AIIMS 2010 Chemistry Question Paper with Answer and Solution

(A) The formula for the angle of banking for a horizontal circular turn is given by $\tan \theta = \frac{v^2}{rg}$.
Given: speed $v = 150 \, m/s$,banking angle $\theta = 12^\circ$,and acceleration due to gravity $g = 10 \, m/s^2$.
Rearranging the formula to solve for the radius $r$: $r = \frac{v^2}{g \tan \theta}$.
Substituting the values: $r = \frac{(150)^2}{10 \times \tan 12^\circ}$.
Using $\tan 12^\circ \approx 0.2125$,we get $r = \frac{22500}{10 \times 0.2125} = \frac{22500}{2.125} \approx 10588 \, m$.
Converting to kilometers: $r \approx 10.6 \, km$.

(A) According to Newton's $3^{rd}$ law of motion,for every action,there is an equal and opposite reaction.
When the horse pulls the wagon,it pushes the ground backward with its feet.
The ground exerts an equal and opposite force on the horse in the forward direction.
This forward force exerted by the ground is what causes the horse to move forward.

(D) According to the law of conservation of linear momentum,the total momentum before collision equals the total momentum after collision.
Let the masses be $m_1 = 0.1 \,kg$ and $m_2 = 0.4 \,kg$,and their velocities be $v_1 = 1 \,m/s$ and $v_2 = -0.1 \,m/s$ (since they move towards each other).
Let $V$ be the velocity of the combined mass after collision.
$m_1 v_1 + m_2 v_2 = (m_1 + m_2) V$
$(0.1)(1) + (0.4)(-0.1) = (0.1 + 0.4) V$
$0.1 - 0.04 = 0.5 V$
$0.06 = 0.5 V$
$V = \frac{0.06}{0.5} = 0.12 \,m/s$
The distance traveled by the combined mass in $t = 10 \,s$ is:
$d = V \times t = 0.12 \,m/s \times 10 \,s = 1.2 \,m$.

(C) The time period of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{l}{g}}$,where $l$ is the length of the pendulum and $g$ is the acceleration due to gravity.
From this formula,it is clear that the time period $T$ is independent of the mass of the bob.
Since the length of the pendulum remains unchanged,the time period will remain the same regardless of the change in mass of the bob.
Therefore,the new time period is $2\, s$.

(D) The induced $e.m.f.$ $(e)$ across a rotating spoke is given by the formula $e = \frac{1}{2} B \omega l^2$.
Given:
$B = 0.4 \ G = 0.4 \times 10^{-4} \ T$
$l = 0.50 \ m$
Frequency $\nu = 120 \ rev/min = \frac{120}{60} \ rev/s = 2 \ rev/s$
Angular velocity $\omega = 2 \pi \nu = 2 \times 3.14 \times 2 = 12.56 \ rad/s$
Substituting the values:
$e = \frac{1}{2} \times (0.4 \times 10^{-4}) \times (12.56) \times (0.5)^2$
$e = 0.2 \times 10^{-4} \times 12.56 \times 0.25$
$e = 0.05 \times 12.56 \times 10^{-4}$
$e = 0.628 \times 10^{-4} \ V = 6.28 \times 10^{-5} \ V$.

(D) The decay rate (activity) is given by the formula: $\left| \frac{dN}{dt} \right| = \lambda N$.
Here,$\lambda$ is the decay constant,which is related to the half-life $T_{1/2}$ by $\lambda = \frac{0.693}{T_{1/2}}$.
Given: $T_{1/2} = 1.2 \times 10^7 \ s$ and $N = 4 \times 10^{15}$ atoms.
Substituting these values:
$\left| \frac{dN}{dt} \right| = \frac{0.693}{1.2 \times 10^7} \times 4 \times 10^{15}$.
$\left| \frac{dN}{dt} \right| = \frac{0.693}{1.2} \times 4 \times 10^8$.
$\left| \frac{dN}{dt} \right| = 0.5775 \times 4 \times 10^8 = 2.31 \times 10^8 \ \text{atoms/s}$.
Thus,the decay rate is approximately $2.3 \times 10^8 \ \text{atoms/s}$.

(A) The enthalpy of formation $(Delta_fH)$ represents the stability of a compound. $A$ more negative value of $Delta_fH$ indicates higher stability.
Given $Delta_fH(HF) = -161 \ kJ$ and $Delta_fH(HCl) = -92 \ kJ$.
Since $-161 < -92$,$HF$ is more stable than $HCl$.
Therefore,the statement '$HCl$ is more stable than $HF$' is incorrect.

(B) When silicon tetrafluoride $(SiF_4)$ reacts with water,it undergoes hydrolysis to form hexafluorosilicic acid $(H_2SiF_6)$ and silicic acid $(H_4SiO_4)$.
The balanced chemical equation is:
$3SiF_4 + 4H_2O \to 2H_2SiF_6 + H_4SiO_4$
Among the given options,$H_4SiO_4$ is the correct product formed.

