KCET 2016 Chemistry Question Paper with Answer and Solution

(A) The most suitable method for introducing alien $DNA$ into a plant cell is $Biolistics$ (also known as the gene gun method).
In this technique,cells are bombarded with high-velocity micro-particles of gold or tungsten coated with $DNA$.
This method is specifically designed for plant cells because they possess a rigid cell wall that makes other methods like micro-injection or lipofection difficult or ineffective.
Micro-injection is typically used for animal cells,while heat shock is commonly used for bacterial transformation.

(D) The given curve is defined by $x=t^{2}+3t-8$ and $y=2t^{2}-2t-5$.
First,we find the derivatives with respect to $t$:
$\frac{dx}{dt} = 2t+3$ and $\frac{dy}{dt} = 4t-2$.
The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t-2}{2t+3}$.
To find the value of $t$ at the point $(2, -1)$,we set $x=2$ and $y=-1$:
For $x=2$: $t^{2}+3t-8=2 \Rightarrow t^{2}+3t-10=0 \Rightarrow (t-2)(t+5)=0$,so $t=2$ or $t=-5$.
For $y=-1$: $2t^{2}-2t-5=-1 \Rightarrow 2t^{2}-2t-4=0 \Rightarrow 2(t^{2}-t-2)=0 \Rightarrow 2(t-2)(t+1)=0$,so $t=2$ or $t=-1$.
The common value of $t$ is $t=2$.
Substituting $t=2$ into the expression for the slope:
$\left. \frac{dy}{dx} \right|_{t=2} = \frac{4(2)-2}{2(2)+3} = \frac{8-2}{4+3} = \frac{6}{7}$.
Thus,the slope of the tangent is $\frac{6}{7}$.

(B) The hybridization of carbon in the given compounds is determined by the number of sigma bonds and lone pairs attached to the carbon atom.
$1$. In diamond,each carbon atom is bonded to four other carbon atoms via single bonds,resulting in $sp^3$ hybridization.
$2$. In graphite,each carbon atom is bonded to three other carbon atoms in a planar hexagonal structure,resulting in $sp^2$ hybridization.
$3$. In ethyne $(C_2H_2)$,each carbon atom is bonded to one hydrogen atom and one other carbon atom via a triple bond,resulting in $sp$ hybridization.
Therefore,the order is $sp^3, sp^2, sp$.

(D) According to Molecular Orbital Theory $(MOT)$,the bond order is calculated as $BO = \frac{1}{2}(N_b - N_a)$.
For $O_2$ $(16 \ e^-)$: $BO = \frac{10-6}{2} = 2$.
For $O_2^+$ $(15 \ e^-)$: $BO = \frac{10-5}{2} = 2.5$.
For $O_2^-$ $(17 \ e^-)$: $BO = \frac{10-7}{2} = 1.5$.
For $O_2^{2-}$ $(18 \ e^-)$: $BO = \frac{10-8}{2} = 1$.
Thus,the increasing order of bond order is $O_2^{2-} < O_2^- < O_2 < O_2^+$.

(A) If the $z$-axis is defined as the internuclear axis (molecular axis),then the $p_{z}$ orbitals overlap head-on to form a $\sigma$-bond.
Since $p_{x}$ and $p_{y}$ orbitals are perpendicular to the $z$-axis,they undergo lateral (sideways) overlap.
This lateral overlap of $p_{x}$ and $p_{y}$ orbitals results in the formation of $\pi$-molecular orbitals.

(D) The electronegativity difference between $H$ and $Cl$ is approximately $0.9$,which is significantly less than $2.1$.
According to the Pauling scale,if the electronegativity difference between two atoms is less than $1.7$ (or $2.1$ in some conventions),the bond formed is predominantly covalent.
In $NaCl$,the electronegativity difference between $Na$ $(0.9)$ and $Cl$ $(3.0)$ is $2.1$,resulting in an ionic bond.

