AIPMT 1997 Chemistry Question Paper with Answer and Solution

(C) According to Kepler's third law of planetary motion,the square of the period of revolution $T$ is proportional to the cube of the semi-major axis $r$ of the orbit: $T^2 \propto r^3$.
Given that the period of revolution of planet $A$ is $8$ times that of planet $B$,we have $T_A = 8 T_B$,or $\frac{T_A}{T_B} = 8$.
Using the relation $\frac{T_A}{T_B} = \left( \frac{r_A}{r_B} \right)^{3/2}$,we substitute the given values:
$8 = \left( \frac{r_A}{r_B} \right)^{3/2}$.
To solve for $\frac{r_A}{r_B}$,we raise both sides to the power of $2/3$:
$\left( 8 \right)^{2/3} = \frac{r_A}{r_B}$.
Since $8 = 2^3$,we have $(2^3)^{2/3} = 2^2 = 4$.
Therefore,$\frac{r_A}{r_B} = 4$,which means the distance of planet $A$ from the sun is $4$ times greater than that of planet $B$.

(B) $CO$ and $CN^{-}$ are isoelectronic species.
Total electrons in $CO = 6 + 8 = 14$.
Total electrons in $CN^{-} = 6 + 7 + 1 = 14$.
Since both have $14$ electrons,they are isoelectronic.

(D) In graphite,each carbon atom is $sp^2$ hybridized and uses $3$ of its $4$ valence electrons to form covalent bonds with $3$ other carbon atoms in a hexagonal layer.
The fourth electron remains in an unhybridized $p$-orbital,which overlaps with adjacent $p$-orbitals to form a delocalized $\pi$-electron cloud.
Therefore,these electrons are spread out between the layers of the structure,allowing graphite to conduct electricity.

(B) The electronic configuration of $N_2$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. Bond order = $(10-4)/2 = 3$.
For $N_2^-$,the extra electron enters the $\pi^* 2p$ orbital. Bond order = $(10-5)/2 = 2.5$. Since bond order decreases,the $N-N$ bond weakens. $N_2^-$ has one unpaired electron,so it is paramagnetic.
The electronic configuration of $O_2$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. Bond order = $(10-6)/2 = 2$.
For $O_2^-$,the extra electron enters the $\pi^* 2p$ orbital. Bond order = $(10-7)/2 = 1.5$. Since bond order decreases,the bond length increases.
Therefore,the statement that the $O-O$ bond order increases is wrong.

(A) . Water is denser than ice because of the hydrogen bonding interactions and the open cage-like structure of ice.

(D) The electronic configuration of $N_2$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$ (Bond order = $3$).
When $N_2$ is converted to $N_2^+$,an electron is removed from the bonding molecular orbital $(\sigma 2p_z)$,resulting in a bond order of $2.5$. Thus,the $N-N$ bond weakens.
$N_2^+$ has $13$ electrons,so its configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^1$. Since it has one unpaired electron,it is paramagnetic.
The electronic configuration of $O_2$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$ (Bond order = $2$).
When $O_2$ is converted to $O_2^+$,an electron is removed from the antibonding molecular orbital $(\pi^*)$,increasing the bond order to $2.5$ and decreasing paramagnetism.
Therefore,the statement that $N_2^+$ becomes diamagnetic is wrong.

(D) The Henderson-Hasselbalch equation is given by $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
For a buffer to be most efficient,the $pH$ of the solution should be as close as possible to the $pK_a$ of the weak acid used,i.e.,$pH \approx pK_a$.
Given the target $pH = 3.58$,we look for an acid with a $pK_a$ value closest to $3.58$.
Comparing the given options:
$(A)$ $pK_a = 3.98$
$(B)$ $pK_a = 4.41$
$(C)$ $pK_a = 2.97$
$(D)$ $pK_a = 3.58$
Since Acetoacetic acid has a $pK_a$ of $3.58$,which is exactly equal to the desired $pH$,it provides the maximum buffer capacity at this $pH$.

