KVPY 2010 Chemistry Question Paper with Answer and Solution

(D)
Total number of electrons in $O_2^{2-} = 18$.
The molecular orbital electronic configuration of $O_2^{2-}$ is:
$\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$.
Bond order is calculated using the formula: $BO = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
Here,$N_b = 10$ and $N_a = 8$.
Therefore,$BO = \frac{10 - 8}{2} = 1$.

(A) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.
Given values are:
Planck's constant $(h) = 6.626 \times 10^{-34} \, J \, s$
Speed of light $(c) = 3 \times 10^8 \, ms^{-1}$
Wavelength $(\lambda) = 1 \, m$
Substituting these values into the formula:
$E = \frac{6.626 \times 10^{-34} \, J \, s \times 3 \times 10^8 \, ms^{-1}}{1 \, m}$
$E = 19.878 \times 10^{-26} \, J$
$E = 1.988 \times 10^{-25} \, J$
Therefore,the correct option is $A$.

(C)
Central atom $Cl$ has $7$ valence electrons,out of which it forms $3$ bond pairs with fluorine and $2$ pairs of electrons remain non-bonded (lone pairs).
According to $VSEPR$ theory,the molecule has $sp^3d$ hybridization with a trigonal bipyramidal electron geometry.
To minimize repulsion,the two lone pairs occupy the equatorial positions,resulting in a $T$-shaped molecular geometry.

(D) The correct answer is $ (D) $.
Friedel-Crafts acylation is an electrophilic aromatic substitution reaction. In this process,an acyl group $( RCO- )$ is introduced into an aromatic ring by reacting the aromatic compound with an acyl chloride or an acid anhydride in the presence of a Lewis acid catalyst,such as $ AlCl_3 $.
The reaction can be represented as:
$ C_6H_6 + CH_3COCl \xrightarrow{AlCl_3} C_6H_5COCH_3 + HCl $

(A) The most stable conformation for $n$-butane is the anti-staggered form,where the two bulky methyl groups are at a dihedral angle of $180^{\circ}$ to each other.
In this conformation,the steric repulsion between the methyl groups is minimized,making it the most stable conformer.
This structure is represented in option $(A)$.

(A) In the given nuclear reaction: ${ }_{90}^{234} Th \longrightarrow{ }_{91}^{234} Pa + X$.
According to the law of conservation of mass number: $234 = 234 + A$,which gives $A = 0$.
According to the law of conservation of atomic number: $90 = 91 + Z$,which gives $Z = -1$.
Thus,the particle $X$ is ${ }_{-1}^0 e$ (a beta particle).
Therefore,the correct option is $A$.

(D) The given reactions are:
$(i) \, CO_2 \rightleftharpoons CO + \frac{1}{2} O_2, \, K_{C_1} = 9.1 \times 10^{-12}$
$(ii) \, H_2O \rightleftharpoons H_2 + \frac{1}{2} O_2, \, K_{C_2} = 7.1 \times 10^{-12}$
We want to find the equilibrium constant $K$ for the reaction:
$CO_2 + H_2 \rightleftharpoons CO + H_2O \quad (iii)$
To obtain reaction $(iii)$,we reverse reaction $(ii)$ and add it to reaction $(i)$:
Reverse of $(ii): \, H_2 + \frac{1}{2} O_2 \rightleftharpoons H_2O, \, K_{C_3} = \frac{1}{K_{C_2}} = \frac{1}{7.1 \times 10^{-12}}$
Adding $(i)$ and reversed $(ii)$:
$CO_2 + H_2 + \frac{1}{2} O_2 \rightleftharpoons CO + \frac{1}{2} O_2 + H_2O$
$CO_2 + H_2 \rightleftharpoons CO + H_2O$
The equilibrium constant $K$ is given by $K = K_{C_1} \times \frac{1}{K_{C_2}} = \frac{9.1 \times 10^{-12}}{7.1 \times 10^{-12}} = 1.28$.

