IIT JEE 2010 Chemistry Question Paper with Answer and Solution

(A) The distance of point $(x_1, y_1, z_1)$ from the plane $Ax + By + Cz - D = 0$ is given by $d = \frac{|Ax_1 + By_1 + Cz_1 - D|}{\sqrt{A^2 + B^2 + C^2}}$.
Given point $P(1, -2, 1)$ and plane $x + 2y - 2z - \alpha = 0$,the distance is $5$.
$5 = \frac{|1(1) + 2(-2) - 2(1) - \alpha|}{\sqrt{1^2 + 2^2 + (-2)^2}} = \frac{|1 - 4 - 2 - \alpha|}{\sqrt{9}} = \frac{|-5 - \alpha|}{3}$.
Since $\alpha > 0$,$|-5 - \alpha| = |5 + \alpha| = 5 + \alpha$.
$5 = \frac{5 + \alpha}{3} \implies 15 = 5 + \alpha \implies \alpha = 10$.
The plane is $x + 2y - 2z - 10 = 0$.
The line passing through $P(1, -2, 1)$ and perpendicular to the plane has direction ratios $(1, 2, -2)$.
The equation of the line is $\frac{x - 1}{1} = \frac{y + 2}{2} = \frac{z - 1}{-2} = k$.
Any point on the line is $(k + 1, 2k - 2, -2k + 1)$.
This point lies on the plane: $(k + 1) + 2(2k - 2) - 2(-2k + 1) = 10$.
$k + 1 + 4k - 4 + 4k - 2 = 10 \implies 9k - 5 = 10 \implies 9k = 15 \implies k = \frac{5}{3}$.
Substituting $k$ back,the foot of the perpendicular is $(\frac{5}{3} + 1, 2(\frac{5}{3}) - 2, -2(\frac{5}{3}) + 1) = (\frac{8}{3}, \frac{4}{3}, -\frac{7}{3})$.

(D) The behavior of a real gas is described by the van der Waals equation: $(P + \frac{a}{V^2})(V - b) = RT$.
$1$. At high temperature $(T)$,the average kinetic energy of the gas molecules is very high,which allows them to overcome the intermolecular forces $(IMF)$.
$2$. At low pressure $(P)$,the volume occupied by the gas is large,meaning the molecules are far apart,making the volume of the molecules $(b)$ negligible compared to the total volume and minimizing the effect of intermolecular forces.
$3$. Therefore,a real gas approaches ideal gas behavior under the conditions of low pressure and high temperature.

(D) The equation of the plane containing the two lines is given by $a(x - 2) + b(y - 3) + c(z - 4) = 0$,where the normal vector $(a, b, c)$ is perpendicular to the direction vectors $(2, 3, 4)$ and $(3, 4, 5)$.
Solving the system $2a + 3b + 4c = 0$ and $3a + 4b + 5c = 0$,we find the cross product: $\vec{n} = (2, 3, 4) \times (3, 4, 5) = (15-16, 12-10, 8-9) = (-1, 2, -1)$.
Thus,the equation of the plane is $-1(x - 2) + 2(y - 3) - 1(z - 4) = 0$,which simplifies to $-x + 2y - z = 0$ or $x - 2y + z = 0$.
Comparing this with the given plane $Ax - 2y + z = d$,we must have $A = 1$ for the planes to be parallel.
The distance between the parallel planes $x - 2y + z = 0$ and $x - 2y + z = d$ is given by $\frac{|d - 0|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \sqrt{6}$.
This gives $\frac{|d|}{\sqrt{6}} = \sqrt{6}$,which implies $|d| = 6$.

(D) In the given two-dimensional square unit cell,there is one atom at the center and four atoms at the corners.
The effective number of atoms $(n)$ is calculated as:
$n = 1_{\text{center}} + 4 \times \frac{1}{4}_{\text{corner}} = 1 + 1 = 2$
The diagonal of the square $(AC)$ is equal to $4r$,where $r$ is the radius of the atom.
$AC = 4r = \sqrt{2}a$,where $a$ is the edge length of the square.
Therefore,$a = \frac{4r}{\sqrt{2}} = 2\sqrt{2}r$.
The packing efficiency ($P$.$E$.) is the ratio of the area occupied by the atoms to the total area of the unit cell:
$P.E. = \frac{n \times \pi r^2}{a^2} \times 100$
$P.E. = \frac{2 \times \pi r^2}{(2\sqrt{2}r)^2} \times 100 = \frac{2\pi r^2}{8r^2} \times 100 = \frac{\pi}{4} \times 100$
$P.E. = \frac{3.14159}{4} \times 100 = 78.54\%$