(B) Since the hydride ion $(H^{-})$ is a much stronger base than the hydroxide ion $(OH^{-})$,it will readily abstract a proton $(H^{+})$ from water.
The reaction is as follows:
$H^{-}_{(aq)} + H_2O_{(l)} \to H_2(g) + OH^{-}_{(aq)}$
In this reaction,$H^{-}$ acts as a Bronsted-Lowry base and $H_2O$ acts as a Bronsted-Lowry acid.

(D) redox reaction involves a change in the oxidation state of elements.
In option $D$,the reaction is $N_2 + O_2 \to 2NO$.
Here,the oxidation number $(O.N.)$ of $N$ increases from $0$ in $N_2$ to $+2$ in $NO$ (oxidation),and the $O.N.$ of $O$ decreases from $0$ in $O_2$ to $-2$ in $NO$ (reduction).
Therefore,this is a redox reaction.

(D) The size of an atom or ion depends on the number of electrons and the effective nuclear charge.
For the same element,the size order is $I^{-} > I > I^{+}$.
$I^{-}$ has $54$ electrons,$I$ has $53$ electrons,and $I^{+}$ has $52$ electrons.
As the number of electrons decreases for the same nuclear charge,the effective nuclear charge per electron increases,leading to a smaller size.
Therefore,the correct order is $I^{-} > I > I^{+}$.

(C) Due to lanthanide contraction,the atomic radii of $Zr$ and $Hf$ are nearly identical.
Lanthanide contraction is explained by the poor shielding effect of $4f$ electrons.
In multi-electron atoms,inner electrons shield outer electrons from the nuclear charge.
The shielding efficiency follows the order $s > p > d > f$.
Because the $4f$ subshell provides poor shielding,the effective nuclear charge experienced by the outer electrons increases,leading to a smaller atomic size.
In the case of $Hf$ $(Z = 72)$,the filling of the $4f$ subshell results in a smaller size than expected,making it comparable to $Zr$ $(Z = 40)$.

(B) Since $H^{-}$ is a stronger base than $OH^{-}$,it will readily accept a proton from water.
$H^{-}$ acts as a Bronsted-Lowry base and $H_2O$ acts as a Bronsted-Lowry acid.
The reaction is: $H^{-}_{(aq)} + H_2O_{(l)} \to OH^{-}_{(aq)} + H_{2(g)}$
This is an acid-base reaction where the stronger base $(H^{-})$ reacts with the acid $(H_2O)$ to form a weaker base $(OH^{-})$ and a weaker acid $(H_2)$.

(A) As we move from top to bottom in a group,the dipole moment of hydrides of the elements of that particular group decreases due to an increase in the size of the central atom and a decrease in electronegativity.
$NH_3$ has the highest dipole moment because nitrogen is the most electronegative element in the group,leading to a greater polarity of the $N-H$ bond.
The order of dipole moment for the given hydrides is:
$NH_3 > PH_3 > AsH_3 > SbH_3$

(C) The reactivity of benzene derivatives towards electrophilic substitution depends on the electron density of the benzene ring.
$I$ (Anisole) has a $-OCH_3$ group,which is an electron-donating group $(EDG)$ via resonance ($+M$ effect),significantly increasing the electron density of the benzene ring.
$II$ (Benzene) has no substituents.
$III$ (Nitrobenzene) has a $-NO_2$ group,which is a strong electron-withdrawing group $(EWG)$ via both inductive $(-I)$ and resonance $(-M)$ effects,significantly decreasing the electron density of the benzene ring.
Since electrophiles are electron-deficient species,they react faster with rings having higher electron density.
Therefore,the order of reactivity is $I > II > III$.