(A) The molecule $PCl_3F_2$ has a trigonal bipyramidal geometry.
According to Bent's rule,more electronegative atoms prefer to occupy the axial positions,while larger and less electronegative atoms prefer the equatorial positions to minimize repulsion.
In $PCl_3F_2$,the $F$ atoms are more electronegative than $Cl$ atoms.
Therefore,the $F$ atoms occupy the axial positions,and the $Cl$ atoms occupy the equatorial positions.
This corresponds to the structure shown in option $A$.

(B) Enantiomers are pairs of stereoisomers that are non-superimposable mirror images of each other. They must contain at least one chiral center.
$1$. Structure $I$ is $pentan-2-ol$ with the $-OH$ group on a wedge.
$2$. Structure $IV$ is $pentan-2-ol$ with the $-OH$ group on a dash (relative to the same carbon chain orientation).
$3$. Comparing $I$ and $IV$,they represent non-superimposable mirror images of each other,both possessing a chiral carbon atom at the $C-2$ position.
$4$. Structures $II$ and $III$ do not contain chiral centers,so they cannot be part of an enantiomeric pair.
Therefore,the correct enantiomeric pair is $I$ and $IV$.

(C) compound is aromatic if it satisfies the following conditions:
$1$. It must be cyclic.
$2$. It must be planar.
$3$. It must have a fully conjugated system of $\pi$ electrons.
$4$. It must follow $H$ückel's rule,i.e.,it must contain $(4n+2) \pi$ electrons,where $n = 0, 1, 2, ...$.
Let us analyze the given options:
- Option $(A)$: The cyclopentadienyl cation has $4 \pi$ electrons. It is anti-aromatic.
- Option $(B)$: The cyclohexadienyl cation is non-planar and non-aromatic.
- Option $(C)$: The tropylium cation (cycloheptatrienyl cation) has $6 \pi$ electrons $(n=1)$. It is cyclic,planar,and fully conjugated. Thus,it is aromatic.
- Option $(D)$: The cyclooctatrienyl cation is non-planar and non-aromatic.
Therefore,the correct option is $(C)$.

(A) The reaction of cyclohexene with bromine in $CCl_4$ in the dark is an electrophilic addition reaction.
Bromine adds across the double bond in an anti-addition manner,resulting in the formation of $trans-1,2-dibromocyclohexane$.
The reaction proceeds through a cyclic bromonium ion intermediate,which is then attacked by the bromide ion from the opposite side,leading to the trans-product.

(A) Bromination of compound $I$ ($2,3$-dimethylbut-$2$-ene) with $Br_2$ yields $2,3$-dibromo-$2,3$-dimethylbutane. This product is a $meso$ compound due to the presence of a plane of symmetry. Therefore,it does not form any enantiomeric pair.
Bromination of compound $II$ ($2$-methylbut-$1$-ene) with $Br_2$ (followed by addition) yields $1,2$-dibromo-$2$-methylbutane. The carbon atom at position $2$ becomes a chiral center $(C^*)$. This chiral compound exists as a pair of enantiomers ($d$ and $l$ forms),which are non-superimposable mirror images of each other. Thus,it produces $1$ enantiomeric pair.
Therefore,the number of enantiomeric pairs produced from $I$ and $II$ are $0$ and $1,$ respectively. The correct option is $(a).$

(A) At equilibrium,the change in Gibbs free energy is $\Delta G^{\circ} = 0 \, kJ \, mol^{-1}$.
Using the relation $\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}$,at equilibrium,we have $0 = \Delta H^{\circ} - T \Delta S^{\circ}$.
Therefore,$T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}}$.
Given $\Delta H^{\circ} = 7.5 \, kJ \, mol^{-1} = 7500 \, J \, mol^{-1}$ and $\Delta S^{\circ} = 25 \, J \, K^{-1} \, mol^{-1}$.
$T = \frac{7500 \, J \, mol^{-1}}{25 \, J \, K^{-1} \, mol^{-1}} = 300 \, K$.
Thus,the values are $0 \, kJ \, mol^{-1}$ and $300 \, K$.