(D) In the given two-dimensional square unit cell,there are four corner atoms and one central atom.
Each corner atom contributes $1/4$ of its area to the unit cell,so $4 \times (1/4) = 1$ atom from the corners.
The central atom contributes $1$ full atom to the unit cell.
Total number of atoms $= 1 + 1 = 2$.
Let the radius of the atom be $r$ and the side of the square be $a$.
The diagonal of the square is $a\sqrt{2} = 4r$,so $a = 4r/\sqrt{2} = 2r\sqrt{2}$.
Packing efficiency $= \frac{\text{Area of 2 atoms}}{\text{Area of square}} \times 100 = \frac{2 \times \pi r^2}{a^2} \times 100$.
Substituting $a = 2r\sqrt{2}$,we get:
Packing efficiency $= \frac{2 \pi r^2}{(2r\sqrt{2})^2} \times 100 = \frac{2 \pi r^2}{8r^2} \times 100 = \frac{\pi}{4} \times 100 \approx 78.54 \%$.

(D) The synthesis of $3-$octyne $(CH_3CH_2C \equiv CCH_2CH_2CH_2CH_3)$ involves the alkylation of an acetylide ion.
First,$1-$butyne $(CH_3CH_2C \equiv CH)$ reacts with sodium amide $(NaNH_2)$ to form the sodium butynide nucleophile: $CH_3CH_2C \equiv CH + NaNH_2 \rightarrow CH_3CH_2C \equiv C^{-}Na^{+} + NH_3$.
Next,this nucleophile performs an $S_N2$ reaction with $1-$bromobutane $(CH_3CH_2CH_2CH_2Br)$: $CH_3CH_2C \equiv C^{-}Na^{+} + CH_3CH_2CH_2CH_2Br \rightarrow CH_3CH_2C \equiv CCH_2CH_2CH_2CH_3 + NaBr$.
Thus,the required reagents are $1-$bromobutane and $1-$butyne.

(B) By definition,the standard molar enthalpy of formation $(\Delta_fH^circ)$ of an element in its most stable state at $298 \ K$ and $1 \ \text{bar}$ pressure is $ZERO$.
$Cl_{2(g)}$ is the most stable form of chlorine at $298 \ K$.
$Br_{2(l)}$ is the most stable form of bromine at $298 \ K$,so $Br_{2(g)}$ does not have a zero enthalpy of formation.
$H_2O_{(g)}$ and $CH_{4(g)}$ are compounds,not elements,so their standard enthalpy of formation is not zero.

(C) The bond dissociation energy of a $C-C$ single bond is approximately $83-100 \ kcal \ mol^{-1}$. Among the given options,$100 \ kcal \ mol^{-1}$ is the closest and most appropriate value.

(B) The structure of $2,2$-dimethylbutane is $CH_3-CH_2-C(CH_3)_2-CH_3$.
For the given Newman projection,the front carbon has two $CH_3$ groups and one $H$ atom,while the back carbon has two $H$ atoms and one group $Y$. The group $X$ is attached to the front carbon.
Considering the $C_2-C_3$ bond axis:
Front carbon $(C_2)$ has two $CH_3$ groups and one $H$ atom. Back carbon $(C_3)$ has two $H$ atoms and one $C_2H_5$ group.
If $X = CH_3$ and $Y = C_2H_5$,this matches the structure.
Considering the $C_1-C_2$ bond axis:
Front carbon $(C_1)$ has three $H$ atoms. Back carbon $(C_2)$ has two $CH_3$ groups and one $C_2H_5$ group.
If $X = H$ and $Y = C_2H_5$,this matches the structure.
Comparing with the options,$X=H$ and $Y=C_2H_5$ (Option $B$) is a valid possibility,and $X=C_2H_5$ and $Y=H$ (Option $C$) is also a valid possibility depending on the rotation and projection chosen. Thus,$B$ and $C$ are correct.

(C) buffer solution is formed by a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid.
$1$. In option $C$,$HNO_3$ (strong acid) reacts with $CH_3COONa$ (salt of weak acid) to form $CH_3COOH$ (weak acid) and $NaNO_3$. If $HNO_3$ is the limiting reagent,the resulting mixture contains $CH_3COOH$ and $CH_3COONa$,which acts as an acidic buffer.
$2$. In option $D$,the mixture contains $CH_3COOH$ (weak acid) and $CH_3COONa$ (conjugate base),which is a classic acidic buffer system.
Therefore,both $C$ and $D$ can form a buffer.

(B) Temporary hardness of water is due to the presence of magnesium and calcium bicarbonates.
$1$. $Ca(OH)_2$ (Clarke's method) is used to remove temporary hardness by converting soluble bicarbonates into insoluble carbonates:
$Ca(HCO_3)_2 + Ca(OH)_2 \longrightarrow 2 CaCO_3 \downarrow + 2 H_2O$
$2$. $Na_2CO_3$ (Washing soda) is used to remove both temporary and permanent hardness by precipitating calcium and magnesium ions as carbonates:
$Ca(HCO_3)_2 + Na_2CO_3 \longrightarrow CaCO_3 \downarrow + 2 NaHCO_3$
Therefore,both $(B)$ and $(C)$ are correct reagents for softening temporary hardness.

