KVPY 2015 Chemistry Question Paper with Answer and Solution

(A) By Stokes' law,$F=6 \pi \eta a v$.
We have,$\eta = \frac{F}{6 \pi a v}$.
Dimensions of viscosity $\eta$ are $[\eta] = [ML^{-1}T^{-1}]$.
The given relation for volume flow rate is $\frac{V}{t} = k \left(\frac{p}{l}\right)^a \eta^b r^c$.
Substituting dimensions: $[L^3 T^{-1}] = [ML^{-2}T^{-2}]^a [ML^{-1}T^{-1}]^b [L]^c$.
$[M^0 L^3 T^{-1}] = [M^{a+b} L^{-2a-b+c} T^{-2a-b}]$.
Equating powers:
$a+b = 0$ $(i)$
$-2a-b+c = 3$ (ii)
$-2a-b = -1$ (iii)
From (iii),$2a+b = 1$. Subtracting $(i)$ from this,$a = 1$.
Substituting $a=1$ in $(i)$,$b = -1$.
Substituting $a=1, b=-1$ in (ii),$-2(1) - (-1) + c = 3 \Rightarrow -2 + 1 + c = 3 \Rightarrow c = 4$.
Thus,$a=1, b=-1, c=4$.

(C) The given boron trihalides are electron-deficient molecules that act as Lewis acids by accepting an electron pair.
The Lewis acid strength is determined by the extent of $ppi-ppi$ back-bonding between the halogen atom and the boron atom.
In $BF_3$,the $2p-2p$ back-bonding is most effective due to the similar size of the orbitals,which reduces the electron deficiency of the boron atom significantly.
As the size of the halogen increases from $F$ to $Cl$ to $Br$,the effectiveness of back-bonding decreases $(2p-2p > 2p-3p > 2p-4p)$.
Therefore,the electron deficiency on boron increases in the order $BF_3 < BCl_3 < BBr_3$,making $BBr_3$ the strongest Lewis acid.

(B) Isoelectronic species are those species which have the same number of electrons.
Total number of electrons in $O^{2-} = 8 + 2 = 10$.
Total number of electrons in $Zn^{2+} = 30 - 2 = 28$.
Total number of electrons in $Mg^{2+} = 12 - 2 = 10$.
Total number of electrons in $K^{+} = 19 - 1 = 18$.
Total number of electrons in $Ni^{2+} = 28 - 2 = 26$.
Thus,$O^{2-}$ is isoelectronic with $Mg^{2+}$.

(B) The bond angle in a molecule is affected by the number of lone pairs and the electronegativity of the central atom.
In $CH_4$ (methane),the central carbon atom has $4$ bonding pairs and $0$ lone pairs,resulting in a tetrahedral geometry with a bond angle of $109.5^{\circ}$.
In $NH_3$ (ammonia),the central nitrogen atom has $3$ bonding pairs and $1$ lone pair. The lone pair-bond pair repulsion reduces the bond angle to $107.1^{\circ}$.
In $H_2O$ (water),the central oxygen atom has $2$ bonding pairs and $2$ lone pairs. The greater lone pair-lone pair repulsion further reduces the bond angle to $104.5^{\circ}$.
Therefore,the bond angles are $109.5^{\circ}, 107.1^{\circ}, 104.5^{\circ}$ respectively.

(D) The reaction of hydrogen peroxide $(H_2O_2)$ with potassium permanganate $(KMnO_4)$ in an alkaline medium is given by the equation:
$2KMnO_4 + 3H_2O_2 \rightarrow 2MnO_2 + 2KOH + 2H_2O + 3O_2$
In this reaction,the product containing manganese is manganese dioxide $(MnO_2)$.
To find the oxidation state of $Mn$ in $MnO_2$,let it be $x$.
Since the oxidation state of oxygen is $-2$,we have:
$x + 2(-2) = 0$
$x - 4 = 0$
$x = +4$
Therefore,the oxidation state of $Mn$ is $+4$.