(D) For the precipitation of $Mg(OH)_2$,the ionic product must exceed the solubility product $(K_{sp})$.
The equilibrium is: $Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$
$K_{sp} = [Mg^{2+}][OH^-]^2 = 1.0 \times 10^{-12}$
Given $[Mg^{2+}] = 0.01 \ M = 10^{-2} \ M$.
Substituting the values: $10^{-2} \times [OH^-]^2 = 10^{-12}$
$[OH^-]^2 = 10^{-10}$
$[OH^-] = 10^{-5} \ M$
$pOH = -\log[OH^-] = -\log(10^{-5}) = 5$
Since $pH + pOH = 14$ at $25^{\circ}C$,
$pH = 14 - 5 = 9$.

(B) In cyclic silicates,two oxygen atoms of each $SiO_4^{4-}$ tetrahedron are shared to form a ring with the general formula $[SiO_3]_n^{2n-}$.
The figure shows a ring consisting of $6$ silicon atoms.
Substituting $n = 6$ into the general formula,we get $[SiO_3]_6^{2(6)-} = [Si_6O_{18}]^{12-}$.
Thus,the cyclic silicate ion is $[Si_6O_{18}]^{12-}$.

(A) To find the enthalpy of formation of $B_2H_{6(g)}$,we use Hess's Law.
We need to obtain the reaction: $2 B_{(s)} + 3 H_{2(g)} \longrightarrow B_2H_{6(g)}$
Using the given equations:
$(ii) \quad 2 B_{(s)} + \frac{3}{2} O_{2(g)} \longrightarrow B_2O_{3(s)}, \quad \Delta H_2^{\circ} = -1273 \, kJ$
$(iv) \times 3 \quad 3 H_{2(g)} + \frac{3}{2} O_{2(g)}$ $\longrightarrow 3 H_2O_{(l)}, \quad 3 \times \Delta H_4^{\circ} = 3 \times (-286) = -858 \, kJ$
$(i) \times 3 \quad 3 H_2O_{(l)}$ $\longrightarrow 3 H_2O_{(g)}, \quad 3 \times \Delta H_1^{\circ} = 3 \times 44 = 132 \, kJ$
$-(iii) \quad B_2O_{3(s)} + 3 H_2O_{(g)}$ $\longrightarrow B_2H_{6(g)} + 3 O_{2(g)}, \quad -\Delta H_3^{\circ} = -(-2035) = 2035 \, kJ$
Summing these equations:
$2 B_{(s)} + 3 H_{2(g)} \longrightarrow B_2H_{6(g)}$
$\Delta H_r^{\circ} = -1273 - 858 + 132 + 2035 = 36 \, kJ$.

(A) The atomic number of silicon $(Si)$ is $14$.
Following the Aufbau principle,the electrons are filled in the order $1 s, 2 s, 2 p, 3 s, 3 p$.
The distribution of $14$ electrons is as follows: $1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^2$.
Thus,the correct electronic configuration is $1 s^2 2 s^2 2 p^6 3 s^2 3 p^2$.

(A) The balanced chemical equation for the thermal decomposition of $CaCO_{3}$ is:
$CaCO_{3}(s) \stackrel{\Delta}{\longrightarrow} CaO(s) + CO_{2}(g)$
From the stoichiometry of the reaction,$1 \ mol$ of $CaCO_{3}$ produces $1 \ mol$ of $CO_{2}$.
The molar mass of $CaCO_{3} = 100 \ g/mol$.
The number of moles of $CaCO_{3}$ in $25 \ g$ is:
$n(CaCO_{3}) = \frac{25 \ g}{100 \ g/mol} = 0.25 \ mol$.
Since $1 \ mol$ of $CaCO_{3}$ yields $1 \ mol$ of $CO_{2}$,$0.25 \ mol$ of $CaCO_{3}$ will yield $0.25 \ mol$ of $CO_{2}$.
The molar mass of $CO_{2} = 12 + (2 \times 16) = 44 \ g/mol$.
The mass of $CO_{2}$ produced is:
$Mass = n \times M = 0.25 \ mol \times 44 \ g/mol = 11 \ g$.