(D) Volume of the gold dispersed in $1 \ L$ of water $= \frac{\text{Mass}}{\text{Density}} = \frac{1.9 \times 10^{-4} \ g}{19 \ g/cm^3} = 1 \times 10^{-5} \ cm^3$.
Radius of gold sol particle $= 10 \ nm = 10 \times 10^{-7} \ cm = 10^{-6} \ cm$.
Volume of one gold sol particle $= \frac{4}{3} \pi r^3 = \frac{4}{3} \times 3.1416 \times (10^{-6} \ cm)^3 \approx 4.19 \times 10^{-18} \ cm^3$.
Number of gold sol particles in $1 \ L$ $(1000 \ cm^3)$ $= \frac{1 \times 10^{-5} \ cm^3}{4.19 \times 10^{-18} \ cm^3} \approx 2.38 \times 10^{12}$.
Since $1 \ L = 10^6 \ mm^3$,the number of particles per $mm^3 = \frac{2.38 \times 10^{12}}{10^6} = 2.38 \times 10^6 \approx 2.4 \times 10^6$.

(B) The hydrolysis of urea $(NH_2CONH_2)$ with water $(H_2O)$ proceeds as follows:
$NH_2CONH_2 + 2H_2O$ $\rightarrow (NH_4)_2CO_3$ $\rightarrow 2NH_3 + H_2CO_3$
$H_2CO_3$ is carbonic acid,which is unstable and decomposes into carbon dioxide $(CO_2)$ and water $(H_2O)$.
Therefore,the primary product of the hydrolysis reaction is carbonic acid.

(B) The dipole moment depends on the molecular geometry and the electronegativity difference between the atoms.
$BF_3$ has a trigonal planar geometry,and the bond dipoles cancel each other out,resulting in a net dipole moment of $0 \ D$.
In $NH_3$ and $NF_3$,both have a trigonal pyramidal geometry with a lone pair on the nitrogen atom.
In $NH_3$,the direction of the bond dipoles $(N-H)$ and the lone pair dipole are in the same direction,reinforcing each other,leading to a high dipole moment of approximately $1.46 \ D$.
In $NF_3$,the bond dipoles $(N-F)$ are in the opposite direction to the lone pair dipole,which partially cancels the net dipole moment,resulting in a lower value of approximately $0.24 \ D$.
$B_2H_6$ (diborane) is a non-polar molecule with a bridged structure,resulting in a net dipole moment of $0 \ D$.
Therefore,$NH_3$ has the highest dipole moment.

(D) According to the law of conservation of linear momentum,the total momentum before collision is equal to the total momentum after collision.
$m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v$
Where $v$ is the common velocity of the two bodies after they stick together.
Given:
$m_{1} = 0.1\, kg$,$m_{2} = 0.4\, kg$
$v_{1} = 1\, m/s$,$v_{2} = -0.1\, m/s$ (since they move towards each other,we take opposite directions).
Substituting the values:
$(0.1 \times 1) + (0.4 \times -0.1) = (0.1 + 0.4) v$
$0.1 - 0.04 = 0.5 v$
$0.06 = 0.5 v$
$v = \frac{0.06}{0.5} = 0.12\, m/s$
Distance covered in $t = 10\, s$ is:
$d = v \times t = 0.12\, m/s \times 10\, s = 1.2\, m$.

(D) The decay rate is given by the formula: $\frac{dN}{dt} = \lambda N$.
First, calculate the decay constant $\lambda$ using the half-life $T_{1/2} = 1.2 \times 10^7 \, s$:
$\lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{1.2 \times 10^7} \, s^{-1}$.
Now, substitute the values of $\lambda$ and $N = 4.0 \times 10^{15}$ atoms into the decay rate formula:
$\frac{dN}{dt} = \left( \frac{0.693}{1.2 \times 10^7} \right) \times (4.0 \times 10^{15})$.
$\frac{dN}{dt} = \frac{0.693 \times 4.0}{1.2} \times 10^{15-7}$.
$\frac{dN}{dt} = \frac{2.772}{1.2} \times 10^8$.
$\frac{dN}{dt} = 2.31 \times 10^8 \, \text{atoms/s}$.
Rounding to the nearest given option, the decay rate is $2.3 \times 10^8 \, \text{atoms/s}$.

(B) The dipole moment of a molecule depends on its geometry and the electronegativity difference between the atoms.
$BF_3$ has a trigonal planar geometry,and due to its symmetry,the individual bond dipoles cancel each other out,resulting in a net dipole moment of $0 \ D$.
$NH_3$ has a trigonal pyramidal geometry with a lone pair on the nitrogen atom. The bond dipoles of the $N-H$ bonds and the lone pair dipole point in the same direction,resulting in a high net dipole moment of approximately $1.47 \ D$.
$NF_3$ also has a trigonal pyramidal geometry,but the bond dipoles of the $N-F$ bonds point in the opposite direction to the lone pair dipole,which partially cancels the net dipole moment,resulting in a lower value of approximately $0.24 \ D$.
$B_2H_6$ (diborane) is a non-polar molecule due to its bridge-bonded structure,resulting in a net dipole moment of $0 \ D$.
Therefore,$NH_3$ has the highest dipole moment.

