and $3$-hydroxypentanoic acid.

(B) $1$. The polysaccharide '$X$' on hydrolysis yields '$Y$'.
$2$. Since '$Y$' gives gluconic acid on treatment with bromine water,'$Y$' must be glucose.
$3$. Cellulose is a polysaccharide composed of $D$-glucose units linked by $\beta-1,4-$glycosidic linkages.
$4$. Starch (amylose and amylopectin) contains $\alpha-$glycosidic linkages.
$5$. Therefore,'$X$' is cellulose.

(D) Broad spectrum antibiotics are effective against a wide range of Gram-positive and Gram-negative bacteria. $Vancomycin$,$Ampicillin$,and $Ofloxacin$ are examples of broad spectrum antibiotics. $Penicillin \ G$ is a narrow spectrum antibiotic,as it is primarily effective against Gram-positive bacteria.

(A) The qualitative analysis of $Ni^{2+}$ ions involves the addition of dimethylglyoxime $(DMG)$ in an ammoniacal (alkaline) medium.
This reaction results in the formation of a bright red precipitate of nickel dimethylglyoximate,$[Ni(DMG)_{2}]$.
The reaction is: $Ni^{2+} + 2DMG \xrightarrow{NH_4OH} [Ni(DMG)_{2}] \downarrow + 2H^+$.
Therefore,the reagent $(X)$ is dimethylglyoxime and the cation $(y^{2+})$ is $Ni^{2+}$.

(C) In an $hcp$ lattice,the number of atoms per unit cell is $6$.
Number of tetrahedral voids $= 2 \times \text{number of atoms} = 2 \times 6 = 12$.
Number of $Y$ atoms $= \frac{2}{3} \times 12 = 8$.
Total number of atoms in the lattice $= 6 (X) + 8 (Y) = 14$.
Percentage of element $X = \frac{6}{14} \times 100 = 42.85 \%$.
Rounding to the nearest integer,we get $43 \%$.

(C) The formula for osmotic pressure is $\pi = C \times R \times T$,where $C$ is the molar concentration in $mol \, L^{-1}$.
Given $\pi = 7.47 \, bar$,$R = 0.083 \, L \, bar \, K^{-1} \, mol^{-1}$,and $T = 300 \, K$.
Substituting the values: $7.47 = C \times 0.083 \times 300$.
$C = \frac{7.47}{0.083 \times 300} = \frac{7.47}{24.9} = 0.3 \, mol \, L^{-1}$.
To find the concentration in $g \, L^{-1}$,multiply the molar concentration by the molar mass of glucose $(180 \, g \, mol^{-1})$:
$\text{Concentration} = 0.3 \, mol \, L^{-1} \times 180 \, g \, mol^{-1} = 54 \, g \, L^{-1}$.

(B) The cell reaction is: $Cu^{2+} + H_{2(g)} \rightarrow Cu_{(s)} + 2H^{+}_{(aq)}$
The standard cell potential $E_{cell}^{0} = E_{cathode}^{0} - E_{anode}^{0} = 0.34 \, V - 0.00 \, V = 0.34 \, V$.
Using the Nernst equation at $298 \, K$:
$E_{cell} = E_{cell}^{0} - \frac{0.0591}{n} \log \frac{[H^{+}]^{2}}{[Cu^{2+}] \cdot P_{H_2}}$
Assuming $P_{H_2} = 1 \, bar$:
$0.576 = 0.34 - \frac{0.0591}{2} \log \frac{[H^{+}]^{2}}{0.01}$
$0.236 = -0.02955 \cdot \log \frac{[H^{+}]^{2}}{10^{-2}}$
$-7.986 = \log [H^{+}]^{2} - \log 10^{-2}$
$-7.986 = 2 \log [H^{+}] + 2$
$-9.986 = 2 \log [H^{+}]$
$\log [H^{+}] = -4.993$
$pH = -\log [H^{+}] = 4.993 \approx 5$.

(B) According to the Arrhenius equation,$\ln \ k = \ln \ A - \frac{E_a}{RT}$.
We are plotting $\ln \ k$ versus $\frac{10^{3}}{T}$.
Rewriting the equation: $\ln \ k = \ln \ A - \left( \frac{E_a}{R \times 10^{3}} \right) \times \left( \frac{10^{3}}{T} \right)$.
Comparing this with the equation of a straight line $y = mx + c$,the slope $m = -\frac{E_a}{R \times 10^{3}}$.
Given the slope is $-18.5$,we have: $-18.5 = -\frac{E_a}{8.31 \times 10^{3}}$.
$E_a = 18.5 \times 8.31 \times 10^{3} \ J \ mol^{-1} = 153735 \ J \ mol^{-1}$.
Converting to $kJ \ mol^{-1}$: $E_a = 153.735 \ kJ \ mol^{-1}$.
Rounding to the nearest integer,we get $154 \ kJ \ mol^{-1}$.