(B)
On moving across the second period of the periodic table,the atomic radii of the elements decrease due to an increase in effective nuclear charge.
The electrons in all shells are pulled closer to the nucleus,thereby making each individual shell smaller.

(B) The correct answer is $B$.
According to the octet rule,atoms of different elements combine with each other to complete their respective octets (i.e.,$8$ electrons in their outermost shell,or $2$ electrons in the case of $H, Li,$ and $Be$ to attain a stable nearest noble gas configuration).
In $BCl_3$,the central Boron $(B)$ atom has only $6$ electrons in its valence shell after forming three covalent bonds with Chlorine atoms.
Since $6 < 8$,$BCl_3$ is an electron-deficient molecule and does not satisfy the octet rule.

(A) The chemical formula $C_4H_7Br$ represents a molecule such as $1$-bromobut-$2$-ene $(CH_3-CH=CH-CH_2Br)$.
To find the total number of covalent bonds,we count all single and multiple bonds in the structure:
$1$. $C-H$ bonds: There are $7$ $C-H$ bonds.
$2$. $C-C$ bonds: There are $2$ single $C-C$ bonds and $1$ double $C=C$ bond (total $3$ bonds).
$3$. $C-Br$ bond: There is $1$ $C-Br$ bond.
Total number of covalent bonds = $7 + 3 + 1 = 11$.
Wait,let us re-examine the provided image structure: The image shows a chain with $12$ numbered bonds. Counting them directly from the provided structure: $1$ $(H-C)$,$2$ $(C-H)$,$3,4$ $(C=C)$,$5,6$ $(C-C)$,$7$ $(C-H)$,$8$ $(C-H)$,$9$ $(C-C)$,$10$ $(C-H)$,$11$ $(C-H)$,$12$ $(C-Br)$.
Total count = $12$.

(D)
As we know,$pH = -\log [H^+]$.
Initial $pH = 2.0$,so initial $[H^+]_i = 10^{-2} \ M$.
Final $pH = 5.0$,so final $[H^+]_f = 10^{-5} \ M$.
The ratio of change is $\frac{[H^+]_f}{[H^+]_i} = \frac{10^{-5}}{10^{-2}} = 10^{-3}$.
Thus,the hydrogen ion concentration decreases by a factor of $1000$ (thousand-fold).

(C) For the $1^{st}$ jar:
Number of moles of $H_2 = \frac{2 \ g}{2 \ g/mol} = 1 \ mol$.
Number of molecules of $H_2 = 1 \times N_A = 6.022 \times 10^{23}$ molecules.
For the $2^{nd}$ jar:
Number of moles of $N_2 = \frac{28 \ g}{28 \ g/mol} = 1 \ mol$.
Number of molecules of $N_2 = 1 \times N_A = 6.022 \times 10^{23}$ molecules.
Since both jars contain $1 \ mol$ of gas,they contain the same number of molecules.

(C) . The types of isomers in the given options are as follows:
$(a)$ $CH_3CH_2CH_2OH$ and $CH_3CH_2OCH_3$: These are functional isomers.
$(b)$ $CH_3CH_2CH_2Cl$ and $CH_3CHClCH_3$: These are positional isomers.
$(c)$ The given structures represent $cis$ and $trans$ forms of but$-2-$ene. They are geometrical isomers,which are a type of stereoisomers.
$(d)$ Methylcyclopentane and cyclohexane: These are ring-chain isomers.
Thus,the correct option is $(c)$.