(D) The balanced chemical equation is: $Ca + 2H_2O \to Ca(OH)_2 + H_2$
From the stoichiometry,$1 \ mol$ of $Ca$ $(40 \ g)$ produces $1 \ mol$ of $H_2$ gas.
$1 \ mol$ of any gas at $STP$ occupies $22400 \ cm^3$.
Given mass of $Ca = 8 \ g$.
Moles of $Ca = \frac{8 \ g}{40 \ g/mol} = 0.2 \ mol$.
Since $1 \ mol$ of $Ca$ produces $1 \ mol$ of $H_2$,$0.2 \ mol$ of $Ca$ will produce $0.2 \ mol$ of $H_2$.
Volume of $H_2$ at $STP = 0.2 \ mol \times 22400 \ cm^3/mol = 4480 \ cm^3$.

(C) The energy of a photon is given by the equation $E = h \nu$,where $E$ is energy,$h$ is Planck's constant,and $\nu$ is frequency.
Given $E = 5.0 \times 10^{-5} \ erg$ and $h = 6.626 \times 10^{-27} \ erg \ sec$ (in $CGS$ units).
Rearranging for frequency: $\nu = \frac{E}{h}$.
$\nu = \frac{5.0 \times 10^{-5} \ erg}{6.626 \times 10^{-27} \ erg \ sec} \approx 7.54 \times 10^{21} \ sec^{-1}$.
Thus,the correct option is $C$.

(D) Ionization potential is the energy required to remove an electron from the outermost shell of an isolated gaseous atom.
As the atomic radius increases,the distance between the nucleus and the outermost electron increases.
This leads to a decrease in the electrostatic force of attraction between the nucleus and the valence electron.
Consequently,less energy is required to remove the electron,meaning the ionization potential decreases with an increase in atomic radii.

(C) The Assertion is correct because electron affinity is the energy released when an electron is added to an isolated gaseous atom,whereas electronegativity is the tendency of a bonded atom to attract shared electrons.
The Reason is incorrect because electron affinity is an experimentally measurable quantity (in $kJ \ mol^{-1}$),whereas electronegativity is a relative scale (like the $Pauling$ scale) and is not directly measurable.

(C) $BF_3$ and $B_2H_6$ are non-polar molecules and have a net dipole moment of $0 \ D$.
In $NF_3$,the highly electronegative $F$ atoms pull the electron density away from the $N$ atom,such that the resultant dipole moment of the three $N-F$ bonds opposes the dipole moment of the lone pair on the $N$ atom.
In $NH_3$,the dipole moments of the three $N-H$ bonds are in the same direction as the dipole moment of the lone pair on the $N$ atom.
Therefore,the bond moments add up in $NH_3$,resulting in the highest dipole moment among the given options.

(C) The Assertion is correct because larger molecules have more loosely held electrons,making them more polarizable.
The Reason is incorrect because polarizability is a property of all molecules,regardless of whether they have a permanent dipole moment or not.
Non-polar molecules (like $H_2$,$O_2$,or $CH_4$) also exhibit polarizability due to induced dipoles.

(B) The inversion temperature $(T_i)$ is the temperature below which a gas cools upon Joule-Thomson expansion.
It is calculated using the formula: $T_i = \frac{2a}{bR}$.
Given:
$a = 0.244 \, atm \, L^2 \, mol^{-2}$
$b = 0.027 \, L \, mol^{-1}$
$R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1}$
Substituting these values into the formula:
$T_i = \frac{2 \times 0.244}{0.027 \times 0.0821} \approx \frac{0.488}{0.0022167} \approx 220.15 \, K$.
Rounding to the nearest integer,we get $220 \, K$.

(B) For an ideal gas,the internal energy $(E)$ is a function of temperature only,i.e.,$E = f(T)$.
Since the process is isothermal,the temperature remains constant $(\Delta T = 0)$.
Therefore,the change in internal energy $(\Delta E)$ is zero.

(C) In the reaction $2Cl_{(g)} \to Cl_{2(g)}$,two moles of gaseous atoms are converted into one mole of a gaseous molecule.
Since the number of moles of gas decreases,the disorder of the system decreases,so $\Delta S$ is negative $(-ve)$.
Bond formation is an exothermic process,which releases energy,therefore $\Delta H$ is negative $(-ve)$.