(B) Optical activity is exhibited by molecules that possess at least one chiral center (a carbon atom bonded to four different groups).
$(a)$ $CH_3-CH_2-CH_2-Br$ ($1-$bromopropane): No chiral center is present,so it is optically inactive.
$(b)$ $CH_3-CH(Br)-CH_2-CH_3$ ($2-$bromobutane): The carbon at position $2$ is bonded to four different groups $(-H, -CH_3, -CH_2CH_3, -Br)$,making it a chiral center. Thus,it exhibits optical activity.
$(c)$ $CH_3-CH_2-CH(Br)-CH_2-CH_3$ ($3-$bromopentane): The central carbon is bonded to two identical ethyl groups,so it is achiral and optically inactive.
$(d)$ Bromocyclohexane: The molecule has a plane of symmetry passing through the $C-Br$ bond,making it achiral and optically inactive.

(A) The difference between the $1^{st}$ and $2^{nd}$ ionisation energies is largest when the removal of the second electron requires breaking a stable,fully-filled shell or subshell configuration.

Electronic configuration	Element/Group
$1s^{2} 2s^{2} 2p^{6}$	Neon (Group $18$)
$1s^{2} 2s^{2} 2p^{6} 3s^{1}$	Sodium (Group $1$)
$1s^{2} 2s^{2} 2p^{6} 3s^{2}$	Magnesium (Group $2$)
$1s^{2} 2s^{2} 2p^{1}$	Boron (Group $13$)

For $1s^{2} 2s^{2} 2p^{6}$ (Neon),the $1^{st}$ ionisation energy is very high due to its stable noble gas configuration. However,the $2^{nd}$ ionisation energy is significantly higher because it involves removing an electron from a stable $2p^{6}$ shell,which is extremely difficult. Thus,the jump between $IE_1$ and $IE_2$ is largest for this configuration.

(A) The electronegativity of a carbon atom is directly proportional to the $s$-character in its hybridised state.
Greater $s$-character leads to higher electronegativity because the electrons are held closer to the nucleus.
The $s$-character in different hybridisations is as follows:
$sp$: $50\% \ s$-character
$sp^{2}$: $33.3\% \ s$-character
$sp^{3}$: $25\% \ s$-character
Therefore,the order of electronegativity is $sp > sp^{2} > sp^{3}$.

(C) The acidic strength of an acid depends on the stability of its conjugate base. Stronger acids have more stable conjugate bases.
The dissociation reactions are:
$HCl \rightleftharpoons H^{+} + Cl^{-}$
$HCOOH \rightleftharpoons H^{+} + HCOO^{-}$
$CH_{3}COOH \rightleftharpoons H^{+} + CH_{3}COO^{-}$
The stability of the conjugate bases follows the order: $Cl^{-} > HCOO^{-} > CH_{3}COO^{-}$.
Therefore,the acidic strength follows the order: $HCl > HCOOH > CH_{3}COOH$.
Since $pH = -\log[H^{+}]$,a higher concentration of $H^{+}$ ions results in a lower $pH$ value.
Thus,the order of $pH$ for $1\, N$ aqueous solutions is: $CH_{3}COOH > HCOOH > HCl$.

(A) The reaction is an acid-catalyzed hydration of an alkene.
$1$. The proton $(H^+)$ attacks the double bond to form the most stable carbocation intermediate.
$2$. The protonation of the alkene leads to a tertiary carbocation at the carbon attached to the cyclohexyl group.
$3$. Water $(H_2O)$ then acts as a nucleophile and attacks this tertiary carbocation.
$4$. Subsequent deprotonation yields the tertiary alcohol as the major product,which corresponds to structure $I$.

(A) The heat of reaction $\Delta H$ is calculated using the formula: $\Delta H = \Sigma BE_{\text{reactants}} - \Sigma BE_{\text{products}}$.
For the formation of $2 \ mol$ of $NF_3$: $N_2 + 3 F_2 \longrightarrow 2 NF_3$.
$\Delta H = [BE_{N \equiv N} + 3 \times BE_{F-F}] - [6 \times BE_{N-F}] = [941 + 3(155)] - [6(272)] = 1406 - 1632 = -226 \ kJ \ mol^{-1}$.
Since this is for $2 \ mol$ of $NF_3$,the heat of formation per mole is $\Delta H_f(NF_3) = -226 / 2 = -113 \ kJ \ mol^{-1}$. However,the question asks for the heat of reaction for the given stoichiometry.
For the formation of $2 \ mol$ of $NCl_3$: $N_2 + 3 Cl_2 \longrightarrow 2 NCl_3$.
$\Delta H = [BE_{N \equiv N} + 3 \times BE_{Cl-Cl}] - [6 \times BE_{N-Cl}] = [941 + 3(242)] - [6(200)] = 1667 - 1200 = +467 \ kJ \ mol^{-1}$.
The values for the reactions are $-226 \ kJ \ mol^{-1}$ and $+467 \ kJ \ mol^{-1}$ respectively.