(C) For the equilibrium: $Fe(OH)_{3(s)} \rightleftharpoons Fe^{3+}_{(aq)} + 3OH^{-}_{(aq)}$
The solubility product expression is: $K_{sp} = [Fe^{3+}][OH^{-}]^3$
Let the initial concentrations be $[Fe^{3+}]_1$ and $[OH^{-}]_1$.
$K_{sp} = [Fe^{3+}]_1 [OH^{-}]_1^3$ ... $(i)$
If the new concentration $[OH^{-}]_2 = \frac{1}{4} [OH^{-}]_1$, then:
$K_{sp} = [Fe^{3+}]_2 [OH^{-}]_2^3 = [Fe^{3+}]_2 (\frac{1}{4} [OH^{-}]_1)^3$
$K_{sp} = [Fe^{3+}]_2 \times \frac{1}{64} [OH^{-}]_1^3$ ... $(ii)$
From $(i)$ and $(ii)$:
$[Fe^{3+}]_1 [OH^{-}]_1^3 = [Fe^{3+}]_2 \times \frac{1}{64} [OH^{-}]_1^3$
$[Fe^{3+}]_2 = 64 \times [Fe^{3+}]_1$
Therefore, the concentration of $Fe^{3+}$ increases by $64$ times.

(D) For reaction $(a)$: $K_1 = \frac{[NO_2]}{[NO][O_2]^{1/2}}$
For reaction $(b)$: $K_2 = \frac{[NO]^2[O_2]}{[NO_2]^2}$
From $(a)$,we can write: $\frac{1}{K_1} = \frac{[NO][O_2]^{1/2}}{[NO_2]}$
Squaring both sides: $\left(\frac{1}{K_1}\right)^2 = \frac{[NO]^2[O_2]}{[NO_2]^2}$
Since the right side is equal to $K_2$,we have $\frac{1}{K_1^2} = K_2$
Therefore,$K_2 = \frac{1}{K_1^2}$.

(C) Let $f(x) = \left(\frac{1}{x}\right)^{x} = x^{-x}$.
Taking the natural logarithm on both sides,we get $\ln f(x) = -x \ln x$.
Differentiating with respect to $x$,we have $\frac{1}{f(x)} f'(x) = -(\ln x + x \cdot \frac{1}{x}) = -(1 + \ln x)$.
Thus,$f'(x) = -f(x)(1 + \ln x)$.
For critical points,set $f'(x) = 0$,which implies $1 + \ln x = 0$,so $\ln x = -1$,which gives $x = e^{-1} = \frac{1}{e}$.
To check for maximum,we observe the sign of $f'(x)$. For $x < \frac{1}{e}$,$f'(x) > 0$,and for $x > \frac{1}{e}$,$f'(x) < 0$. Thus,$x = \frac{1}{e}$ is a point of local maxima.
The maximum value is $f\left(\frac{1}{e}\right) = \left(\frac{1}{1/e}\right)^{1/e} = e^{1/e}$.

(C) Let $ y = \left(\frac{1}{x}\right)^{x} $. Taking the natural logarithm on both sides,we get $ \ln y = x \ln \left(\frac{1}{x}\right) = -x \ln x $.
Differentiating with respect to $ x $,we have $ \frac{1}{y} \frac{dy}{dx} = - (1 \cdot \ln x + x \cdot \frac{1}{x}) = -(1 + \ln x) $.
Thus,$ \frac{dy}{dx} = -y(1 + \ln x) $.
Setting $ \frac{dy}{dx} = 0 $,we get $ 1 + \ln x = 0 $,which implies $ \ln x = -1 $,so $ x = e^{-1} = \frac{1}{e} $.
To check for the maximum,we find the second derivative $ \frac{d^2y}{dx^2} = -\frac{dy}{dx}(1 + \ln x) - y(\frac{1}{x}) $.
At $ x = \frac{1}{e} $,$ \ln x = -1 $,so $ \frac{d^2y}{dx^2} = 0 - y(e) = -y e < 0 $. Since the second derivative is negative,the function has a maximum at $ x = \frac{1}{e} $.
The maximum value is $ y = \left(\frac{1}{1/e}\right)^{1/e} = e^{1/e} $.

(C) When two dice are thrown simultaneously,the total number of possible outcomes is $6 \times 6 = 36$.
We are looking for the outcomes where the sum of the numbers on the two dice is $5$.
The favorable outcomes are $(1, 4), (4, 1), (2, 3), (3, 2)$.
There are $4$ such favorable outcomes.
Therefore,the probability of obtaining a total score of $5$ is $P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{36} = \frac{1}{9}$.