(D) The balanced chemical equations are:
$(i)$ $3 \,Cu + 8 \,HNO_3 \rightarrow 3 \,Cu(NO_3)_2 + 2 \,NO + 4 \,H_2O$
$(ii)$ $2 \,CuI_2 \rightarrow Cu_2I_2 + I_2$
$(iii)$ $2 \,Na_2S_2O_3 + I_2 \rightarrow Na_2S_4O_6 + 2 \,NaI$
$(a)$ The coefficients are: $a=3, b=8, c=3, d=2, e=4$.
$(b)$ The coefficients are: $f=2, g=1, h=1$.
$(c)$ The coefficients are: $i=2, j=1, k=1, l=2$.
$(d)$ Moles of $I_2 = \frac{2.54 \,g}{254 \,g/mol} = 0.01 \,mol$.
From equation $(ii)$,$1 \,mol$ of $I_2$ is produced from $2 \,mol$ of $CuI_2$,which corresponds to $2 \,mol$ of $Cu$.
Therefore,moles of $Cu = 2 \times 0.01 \,mol = 0.02 \,mol$.
Mass of $Cu = 0.02 \,mol \times 63.5 \,g/mol = 1.27 \,g$.
Percentage of $Cu = \frac{1.27 \,g}{2.0 \,g} \times 100 = 63.5 \%$.

(A) Given $1 \, kcal = 4.184 \, kJ$. Therefore,$2500 \, kcal = 2500 \times 4.184 \, kJ = 10460 \, kJ$. The energy requirement is $10460 \, kJ$.
$(b)$ Molar mass of sucrose $(C_{12}H_{22}O_{11})$ = $(12 \times 12) + (22 \times 1) + (11 \times 16) = 342 \, g/mol$.
Energy released per mole of sucrose = $5.6 \times 10^6 \, J = 5600 \, kJ/mol$.
Moles of sucrose required = $\frac{10460 \, kJ}{5600 \, kJ/mol} \approx 1.8679 \, mol$.
Mass of sucrose = $1.8679 \, mol \times 342 \, g/mol \approx 638.82 \, g$.
From the equation,$1 \, mol$ of sucrose produces $12 \, mol$ of $CO_2$.
Moles of $CO_2$ produced = $1.8679 \times 12 = 22.4148 \, mol$.
Volume of $CO_2$ at $STP$ = $22.4148 \, mol \times 22.4 \, L/mol \approx 502.09 \, L$.

(D) Initially,the total weight of the watermelon is $1000\,kg$.
Since the water content is $99\%$,the weight of the solid part (non-water) is $1\% \text{ of } 1000\,kg = 0.01 \times 1000 = 10\,kg$.
Let the new total weight of the watermelon at the destination be $x\,kg$.
At the destination,the water content is $98\%$,which means the solid part is $2\% \text{ of } x$.
Since the weight of the solid part remains constant,we have:
$0.02 \times x = 10\,kg$.
$x = \frac{10}{0.02} = 500\,kg$.
The reduction in weight is $1000\,kg - 500\,kg = 500\,kg$.

(D) Let $n$ be a natural number such that $\gcd(n, 10) = 1$. Consider the sequence of $n+1$ numbers $m_k = \sum_{i=0}^{k-1} 10^i = \frac{10^k-1}{9}$ for $k = 1, 2, \dots, n+1$. By the Pigeonhole Principle,at least two of these numbers,say $m_i$ and $m_j$ with $i < j$,must have the same remainder when divided by $n$. Thus,$n$ divides $m_j - m_i = 11\dots100\dots0 = 11\dots1 \times 10^i$. Since $\gcd(n, 10) = 1$,it follows that $\gcd(n, 10^i) = 1$,so $n$ must divide $m_{j-i}$,which consists only of digits $1$.
$(b)$ Let the rational number be $\frac{p}{q}$. We can write $q = 2^r \cdot 5^s \cdot t$,where $\gcd(t, 10) = 1$. From part $(a)$,there exists $m$ consisting only of $1$'s such that $t \mid m$. Let $m = \frac{10^c-1}{9}$. Then $9m = 10^c-1$. Since $t \mid m$,we have $t \mid (10^c-1)$. Let $10^c-1 = kt$. We can choose $b = \max(r, s)$. Then $10^b$ is divisible by $2^r$ and $5^s$. Thus,$10^b(10^c-1)$ is divisible by $2^r \cdot 5^s \cdot t = q$. Therefore,$\frac{p}{q} = \frac{p \cdot \frac{10^b(10^c-1)}{q}}{10^b(10^c-1)} = \frac{a}{10^b(10^c-1)}$ for some integer $a$.