(B) The structure of the cobalt-carbonyl complex $[Co_{2}(CO)_{8}]$ consists of two $Co$ atoms bonded to each other.
$1$. Number of $Co-Co$ bonds $(X)$: There is $1$ direct metal-metal bond between the two cobalt atoms. So,$X = 1$.
$2$. Number of terminal $CO$ ligands $(Y)$: In the solid-state structure of $[Co_{2}(CO)_{8}]$,there are $6$ terminal $CO$ ligands (each $Co$ atom is bonded to $3$ terminal $CO$ groups) and $2$ bridging $CO$ ligands. So,$Y = 6$.
$3$. Calculation: $X + Y = 1 + 6 = 7$.

(D) Step $1$: Calculate the milliequivalents $(m_{eq})$ of $NaOH$ used for back titration: $m_{eq} = 30.0 \ mL \times 0.25 \ N = 7.5 \ m_{eq}$.
Step $2$: Calculate the total $m_{eq}$ of $H_2SO_4$ taken: $m_{eq} = 50.0 \ mL \times 0.5 \ N = 25.0 \ m_{eq}$.
Step $3$: Calculate the $m_{eq}$ of $H_2SO_4$ reacted with $NH_3$: $m_{eq} = 25.0 - 7.5 = 17.5 \ m_{eq}$.
Step $4$: Since $m_{eq}$ of $NH_3 = m_{eq}$ of $H_2SO_4$ reacted,$m_{eq}$ of $NH_3 = 17.5 \ m_{eq}$.
Step $5$: Calculate the mass of nitrogen: $\text{Mass of } N = 17.5 \times 10^{-3} \times 14 \ g = 0.245 \ g$.
Step $6$: Calculate the percentage of nitrogen: $\% N = (0.245 / 0.166) \times 100 \approx 147.59 \%$. Rounding to the nearest integer,we get $148 \%$.

(D) The formula for freezing point depression is $\Delta T_f = i \cdot K_f \cdot m$.
First,calculate the molality $(m)$: $m = \frac{0.7 \ g}{93 \ g \ mol^{-1}} \times \frac{1000 \ g \ kg^{-1}}{42.0 \ g} = 0.1792 \ mol \ kg^{-1}$.
Using $\Delta T_f = 0.2 \ K$ and $K_f = 1.86 \ K \ kg \ mol^{-1}$,we find the van't Hoff factor $(i)$:
$0.2 = i \cdot 1.86 \cdot 0.1792 \implies i = \frac{0.2}{1.86 \cdot 0.1792} \approx 0.60$.
For the association reaction $2A \rightleftharpoons A_2$,the van't Hoff factor is $i = 1 - \alpha + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}$.
Equating $1 - \frac{\alpha}{2} = 0.60$,we get $\frac{\alpha}{2} = 0.40$,so $\alpha = 0.80$.
The percentage association is $80 \%$.

(B) The correct matches are:
$a$. Zymase is an enzyme obtained from $ii$. Yeast.
$b$. Diastase is an enzyme obtained from $iii$. Malt.
$c$. Urease is an enzyme obtained from $iv$. Soyabean.
$d$. Pepsin is an enzyme found in $i$. Stomach.
Therefore,the correct sequence is $a-ii, b-iii, c-iv, d-i$.

(A) In electrolytic refining,the impure metal (blister copper) is made the anode.
During the process,the impure copper dissolves into the electrolyte,while precious metals like $Au$ and $Pt$ do not dissolve and settle at the bottom of the anode as anode mud.
Statement $I$ is true because precious metals are recovered from the anode mud,not deposited as pure metal on the cathode.
Statement $II$ is true because blister copper is indeed used as the anode in the electrolytic refining of copper.
Therefore,both statements are true.

(A) The standard reduction potentials $(E^{\circ})$ for the given pairs are:
$E^{\circ}_{Cl_2 / Cl^{-}} = +1.36 \, V$
$E^{\circ}_{Ag^{+} / Ag} = +0.80 \, V$
$E^{\circ}_{I_2 / I^{-}} = +0.54 \, V$
$E^{\circ}_{Na^{+} / Na} = -2.71 \, V$
$E^{\circ}_{Li^{+} / Li} = -3.05 \, V$
Comparing these values,the order is $1.36 > 0.80 > 0.54 > -2.71 > -3.05$.
Therefore,the correct order is $A > C > B > D > E$.