(A) The given structure is $CH_3-CH=CH-CH_2-Br$.
According to $IUPAC$ nomenclature rules,the principal functional group or the multiple bond should get the lowest possible number.
Here,the double bond is present,so we number the chain starting from the end that gives the double bond the lowest locant.
Numbering from the right side: $CH_3(4)-CH(3)=CH(2)-CH_2(1)-Br$.
The double bond starts at position $2$ and the bromine atom is at position $1$.
Therefore,the $IUPAC$ name is $1-$bromobut$-2-$ene.

(B) miscible mixture of $C_6H_6 + CHCl_3$ can be separated by distillation.
The boiling point of benzene $(C_6H_6)$ is $80^{\circ} C$ and that of chloroform $(CHCl_3)$ is $61.2^{\circ} C$.
Since there is a significant difference in their boiling points,the mixture can be separated by distillation,where the component with the lower boiling point $(CHCl_3)$ evaporates first and is collected separately.

(B) The nitration of benzene involves the generation of the nitronium ion $(NO_2^{+})$ as the active electrophile.
Step $1$: Sulfuric acid $(H_2SO_4)$ acts as a proton donor and nitric acid $(HNO_3)$ acts as a base. The reaction is: $HNO_3 + H_2SO_4 \rightleftharpoons H_2NO_3^{+} + HSO_4^{-}$.
Step $2$: The protonated nitric acid $(H_2NO_3^{+})$ loses a water molecule to form the nitronium ion: $H_2NO_3^{+} \rightarrow NO_2^{+} + H_2O$.
This nitronium ion $(NO_2^{+})$ then attacks the benzene ring to form nitrobenzene.

(D) The given circuit consists of two $NAND$ gates acting as $NOT$ gates,followed by a $NAND$ gate.
Let the inputs be $A$ and $B$.
The output of the first $NAND$ gate is $\overline{A \cdot A} = \bar{A}$.
The output of the second $NAND$ gate is $\overline{B \cdot B} = \bar{B}$.
These outputs are fed into the final $NAND$ gate.
The final output $Y$ is given by:
$Y = \overline{\bar{A} \cdot \bar{B}}$
Using De Morgan's law,$\overline{\bar{A} \cdot \bar{B}} = \overline{\bar{A}} + \overline{\bar{B}} = A + B$.
Thus,$Y = A + B$,which represents the $OR$ operation.
Therefore,the correct option is $D$.

(A) Molar mass of $CO_2 = 12 + (2 \times 16) = 44 \ g/mol$.
Number of moles of $CO_2 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{4.4 \ g}{44 \ g/mol} = 0.1 \ mol$.
$1$ molecule of $CO_2$ contains $2$ oxygen atoms.
Therefore,$1$ mole of $CO_2$ contains $2 \times 6.022 \times 10^{23}$ oxygen atoms.
Number of oxygen atoms in $0.1 \ mol$ of $CO_2 = 0.1 \times 2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{23}$ oxygen atoms.
Thus,the number of oxygen atoms is approximately $1.2 \times 10^{23}$.

(C) To find the empirical formula,we calculate the relative number of moles of each element:

Element	Percentage (%) / Atomic Mass	Relative Moles	Simplest Ratio
$C$	$40 / 12$	$3.33$	$3.33 / 3.33 = 1$
$H$	$13.33 / 1$	$13.33$	$13.33 / 3.33 \approx 4$
$N$	$46.67 / 14$	$3.33$	$3.33 / 3.33 = 1$

The empirical formula is $CH_{4}N$.

(A) Based on the phase diagram,the liquid phase is bounded by the triple point at the lower temperature limit and the critical point at the upper temperature limit. Below the triple point,the substance exists as a solid or vapor. Above the critical temperature $(T_c)$,the substance exists as a supercritical fluid,where the distinction between liquid and gas disappears. Therefore,a liquid can exist only between the triple point and the critical point.

(A) The energy of an electron in the $n^{\text{th}}$ Bohr orbit of a hydrogen atom is given by the formula:
$E_n = -\frac{13.6 \times Z^2}{n^2} \text{ eV}$
For a hydrogen atom,the atomic number $Z = 1$.
Substituting $Z = 1$ into the formula,we get:
$E_n = -\frac{13.6 \times (1)^2}{n^2} \text{ eV} = -\frac{13.6}{n^2} \text{ eV}$.