(A) The complex $[Co(dien)Cl_3]$ contains a tridentate ligand,diethylenetriamine $(dien)$.
This complex exists in two geometrical isomeric forms: $fac$ (facial) and $mer$ (meridional) isomers.
In the $fac$-isomer,the three donor nitrogen atoms of the $dien$ ligand and the three $Cl^-$ ions occupy the corners of the same face of the octahedron.
In the $mer$-isomer,the three donor nitrogen atoms of the $dien$ ligand and the three $Cl^-$ ions are arranged in a meridional plane.
Therefore,the total number of isomers is $2$.

(C)
$\pi$-acid ligands are those ligands that possess empty $\pi^*$ (anti-bonding) molecular orbitals,which can accept electron density from the filled $d$-orbitals of the central metal atom through back-bonding.
Among the given options,$CN^{-}$ acts as a $\pi$-acid ligand because it has empty $\pi^*$ orbitals available for back-donation from the metal.

(B)
As given in the question,the concentration of a substance becomes one-half of its original value,which refers to the half-life period $(t_{1/2})$.
The half-life period is the time required to reduce the initial concentration to half of its value.
For a $zero$ order reaction,$t_{1/2} = \frac{[A]_0}{2k}$.
For a $first$ order reaction,$t_{1/2} = \frac{0.693}{k}$.
For a $second$ order reaction,$t_{1/2} = \frac{1}{k[A]_0}$.
For a $third$ order reaction,$t_{1/2} = \frac{1}{2k[A]_0^2}$.
Here,$[A]_0$ is the initial concentration.
Since the half-life of a $first$ order reaction is independent of the initial concentration,the correct option is $(b)$.

(A) The acidic strength of a compound is directly proportional to the stability of its conjugate base. More stable the conjugate base,stronger is the acid.
$I$: Benzyl alcohol $(C_6H_5CH_2OH)$ forms a conjugate base where the negative charge is on the oxygen atom and is not resonance-stabilized by the benzene ring.
$II$: Benzoic acid $(C_6H_5COOH)$ forms a carboxylate ion $(C_6H_5COO^-)$ which is resonance-stabilized by two oxygen atoms.
$III$: $p$-Cresol $(CH_3-C_6H_4-OH)$ forms a phenoxide ion which is resonance-stabilized,but the electron-donating $+I$ effect of the $-CH_3$ group destabilizes it compared to benzoic acid.
$IV$: Benzenesulfonic acid $(C_6H_5SO_3H)$ forms a sulfonate ion $(C_6H_5SO_3^-)$ where the negative charge is delocalized over three highly electronegative oxygen atoms,making it the most stable conjugate base.
Comparing the stability of conjugate bases:
$I$ (least stable) $< III < II < IV$ (most stable).
Therefore,the order of acidity is $I < III < II < IV$.

(C) The correct option is $(C)$.
When a concentrated solution of $CuSO_4$ is mixed with a dilute solution of $CuSO_4$,the process is essentially a dilution process.
During dilution,the randomness of the solute particles increases as they spread out in a larger volume,which leads to an increase in entropy $(\Delta S > 0)$.
Since the mixing of two solutions of the same solute at the same temperature involves negligible heat change (ideal mixing approximation),the enthalpy change $(\Delta H)$ is approximately zero.

(B)
Increasing the temperature increases the kinetic energy of the reactant molecules.
This leads to an increase in the number of collisions,the average energy of collisions,and the average velocity of the reactant molecules.
However,the activation energy $(E_a)$ is a characteristic property of the reaction pathway and is independent of temperature.
Therefore,increasing the temperature does not increase the activation energy.

(C) The correct option is $C$.
$A$ face-centered cubic $(fcc)$ unit cell contains atoms at all the corners and at the center of all faces of the cube.
Thus,the number of atoms in $fcc = (\frac{1}{8} \times 8) + (\frac{1}{2} \times 6) = 1 + 3 = 4$.
$A$ body-centered cubic $(bcc)$ unit cell has an atom at each of its corners and also one atom at its body center.
Thus,the number of atoms in $bcc = (\frac{1}{8} \times 8) + 1 = 1 + 1 = 2$.