(A) The given titre values are $25.2 \ mL$,$25.25 \ mL$,and $25.0 \ mL$.
First,calculate the average: $\text{Average} = \frac{25.2 + 25.25 + 25.0}{3} = \frac{75.45}{3} = 25.15 \ mL$.
According to the rules of significant figures in addition/subtraction,the result should be reported to the same number of decimal places as the measurement with the fewest decimal places.
Here,$25.2$ and $25.0$ have one decimal place,so the sum $75.45$ should be rounded to one decimal place,which is $75.5$.
Then,$75.5 / 3 = 25.166...$,which rounds to $25.2$.
The value $25.2$ has $3$ significant figures.

(B) The molecular formula $C_4H_6$ corresponds to a degree of unsaturation (double bond equivalent) of $4 - (6/2) + 1 = 2$.
For cyclic isomers,we consider structures with one ring and one double bond,or two rings.
The possible cyclic isomers are:
$1$. Cyclobutene
$2$. $1-$Methylcyclopropene
$3$. $3-$Methylcyclopropene
$4$. Methylenecyclopropane
$5$. Bicyclobutane
Thus,there are a total of $5$ cyclic isomers.

(A) An aqueous solution turns red litmus paper blue if it is basic in nature.
$1$. $KCN$: Salt of strong base $(KOH)$ and weak acid $(HCN)$,so it is basic.
$2$. $K_2SO_4$: Salt of strong base $(KOH)$ and strong acid $(H_2SO_4)$,so it is neutral.
$3$. $(NH_4)_2C_2O_4$: Salt of weak base $(NH_4OH)$ and weak acid $(H_2C_2O_4)$,so it is weakly acidic/neutral.
$4$. $NaCl$: Salt of strong base $(NaOH)$ and strong acid $(HCl)$,so it is neutral.
$5$. $Zn(NO_3)_2$: Salt of weak base $(Zn(OH)_2)$ and strong acid $(HNO_3)$,so it is acidic.
$6$. $FeCl_3$: Salt of weak base $(Fe(OH)_3)$ and strong acid $(HCl)$,so it is acidic.
$7$. $K_2CO_3$: Salt of strong base $(KOH)$ and weak acid $(H_2CO_3)$,so it is basic.
$8$. $NH_4NO_3$: Salt of weak base $(NH_4OH)$ and strong acid $(HNO_3)$,so it is acidic.
$9$. $LiCN$: Salt of strong base $(LiOH)$ and weak acid $(HCN)$,so it is basic.
The compounds that are basic are $KCN$,$K_2CO_3$,and $LiCN$. The total number is $3$.

(B) The molecule $BrF_5$ has a square pyramidal geometry with one lone pair on the central $Br$ atom.
Due to the strong repulsion between the lone pair and the bonding pairs,the equatorial $F$ atoms are pushed downwards.
Consequently,the bond angles between the axial $F$ and the equatorial $F$ atoms are reduced from $90^{\circ}$ to approximately $84.8^{\circ}$.
Therefore,there are no $90^{\circ}$ $F-Br-F$ bond angles in the molecule.

(C) The mineral Beryl has the chemical formula $Be_3 Al_2 Si_6 O_{18}$.
To determine $n$ using the principle of charge neutrality (sum of oxidation states $= 0$):
$Be = +2$,$Al = +3$,$Si = +4$,$O = -2$
$n(+2) + 2(+3) + 6(+4) + 18(-2) = 0$
$2n + 6 + 24 - 36 = 0$
$2n - 6 = 0$
$2n = 6$
$n = 3$

(D) Since $A, B, C$ are in arithmetic progression,$2B = A + C$.
Given $A + B + C = 180^{\circ}$,we have $3B = 180^{\circ}$,so $B = 60^{\circ}$.
Using the sine rule,$\frac{a}{\sin A} = \frac{c}{\sin C} = 2R$,so $a = 2R \sin A$ and $c = 2R \sin C$.
The expression becomes $\frac{\sin A}{\sin C} \sin 2C + \frac{\sin C}{\sin A} \sin 2A$.
$= \frac{\sin A}{\sin C} (2 \sin C \cos C) + \frac{\sin C}{\sin A} (2 \sin A \cos A) = 2 \sin A \cos C + 2 \sin C \cos A$.
$= 2 \sin(A + C) = 2 \sin(180^{\circ} - B) = 2 \sin B$.
Since $B = 60^{\circ}$,the value is $2 \sin 60^{\circ} = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$.