(B) For an adiabatic process,the change in internal energy $\Delta U$ is equal to the work done $w$ on the system. According to the first law of thermodynamics,$\Delta U = q + w$. Since the process is adiabatic,$q = 0$,so $\Delta U = w$.
Given that the gas expands,work is done by the system,so $w = -3000\, J$.
Using the relation $\Delta U = nC_V\Delta T$,where $n = 1\, mol$ and $C_V = 20\, J/(mol\cdot K)$:
$-3000\, J = 1\, mol \times 20\, J/(mol\cdot K) \times (T_f - T_i)$.
Initial temperature $T_i = 27 + 273 = 300\, K$.
$-3000 = 20 \times (T_f - 300)$.
$-150 = T_f - 300$.
$T_f = 300 - 150 = 150\, K$.

(A) The enthalpy of formation $(\Delta_f H^\circ)$ represents the energy released or absorbed during the formation of a compound from its elements.
More negative (lower) enthalpy of formation indicates higher stability of the compound.
Given $\Delta_f H^\circ (HF) = -161 \ kJ$ and $\Delta_f H^\circ (HCl) = -92 \ kJ$.
Since $-161 \ kJ < -92 \ kJ$,$HF$ is more stable than $HCl$.
Therefore,the statement '$HCl$ is more stable than $HF$' is incorrect.

(C) The dissociation of the salt is given by: $AB \rightleftharpoons A^{+} + B^{-}$
The solubility product expression is: $K_{sp} = [A^{+}][B^{-}]$
Precipitation occurs when the ionic product exceeds the solubility product: $[A^{+}][B^{-}] > K_{sp}$
Given $[A^{+}] = 10^{-3} \ M$ and $K_{sp} = 1 \times 10^{-8}$,we substitute these values:
$(10^{-3})[B^{-}] > 1 \times 10^{-8}$
$[B^{-}] > \frac{1 \times 10^{-8}}{10^{-3}}$
$[B^{-}] > 1 \times 10^{-5} \ M$

(A) $1$. For $S_8$: Since it is an elemental form of sulphur,the oxidation state is $0$.
$2$. For $S_2F_2$: Let the oxidation state of $S$ be $x$. The oxidation state of $F$ is $-1$. Thus,$2x + 2(-1) = 0$,which gives $2x = 2$,so $x = +1$.
$3$. For $H_2S$: Let the oxidation state of $S$ be $x$. The oxidation state of $H$ is $+1$. Thus,$2(+1) + x = 0$,which gives $2 + x = 0$,so $x = -2$.
Therefore,the oxidation states are $0, +1, -2$.

(A) In an alkaline medium,$Na_2CO_3$ reacts with $SO_2$ to form $Na_2SO_3$ (sodium sulphite).
The chemical reaction is as follows:
$Na_2CO_3 + SO_2 \rightarrow Na_2SO_3 + CO_2$
When $SO_2$ reacts with caustic alkalies or carbonates in an alkaline medium,it forms sulphites $(Na_2SO_3)$. In an acidic medium,it would form bisulphites $(NaHSO_3)$.

(B) When $SiF_4$ reacts with water,it undergoes hydrolysis to form silicic acid $(H_4SiO_4)$ and hydrofluoric acid $(HF)$.
The balanced chemical equation is:
$SiF_4 + 4H_2O \to H_4SiO_4 + 4HF$

(B) The reaction of an alkene with diborane $(B_2H_6)$ followed by oxidation with alkaline hydrogen peroxide $(H_2O_2/OH^-)$ is known as hydroboration-oxidation.
This reaction results in the anti-Markovnikov addition of water across the double bond,yielding a primary alcohol.
$6R-CH=CH_2 + B_2H_6$ $\rightarrow 2(R-CH_2-CH_2)_3B$ $\xrightarrow{H_2O_2/OH^-} 6R-CH_2-CH_2-OH + 2H_3BO_3$.

(A) Benzene is a non-polar solvent.
Butter consists of fats and oils,which are organic compounds with low polarity.
According to the principle of 'like dissolves like',non-polar solutes dissolve in non-polar solvents.
Therefore,butter dissolves in benzene,making the assertion correct and the reason a valid explanation.

(A) catalyst provides an alternate reaction pathway with a lower activation energy $(E_a)$.
It speeds up both the forward and backward reactions to the same extent.
Since the rate of both reactions increases equally,the equilibrium position remains unchanged,and the equilibrium constant $(K_{eq})$ is not altered.

(C) The stability of carbon tetrahalides depends on the bond dissociation energy of the $C-X$ bond.
As the size of the halogen atom increases from $F$ to $I$,the bond length increases and the bond strength decreases.
Therefore,the stability of carbon tetrahalides decreases in the order: $CF_4 > CCl_4 > CBr_4 > CI_4$.
Thus,$CF_4$ is the most stable halide among the given options.