(A) The thermal decomposition of lead nitrate is given by:
$2Pb(NO_3)_2 \xrightarrow{673 \ K} 2PbO + 4NO_2 + O_2$
Here,$A$ is $NO_2$.
On dimerisation,$NO_2$ forms $N_2O_4$ $(B)$:
$2NO_2 \rightleftharpoons N_2O_4$
The structure of $N_2O_4$ consists of two $NO_2$ units linked by a $N-N$ bond. The structure is $O_2N-NO_2$. It does not contain any bridged oxygen atoms (i.e.,no $N-O-N$ linkage).
Therefore,the number of bridged oxygen atoms in $B$ is $0$.

(B) The spin-only magnetic moment $(\mu)$ is given by the formula $\mu = \sqrt{n(n+2)} \ BM$, where $n$ is the number of unpaired electrons.
To find the lowest magnetic moment, we need to identify the ion with the minimum number of unpaired electrons $(n)$.
$1$. $V^{2+}$ $(Z=23)$: Electronic configuration is $[Ar] 3d^3$. Number of unpaired electrons $(n)$ = $3$.
$2$. $Ni^{2+}$ $(Z=28)$: Electronic configuration is $[Ar] 3d^8$. Number of unpaired electrons $(n)$ = $2$.
$3$. $Cr^{2+}$ $(Z=24)$: Electronic configuration is $[Ar] 3d^4$. Number of unpaired electrons $(n)$ = $4$.
$4$. $Fe^{2+}$ $(Z=26)$: Electronic configuration is $[Ar] 3d^6$. Number of unpaired electrons $(n)$ = $4$.
Since $Ni^{2+}$ has the lowest number of unpaired electrons $(n=2)$, it has the lowest spin-only magnetic moment value.

(B) The reaction involves the formation of a Grignard reagent from $A$ followed by its reaction with methyl benzoate to form triphenylmethanol.
$1$. $A + Mg \xrightarrow{THF} A-MgBr$
$2$. The Grignard reagent $(PhMgBr)$ attacks the carbonyl carbon of methyl benzoate $(PhCOOCH_3)$.
$3$. This leads to the formation of a ketone intermediate,benzophenone $(Ph_2CO)$,which further reacts with another equivalent of $PhMgBr$ to form the final product,triphenylmethanol $(Ph_3COH)$.
Since the product contains three phenyl groups attached to the central carbon,and one comes from the methyl benzoate,the other two must come from the Grignard reagent. Thus,$A$ must be bromobenzene $(C_6H_5Br)$.

(C) Acetophenone is $C_6H_5COCH_3$.
$(A)$ $C_6H_5CHO + CH_3MgBr \rightarrow C_6H_5CH(OH)CH_3$. Oxidation with $Na_2Cr_2O_7/H^+$ gives $C_6H_5COCH_3$ (Acetophenone).
$(B)$ $CH_3CHO + C_6H_5MgBr \rightarrow CH_3CH(OH)C_6H_5$. Oxidation with $PCC$ gives $CH_3COC_6H_5$ (Acetophenone).
$(C)$ Esters react with $2$ equivalents of Grignard reagent to form tertiary alcohols. $C_6H_5COOC_2H_5 + 2CH_3MgBr \rightarrow C_6H_5C(OH)(CH_3)_2$. This does not give acetophenone.
$(D)$ Acid chlorides react with organocadmium reagents (formed in situ from $2RMgX + CdCl_2$) to give ketones. $2C_6H_5COCl + (CH_3)_2Cd \rightarrow 2C_6H_5COCH_3$. This gives acetophenone.

(C) Enamine formation is a nucleophilic addition-elimination reaction between a carbonyl compound (having at least one $\alpha$-hydrogen) and a secondary amine.
In the case of $Di-tert-butyl \ ketone$ $((t-Bu)_2CO)$,the carbonyl carbon is highly sterically hindered by two bulky $tert-butyl$ groups.
Due to this extreme steric hindrance,the secondary amine cannot approach the carbonyl carbon to initiate the nucleophilic attack,thus preventing the formation of an enamine.

(B) Dettol is a mixture of Chloroxylenol and Terpineol.
Chloroxylenol (Compound '$A$') contains a benzene ring,which has $6$ $\pi$ electrons.
Terpineol (Compound '$B$') contains one double bond,which has $2$ $\pi$ electrons.
Therefore,compound '$B$' is Terpineol.