(D) Tautomerism requires the presence of at least one $\alpha$-hydrogen atom attached to a carbon atom adjacent to a functional group that can undergo tautomeric rearrangement.
In the case of nitro compounds $(RCH_2NO_2)$,the $\alpha$-hydrogen atom is acidic and can migrate to the oxygen atom of the nitro group to form an aci-nitro form.
The equilibrium is represented as:
$R-CH_2-N^+(O)O^- \rightleftharpoons R-CH=N^+(OH)O^-$
Since $(CH_3)_3CNO$,$(CH_3)_2NH$,and $R_3CNO_2$ lack an $\alpha$-hydrogen atom,they do not exhibit tautomerism.
Therefore,the correct option is $D$.

(A) The correct answer is $A$. Branched chain hydrocarbons are more desirable in commercial gasoline because they have higher octane ratings compared to their straight-chain isomers,which helps in reducing engine knocking.

(A) The reaction $2RCOOK + 2H_2O \xrightarrow{\text{Electrolysis}} R-R + 2CO_2 + 2KOH + H_2$ is known as the Kolbe's electrolysis method.
This method is used for the preparation of alkanes (hydrocarbons) with an even number of carbon atoms.
It proceeds via a free radical mechanism and provides good yields of the symmetric alkane product $(R-R)$.

(B) The reaction of ethene $(CH_2=CH_2)$ with hypochlorous acid $(HOCl)$ follows an electrophilic addition mechanism to form ethylene chlorohydrin $(Cl-CH_2-CH_2-OH)$,which is molecule $M$.
Ethylene chlorohydrin then undergoes hydrolysis in the presence of a mild base like aqueous sodium bicarbonate $(aq. NaHCO_3)$,which acts as reagent $R$,to produce ethylene glycol $(HO-CH_2-CH_2-OH)$.

(D) In ethyne $(CH \equiv CH)$,the carbon-carbon triple bond consists of $1$ $\sigma$ bond and $2$ $\pi$ bonds.
The cylindrical shape of the alkyne molecule arises due to the presence of these two $\pi$ bonds,which are formed by the sideways overlap of two sets of $p$-orbitals perpendicular to each other.

(C) Strength = $\text{Normality} \times \text{Equivalent Weight (EW)}$ of $H_2O_2$
$= 1.5 \ N \times 17 \ g \ eq^{-1} = 25.5 \ g \ L^{-1}$
Decomposition reaction: $2H_2O_2 \rightarrow 2H_2O + O_2$
$2 \times 34 \ g = 68 \ g$ of $H_2O_2$ produces $22.4 \ L$ of $O_2$ at $STP$.
Volume strength = $\frac{11.2 \times \text{Normality}}{2} = 5.6 \times \text{Normality}$
$= 5.6 \times 1.5 = 8.4$
Thus,the volume strength of $1.5 \ N \ H_2O_2$ solution is $8.4$.

(B) The correct answer is $B$.
Opiate narcotics are drugs that bind to specific opioid receptors in our central nervous system and gastrointestinal tract.
They act as central nervous system depressants,which help in relieving intense pain and inducing sleep.
Morphine is a well-known opiate extracted from the latex of the poppy plant,$Papaver$ $somniferum$.
Other examples include heroin,pethidine,and methadone.

(A) The dipole moment of hydrides of group $15$ elements depends on the electronegativity difference between the central atom and hydrogen,as well as the lone pair contribution.
As we move down the group from $N$ to $Sb$,the electronegativity of the central atom decreases,which reduces the polarity of the $M-H$ bond.
Additionally,the bond angle decreases down the group,which also affects the resultant dipole moment.
$NH_3$ has the highest electronegativity difference and a significant lone pair contribution,resulting in the highest dipole moment of $1.47 \ D$.