(A) Let the initial concentration of $X$ and $Z$ be $C$ and $\alpha$ be the degree of dissociation.
For reaction $X \rightleftharpoons 2Y$:
Initial concentration: $C$ $0$
Equilibrium concentration: $C(1-\alpha)$ $2C\alpha$
$K_{1} = \frac{[2C\alpha]^2}{C(1-\alpha)} = \frac{4C^2\alpha^2}{C(1-\alpha)} = \frac{4C\alpha^2}{1-\alpha}$
For reaction $Z \rightleftharpoons P + Q$:
Initial concentration: $C$ $0$ $0$
Equilibrium concentration: $C(1-\alpha)$ $C\alpha$ $C\alpha$
$K_{2} = \frac{[C\alpha][C\alpha]}{C(1-\alpha)} = \frac{C^2\alpha^2}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha}$
Therefore,the ratio $\frac{K_{1}}{K_{2}} = \frac{4C\alpha^2 / (1-\alpha)}{C\alpha^2 / (1-\alpha)} = 4$.

(C) The molecular formula $C_3H_2Cl_2$ corresponds to a degree of unsaturation of $2$ $(3 - 2/2 + 2/2 = 2)$.
For cyclic isomers,we consider cyclopropene derivatives:
$1$. $3,3$-dichlorocyclopropene
$2$. $1,3$-dichlorocyclopropene (This molecule has a chiral center at the $C_1$ position,so it exists as a pair of enantiomers).
$3$. $1,2$-dichlorocyclopropene
Thus,the total number of cyclic isomers is $4$ ($3,3$-dichlorocyclopropene,$1,3$-dichlorocyclopropene (enantiomer $I$),$1,3$-dichlorocyclopropene (enantiomer $II$),and $1,2$-dichlorocyclopropene).
However,based on the provided options and standard interpretation of such problems,the count of distinct structural isomers is $3$,but including optical isomers,the total is $4$.

(A) According to the ideal gas equation,$pV = nRT$.
For $1 \ mole$ of gas,$n = 1$,so $pV = RT$ or $p = \frac{RT}{V}$.
Taking $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.
For point $X$: $V = 50 \ L, T = 200 \ K$.
$p_X = \frac{0.0821 \times 200}{50} = 0.3284 \ atm \approx 0.328 \ atm$.
For point $Y$: $V = 50 \ L, T = 500 \ K$.
$p_Y = \frac{0.0821 \times 500}{50} = 0.821 \ atm \approx 0.820 \ atm$.
For point $Z$: $V = 20 \ L, T = 200 \ K$.
$p_Z = \frac{0.0821 \times 200}{20} = 0.821 \ atm \approx 0.820 \ atm$.
Thus,the pressures at $X, Y, Z$ are $0.328 \ atm, 0.820 \ atm, 0.820 \ atm$ respectively.

(A) The chemical formula of ammonium sulphate is $(NH_4)_2SO_4$.
The molar mass of $(NH_4)_2SO_4 = 2 \times (14 + 4 \times 1) + 32 + 4 \times 16 = 2 \times 18 + 32 + 64 = 36 + 96 = 132 \ g/mol$.
The mass of nitrogen in one mole of $(NH_4)_2SO_4$ is $2 \times 14 = 28 \ g$.
The percentage of nitrogen by mass $= (\frac{28}{132}) \times 100 \approx 21.2 \ \%$.
Thus,the value is closest to $21 \ \%$.

(C) The correct answer is $C$.
According to Mendeleev's periodic law,the physical and chemical properties of the elements are a periodic function of their atomic mass.

(D) . The maximum number of electrons that can be accommodated in a subshell is given by the formula $2(2l + 1)$.
For the given azimuthal quantum number $l = 4$,the calculation is:
Maximum number of electrons $= 2(2 \times 4 + 1) = 2(8 + 1) = 2 \times 9 = 18$.