(B) For a gaseous reaction,the half-life $t_{1/2}$ is related to the initial pressure $P_0$ by the relation: $t_{1/2} \propto \frac{1}{(P_0)^{n-1}}$,where $n$ is the order of the reaction.
Given:
$P_1 = 55.5 \ kPa$,$t_1 = 340 \ s$
$P_2 = 27.8 \ kPa$,$t_2 = 170 \ s$
Using the ratio formula:
$\frac{t_1}{t_2} = \left(\frac{P_2}{P_1}\right)^{n-1}$
Substituting the values:
$\frac{340}{170} = \left(\frac{27.8}{55.5}\right)^{n-1}$
$2 = \left(\frac{1}{2}\right)^{n-1}$
$2^1 = (2^{-1})^{n-1}$
$2^1 = 2^{-(n-1)}$
$1 = -(n-1)$
$1 = -n + 1$
$n = 0$
Thus,the order of the reaction is $0$.

(B) The reduction reaction is: $Fe^{3+} + 3e^{-} \longrightarrow Fe$
From the stoichiometry,$3 \ mol$ of electrons are required to deposit $1 \ mol$ of $Fe$ $(56 \ g)$.
Total charge required for $0.3482 \ g$ of $Fe$ is:
$Q = \frac{3 \times 96500 \times 0.3482}{56} \approx 1800 \ C$
Using the formula $Q = I \times t$ (where $t$ is in seconds):
$1800 = 1.5 \times t$
$t = \frac{1800}{1.5} = 1200 \ s$
Converting time to minutes:
$x = \frac{1200}{60} = 20 \ min$

(C) The inner-orbital complexes are $K_{3}[Fe(CN)_{6}]$ and $[Co(NH_{3})_{6}]Cl_{3}$.
$K_{3}[Fe(CN)_{6}]$ contains $Fe^{3+}$ $(3d^{5})$,which forms a $d^{2}sp^{3}$ complex with $1$ unpaired electron. Its spin-only magnetic moment is $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.732 \ B.M.$
$[Co(NH_{3})_{6}]Cl_{3}$ contains $Co^{3+}$ $(3d^{6})$,which forms a $d^{2}sp^{3}$ complex with $0$ unpaired electrons. Its spin-only magnetic moment is $\mu = 0 \ B.M.$
$CN^{-}$ is a stronger field ligand than $NH_{3}$,so $K_{3}[Fe(CN)_{6}]$ has a larger crystal field splitting energy $(\Delta_{0})$ than $[Co(NH_{3})_{6}]Cl_{3}$.
Since $\Delta_{0} = \frac{hc}{\lambda}$,a larger $\Delta_{0}$ corresponds to a shorter wavelength $(\lambda)$ of absorbed light.
Therefore,$K_{3}[Fe(CN)_{6}]$ is the complex that absorbs light at the shortest wavelength.
Its spin-only magnetic moment is $1.732 \ B.M.$,which rounds to $2$ to the nearest integer.

(C) The monomer unit of Novolac is $o$-hydroxybenzyl alcohol (saligenin),which has the chemical formula $C_7H_8O_2$.
The molecular mass of the monomer unit is $(7 \times 12) + (8 \times 1) + (2 \times 16) = 84 + 8 + 32 = 124 \ g/mol$.
Novolac is formed by the condensation polymerization of phenol and formaldehyde,where water molecules are eliminated.
For a polymer chain consisting of $n$ monomer units,$(n-1)$ molecules of water are removed.
The mass of the polymer is given by: $Mass_{polymer} = (n \times Mass_{monomer}) - ((n-1) \times Mass_{water})$.
Substituting the given values: $963 = (n \times 124) - ((n-1) \times 18)$.
$963 = 124n - 18n + 18$.
$963 - 18 = 106n$.
$945 = 106n$.
$n = 945 / 106 \approx 8.91$.
Rounding to the nearest integer,the number of monomer units $n = 9$.

(D) The Biuret test is a chemical test used for detecting the presence of peptide bonds.
It gives a positive result for compounds containing at least two peptide linkages $(-CO-NH-)$.
$1$. Glycine: An amino acid,contains no peptide linkage.
$2$. Glycylalanine: $A$ dipeptide,contains only one peptide linkage.
$3$. Tripeptide: Contains two peptide linkages.
$4$. Biuret: Contains two peptide linkages $(NH_2-CO-NH-CO-NH_2)$.
Therefore,only the Tripeptide and Biuret give a positive Biuret test.
The total count is $2$.