(B) Since the $H^-$ ion is a stronger base than the $OH^-$ ion,it acts as a Bronsted-Lowry base and accepts a proton $(H^+)$ from water.
The reaction is: $H^-_{(aq)} + H_2O_{(l)} \to OH^-_{(aq)} + H_{2(g)}$

(A) For $CuS$ ($1:1$ type salt): $K_{sp} = s^2 \implies s = \sqrt{K_{sp}} = \sqrt{10^{-31}} = 10^{-15.5}$.
For $Ag_2S$ ($2:1$ type salt): $K_{sp} = 4s^3 \implies s = (K_{sp}/4)^{1/3} = (10^{-44}/4)^{1/3} \approx 0.63 \times 10^{-14.6} \approx 10^{-15.2}$.
For $HgS$ ($1:1$ type salt): $K_{sp} = s^2 \implies s = \sqrt{K_{sp}} = \sqrt{10^{-54}} = 10^{-27}$.
Comparing the values: $10^{-15.2} > 10^{-15.5} > 10^{-27}$.
Therefore,the order of solubility is $Ag_2S > CuS > HgS$.

(D) Anti or completely staggered.
The anti-staggered conformation is the most stable because it minimizes the steric repulsion (van der Waals strain) between the two bulky methyl groups,which are placed at a dihedral angle of $180^{\circ}$.

(A) The human kidney consists of approximately one million complex tubular structures called nephrons.
Each nephron is the structural and functional unit of the kidney.
It consists of two major parts: the glomerulus and the renal tubule.
Nephridia are excretory organs found in annelids like earthworms,not humans.
Pyramids are conical masses in the renal medulla,and Henle's loop is just a part of the nephron.

(C) Silverfish $(Lepisma)$,scorpion $(Palamnaeus)$,crab $(Cancer)$,and honeybee $(Apis)$ all belong to the phylum $Arthropoda$.
The most characteristic feature of all organisms in the phylum $Arthropoda$ is the presence of jointed appendages ($arthros$ = joint,$poda$ = foot).
Therefore,the common feature among these organisms is the presence of jointed appendages.

(B) Given: Voltage drop across the diode $(V_D) = 0.5 \, V$. Maximum power rating of the diode $(P) = 100 \, mW = 100 \times 10^{-3} \, W$. Source voltage $(V_S) = 1.5 \, V$.
To find the maximum current $(I_{max})$ that the diode can handle:
$P = V_D \times I_{max}$
$I_{max} = \frac{P}{V_D} = \frac{100 \times 10^{-3} \, W}{0.5 \, V} = 0.2 \, A$.
Applying Kirchhoff's Voltage Law $(KVL)$ to the circuit:
$V_S = I_{max} \times R + V_D$
$1.5 \, V = (0.2 \, A) \times R + 0.5 \, V$
$1.5 - 0.5 = 0.2 \times R$
$1.0 = 0.2 \times R$
$R = \frac{1.0}{0.2} = 5 \, \Omega$.
Therefore,the value of the series resistor $R$ is $5 \, \Omega$.

(A) The dipole moment of the hydrides of group $15$ elements decreases down the group.
This is because the electronegativity of the central atom decreases from $N$ to $Bi$.
Since $N$ is the most electronegative element among them,$NH_3$ has the highest dipole moment.

(B) The maximum power rating of the diode is $P = 100 \, mW = 0.1 \, W$.
The voltage drop across the diode is $V_d = 0.5 \, V$.
The maximum current $I$ that can flow through the diode is given by $I = \frac{P}{V_d} = \frac{0.1 \, W}{0.5 \, V} = 0.2 \, A$.
According to Kirchhoff's voltage law for the circuit,the total voltage of the battery $V_b = 1.5 \, V$ is equal to the sum of the voltage drop across the resistor $V_R$ and the voltage drop across the diode $V_d$.
$V_b = V_R + V_d$
$1.5 \, V = I \cdot R + 0.5 \, V$
$1.5 \, V = (0.2 \, A) \cdot R + 0.5 \, V$
$1.0 \, V = 0.2 \cdot R$
$R = \frac{1.0}{0.2} \, \Omega = 5 \, \Omega$.