(C) The correct option is $C$.
Electron-donating substituents tend to decrease the acidic strength,while electron-withdrawing substituents tend to increase the acidic strength of substituted benzoic acids relative to benzoic acid.
$OCH_3$ exerts a $+M$ effect,which destabilizes the conjugate base of compound $2$ and hence decreases the acidity.
$NO_2$ exerts a $-M$ effect,which stabilizes the conjugate base of compound $3$ and hence increases the acidity.
Therefore,the correct order of acidity of the given compounds is $3 > 1 > 2$.

(D) The reaction of but$-2-$ene $(CH_3-CH=CH-CH_3)$ with acidic $KMnO_4$ is an oxidative cleavage reaction.
Acidic $KMnO_4$ is a strong oxidizing agent that breaks the carbon-carbon double bond.
Since but$-2-$ene is a symmetrical alkene,the oxidative cleavage of the double bond results in the formation of two molecules of acetic acid $(CH_3COOH)$.
The reaction is: $CH_3-CH=CH-CH_3 + [O] \xrightarrow{KMnO_4/H^+} 2CH_3COOH$.

(B) The correct option is $(b)$.
When baking soda $(NaHCO_3)$ is mixed with vinegar (which contains acetic acid,$CH_3COOH$),a chemical reaction occurs that produces sodium acetate,water,and carbon dioxide gas.
The balanced chemical equation is:
$CH_3COOH_{(aq)} + NaHCO_{3(s)} \longrightarrow CH_3COONa_{(aq)} + H_2O_{(l)} + CO_{2(g)}$
Thus,the gas released is carbon dioxide $(CO_2)$.

(D) . Alkali metals have the highest tendency to form ionic bonds because they possess low ionization energy.
The general electronic configuration of an alkali metal is $ns^1$.
Among the given configurations,$1 s^2 2 s^2 2 p^6 3 s^1$ corresponds to $Na$ (Sodium),which is an alkali metal and thus readily forms an ionic bond.

(C)
Given,
Normality of $H_2SO_4 = 1 \, N$
Avogadro's number $= A_0$
Volume of $H_2SO_4 = 200 \, mL = 0.2 \, L$
Normality $=$ Basicity $\times$ Molarity
For $H_2SO_4$,basicity $= 2$
$\therefore 1 = 2 \times M \implies M = 0.5 \, mol/L$
Number of moles of $H_2SO_4 = M \times V(L) = 0.5 \times 0.2 = 0.1 \, mol$
Since each molecule of $H_2SO_4$ contains $1$ atom of $S$,the number of moles of $S$ atoms $= 0.1 \, mol$
Number of $S$ atoms $= 0.1 \times A_0 = \frac{A_0}{10}$

(C) The reduction half-reaction for the dichromate ion in an acidic medium is given by:
$Cr_2O_7^{2-} + 14 H^+ + 6 e^- \rightarrow 2 Cr^{3+} + 7 H_2O$
In this reaction,the oxidation state of $Cr$ changes from $+6$ to $+3$.
Since there are two $Cr$ atoms in $Cr_2O_7^{2-}$,the total change in oxidation state is $2 \times (6 - 3) = 6$.
Therefore,$6$ electrons are required to reduce $Cr_2O_7^{2-}$ to $2 Cr^{3+}$.

(D) From the given graph:
Volume of gas at $0^{\circ}C$ is $V_1 = 250 \ cm^3$.
Volume of gas at $300^{\circ}C$ is $V_2 = 500 \ cm^3$.
Therefore,the ratio of the volumes is:
$\frac{V_2}{V_1} = \frac{500}{250} = 2$.
Thus,the volume of the gas at $300^{\circ}C$ is larger than that at $0^{\circ}C$ by a factor of $2$.

(C) The chemical reaction is: $2Xe + 4F_2 \longrightarrow XeF_2 + XeF_6$
Initial moles of $Xe = \frac{262}{131} = 2 \ mol$.
Initial moles of $F_2 = \frac{152}{38} = 4 \ mol$.
Let $x$ be the moles of $XeF_2$ and $y$ be the moles of $XeF_6$ formed.
According to the law of conservation of mass (atoms):
For $Xe$: $x + y = 2$
For $F$: $2x + 6y = 2 \times 4 = 8$
Solving the equations: $x + y = 2$ and $x + 3y = 4$.
Subtracting the first from the second: $2y = 2$,so $y = 1$.
Substituting $y = 1$ into $x + y = 2$,we get $x = 1$.
Thus,the molar ratio $XeF_2 : XeF_6$ is $1 : 1$.