(C) Let the equation of the required plane be $ax+by+cz=0$. Since it contains the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$,the normal vector $(a, b, c)$ is perpendicular to the direction vector $(2, 3, 4)$.
Thus,$2a+3b+4c=0$ ............$(i)$
Next,consider the plane containing the lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$. The normal vector $(a', b', c')$ of this plane is the cross product of the direction vectors $(3, 4, 2)$ and $(4, 2, 3)$:
$(a', b', c') = (3, 4, 2) \times (4, 2, 3) = (12-4, 8-9, 6-16) = (8, -1, -10)$.
Since our required plane is perpendicular to this plane,its normal vector $(a, b, c)$ must be perpendicular to the normal vector $(8, -1, -10)$ of the second plane.
Thus,$8a-b-10c=0$ ............$(ii)$
Solving $(i)$ and $(ii)$ using cross multiplication:
$\frac{a}{(3)(-10) - (4)(-1)} = \frac{b}{(4)(8) - (2)(-10)} = \frac{c}{(2)(-1) - (3)(8)}$
$\frac{a}{-30+4} = \frac{b}{32+20} = \frac{c}{-2-24}$
$\frac{a}{-26} = \frac{b}{52} = \frac{c}{-26}$
Dividing by $-26$,we get the ratios $a:b:c = 1:-2:1$.
Substituting these into $ax+by+cz=0$,we get $x-2y+z=0$.

(B) Using the Law of Cosines for $\angle C$:
$\cos(C) = \frac{a^2 + b^2 - c^2}{2ab}$
$\cos(\frac{\pi}{6}) = \frac{(x^2+x+1)^2 + (x^2-1)^2 - (2x+1)^2}{2(x^2+x+1)(x^2-1)}$
$\frac{\sqrt{3}}{2} = \frac{(x^4+x^2+1+2x^3+2x^2+2x) + (x^4-2x^2+1) - (4x^2+4x+1)}{2(x^2+x+1)(x^2-1)}$
$\frac{\sqrt{3}}{2} = \frac{2x^4+2x^3-3x^2-2x+1}{2(x^2+x+1)(x^2-1)}$
Since $(x^2+x+1)(x^2-1) = x^4+x^3-x-1$,the equation simplifies to:
$\sqrt{3}(x^4+x^3-x-1) = 2x^4+2x^3-3x^2-2x+1$
Solving this quadratic in $x$ leads to $x = 1+\sqrt{3}$ as the valid solution,since side lengths must be positive ($x > 1$ for $b > 0$).

(A) The electronic configuration of $B_2$ ($10$ electrons) is $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \pi_{2p_x}^2$.
Bond order is calculated as $\frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2}$.
Bond order = $\frac{6 - 4}{2} = 1$.
Since all electrons are paired in the $\pi_{2p_x}$ orbital under the assumption that Hund's rule is violated,the molecule is diamagnetic.

(D) To determine the shape,we calculate the hybridization and number of lone pairs for each species:
$1$. $SO_3$: The central $S$ atom has $3$ bond pairs and $0$ lone pairs. It has $sp^2$ hybridization and a trigonal planar shape.
$2$. $BrF_3$: The central $Br$ atom has $3$ bond pairs and $2$ lone pairs. It has $sp^3d$ hybridization and a $T$-shaped geometry.
$3$. $SiO_3^{2-}$: The central $Si$ atom has $3$ bond pairs and $0$ lone pairs. It has $sp^2$ hybridization and a trigonal planar shape.
$4$. $OSF_2$ (Thionyl fluoride): The central $S$ atom has $3$ bond pairs (one double bond to $O$ and two single bonds to $F$) and $1$ lone pair. The total steric number is $4$ ($3$ bond pairs + $1$ lone pair),which corresponds to $sp^3$ hybridization. Due to the presence of one lone pair,the geometry is pyramidal.

(B) The elements listed are $O, Cl, F, N, P, Sn, Tl, Na, Ti$.
$F$ (Fluorine) always shows $-1$ oxidation state.
$Na$ (Sodium) always shows $+1$ oxidation state.
Other elements show variable oxidation states:
$O$ $(-2, -1, +2)$,$Cl$ $(-1, +1, +3, +5, +7)$,$N$ ($-3$ to $+5$),$P$ ($-3$ to $+5$),$Sn$ $(+2, +4)$,$Tl$ $(+1, +3)$,$Ti$ $(+2, +3, +4)$.
Thus,only $2$ elements ($F$ and $Na$) show a single non-zero oxidation state.

(D) diprotic acid is an acid that can donate two protons ($H^+$ ions) per molecule.
$1$. $H_3PO_4$: Triprotic acid.
$2$. $H_2SO_4$: Diprotic acid.
$3$. $H_3PO_3$: Diprotic acid (contains two $P-OH$ bonds).
$4$. $H_2CO_3$: Diprotic acid.
$5$. $H_2S_2O_7$: Diprotic acid.
$6$. $H_3BO_3$: Monoprotic acid (Lewis acid).
$7$. $H_3PO_2$: Monoprotic acid (contains one $P-OH$ bond).
$8$. $H_2CrO_4$: Diprotic acid.
$9$. $H_2SO_3$: Diprotic acid.
The diprotic acids are: $H_2SO_4, H_3PO_3, H_2CO_3, H_2S_2O_7, H_2CrO_4, H_2SO_3$.
Total count = $6$.