(B) When excess $6 \ M \ NH_3$ is added to the solution containing $Fe^{3+}$,$Zn^{2+}$,and $Cu^{2+}$:
$1$. $Fe^{3+}$ reacts with $NH_3$ and $H_2O$ to form a brown precipitate of $Fe(OH)_3$,which is insoluble in excess $NH_3$.
$2$. $Cu^{2+}$ reacts with excess $NH_3$ to form a deep blue soluble complex $[Cu(NH_3)_4]^{2+}$.
$3$. $Zn^{2+}$ reacts with excess $NH_3$ to form a colorless soluble complex $[Zn(NH_3)_4]^{2+}$.
Thus,$Fe^{3+}$ is separated as a precipitate,while $Cu^{2+}$ and $Zn^{2+}$ remain in the solution.

(D) Chloropicrin,also known as trichloronitromethane $(CCl_3NO_2)$,is prepared by the reaction of nitric acid $(HNO_3)$ with chloroform $(CHCl_3)$.
The chemical equation is: $CHCl_3 + HNO_3 \rightarrow CCl_3NO_2 + H_2O$.

(D) The reduction in the force of contraction of a muscle after prolonged stimulation is called muscle fatigue.
During strenuous physical activity,the demand for $ATP$ increases,and when the supply of $O_{2}$ is insufficient for aerobic respiration,muscle cells perform anaerobic respiration (glycolysis).
This process results in the accumulation of lactic acid in the muscle tissues.
The accumulation of lactic acid is the primary cause of the sensation of fatigue in the muscles.

(C) When $6 \ M \ NaOH$ is added in excess to a solution containing $Fe^{3+}$,$Zn^{2+}$,and $Cu^{2+}$:
$1$. $Fe^{3+}$ reacts with $OH^-$ to form a reddish-brown precipitate of $Fe(OH)_3$,which is insoluble in excess $NaOH$.
$2$. $Zn^{2+}$ reacts with $OH^-$ to form $Zn(OH)_2$,which dissolves in excess $NaOH$ to form the soluble complex $[Zn(OH)_4]^{2-}$.
$3$. $Cu^{2+}$ reacts with $OH^-$ to form a blue precipitate of $Cu(OH)_2$,which is insoluble in excess $NaOH$.
However,in the context of standard qualitative analysis,$Fe^{3+}$ is separated as $Fe(OH)_3$ from the others. Given the options,$6 \ M \ NaOH$ is the correct reagent to distinguish $Fe^{3+}$ (as a precipitate) from the amphoteric $Zn^{2+}$ (as a soluble complex).

(B) When chloroform is treated with concentrated nitric acid,its hydrogen is replaced by a nitro group.
$CHCl_{3} + HNO_{3} \to CCl_{3}NO_{2} + H_{2}O$
(Chloropicrin)

(B) Given mass of $NaOH$ $(W_{NaOH})$ = $24.5 \ g$
Molar mass of $NaOH$ $(M_{NaOH})$ = $40.0 \ g \ mol^{-1}$
Number of moles of $NaOH$ $(n)$ = $\frac{W_{NaOH}}{M_{NaOH}} = \frac{24.5}{40.0} = 0.6125 \ mol$
Volume of solution $(V)$ = $1 \ L$
Molarity $(M)$ = $\frac{n}{V(L)} = \frac{0.6125 \ mol}{1 \ L} = 0.6125 \ M$

(A) The acidity of metal complexes depends on the charge-to-size ratio (ionic potential) of the central metal ion.
$Al^{3+}$ has a higher charge $(+3)$ and a smaller ionic radius compared to $Mg^{2+}$ $(+2)$.
This results in a higher charge density for $Al^{3+}$,which polarizes the $O-H$ bond in the coordinated water molecules more effectively,facilitating the release of $H^+$ ions.
Therefore,$[Al(H_2O)_6]^{3+}$ is a stronger acid than $[Mg(H_2O)_6]^{2+}$.
Both the Assertion and the Reason are correct,and the Reason correctly explains the Assertion.

(C) The Assertion is true because $CaCl_2$ reacts with alcohols and $NH_3$ to form addition compounds (e.g.,$CaCl_2 \cdot 4C_2H_5OH$ and $CaCl_2 \cdot 8NH_3$),making it unsuitable as a drying agent for these substances.
The Reason is false because $CaCl_2$ is actually a very effective and commonly used desiccant for many other gases and solvents.

(C) The correct answer is $(c)$.
Diethyl ether $(CH_3CH_2OCH_2CH_3)$ is resistant to nucleophilic attack by $OH^{-}$ ions because it lacks an electrophilic carbon atom that can be attacked by a nucleophile.
In contrast,carboxylic acid derivatives such as methyl acetate $(CH_3COOCH_3)$ and acetamide $(CH_3CONH_2)$,as well as nitriles like acetonitrile $(CH_3CN)$,contain a carbonyl or cyano carbon atom which is electrophilic and susceptible to nucleophilic attack.
Ethers are generally inert towards bases because the alkoxide ion $(RO^{-})$ is a poor leaving group.