(A) Step $1$: Ethanol $(CH_3CH_2OH)$ undergoes dehydration in the presence of concentrated sulphuric acid $(H_2SO_4)$ at $170^{\circ} C$ to produce ethene $(CH_2=CH_2)$ gas.
$CH_3CH_2OH \xrightarrow{Conc. H_2SO_4, 170^{\circ} C} CH_2=CH_2 + H_2O$
Step $2$: The ethene gas is then treated with bromine $(Br_2)$ in carbon tetrachloride $(CCl_4)$,which is an electrophilic addition reaction,resulting in the formation of $1,2$-dibromoethane.
$CH_2=CH_2 + Br_2 \xrightarrow{CCl_4} CH_2Br-CH_2Br$
Therefore,the major product is $1,2$-dibromoethane.

(C) The balanced chemical equation for the combustion of $C_4H_8$ is:
$C_4H_{8(g)} + 6O_{2(g)} \longrightarrow 4CO_{2(g)} + 4H_2O_{(l)}$
According to the stoichiometry of the reaction,$1 \, mole$ of $C_4H_8$ reacts with $6 \, moles$ of $O_2$.
At $STP$,$1 \, mole$ of any gas occupies $22.4 \, L$.
Given $22.4 \, L$ of $C_4H_8$ corresponds to $1 \, mole$.
Therefore,the volume of $O_2$ required is $6 \times 22.4 \, L = 134.4 \, L$.

(A) The reaction involves the acid-catalyzed hydration of $1$-methylcyclopentene.
In the presence of an acid catalyst $(H^+)$,water adds across the double bond of the alkene.
This reaction proceeds via an electrophilic addition mechanism,where the proton $(H^+)$ first adds to the less substituted carbon of the double bond to form the more stable carbocation intermediate.
For $1$-methylcyclopentene,the proton adds to the $CH$ group,resulting in a tertiary carbocation at the carbon atom already bearing the methyl group.
Subsequently,a water molecule attacks this tertiary carbocation,followed by deprotonation to yield $1$-methylcyclopentan-$1$-ol.
This follows the Markownikoff rule,which states that the nucleophile $(-OH)$ attaches to the more substituted carbon atom.
Therefore,the major product is structure $I$.

(B) Eubacteria,also known as 'true bacteria',are characterized by a rigid cell wall and,if motile,a flagellum.
$(2)$ $Anabaena$ is a cyanobacterium,which is a type of eubacteria.
$(3)$ Many cyanobacteria (eubacteria) perform $N_2$ fixation.
$(7)$ Cyanobacteria often possess a gelatinous sheath surrounding their colonies.
$(8)$ $Heterocyst$ is a specialized cell found in some filamentous cyanobacteria (eubacteria) for nitrogen fixation.
Methanogens $(1)$ and Thermoacidophiles $(4)$ are Archaebacteria.
Diatoms $(6)$ are Protists (Eukaryotes).
Therefore,the correct set associated with Eubacteria is $(2, 3, 7, 8)$.

(C) The temperature dependence of the rate of a chemical reaction is expressed by the Arrhenius equation: $k = A e^{-E_{a} / (RT)}$.
When $T$ is very high,the factor $E_{a} / (RT)$ approaches $0$.
Therefore,$k = A e^{0} = A \times 1 = A$.
Thus,the rate constant of a chemical reaction at a very high temperature will approach the Arrhenius frequency factor $(A)$.

(B) The reducing strength of a metal is inversely proportional to its standard reduction potential. $A$ more negative standard reduction potential indicates a stronger reducing agent.
Comparing the given values:
$Cr^{3+}/Cr = -0.74 \ V$
$Pb^{2+}/Pb = -0.13 \ V$
$Cu^{2+}/Cu = +0.34 \ V$
$Ag^{+}/Ag = +0.80 \ V$
The order of increasing standard reduction potential is $Cr < Pb < Cu < Ag$. Therefore,the order of decreasing reducing strength is $Cr > Pb > Cu > Ag$.