(B) $1.$ For $S_1$ (spherically symmetric,$\ell = 0$):
$\text{Radial node} = n - \ell - 1 = 1$ $\Rightarrow n - 0 - 1 = 1$ $\Rightarrow n = 2.$
Thus,state $S_1$ is $2s$.
For $S_2$,energy $E_{S_2} = E_H(\text{ground}) = -13.6 \ eV$.
$E_{S_2} = \frac{-13.6 \times Z^2}{n^2} = -13.6 \ eV$ $\Rightarrow \frac{3^2}{n^2} = 1$ $\Rightarrow n = 3.$
Given $S_2$ has one radial node: $n - \ell - 1 = 1$ $\Rightarrow 3 - \ell - 1 = 1$ $\Rightarrow \ell = 1$.
$2.$ Energy of $S_1$ $(n=2, Z=3)$: $E_{S_1} = \frac{-13.6 \times 3^2}{2^2} = -13.6 \times 2.25 \ eV$.
In units of $E_H(\text{ground}) = -13.6 \ eV$,the energy is $2.25$.
$3.$ For $S_2$,$\ell = 1$ (as calculated above).

(B) The fundamental frequency of a pipe closed at one end is given by $f_P = \frac{v}{4L_P}$,where $v = 320 \ ms^{-1}$ and $L_P = 0.8 \ m$.
$f_P = \frac{320}{4 \times 0.8} = \frac{320}{3.2} = 100 \ Hz$.
The frequency of a string vibrating in its second harmonic is $f_S = \frac{2}{2L_S} \sqrt{\frac{T}{\mu}} = \frac{1}{L_S} \sqrt{\frac{T}{\mu}}$,where $L_S = 0.5 \ m$ and $T = 50 \ N$.
Since the string resonates with the pipe,$f_S = f_P = 100 \ Hz$.
$100 = \frac{1}{0.5} \sqrt{\frac{50}{\mu}} \implies 100 = 2 \sqrt{\frac{50}{\mu}} \implies 50 = \sqrt{\frac{50}{\mu}}$.
Squaring both sides: $2500 = \frac{50}{\mu} \implies \mu = \frac{50}{2500} = 0.02 \ kg/m$.
The mass of the string $M = \mu \times L_S = 0.02 \times 0.5 = 0.01 \ kg = 10 \ g$.

(A) The given structure represents a disaccharide where ring $(a)$ is a six-membered pyranose ring (glucose unit) and ring $(b)$ is a five-membered furanose ring (fructose unit).
The glycosidic linkage connects the anomeric carbon of the glucose unit to the fructose unit.
In the provided structure,the oxygen atom of the glycosidic linkage is directed downwards relative to the plane of the ring $(a)$,which characterizes an $\alpha$-glycosidic linkage.
Therefore,ring $(a)$ is a pyranose ring with an $\alpha$-glycosidic linkage.

(D) The reaction of anisole $(C_6H_5OCH_3)$ with $HBr$ involves the protonation of the ether oxygen atom by the acid.
This is followed by the nucleophilic attack of the bromide ion $(Br^-)$ on the less sterically hindered alkyl group.
Since the phenyl group is attached to the oxygen,the $C-O$ bond between the phenyl ring and oxygen has partial double bond character due to resonance,making it stronger and harder to break.
Therefore,the $Br^-$ ion attacks the methyl group $(CH_3)$,resulting in the formation of phenol $(C_6H_5OH)$ and methyl bromide $(CH_3Br)$.

(A) The Arrhenius equation is given by $k = Ae^{-E_a / RT}$.
As the temperature $T$ increases,the term $e^{-E_a / RT}$ increases exponentially.
Therefore,the rate constant $k$ increases exponentially with an increase in temperature $T$.
This behavior is represented by the exponential curve shown in option $A$.

(C) Ethylenediaminetetraacetic acid $(EDTA)$ is a hexadentate ligand.
Its structure consists of an ethylenediamine backbone $(H_2N-CH_2-CH_2-NH_2)$ where the four hydrogen atoms on the nitrogen atoms are replaced by four acetic acid groups $(CH_2COOH)$.
Therefore,the structure is $(HOOCCH_2)_2N-CH_2-CH_2-N(CH_2COOH)_2$.
This corresponds to option $C$.

(B) Ionization isomers are coordination compounds that produce different ions in solution.
In the complex $[Cr(H_2O)_4Cl(NO_2)]Cl$,the $Cl^-$ ion is present in the ionization sphere.
To form an ionization isomer,the $Cl^-$ ion in the ionization sphere is exchanged with the $NO_2^-$ ligand inside the coordination sphere.
Thus,the isomer is $[Cr(H_2O)_4Cl_2](NO_2)$.