(D) The rules for quantum numbers are as follows:
$1$. The principal quantum number $n$ can be any positive integer.
$2$. The azimuthal quantum number $l$ can have values from $0$ to $n-1$.
$3$. The magnetic quantum number $m_l$ can have values from $-l$ to $+l$.
$4$. The spin quantum number $s$ can only be $+1/2$ or $-1/2$.
Evaluating the options:
- In option $D$,for $n = 3$,the maximum value of $l$ is $2$. If $l = 2$,the possible values for $m_l$ are $-2, -1, 0, +1, +2$.
- Therefore,$m_l = -3$ is not permissible for $l = 2$.
- Additionally,the spin quantum number $s$ must be $\pm 1/2$. The original options provided $s = \pm 2$,which is physically impossible.

(D) For the reaction: $H_{2(g)} + Br_{2(g)} \rightarrow 2HBr_{(g)}$
$\Delta H^{\circ}_{reaction} = \Sigma \Delta H_{\text{bonds broken}} - \Sigma \Delta H_{\text{bonds formed}}$
$\Delta H^{\circ}_{reaction} = [BE(H-H) + BE(Br-Br)] - [2 \times BE(H-Br)]$
$\Delta H^{\circ}_{reaction} = (433 + 192) - (2 \times 364) \ kJ \ mol^{-1}$
$\Delta H^{\circ}_{reaction} = 625 - 728 \ kJ \ mol^{-1}$
$\Delta H^{\circ}_{reaction} = -103 \ kJ \ mol^{-1}$

(A) The $3^{rd}$ law of thermodynamics states that at absolute zero temperature (i.e.,$0 \ K$),the entropy of a perfectly crystalline solid is zero. This is because there is no thermal motion and perfect order in the crystal lattice.

(C) The acidity of a compound depends on the stability of its conjugate base.
$o$-Nitrophenol is the most acidic among the given options because the $-NO_2$ group is a strong electron-withdrawing group ($-I$ and $-M$ effect).
This group stabilizes the phenoxide ion formed after the loss of a proton $(H^+)$ by dispersing the negative charge through resonance.
In contrast,$-CH_3$ (in $o$-cresol) is an electron-donating group ($+I$ and hyperconjugation),which destabilizes the phenoxide ion,making it less acidic than phenol.
$Cl-CH_2-CH_2-OH$ is an aliphatic alcohol,which is significantly less acidic than phenolic compounds.

(D) The reaction for chlorobenzene to phenol is: $C_6H_5Cl \xrightarrow[(ii) dil. HCl]{(i) 6-8\% NaOH, 623 K, 300 atm} C_6H_5OH$.
The presence of the $-NO_2$ group at $o$- and $p$- positions withdraws electron density from the benzene ring due to its strong $-I$ and $-M$ effects.
This reduction in electron density facilitates the nucleophilic attack of $OH^-$ on the haloarene.
Furthermore,the intermediate carbanion formed is stabilized through resonance by the $-NO_2$ group at the $o$- and $p$- positions,which delocalizes the negative charge.

(A) The Reimer-Tiemann reaction involves the treatment of phenol with $CHCl_3$ in the presence of aqueous $NaOH$.
In this reaction,$CHCl_3$ reacts with $OH^-$ to generate the electrophile,dichlorocarbene $(:CCl_2)$.
The dichlorocarbene then attacks the phenoxide ion to form an intermediate,which upon hydrolysis yields salicylaldehyde.

(A) The dehydration of ethanol to form ethoxyethane (diethyl ether) is an intermolecular dehydration reaction.
When an excess of ethanol is heated with concentrated $H_2SO_4$ at $413\ K$ $(140^{\circ}C)$,two molecules of ethanol undergo condensation to form ethoxyethane and water.
The reaction is: $2CH_3CH_2OH \xrightarrow{conc. H_2SO_4, 413\ K} CH_3CH_2-O-CH_2CH_3 + H_2O$.
Heating at $443\ K$ leads to intramolecular dehydration,forming ethene $(CH_2=CH_2)$.

(B) This reaction is an example of the Cannizzaro reaction,which occurs in aldehydes lacking $\alpha$-hydrogen atoms.
$2 C_6H_5CHO + NaOH (conc.) \rightarrow C_6H_5COONa (A) + C_6H_5CH_2OH (B)$
Product $A$ is Sodium benzoate,which is commonly used as a food preservative.
Product $B$ is Phenyl methanol (also known as Benzyl alcohol),which is an aromatic hydroxy compound where the $OH$ group is attached to an $sp^3$ hybridized carbon atom $(-CH_2-)$ adjacent to the benzene ring.
Therefore,the products $A$ and $B$ are Sodium benzoate and Phenyl methanol.