(A) The reaction involves the free radical polymerization of styrene $(C_6H_5CH=CH_2)$ in the presence of dibenzoyl peroxide as an initiator.
During polymerization,the double bond of the vinyl group breaks,and the monomer units link together to form a long chain.
The structure of the repeating unit in polystyrene is $[-CH_2-CH(C_6H_5)-]_n$.
Comparing this with the given options,structure $I$ represents the correct polymer structure.

(B) The reaction between a tertiary alkyl halide,$(CH_{3})_{3}CCl$,and a strong base,$CH_{3}CH_{2}ONa$,in ethanol primarily undergoes an $E2$ elimination reaction.
Since $(CH_{3})_{3}CCl$ is a tertiary halide,steric hindrance makes substitution difficult.
The strong base $CH_{3}CH_{2}O^-$ abstracts a proton from the $\beta$-carbon,leading to the formation of an alkene.
The reaction is: $(CH_{3})_{3}CCl + CH_{3}CH_{2}ONa \rightarrow CH_{2}=C(CH_{3})_{2} + CH_{3}CH_{2}OH + NaCl$.
The major product is $2$-methylpropene,which corresponds to option $B$.

(D) The correct answer is $D$.
When $H_2S$ gas is passed through an acidic solution,only group $II$ cations precipitate as their sulfides because the low concentration of $S^{2-}$ ions (due to the common ion effect of $H^+$) is sufficient to exceed the solubility product $(K_{sp})$ of group $II$ sulfides.
Among the given ions,$Cu^{2+}$ and $Pb^{2+}$ belong to group $II$.
$Al^{3+}$ belongs to group $III$ and $Ni^{2+}$ belongs to group $IV$,which do not precipitate in acidic medium.
The reactions are:
$Cu^{2+} + H_2S \longrightarrow CuS(s) + 2H^+$
$Pb^{2+} + H_2S \longrightarrow PbS(s) + 2H^+$

(B) The most abundant transition metal in the human body is iron $(Fe)$.
Iron is a crucial component of haemoglobin.
It is the iron-containing oxygen transport metalloprotein found in red blood cells.
Haemoglobin in the blood carries oxygen from the lungs to the rest of the body.

(A) Molar conductivity at infinite dilution $(\Lambda_{m}^{\circ})$ depends on the mobility of ions in the solution.
$HCl$ is a strong acid and dissociates completely into $H^{ }$ and $Cl^{-}$ ions. $H^{ }$ ions have the highest ionic mobility.
$NaCl$ is a strong electrolyte,dissociating into $Na^{ }$ and $Cl^{-}$ ions.
$CH_{3}COONa$ is a strong electrolyte,dissociating into $CH_{3}COO^{-}$ and $Na^{ }$ ions.
$CH_{3}COOH$ is a weak acid and dissociates only partially,resulting in fewer ions in the solution compared to strong electrolytes.
Since $HCl$ provides the highly mobile $H^{ }$ ion,it has the highest molar conductivity.
$NaCl$ has higher conductivity than $CH_{3}COONa$ because the $Cl^{-}$ ion has higher mobility than the $CH_{3}COO^{-}$ ion.
$CH_{3}COOH$ has the lowest molar conductivity due to its weak dissociation.
Thus,the correct order is $HCl > NaCl > CH_{3}COONa > CH_{3}COOH$.

(C) The spin-only magnetic moment of a complex is calculated using the formula: $\mu = \sqrt{n(n+2)} \, BM$,where $n$ is the number of unpaired electrons.
The oxidation state of $Z$ in $[ZCl_{4}]^{2-}$ is $+2$.
The spin-only magnetic moments for the given metals in the $+2$ oxidation state are:
$(a)$ $Mn^{2+}$: Electronic configuration is $3d^{5}$,$n=5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \, BM$.
$(b)$ $Ni^{2+}$: Electronic configuration is $3d^{8}$,$n=2$,$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \, BM$.
$(c)$ $Co^{2+}$: Electronic configuration is $3d^{7}$,$n=3$,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, BM$.
$(d)$ $Cu^{2+}$: Electronic configuration is $3d^{9}$,$n=1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \, BM$.
Since the given magnetic moment is $3.87 \, BM$,$Z$ must be $Co$.