(A) The reaction of phenol with $NaOH(aq)$ forms the phenoxide ion,which is highly activated towards electrophilic aromatic substitution.
$Br_2$ acts as an electrophile.
The phenoxide ion undergoes electrophilic attack by $Br_2$ at the ortho and para positions.
Intermediate $C$ represents the $p$-bromophenoxide ion.
Intermediate $A$ represents the $o$-bromophenoxide ion.
Intermediate $B$ represents a dienone intermediate formed during the bromination process.
Thus,the intermediates involved in the reaction pathway are $A$,$B$,and $C$.

(C) Intensive properties are independent of the amount of matter present in the system.
$1$. Molar conductivity is an intensive property because it is defined as the conductivity per mole of electrolyte.
$2$. Electromotive force $(EMF)$ is an intensive property because it is a potential difference independent of the size of the cell.
$3$. Resistance $(R)$ is an extensive property because it depends on the dimensions and amount of material.
$4$. Heat capacity $(C)$ is an extensive property because it depends on the total amount of substance.
Therefore,$A$ and $B$ are intensive properties.

(B) $1.$ The cell reaction is $M_{(s)} + M^{+}_{(aq, 0.05 \ M)} \rightarrow M^{+}_{(aq, 1 \ M)} + M_{(s)}$.
According to the Nernst equation,$E_{cell} = E^{\circ}_{cell} - \frac{RT}{nF} \ln Q$. For a concentration cell,$E^{\circ}_{cell} = 0$.
$E_{cell} = 0 - \frac{0.0591}{1} \log \frac{0.05}{1} = -0.0591 \times \log(5 \times 10^{-2}) = -0.0591 \times (-1.301) \approx +0.077 \ V = 77 \ mV$ (given as $70 \ mV$).
Since $E_{cell} > 0$,the reaction is spontaneous,so $\Delta G < 0$.
$2.$ New concentration of $M^{+}$ is $0.0025 \ M$.
$E_{cell} = -0.0591 \log \frac{0.0025}{1} = -0.0591 \times \log(2.5 \times 10^{-3}) = -0.0591 \times (-2.602) \approx 0.153 \ V \approx 140 \ mV$ (using the constant derived from the first part: $70 \ mV = -k \log(0.05) \Rightarrow k = 70 / 1.301 \approx 53.8$. Then $E_{new} = 53.8 \times 2.602 \approx 140 \ mV$).

(C) $1.$ Partial roasting of chalcopyrite: $2CuFeS_2 + O_2 \rightarrow Cu_2S + 2FeS + SO_2 \uparrow$. Further roasting: $2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2 \uparrow$ and $2FeS + 3O_2 \rightarrow 2FeO + 2SO_2 \uparrow$. Thus,the products are $Cu_2O$ and $FeO$.
$2.$ Iron is removed as slag by adding silica $(SiO_2)$: $FeO + SiO_2 \rightarrow FeSiO_3$ (slag).
$3.$ In self-reduction: $2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2 \uparrow$. Here,the sulfur in $Cu_2S$ (as $S^{2-}$) is oxidized to $SO_2$ (where $S$ is $+4$),making $S^{2-}$ the reducing agent.

(B) For a zero-order reaction,the rate constant $k$ is given by the expression $k = \frac{[R]_0 - [R]_t}{t}$.
Using the data points:
$1$. For $t = 0.05 \text{ min}$,$[R] = 0.75 \text{ M}$:
$k_1 = \frac{1.0 - 0.75}{0.05} = \frac{0.25}{0.05} = 5 \text{ M min}^{-1}$.
$2$. For $t = 0.12 \text{ min}$,$[R] = 0.40 \text{ M}$:
$k_2 = \frac{1.0 - 0.40}{0.12} = \frac{0.60}{0.12} = 5 \text{ M min}^{-1}$.
$3$. For $t = 0.18 \text{ min}$,$[R] = 0.10 \text{ M}$:
$k_3 = \frac{1.0 - 0.10}{0.18} = \frac{0.90}{0.18} = 5 \text{ M min}^{-1}$.
Since the rate constant $k$ is constant for all time intervals,the reaction follows zero-order kinetics.

(D) The nuclear fission reaction is represented as: ${}_{92}^{235} U + {}_{0}^{1} n \rightarrow {}_{54}^{142} Xe + {}_{38}^{90} Sr + x({}_{0}^{1} n)$.
By balancing the mass number: $235 + 1 = 142 + 90 + x$.
$236 = 232 + x$.
$x = 236 - 232 = 4$.
Therefore,the number of neutrons emitted is $4$.

(A) In the given structure of lysine,the basic groups are the amino group $(-NH_2)$ and the carboxylate ion $(-COO^-)$,which can act as a base by accepting a proton.
Therefore,there are $2$ basic groups present in this form of lysine.