(C) Due to lanthanoid contraction,the atomic radii of $Zr$ $(160 \ pm)$ and $Hf$ $(159 \ pm)$ are almost the same. This occurs because the $4f$ orbitals are filled before the $5d$ series,and the poor shielding effect of $4f$ electrons leads to a greater effective nuclear charge,which offsets the expected increase in size down the group.

(B) Since $H^{-}$ is a stronger base than $OH^{-}$,it will abstract a proton $(H^{+})$ from water.
The reaction is: $H^{-}_{(aq)} + H_2O_{(l)} \to OH^{-}_{(aq)} + H_{2(g)}$.
In this reaction,$H^{-}$ acts as a Bronsted-Lowry base and water acts as an acid.

(D) The electronic configuration of $N_2$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. Bond order = $(10-4)/2 = 3$.
In $N_2^+$,one electron is removed from the bonding molecular orbital $(\sigma 2p_z)$,so the configuration becomes $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^1$. Bond order = $(9-4)/2 = 2.5$. Since the bond order decreases,the $N-N$ bond weakens and $N_2^+$ is paramagnetic.
For $O_2$,the configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. Bond order = $(10-6)/2 = 2$.
In $O_2^+$,one electron is removed from the antibonding molecular orbital $(\pi^*)$,so the configuration becomes $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Bond order = $(10-5)/2 = 2.5$. The bond order increases,and paramagnetism decreases as the number of unpaired electrons reduces from $2$ to $1$.
Therefore,the statement that $N_2^+$ becomes diamagnetic is wrong.

(A) Electrophilic substitution reaction is favored by the presence of electron-donating groups on the benzene ring,which increase the electron density and activate the ring. Conversely,electron-withdrawing groups decrease the electron density and deactivate the ring.
$I$: Anisole $(C_6H_5OCH_3)$ contains a $-OCH_3$ group,which is a strong electron-donating group due to the resonance effect ($+R$ effect),making it highly reactive.
$II$: Benzene $(C_6H_6)$ has no substituent,serving as the reference.
$III$: Nitrobenzene $(C_6H_5NO_2)$ contains a $-NO_2$ group,which is a strong electron-withdrawing group due to the $-R$ and $-I$ effects,making it the least reactive.
Therefore,the order of reactivity is $I > II > III$.

(D) The centre of mass of a system of particles is defined by the position vector $\vec{R} = \frac{\sum m_i \vec{r}_i}{\sum m_i}$.
From this formula,it is evident that the centre of mass depends on the individual masses $(m_i)$ and their respective position vectors $(\vec{r}_i)$.
The relative distance between particles is inherently determined by their positions.
However,the centre of mass is a geometric property of the distribution of mass and is independent of any external or internal forces acting on the system.

(A) The position of the centre of mass $(R_{cm})$ of a system of particles is defined by the formula:
$R_{cm} = \frac{\sum m_i r_i}{\sum m_i}$
From this formula,it is evident that the centre of mass depends on the masses $(m_i)$ of the particles and their respective positions $(r_i)$.
The internal forces acting between particles in a system cancel each other out due to Newton's third law,and external forces affect the motion of the centre of mass,but the definition of the centre of mass itself is purely a geometric property based on mass distribution.
Therefore,the centre of mass does not depend on the forces acting on the particles.

(B) The electronic configuration of $La$ $(Z=57)$ is $[Xe] 5d^1 6s^2$.
Following this,the $4f$ orbitals are filled.
Europium ($Eu$,$Z=63$) has a stable half-filled $4f$ subshell with the configuration $[Xe] 4f^7 6s^2$.
For Gadolinium ($Gd$,$Z=64$),the next electron enters the $5d$ orbital instead of the $4f$ orbital to maintain the stability of the half-filled $4f^7$ configuration.
Therefore,the electronic configuration of $Gd$ $(Z=64)$ is $[Xe] 4f^7 5d^1 6s^2$.