(D) When $\alpha-D-(+)$-glucose is dissolved in water,it undergoes mutarotation.
This process involves the conversion of $\alpha-D-(+)$-glucose into an open-chain structure,which then cyclizes to form $\beta-D-(+)$-glucose.
At equilibrium,the solution contains a mixture of both $\alpha-D-(+)$-glucose (approx. $36\%$) and $\beta-D-(+)$-glucose (approx. $64\%$),with only a negligible amount of the open-chain form.
Therefore,the major constituents present in the solution are $\alpha-D-(+)$-glucose and $\beta-D-(+)$-glucose.

(C) The correct answer is $C$ $(III)$.
Step $1$: Aniline reacts with $NaNO_{2} +$ dil. $HCl$ at $0-5^{\circ} C$ to form benzene diazonium chloride $(C_{6}H_{5}N_{2}^{+}Cl^{-})$.
Step $2$: The benzene diazonium chloride then reacts with $CuCN$ (Sandmeyer reaction) to replace the diazonium group with a cyano group $(-CN)$,yielding cyanobenzene (benzonitrile) and releasing $N_{2}$ gas.
Structure $III$ represents cyanobenzene.

(A) Schottky defect is a type of vacancy defect in ionic solids,where an equal number of cations and anions are missing from their lattice sites to maintain electrical neutrality.This defect leads to a decrease in the density of the crystal.It is typically observed in ionic compounds where the cations and anions have similar sizes,such as $NaCl$ and $AgBr$.

(C) The oxidation state of $Mn$ in $[MnBr_4]^{2-}$ is $+2$.
The electronic configuration of $Mn^{2+}$ is $[Ar] 3d^{5} 4s^{0}$.
Since $Br^{-}$ is a weak field ligand,pairing of electrons will not occur.
As the hybridisation of $Mn$ in $[MnBr_4]^{2-}$ is $sp^{3}$,its geometry is tetrahedral and it contains $5$ unpaired electrons.
$Mn^{2+}$ (Ground state): $3d$ $(1|1|1|1|1)$ $4s$ ( ) $4p$ ( ) ( )
$[MnBr_4]^{2-}$: $3d$ $(1|1|1|1|1)$ $4s$ ( ) $4p$ ( ) ( ) ( ) $\rightarrow sp^{3}$ hybridisation

(A) The cell reaction is: $Zn_{(s)} + Cu^{2+}_{(aq)} \longrightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$
According to the Nernst equation:
$E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$
When the cell is completely discharged,$E_{cell} = 0$ and $n = 2$.
Substituting the values:
$0 = 1.10 - \frac{0.0591}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$
$1.10 = \frac{0.0591}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$
$\log \frac{[Zn^{2+}]}{[Cu^{2+}]} = \frac{1.10 \times 2}{0.0591} = \frac{2.20}{0.0591} \approx 37.225$
Thus,the value is closest to $37.3$.

(C) The correct option is $C$.
Bromobenzene reacts with $Mg$ in the presence of dry ether to form a Grignard reagent (phenylmagnesium bromide).
This Grignard reagent then reacts with $CO_{2}$ to form an intermediate,which upon hydrolysis with $\text{dil. } HCl$ gives benzoic acid.

(B) The organic compound with molecular formula $C_2H_6O$ is ethanol $(CH_3CH_2OH)$,as it undergoes oxidation to form a carboxylic acid.
$CH_3CH_2OH \xrightarrow{K_2Cr_2O_7 / H_2SO_4} CH_3COOH$ $(X)$
To determine the molecular formula of $X$,we calculate its empirical formula based on the given percentage composition:

Element	Percentage	Moles $(Percentage / Atomic Mass)$	Simplest Ratio
$C$	$40\%$	$40 / 12 = 3.33$	$3.33 / 3.33 = 1$
$H$	$6.7\%$	$6.7 / 1 = 6.7$	$6.7 / 3.33 = 2$
$O$	$53.3\%$	$53.3 / 16 = 3.33$	$3.33 / 3.33 = 1$

The empirical formula is $CH_2O$. The empirical formula mass is $12 + (2 \times 1) + 16 = 30$.
The molar mass of acetic acid $(CH_3COOH)$ is $60$.
$n = \frac{\text{Molecular Mass}}{\text{Empirical Formula Mass}} = \frac{60}{30} = 2$.
Therefore,the molecular formula is $(CH_2O)_2 = C_2H_4O_2$.