(B) For a first order reaction $R \longrightarrow P$,the rate law is given by: $\text{Rate} = -\frac{d[R]}{dt} = k[R]$.
Rearranging the terms,we get: $\frac{d[R]}{[R]} = -k dt$.
Integrating both sides: $\int \frac{d[R]}{[R]} = -\int k dt \Rightarrow \ln[R] = -kt + C$.
At $t = 0$,$[R] = [R_0]$,so $\ln[R_0] = C$.
Substituting $C$ back into the equation: $\ln[R] = -kt + \ln[R_0]$.
Rearranging: $\ln \frac{[R]}{[R_0]} = -kt$.
Taking the exponential of both sides: $\frac{[R]}{[R_0]} = e^{-kt}$.
Therefore,the concentration at time $t$ is: $[R] = [R_0] e^{-kt}$.

(A) For the reaction,$2 NO_{2(g)} \longrightarrow 2 NO_{(g)} + O_{2(g)}$,the rate expression is given by:
$r = -\frac{1}{2} \frac{d[NO_2]}{dt} = +\frac{1}{2} \frac{d[NO]}{dt} = +\frac{d[O_2]}{dt}$.
$1$. Curve $Z$ represents the reactant $NO_2$ because its concentration decreases with time.
$2$. Curves $X$ and $Y$ represent the products $NO$ and $O_2$ because their concentrations increase with time.
$3$. From the stoichiometry,the rate of formation of $NO$ is twice the rate of formation of $O_2$ (i.e.,$\frac{d[NO]}{dt} = 2 \frac{d[O_2]}{dt}$).
$4$. Therefore,the concentration of $NO$ increases faster than $O_2$,meaning curve $X$ corresponds to $NO$ and curve $Y$ corresponds to $O_2$.
Thus,$X = NO, Y = O_2, Z = NO_2$.

(C) The molar mass of oxalic acid dihydrate $(H_2C_2O_4 \cdot 2H_2O)$ is $126 \ g/mol$. The equivalent weight is $126/2 = 63 \ g/eq$.
Initial normality $(N_1)$ of the $100 \ mL$ solution: $N_1 = \frac{2.52 \ g}{63 \ g/eq \times 0.1 \ L} = 0.4 \ N$.
Using the dilution formula $N_1V_1 = N_2V_2$: $0.4 \ N \times 10 \ mL = N_2 \times 500 \ mL$.
$N_2 = \frac{0.4 \times 10}{500} = 0.008 \ N$.
Amount of oxalic acid in $mg/mL$ in the final solution:
Mass of oxalic acid in $10 \ mL$ aliquot $= \frac{2.52 \ g}{100 \ mL} \times 10 \ mL = 0.252 \ g = 252 \ mg$.
Concentration $= \frac{252 \ mg}{500 \ mL} = 0.504 \ mg/mL$.

(A) In reaction $(I)$,the $OH$ group of the $CH_2OH$ side chain is protonated by $HBr$ to form a good leaving group $(H_2O^+)$. This leaves as water to form a resonance-stabilized benzylic carbocation,which is then attacked by $Br^-$ to yield $3-(bromomethyl)phenol$.
In reaction $(II)$,the ether oxygen is protonated by $HBr$. The $C-O$ bond cleavage occurs such that the methyl group is released as $CH_3Br$ because the $C-O$ bond involving the phenyl ring has partial double bond character due to resonance,making it stronger and harder to break. Thus,the reaction yields benzene-$1,3-diol$ (resorcinol) and $CH_3Br$.

(C) The correct option is $C$.
Given, atomic radius of the metal, $r = 141.4 \, pm$.
For a face-centred cubic $(fcc)$ structure, the relationship between the edge length $a$ and the atomic radius $r$ is given by $r = \frac{a}{2\sqrt{2}}$.
Rearranging for $a$, we get $a = 2\sqrt{2}r$.
Substituting the values: $a = 2 \times 1.414 \times 141.4 = 400 \, pm$.
The volume of the unit cell is $V = a^3$.
$V = (400 \, pm)^3 = 64,000,000 \, pm^3 = 64 \times 10^6 \, pm^3$.
This can be expressed as $6.4 \times 10^7 \, pm^3$.