(A) The starting material is $1,2-dimethylenecyclohexane$ derivative,which upon ozonolysis ($O_3$ followed by $Zn, H_2O$) yields $Y$,which is $cyclodecane-1,6-dione$.
$Y$ $(cyclodecane-1,6-dione)$ has two equivalent carbonyl groups and equivalent $\alpha$-hydrogens.
Intramolecular aldol condensation of $cyclodecane-1,6-dione$ involves the formation of an enolate at one of the $\alpha$-positions,followed by nucleophilic attack on the other carbonyl group.
Due to the symmetry of the molecule,the attack can occur to form a $5$-membered ring (via a $5$-exo-trig cyclization) or a $7$-membered ring (via a $7$-endo-trig cyclization).
According to Baldwin's rules and thermodynamic stability,the formation of the $5$-membered ring is highly favored over the $7$-membered ring.
Thus,only $1$ major intramolecular aldol condensation product is formed.

(B) Compounds that are acidic enough to react with aqueous $NaOH$ (a base) to form water-soluble salts are soluble in aqueous $NaOH$.
$1$. Cyclohexanecarboxylic acid (contains $-COOH$ group,acidic).
$2$. Phenol (contains phenolic $-OH$ group,acidic).
$3$. $4$-(Dimethylamino)phenol (contains phenolic $-OH$ group,acidic).
$4$. $1$-Naphthoic acid (contains $-COOH$ group,acidic).
Other compounds like $N,N$-dimethylaniline,$2$-ethoxybenzyl alcohol,nitrobenzene,and $1,2$-diethylbenzene are not sufficiently acidic to react with $NaOH$.
Therefore,the total number of compounds soluble in aqueous $NaOH$ is $4$.

(A) For compound $P$ ($4$-hydroxybenzoic acid),the $-OH$ group is a strong activating group and ortho/para directing. Since the para position is occupied by $-COOH$,nitration occurs at the ortho position to the $-OH$ group.
For compound $Q$ ($1$-methoxy-$4$-methylbenzene),both $-OCH_3$ and $-CH_3$ are activating groups. $-OCH_3$ is a stronger activating group than $-CH_3$. Therefore,nitration occurs ortho to the $-OCH_3$ group.
For compound $S$ (phenyl benzoate),the ester group $-COO-$ is deactivating and meta-directing for the benzoyl ring,but the phenoxy ring is activated by the oxygen atom. The oxygen atom of the ester group is ortho/para directing. Due to steric hindrance,the para position of the phenoxy ring is the major site for electrophilic substitution.

(D) In the given two-dimensional square unit cell,the atoms are present at the four corners and one at the center of the square.
Let the side length of the square be $a$ and the radius of each atom be $r$.
The diagonal of the square is $d = a \sqrt{2}$.
Along the diagonal,the atoms are in contact,so $d = 4r$.
Therefore,$a \sqrt{2} = 4r$,which gives $a = 2 \sqrt{2} r$.
The number of atoms per unit cell is $Z = 4 \times (1/4) + 1 = 2$.
The area of the unit cell is $a^2 = (2 \sqrt{2} r)^2 = 8r^2$.
The area occupied by the atoms is $Z \times \pi r^2 = 2 \pi r^2$.
Packing efficiency = $\frac{\text{Area occupied by atoms}}{\text{Total area of unit cell}} \times 100$
Packing efficiency = $\frac{2 \pi r^2}{8 r^2} \times 100 = \frac{\pi}{4} \times 100 \approx 0.7854 \times 100 = 78.54 \%$.

(B) The spin-only magnetic moment $\mu$ is given by $\sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons. For $\mu = 2.82 \ B.M.$,$n=2$.
$(A)$ $Ni(CO)_4$: $Ni$ is in $0$ oxidation state. $Ni(0) = 3d^8 4s^2$. $CO$ is a strong field ligand,causing pairing of electrons. All electrons are paired $(n=0)$.
$(B)$ $\left[NiCl_4\right]^{2-}$: $Ni$ is in $+2$ oxidation state. $Ni^{2+} = 3d^8$. $Cl^-$ is a weak field ligand,so no pairing occurs. The $3d$ orbitals have $2$ unpaired electrons $(n=2)$. Thus,$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.82 \ B.M.$
$(C)$ $Ni\left(PPh_3\right)_4$: $Ni$ is in $0$ oxidation state. $Ni(0) = 3d^8 4s^2$. $PPh_3$ is a strong field ligand,causing pairing. All electrons are paired $(n=0)$.
$(D)$ $\left[Ni(CN)_4\right]^{2-}$: $Ni$ is in $+2$ oxidation state. $Ni^{2+} = 3d^8$. $CN^-$ is a strong field ligand,causing pairing. All electrons are paired $(n=0)$.