(A) $MnO_2$,when fused with $KOH$ and oxidised in air,gives a dark green compound $X$,which is potassium manganate $(K_2MnO_4)$.
In acidic solution,$K_2MnO_4$ undergoes disproportionation to give an intense purple solution of potassium permanganate $Y$ $(KMnO_4)$ and $MnO_2$.
The chemical reactions are:
$2MnO_2 + 4KOH + O_2 \longrightarrow 2K_2MnO_4 + 2H_2O$
$3K_2MnO_4 + 4H^{+} \longrightarrow 2KMnO_4 + MnO_2 + 2H_2O + 4K^{+}$
Thus,$X$ is $K_2MnO_4$ and $Y$ is $KMnO_4$.

(B) $Al$ $(X)$ is an amphoteric metal that dissolves in both dilute $HCl$ and dilute $NaOH$ to liberate $H_2$ gas.
$2Al + 6HCl \longrightarrow 2AlCl_3 + 3H_2$
$2Al + 2NaOH + 2H_2O \longrightarrow 2NaAlO_2 + 3H_2$
When $NH_4Cl$ and excess $NH_4OH$ are added to the $AlCl_3$ solution,$Al(OH)_3$ $(Y)$ is precipitated.
$AlCl_3 + 3NH_4OH \longrightarrow Al(OH)_3 \downarrow + 3NH_4Cl$
Similarly,adding $NH_4Cl$ to the $NaAlO_2$ solution (formed by $Al$ in $NaOH$) also precipitates $Al(OH)_3$ $(Y)$ due to the common ion effect and hydrolysis of the aluminate ion.
$NaAlO_2 + NH_4Cl + H_2O \longrightarrow Al(OH)_3 \downarrow + NaCl + NH_3$
Thus,$(X)$ is $Al$ and $(Y)$ is $Al(OH)_3$.

(A) The correct option is $A$.
The reaction is $2ZnS_{(s)} + 3O_{2(g)} \xrightarrow{\Delta} 2ZnO_{(s)} + 2SO_{2(g)}$.
This process is known as roasting,where the sulphide ore is heated in a regular supply of air to convert it into its oxide form at a temperature below the melting point of the metal.

(B) The molecule with the formula $C_{12}O_9$ is known as mellitic anhydride.
Its structure consists of a benzene ring fused with three cyclic anhydride groups.
Therefore,the functional group present in the molecule is an anhydride group.

(A) The correct option is $A$.
When acetic acid $(CH_3COOH)$ reacts with ethanol $(C_2H_5OH)$ in the presence of an acid catalyst like hydrochloric acid $(HCl)$,an esterification reaction occurs.
This reaction produces ethyl acetate $(CH_3COOC_2H_5)$,which is a sweet-smelling compound,along with water $(H_2O)$.
The chemical equation is: $CH_3COOH + C_2H_5OH \xrightarrow{HCl} CH_3COOC_2H_5 + H_2O$.

(A) .
Metals that are placed below hydrogen in the electrochemical reactivity series cannot displace hydrogen from dilute acids.
Among the given metals,$Cu$ (copper) is less reactive than hydrogen $(H)$,meaning its standard reduction potential is higher than that of hydrogen.
Therefore,$Cu$ does not react with $HCl$ to produce $H_2$ gas.

(B)
The isomeric ethers with the molecular formula $C_4H_{10}O$ are:
$1. CH_3-O-CH_2-CH_2-CH_3$ ($1$-methoxypropane)
$2. CH_3-CH_2-O-CH_2-CH_3$ (ethoxyethane)
$3. CH_3-O-CH(CH_3)_2$ ($2$-methoxypropane)
Thus,there are $3$ isomeric ethers with the molecular formula $C_4H_{10}O$.

(A) The reaction at the cathode is: $Ag^+ + e^- \rightarrow Ag(s)$.
According to Faraday's law of electrolysis,the mass $W$ deposited is given by: $W = \frac{M \times I \times t}{n \times F}$.
Given: $M = 108 \, g/mol$,$I = 0.5 \, A$,$t = 1 \, h = 3600 \, s$,$n = 1$ (for $Ag^+$),and $F \approx 96500 \, C/mol$.
Substituting the values: $W = \frac{108 \times 0.5 \times 3600}{1 \times 96500}$.
$W = \frac{194400}{96500} \approx 2.014 \, g$.
Thus,the amount is closest to $2 \, g$.