(A) $1$. The oxidation product of the organic compound $(X)$ reacts with phenyl hydrazine,which indicates that the product is a carbonyl compound (aldehyde or ketone).
$2$. The product does not give the silver mirror test (Tollens' test),which confirms that it is a ketone,not an aldehyde.
$3$. Secondary alcohols are oxidized by acidified $K_2Cr_2O_7$ to form ketones.
$4$. Among the given options,$2-$propanol is a secondary alcohol. Its oxidation gives acetone (propanone),which is a ketone.
$5$. Acetone reacts with phenyl hydrazine to form acetone phenylhydrazone but does not give the silver mirror test.
$6$. Therefore,the compound $X$ is $2-$propanol.

(B) The given reaction sequence is:
$CH_3CN$ $\xrightarrow{\text{Reduction}} CH_3CH_2NH_2$ $\xrightarrow{HNO_2} CH_3CH_2OH$
Here,$A$ is $CH_3CN$ (ethane nitrile),$B$ is $CH_3CH_2NH_2$ (ethylamine),and the final product is ethanol.
Therefore,the compound $A$ is ethane nitrile.

(A) The reaction sequence is as follows:
$1$. $A$ is Nitrobenzene $(C_6H_5NO_2)$.
$2$. Reduction of $A$ gives $B$,which is Aniline $(C_6H_5NH_2)$.
$3$. Aniline reacts with trichloromethane $(CHCl_3)$ and caustic potash $(KOH)$ to form Phenyl isocyanide $(C_6H_5NC)$ as compound $C$ (Carbylamine reaction).
$4$. Catalytic reduction of $C$ $(C_6H_5NC)$ gives $N$-methylbenzenamine $(C_6H_5NHCH_3)$.
Thus,compound $A$ is Nitrobenzene.

(B) Glucose gives a positive Fehling's solution test.
This is because glucose is a reducing sugar,as it contains a free aldehyde group $(-CHO)$ in its open-chain structure,which can reduce $Cu^{2+}$ ions to $Cu^+$ ions.

(A) In $DNA$,the nitrogenous bases pair specifically through hydrogen bonding.
Adenine $(A)$ always pairs with thymine $(T)$ via two hydrogen bonds.
Guanine $(G)$ always pairs with cytosine $(C)$ via three hydrogen bonds.
Therefore,the correct base pairing is $A-T$ and $G-C$.

(D) The reaction sequence is as follows:
$1$. $CH_3COOH + PCl_5 \rightarrow CH_3COCl$ (Product $A$ is acetyl chloride).
$2$. $CH_3COCl + C_6H_6 \xrightarrow{Anhyd. AlCl_3} CH_3COC_6H_5$ (Product $B$ is acetophenone,a Friedel-Crafts acylation reaction).
$3$. $CH_3COC_6H_5 + CH_3MgBr \rightarrow (CH_3)_2C(OH)C_6H_5$ (Product $C$ is $2-$phenylpropan$-2-$ol,formed by the nucleophilic addition of the Grignard reagent to the ketone).

(B) For the given reaction, $3 \,A \rightarrow 2 \,B$.
According to the rate law expression, the rate of reaction is defined as:
Rate $= -\frac{1}{3} \frac{d[A]}{d t} = +\frac{1}{2} \frac{d[B]}{d t}$.
To find the rate of appearance of $B$, which is $+\frac{d[B]}{d t}$, we multiply both sides by $2$:
$+\frac{d[B]}{d t} = -\frac{2}{3} \frac{d[A]}{d t}$.

(B) For the reaction $m A \rightarrow x B$,the rate law is given by $r = k[A]^{2}$.
If the concentration of $A$ is doubled,the new concentration becomes $[A]' = 2[A]$.
The new rate $r'$ is given by $r' = k[A]'{}^{2} = k(2[A])^{2}$.
$r' = 4k[A]^{2}$.
Since $r = k[A]^{2}$,we have $r' = 4r$.
Therefore,the reaction rate will be quadrupled.

(A) For a $1^{st}$ order reaction,the amount remaining after $n$ half-lives is given by the formula: $[A] = [A]_0 \times (1/2)^n$.
Number of half-lives $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{240 \ minutes}{60 \ minutes} = 4$.
Percentage remaining $= (1/2)^n \times 100 = (1/2)^4 \times 100$.
Percentage remaining $= \frac{1}{16} \times 100 = 6.25 \%$.