(B) Weight of solute $(w) = 1 \ g$
Weight of solvent $(W) = 75 \ g$
Boiling point of solution $= 100.114 \ ^oC$
Boiling point of solvent $= 100 \ ^oC$
$\Delta T = 100.114 - 100 = 0.114 \ ^oC$
Molecular weight of solute $(m) = 60.1$
Boiling point elevation constant $(K_b) = ?$
The formula for boiling point elevation is $\Delta T = K_b \times m_{olality} = K_b \times \frac{w \times 1000}{m \times W}$
Rearranging for $K_b$: $K_b = \frac{\Delta T \times m \times W}{w \times 1000}$
$K_b = \frac{0.114 \times 60.1 \times 75}{1 \times 1000}$
$K_b = \frac{513.855}{1000} \approx 0.513 \ ^oC \ kg \ mol^{-1}$

(C) The relationship between standard cell potential $(E_{cell}^o)$ and equilibrium constant $(K_c)$ is given by:
$E_{cell}^o = \frac{0.0591 \ V}{n} \log K_c$
Given:
$E_{cell}^o = 0.295 \ V$
$n = 2$
Substituting the values:
$0.295 = \frac{0.0591}{2} \log K_c$
$\log K_c = \frac{0.295 \times 2}{0.0591} = \frac{0.59}{0.0591} \approx 10$
$K_c = 10^{10}$

(B) Given: Resistance $R = 31.6\,\Omega$,Cell constant $G^* = 0.367\,cm^{-1}$,Molarity $M = 0.5\,M$.
Conductance $C = \frac{1}{R} = \frac{1}{31.6} \approx 0.0316\,S$.
Specific conductance $\kappa = C \times G^* = 0.0316\,S \times 0.367\,cm^{-1} \approx 0.0116\,S\,cm^{-1}$.
Molar conductance $\Lambda_m = \frac{\kappa \times 1000}{M} = \frac{0.0116 \times 1000}{0.5} = \frac{11.6}{0.5} = 23.2\,S\,cm^2\,mol^{-1}$.

(C) The rate of reaction is defined as the change in concentration of a reactant or product per unit time. Thus,the Assertion is correct.
However,the rate of reaction generally depends on the concentration of the reactants. As the reaction proceeds,the concentration of reactants decreases,which leads to a decrease in the rate of reaction. Therefore,the rate of reaction does not remain constant. Thus,the Reason is incorrect.

(D) Volume of gold present in solution = $\frac{\text{Mass of gold}}{\text{Density of gold}} = \frac{1.9 \times 10^{-4} \ g}{19 \ g/cm^3} = 1.0 \times 10^{-5} \ cm^3$.
For spherical particles of gold with radius $r = 10 \ nm = 10 \times 10^{-7} \ cm = 10^{-6} \ cm$,the volume of each particle = $\frac{4}{3} \pi r^3 = \frac{4}{3} \times 3.14 \times (10^{-6} \ cm)^3 = 4.186 \times 10^{-18} \ cm^3$.
Number of gold particles present in $1 \ L$ $(1000 \ cm^3)$ = $\frac{\text{Total volume of gold}}{\text{Volume of one particle}} = \frac{1.0 \times 10^{-5} \ cm^3}{4.186 \times 10^{-18} \ cm^3} \approx 2.389 \times 10^{12} \text{ particles}$.
Since $1 \ L = 1000 \ cm^3 = 10^6 \ mm^3$,the number of particles per $mm^3$ = $\frac{2.389 \times 10^{12}}{10^6} \approx 2.4 \times 10^6 \text{ particles/mm}^3$.

(A) According to the Hardy-Schulze law,the flocculating value is inversely proportional to the valency of the coagulating ion,i.e.,$\text{Flocculating value} \propto \frac{1}{z}$,where $z$ is the valency of the coagulating ion.
$Fe(OH)_3$ is a positively charged sol,so it is coagulated by negative ions.
The negative ions present in the given electrolytes are:
$A) Cl^-$ (valency = $1$)
$B) S^{2-}$ (valency = $2$)
$C) PO_4^{3-}$ (valency = $3$)
$D) SO_4^{2-}$ (valency = $2$)
Since $Cl^-$ has the lowest valency $(z=1)$,it will have the maximum flocculation value. Therefore,$NaCl$ has the maximum flocculation value.

(D) Activated charcoal is a very good adsorbent of gases due to its high surface area and porous structure.
In contrast,anhydrous $CaCl_2$,$Fe(OH)_3$,and conc. $H_2SO_4$ are primarily used as dehydrating agents to remove moisture.

(D) Among the hydrides of group $16$ elements,the boiling point of $H_2O$ is the highest due to strong intermolecular hydrogen bonding.
For the remaining hydrides ($H_2S$,$H_2Se$,$H_2Te$),the boiling point increases with the increase in molecular mass and size,which leads to stronger van der Waals forces.
Thus,the order is $H_2O > H_2Te > H_2Se > H_2S$.