(D) The formula for molar mass $(M_2)$ using freezing point depression is: $M_2 = \frac{k_f \times w_2 \times 1000}{\Delta T_f \times w_1}$
Given: $w_2 = 8.0 \, g$,$w_1 = 92 \, g$,$\Delta T_f = 0.925 \, ^{\circ}C$,$k_f = 1.85 \, ^{\circ}C \, kg \, mol^{-1}$.
Substituting the values: $M_2 = \frac{1.85 \times 8.0 \times 1000}{0.925 \times 92} = \frac{14800}{85.1} \approx 173.9 \, g \, mol^{-1}$.
Rounding to the nearest provided option,the molar mass is $160 \, g \, mol^{-1}$.

(B) The oxidation state of $Fe$ in $K_3[Fe(CN)_6]$ is $+3$. Thus,the electronic configuration for $Fe^{3+}$ is $[Ar] 3d^5$.
Since $CN^-$ is a strong field ligand,it causes pairing of electrons in the $d$-orbitals.
The configuration in the crystal field is $t_{2g}^5 e_g^0$.
$CFSE = (-0.4 \times n(t_{2g}) + 0.6 \times n(e_g)) \Delta_o + mP$
$CFSE = (-0.4 \times 5 + 0.6 \times 0) \Delta_o + 2P = -2.0 \Delta_o + 2P$.
However,considering only the crystal field splitting energy contribution:
$CFSE = -0.4 \times 5 = -2.0 \Delta_o$.
Number of unpaired electrons $(n) = 1$.
Magnetic moment,$\mu = \sqrt{n(n+2)} = \sqrt{1(1+2)} = \sqrt{3} BM$.
Therefore,the correct option is $B$.

(D)
Boiling point elevation is a colligative property,which depends on the number of solute particles in the solution.
For the same molar concentration $(0.01 \, M)$,the solution with the highest van't Hoff factor $(i)$ will show the greatest elevation in boiling point.
$1$. Sucrose $(C_{12}H_{22}O_{11})$: Non-electrolyte,$i = 1$.
$2$. $NaCl$: Dissociates as $NaCl \rightarrow Na^{+} + Cl^{-}$,$i = 2$.
$3$. $CaCl_2$: Dissociates as $CaCl_2 \rightarrow Ca^{2+} + 2Cl^{-}$,$i = 3$.
Since $CaCl_2$ produces the maximum number of particles ($3$ ions per formula unit),it will have the highest boiling point.

(A) .
On heating $MnO_2$ with conc. $HCl$,$MnCl_2$ and $H_2O$ are produced along with the evolution of $Cl_2$ gas.
The balanced chemical equation is:
$MnO_2 + 4HCl \longrightarrow MnCl_2 + 2H_2O + Cl_2$

(D) The solubility of organic compounds in $1 \ N \ NaOH$ (a base) and $1 \ N \ HCl$ (an acid) depends on their acidic or basic nature.
$(i)$ Compound $A$ is neutral as it does not react with either acid or base. Thus,$A = III$ $(C_6H_5CH_2CH_3)$.
$(ii)$ Compound $B$ is acidic as it dissolves in $NaOH$ but not $HCl$. Thus,$B = II$ $(C_6H_5CH_2COOH)$.
$(iii)$ Compound $C$ is amphoteric as it dissolves in both $NaOH$ and $HCl$. Thus,$C = IV$ $(C_6H_5CH_2CH(NH_2)COOH)$.
$(iv)$ Compound $D$ is basic as it dissolves in $HCl$ but not $NaOH$. Thus,$D = I$ $(C_6H_5CH_2NH_2)$.
$(b)$ The compound with the highest solubility in distilled water is $IV$ $(C_6H_5CH_2CH(NH_2)COOH)$ because it exists as a zwitterion $(C_6H_5CH_2CH(NH_3^+)COO^-)$ in water,which is highly polar.