(A) According to the Arrhenius equation,$\ln \frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$.
By measuring the rate constants $k_1$ and $k_2$ at two different temperatures $T_1$ and $T_2$,the activation energy $E_a$ can be calculated.

(A) Equanil is a well-known tranquilizer used to treat stress,depression,and hypertension.

(A) $1$. Identify the ligands: $en$ is ethane-$1,2$-diamine,$Cl^-$ is chlorido,and $ONO^-$ is nitro-$O$.
$2$. Determine the oxidation state of the central metal atom $(Co)$: Let the oxidation state be $x$. The complex is $[Co(en)_{2}(ONO)Cl]Cl$. The charge on $en$ is $0$,$ONO$ is $-1$,$Cl$ is $-1$,and the outer $Cl$ is $-1$. Thus,$x + 2(0) + (-1) + (-1) = +1$,so $x = +3$.
$3$. Arrange ligands alphabetically: Chlorido comes before ethane-$1,2$-diamine,which comes before nitro-$O$.
$4$. Combine the parts: Chloridobis(ethane-$1,2$-diamine)nitro-$O$-cobalt$(III)$ chloride.

(B) The complex $[CoCl_{2}(en)(NH_{3})_{2}]^{+}$ has the general formula $[M(AA)a_{2}b_{2}]$,where $M = Co$,$AA = en$,$a = Cl$,and $b = NH_{3}$.
For this type of complex,there are $3$ geometrical isomers possible:
$1$. Trans-isomer: Both $Cl$ atoms are trans to each other,and both $NH_{3}$ molecules are trans to each other.
$2$. Cis-isomer $(I)$: $Cl$ atoms are cis to each other,and $NH_{3}$ molecules are cis to each other.
$3$. Cis-isomer $(II)$: $Cl$ atoms are cis to each other,and $NH_{3}$ molecules are trans to each other.
Thus,a total of $3$ geometrical isomers are possible.

(A) The crystal field splitting energy $(\Delta)$ depends on the nature of the ligands.
According to the spectrochemical series,the field strength of ligands increases in the order: $I^{-} < Br^{-} < SCN^{-} < Cl^{-} < S^{2-} < F^{-} < OH^{-} < C_2O_4^{2-} < H_2O < NCS^{-} < EDTA^{4-} < NH_3 < en < CN^{-} < CO$.
Comparing the given options with this series,the correct sequence is $SCN^{-} < F^{-} < CN^{-} < CO$.

(B) $CO$ is a stronger ligand than $Cl^{-}$ because $CO$ acts as a $\pi$-acid (or $\pi$-acceptor) ligand.
It possesses vacant antibonding molecular orbitals that can accept electron density from the metal $d$-orbitals,forming a synergistic bond (back-bonding),which stabilizes the metal-ligand complex.

(A) The magnetic moment of the metal ions is determined by the number of unpaired electrons $(n)$ using the formula $\mu_{spin} = \sqrt{n(n+2)} \ BM$.
$1$. For $Mn^{2+}$ $(3d^5)$: $n = 5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \ BM \approx 5.92 \ BM$.
$2$. For $Cu^{2+}$ $(3d^9)$: $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \ BM \approx 1.73 \ BM$.
$3$. For $Sc^{2+}$ $(3d^1)$: $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \ BM \approx 1.73 \ BM$.
$4$. For $Cu^{+}$ $(3d^{10})$: $n = 0$,$\mu = 0 \ BM$.
Since $Mn^{2+}$ has the highest number of unpaired electrons $(n=5)$,it exhibits the maximum paramagnetic behaviour.

(A) The electronic configuration of $Mn^{2+}$ is $[Ar] 3d^5$. This configuration is half-filled,which provides extra stability.
The electronic configuration of $Fe^{2+}$ is $[Ar] 3d^6$.
Oxidation to the $+3$ state involves the removal of an electron from the $d$-subshell. For $Mn^{2+}$,removing an electron disrupts the stable half-filled $d^5$ configuration,requiring a very high third ionization energy.
For $Fe^{2+}$,removing an electron from the $3d^6$ configuration results in a stable $3d^5$ configuration,which requires less energy.
Therefore,$Mn^{2+}$ is more resistant to oxidation to the $+3$ state compared to $Fe^{2+}$.