(B) The Assertion is correct because nitrogen lacks $d-$orbitals in its valence shell,preventing it from expanding its octet to form $NX_5$ type compounds.
The Reason is also correct because phosphorus $(2.19)$ has a lower electronegativity than nitrogen $(3.04)$.
However,the inability of nitrogen to form pentahalides is due to the absence of $d-$orbitals,not due to its electronegativity. Therefore,the Reason is not the correct explanation of the Assertion.

(B) Copper sulphide $(CuS)$ is black in colour.
$Cu^{2+}$ is placed in group $II$ of inorganic qualitative analysis.
It is precipitated in the form of sulphide by passing $H_2S$ gas in the presence of dilute $HCl$.
The reaction is: $Cu^{2+} + H_2S \to \underset{\text{black}}{CuS} + 2H^{+}$

(B) The ionic radii of trivalent lanthanides decrease progressively with an increase in atomic number. This phenomenon is known as lanthanide contraction. Therefore,the statement that ionic radii increase with atomic number is false.

(C) The electronic configuration of $Eu^{2+}$ is $[Xe] 4f^7$,which is a half-filled $f$-orbital configuration,providing extra stability.
The electronic configuration of $Ce^{2+}$ is $[Xe] 4f^1 5d^1$,which is less stable.
Therefore,the Assertion is correct.
Cerium salts are generally not used as catalysts in petroleum cracking; rather,zeolites or other specific catalysts are used. Thus,the Reason is incorrect.

(A) The liquation method is used for the purification of metals that have low melting points $(m.p.)$ compared to their impurities.
Lead $(Pb)$,tin $(Sn)$,and bismuth $(Bi)$ are metals with relatively low melting points.
In this process,the crude metal is placed on a sloping hearth and heated.
The metal melts and flows down,leaving behind the infusible impurities.
Thus,both the Assertion and the Reason are correct,and the Reason correctly explains why these metals are purified by liquation.

(D) The magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
Given $\mu = 4.5 \ BM$,we have $\sqrt{n(n+2)} \approx 4.5$,which implies $n(n+2) \approx 20.25$. Solving for $n$,we get $n \approx 4$.
For $Co$ in $+3$ oxidation state $(3d^6)$,if the ligand $L$ is weak,the electrons will be arranged as shown in the diagram,resulting in $4$ unpaired electrons.
Thus,$Co$ must be in $+3$ oxidation state.

(C) $AgCl$ is insoluble in water,conc. $HCl$,and $CCl_4$.
It dissolves in $NH_4OH$ solution due to the formation of a soluble complex salt.
The reaction is: $AgCl + 2NH_4OH \to [Ag(NH_3)_2]Cl + 2H_2O$.
The product formed is Diamminesilver$(I)$ chloride.

(A) The complex is $Hg[Co(CNS)_4]$.
In this complex,the cation is mercury $(Hg^{2+})$ and the anion is the coordination entity $[Co(CNS)_4]^{2-}$.
The ligand $CNS^-$ is named as thiocyanato-$S$ or simply thiocyanato.
Since there are $4$ such ligands,it is named as tetrathiocyanato.
The central metal atom is cobalt,and since it is in an anionic complex,it is named as cobaltate.
The oxidation state of $Co$ is calculated as: $x + 4(-1) = -2$,which gives $x = +2$.
Thus,the $IUPAC$ name is mercury tetrathiocyanatocobaltate $(II)$.

(D) The statement in option $D$ is incorrect.
In metal carbonyls,the metal-carbon bond possesses both $\sigma$ and $\pi$ character.
Option $A$ is correct: The $\sigma$-bond is formed by the donation of a lone pair of electrons from the carbonyl carbon to the vacant orbital of the metal.
Option $B$ is correct: The $\pi$-bond is formed by the back-donation of electrons from a filled $d$-orbital of the metal into the vacant antibonding $\pi^*$-orbital of the carbon monoxide ligand.
Option $C$ is correct: This back-donation creates a synergic effect that strengthens the $M-C$ bond while weakening the $C-O$ bond.

(C) The Assertion is correct because a chelating ligand requires donor atoms positioned to form stable,strain-free rings (typically $5$ or $6$ membered) with the metal ion.
The Reason is incorrect because hydrazine $(H_2N-NH_2)$ acts as a monodentate ligand. Coordination by hydrazine would result in a $3$-membered ring,which is highly unstable due to significant angle strain,thus it does not act as a chelating ligand.

(C) The Assertion is true because $S_{N}2$ reactions involve the attack of a nucleophile from the side opposite to the leaving group,leading to the Walden inversion (inversion of configuration).
However,the Reason is false. Aryl halides are extremely unreactive towards nucleophilic substitution reactions under ordinary conditions due to the partial double bond character of the $C-X$ bond caused by resonance. Therefore,they do not undergo $S_{N}2$ reactions with $KOH$ to form alcohols.