(C) The reaction proceeds in two steps:
$1$. The first step is the Hoffmann bromamide degradation of $4$-methylbenzamide using $NaOH/Br_2$,which converts the amide group $(-CONH_2)$ into a primary amine group $(-NH_2)$,yielding $p$-toluidine ($4$-methylaniline).
$2$. The second step is the acylation of the resulting $p$-toluidine with benzoyl chloride $(C_6H_5COCl)$. The lone pair on the nitrogen atom of the amine attacks the carbonyl carbon of benzoyl chloride,leading to the formation of $N$-($4$-methylphenyl)benzamide as the final product $T$.

(A) The density of silver is $d = 10.5 \ g \ cm^{-3}$.
The number of atoms per unit volume is given by $n = \frac{d \times N_A}{M} = \frac{10.5 \times 6.022 \times 10^{23}}{108} \approx 5.856 \times 10^{22} \text{ atoms } cm^{-3}$.
The number of atoms per unit area is approximately $n^{2/3} = (5.856 \times 10^{22})^{2/3} \approx 1.5 \times 10^{15} \text{ atoms } cm^{-2}$.
Given area $= 10^{-12} \ m^2 = 10^{-8} \ cm^2$.
Number of atoms $= (1.5 \times 10^{15} \text{ atoms } cm^{-2}) \times (10^{-8} \ cm^2) = 1.5 \times 10^7$.
Comparing with $y \times 10^x$,we get $x = 7$.

(A) The reaction sequence is an aldol condensation followed by cyanohydrin formation and subsequent hydrolysis/cyclization.
$1.$ The reaction between isobutyraldehyde $(P)$ and formaldehyde $(Q)$ in the presence of aqueous $K_2CO_3$ is a cross-aldol condensation. $P$ is $(CH_3)_2CHCHO$ and $Q$ is $HCHO$. Thus,$P$ and $Q$ are $(B)$ and $(A)$ respectively.
$2.$ The aldol product $R$ formed is $3-hydroxy-2,2-dimethylpropanal$,which corresponds to structure $(B)$.
$3.$ Treatment of $R$ with $HCN$ yields the cyanohydrin $S$,which is $2,2-dimethyl-3-hydroxy-4-hydroxybutanenitrile$ derivative,corresponding to structure $(D)$.
Therefore,the correct sequence is $(B, B, D)$ for $(P, Q)$,$R$,and $S$ respectively. However,based on the provided options,the closest match is $(B, A, D)$.

(C) . This is an electrophilic aromatic substitution reaction,specifically a diazo coupling reaction.
$B$. This is the Pinacol-Pinacolone rearrangement,which proceeds via a carbocation intermediate.
$C$. This is a nucleophilic addition reaction of a hydride to a ketone,resulting in a chiral alcohol. Since the carbonyl carbon is planar,the hydride can attack from either side,producing a racemic mixture.
$D$. This is an intramolecular nucleophilic substitution reaction where the thiolate ion displaces the chloride ion.

(A) $(A). (CH_3)_2 SiCl_2 + 2 H_2O \rightarrow (CH_3)_2 Si(OH)_2 + 2 HCl$. The product $(CH_3)_2 Si(OH)_2$ undergoes polymerization to form silicones. Thus,it matches $(p)$ and $(s)$.
$(B). 3 XeF_4 + 6 H_2O \rightarrow XeO_3 + 2 Xe + 12 HF + 1.5 O_2$. This reaction involves hydrogen halide $(HF)$ formation $(p)$,redox reaction $(q)$,reaction with glass (due to $HF$ produced) $(r)$,and $O_2$ formation $(t)$. Thus,it matches $(p, q, r, t)$.
$(C). Cl_2 + H_2O \rightarrow HCl + HOCl$. Further,$2 HOCl \rightarrow 2 HCl + O_2$. This involves hydrogen halide $(HCl)$ formation $(p)$,redox reaction $(q)$,and $O_2$ formation $(t)$. Thus,it matches $(p, q, t)$.
$(D). VCl_5 + H_2O \rightarrow VOCl_3 + 2 HCl$. This involves hydrogen halide $(HCl)$ formation $(p)$. Thus,it matches $(p)$.

(D) The rate of radioactive decay is given by $|dN/dt| = \lambda N = \lambda N_0 e^{-\lambda t}$.
Taking the natural logarithm on both sides: $\ln|dN/dt| = \ln(\lambda N_0) - \lambda t$.
Comparing this with the equation of a straight line $y = mx + c$,the slope of the graph is $-\lambda$.
From the given graph (assuming slope $= -0.5$),we get $\lambda = 0.5 \ \text{year}^{-1}$.
The half-life is $t_{1/2} = \ln(2) / \lambda \approx 0.693 / 0.5 = 1.386 \ \text{years}$.
After $t = 4.16 \ \text{years}$,the number of nuclei remaining is $N = N_0 / p$.
Since $4.16 / 1.386 = 3$,the time elapsed is $3 \ t_{1/2}$.
Thus,$N = N_0 / 2^3 = N_0 / 8$.
Therefore,$p = 8$.