(A) The brown compound of manganese is $MnO_2$ $(A)$. When $MnO_2$ reacts with $HCl$,it produces chlorine gas $(B)$:
$MnO_2 (A) + 4 HCl \rightarrow MnCl_2 + 2 H_2O + Cl_2 (B)$
When excess chlorine gas $(B)$ reacts with ammonia $(NH_3)$,it forms nitrogen trichloride $(C)$,which is an explosive compound:
$3 Cl_2 (B) + NH_3 \rightarrow NCl_3 (C) + 3 HCl$
Thus,the compounds $A$,$B$,and $C$ are $MnO_2$,$Cl_2$,and $NCl_3$ respectively.

(D) In a galvanic cell,oxidation occurs at the anode,where electrons are lost. Reduction occurs at the cathode,where electrons are gained. Therefore,the statement that the electrode at which electrons are lost is called the cathode is incorrect; it is actually called the anode.

(C) In secondary cells,electrode reactions can be reversed by an external electric energy source.
Therefore,these cells can be recharged by passing electric current and used again and again.
The current is passed in the direction opposite to that of the flow of current generated by the cell during discharge.

(C) The balanced chemical equation for the reduction of dichromate ions is:
$Cr_{2}O_{7}^{2-} + 14 H^{+} + 6 e^{-} \rightarrow 2 Cr^{3+} + 7 H_{2}O$
From the stoichiometry of the balanced equation,$1 \ mol$ of $Cr_{2}O_{7}^{2-}$ ions requires $6 \ mol$ of electrons for reduction to $Cr^{3+}$ ions.
Since $1 \ mol$ of electrons carries a charge of $1 \ F$,the total charge required is $6 \ F$.

(A) The Van-Arkel method is a technique used for refining metals like $Zr$ and $Ti$.
In this process,the crude metal is heated in an evacuated vessel with iodine to form a volatile metal iodide,such as $ZrI_4$.
This volatile compound is then decomposed on a tungsten filament at a high temperature $(1800 \ K)$ to obtain the pure metal.
This method is specifically effective for removing all oxygen and nitrogen impurities present in the metal.

(A) Copper matte is obtained during the extraction of copper from copper pyrites $(CuFeS_{2})$.
It is a molten mixture consisting mainly of copper$(I)$ sulphide $(Cu_{2}S)$ and iron$(II)$ sulphide $(FeS)$.

(A) When $Al_2O_3$ is leached from bauxite using concentrated $NaOH$ solution,the complex formed is $Na[Al(OH)_4]$.
The chemical reaction involved in the leaching process of bauxite ore is:
$Al_2O_3 \cdot 2H_2O + 2NaOH + 3H_2O \xrightarrow{473-523 \ K} 2Na[Al(OH)_4]$

(B) Dehydrohalogenation is a $\beta$-elimination reaction where an alkyl halide reacts with a strong base to form an alkene.
The rate of this reaction depends on the strength of the $C-X$ bond.
The bond dissociation energy follows the order $C-F > C-Cl > C-Br > C-I$.
Since the $C-I$ bond is the weakest,it is the easiest to break,making the alkyl iodide the most reactive.
Therefore,the order of reactivity is $R-I > R-Br > R-Cl > R-F$.

(D) The reaction sequence is as follows:
$1$. $\text{Ethanol} (CH_3 CH_2 OH)$ reacts with $PCl_5$ to form $X$,which is chloroethane $(CH_3 CH_2 Cl)$.
$2$. Chloroethane $(CH_3 CH_2 Cl)$ reacts with alcoholic $KOH$ (dehydrohalogenation) to form $Y$,which is ethene $(CH_2 = CH_2)$.
$3$. Ethene $(CH_2 = CH_2)$ reacts with $H_2SO_4$ at room temperature followed by hydrolysis $(H_2O, \Delta)$ to form $Z$,which is ethanol $(CH_3 CH_2 OH)$.
Therefore,the product $Z$ is $CH_3 CH_2 OH$.

(D) The electronic configuration of $N$ $(Z=7)$ is $1s^{2} 2s^{2} 2p^{3}$.
Since the principal quantum number $n=2$,it does not have $d$-orbitals available.
Due to the stable half-filled $2p$ subshell,its ionization enthalpy is very high.
Nitrogen is the third most electronegative element in the periodic table.
Due to its small size,nitrogen has a unique ability to form $p\pi-p\pi$ multiple bonds with itself (forming $N \equiv N$).