(A) The Assertion is correct because the presence of an electron-withdrawing $-NO_2$ group at the $para-$position stabilizes the carbanion intermediate formed during nucleophilic aromatic substitution.
The Reason is also correct. Chlorobenzene typically requires harsh conditions and proceeds via an elimination-addition mechanism (benzyne intermediate) due to the lack of electron-withdrawing groups. In contrast,$4-$nitrochlorobenzene undergoes nucleophilic aromatic substitution via an addition-elimination mechanism (Meisenheimer complex) because the $-NO_2$ group strongly activates the ring towards nucleophilic attack.
Since the mechanism described in the Reason explains why $4-$nitrochlorobenzene is more reactive,the Reason is the correct explanation of the Assertion.

(C) The Reimer-Tiemann reaction of phenol with $CHCl_3$ in $NaOH$ proceeds through the formation of a dichlorocarbene $(:CCl_2)$ intermediate to yield salicylaldehyde. However,when phenol reacts with $CCl_4$ in $NaOH$,the reaction proceeds via a different mechanism involving a trichloromethyl anion intermediate to yield salicylic acid. Therefore,the assertion is correct,but the reason is incorrect because dichlorocarbene is not the intermediate when $CCl_4$ is used.

(A) Phenol is an effective germicide.
Since phenyl is a derivative of phenol,it inherits the germicidal properties of phenol.
Therefore,both the Assertion and the Reason are correct,and the Reason is the correct explanation for the Assertion.

(B) $Assertion$ is correct because $tert$-butyl bromide is a $3^\circ$ alkyl halide,which undergoes $E2$ elimination rather than $S_N2$ substitution when treated with a strong base like sodium methoxide. The major product is isobutylene $(CH_3-C(CH_3)=CH_2)$.
$Reason$ is also correct because sodium methoxide $(CH_3ONa)$ is indeed a strong nucleophile,but its primary role in this specific reaction with a bulky $3^\circ$ substrate is that of a strong base.
However,the reason for the failure of the reaction is the steric hindrance of the $3^\circ$ alkyl halide,not just the nucleophilicity of the methoxide ion. Thus,the Reason is not the correct explanation for the Assertion.

(C) Aldol condensation is a reaction given by aldehydes or ketones that contain at least one $\alpha-$ hydrogen atom.
In the molecule $(CH_3)_3CCHO$ ($2$,$2$-dimethylpropanal),the carbonyl carbon is attached to a tertiary butyl group $(CH_3)_3C-$.
There is no hydrogen atom attached to the $\alpha-$ carbon (the carbon adjacent to the carbonyl group).
Due to the absence of $\alpha-$ hydrogen atoms,it cannot form an enolate ion,and therefore,it does not undergo aldol condensation.

(A) The decarboxylation of $\beta$-keto carboxylic acids occurs via a cyclic transition state involving the carbonyl oxygen of the ketone and the hydrogen of the carboxylic acid group.
This process releases $CO_2$ and initially forms an enol intermediate.
The enol is unstable and rapidly undergoes tautomerization to form the more stable ketone.
Therefore,both the Assertion and the Reason are correct,and the Reason provides the correct explanation for the decarboxylation process.

(B) The hydrolysis of urea $(NH_2CONH_2)$ in the presence of an acid,base,or the enzyme urease produces ammonia and carbon dioxide.
$NH_2CONH_2 + H_2O \rightarrow 2NH_3 + CO_2$
Subsequently,the carbon dioxide reacts with water to form carbonic acid:
$CO_2 + H_2O \rightarrow H_2CO_3$ (Carbonic acid)
Therefore,the final product of complete hydrolysis is carbonic acid.

(C) The Assertion is correct. Aniline is highly reactive and prone to oxidation; therefore,the $-NH_2$ group is protected by acetylation with acetic anhydride to form acetanilide.
The Reason is incorrect. The acetyl group $(-COCH_3)$ is an electron-withdrawing group due to resonance with the lone pair of nitrogen,which decreases the electron density on the benzene ring,thereby moderating the reactivity of the ring and preventing the formation of oxidation products.

(C) Mifepristone is a synthetic steroid that acts as an antagonist to progesterone and is commonly used as a 'morning after pill' to prevent pregnancy after unprotected intercourse.

(C) The Assertion is true because Penicillin is a well-known antibiotic used to treat bacterial infections.
The Reason is false because drugs that act on the central nervous system to reduce anxiety are known as tranquilizers,not antibiotics.

(C) Nessler's reagent,$K_2[HgI_4]$,is used for the detection and quantitative determination of ammonia (or $NH_4^+$) in solution.
It reacts with ammonia to form a brown precipitate of $Hg_2NI \cdot H_2O$ (often represented as $IHg-NH_2-Hg-O-I$ or similar structures),known as the iodide of Millon's base.
The reaction is: $NH_4^+ + 2[HgI_4]^{2-} + 4OH^- \rightarrow Hg_2NI \cdot H_2O + 7I^- + 3H_2O$.