(C) sol is a colloidal system where solid particles are dispersed in a liquid medium. Sulphur sol is a lyophobic colloid formed by the aggregation of a large number of $S_8$ molecules. Therefore,it contains large aggregates of $S$ molecules.

(D) Fluorine is the most electronegative element in the periodic table,which makes most of its reactions highly exothermic.
It forms only one oxoacid,which is $HOF$ (hypofluorous acid).
Due to the small size of the fluorine atom,the lone pairs on the two fluorine atoms in $F_2$ experience strong inter-electronic repulsion,resulting in a very low $F-F$ bond dissociation enthalpy.
Therefore,the statement that fluorine has a high $F-F$ bond dissociation enthalpy is incorrect.

(B) The molecule $XeF_6$ has $7$ electron pairs around the central $Xe$ atom ($6$ bond pairs and $1$ lone pair),which corresponds to $sp^3d^3$ hybridization.
According to $VSEPR$ theory,the presence of one lone pair on the $Xe$ atom causes distortion in the regular octahedral geometry.
Therefore,the shape of $XeF_6$ is described as distorted octahedral.

(B) The statement in option $B$ is incorrect because natural rubber (cis$-1,4-$polyisoprene) has a 'cis' configuration at every double bond,not a 'trans' configuration. Gutta-percha is the 'trans' isomer of natural rubber.
Buna-$S$ is a co-polymer of $1,3$-butadiene and styrene,which is correct.
Vulcanisation makes rubber harder and stronger by forming cross-links between polymer chains.
Natural rubber is indeed a $1,4$-polymer of isoprene ($2$-methyl-$1,3$-butadiene).

(A) $Nylon-6,6$ is a polymer containing amide linkages,therefore,it is a polyamide.
It is formed by the condensation polymerization of hexamethylene diamine and adipic acid:
$n NH_2(CH_2)_6NH_2 + n HOOC(CH_2)_4COOH \xrightarrow{\text{Polymerization}} [NH-(CH_2)_6-NH-CO-(CH_2)_4-CO]_n + 2n H_2O$

(B) In a crystal lattice,an atom located at the edge center is shared by $4$ adjacent unit cells.
Therefore,the contribution of a particle at the edge center to a single unit cell is $1/4$.

(B) Schottky defect arises when an equal number of cations and anions are missing from their normal lattice sites to maintain electrical neutrality. This defect leads to a decrease in the density of the crystal.

(B) Colligative properties are properties of solutions that depend only on the number of solute particles present in a given amount of solvent,not on their identity.
Common colligative properties include:
$1$. Relative lowering of vapour pressure
$2$. Elevation in boiling point
$3$. Depression in freezing point
$4$. Osmotic pressure
Optical activity is a property related to the interaction of a substance with plane-polarized light and is not dependent on the number of solute particles.
Therefore,optical activity is not a colligative property.

(A) When an electrolyte undergoes dissociation in a solution,it splits into multiple ions or particles.
This increase in the total number of solute particles in the solution leads to an increase in the observed colligative properties compared to the calculated values.
Since the van't Hoff factor $(i)$ is defined as the ratio of observed colligative property to the calculated colligative property,for dissociation,$i > 1$.

(A) The osmotic pressure of a solution is given by the equation $\Pi = CRT$,where $\Pi$ is the osmotic pressure,$C$ is the molar concentration,$R$ is the gas constant,and $T$ is the absolute temperature.
Since $\Pi$ is directly proportional to the temperature $T$ $(\Pi \propto T)$,increasing the temperature of the solution will result in an increase in the osmotic pressure.

(D) $ \text{Physisorption} $ is an exothermic process,meaning the enthalpy of adsorption $( \Delta H_{\text{adsorption}} )$ is negative $( -Ve )$.
It involves weak $ \text{van der Waals} $ forces,resulting in a low enthalpy change,typically in the range of $ 20-40 \ kJ \ mol^{-1} $.
Since it is an exothermic process,the correct statement is that $ \Delta H_{\text{adsorption}} $ is low and negative $( -Ve )$,making option $ D $ the incorrect statement.

(D) Reactions in $Zeolite$ catalysts are shape-selective in nature.
These reactions depend on the pore structure of the catalyst,the size of the apertures,and the size of the cavities within the $Zeolite$ framework.