(A) The correct order for ionic radii of $M^{3+}$ ions is $Sc^{3+} > Cr^{3+} > Mn^{3+} > Fe^{3+}$.
Option $A$ is incorrect because the given order $Sc^{3+} > Cr^{3+} > Fe^{3+} > Mn^{3+}$ does not match the actual trend.
Option $B$ is correct as density generally increases across the $3d$ series.
Option $C$ is correct as ionic radii decrease with increasing atomic number for $M^{2+}$ ions,though there are minor variations.
Option $D$ is correct as basic nature depends on the oxidation state and electronegativity of the metal.

(D) Calcium is obtained by the electrolysis of a fused mixture of calcium chloride $(CaCl_2)$ with about $16 \%$ calcium fluoride $(CaF_2)$ in a graphite crucible.
$CaF_2$ is added to lower the melting point of the mixture and to increase the electrical conductivity.

(A) Hypophosphorus acid is $H_3PO_2$.
In its structure,the central phosphorus atom is bonded to one oxygen atom by a double bond $(P=O)$,one hydroxyl group $(-OH)$,and two hydrogen atoms directly attached to the phosphorus atom $(P-H)$.
This structure is represented by option $A$.

(C) The reaction shows the nucleophilic substitution of a bromide ion by a hydroxide ion.
The product shows an inversion of configuration at the chiral carbon atom,which is a characteristic feature of the $S_N2$ mechanism.
In an $S_N2$ reaction (Bimolecular Nucleophilic Substitution),the nucleophile attacks from the side opposite to the leaving group,leading to Walden inversion.

(A) The freezing point depression is given by the formula $\Delta T_f = i \times K_f \times m$.
Since the molality $(m)$ and the solvent $(K_f)$ are the same for all solutions,the freezing point depends on the van't Hoff factor $(i)$.
$Al_2(SO_4)_3$ dissociates as $2Al^{3+} + 3SO_4^{2-}$,giving $i = 5$.
$C_5H_{10}O_5$ is a non-electrolyte,$i = 1$.
$KI$ dissociates as $K^+ + I^-$,giving $i = 2$.
$C_{12}H_{22}O_{11}$ is a non-electrolyte,$i = 1$.
Since $Al_2(SO_4)_3$ has the highest van't Hoff factor $(i=5)$,it will show the maximum depression in freezing point,resulting in the lowest freezing point.

(B) The coordination number of $8$ is characteristic of the body-centred cubic $(BCC)$ lattice structure.
In a $BCC$ unit cell,each atom at the center is surrounded by $8$ nearest neighbors,and each corner atom is also surrounded by $8$ nearest neighbors.
Therefore,the crystal class is body-centred cube.

(C) The number of atoms in $1 \ g$ of $_{92}U^{235}$ is given by $\frac{1 \ g}{235 \ g/mol} \times 6.023 \times 10^{23} \ \text{atoms/mol} \approx 2.563 \times 10^{21} \ \text{atoms}$.
Energy released per atom is $3.20 \times 10^{-11} \ J$.
Total energy released $= (2.563 \times 10^{21} \ \text{atoms}) \times (3.20 \times 10^{-11} \ J/\text{atom}) = 8.2016 \times 10^{10} \ J$.
Converting to $kJ$: $8.2016 \times 10^{10} \ J = 8.2016 \times 10^7 \ kJ \approx 8.21 \times 10^7 \ kJ$.

(A) Let the rate law be $Rate = k [A]^x [B_2]^y$.
From experiment $(1)$ and $(2)$,$[A]_0$ is constant $(0.50 \ M)$ and $[B_2]_0$ is doubled ($0.50 \ M$ to $1.00 \ M$). The rate increases from $1.6 \times 10^{-4}$ to $3.2 \times 10^{-4}$ (doubled).
Therefore,$2^y = 2$,which implies $y = 1$.
From experiment $(2)$ and $(3)$,$[B_2]_0$ is constant $(1.00 \ M)$ and $[A]_0$ is doubled ($0.50 \ M$ to $1.00 \ M$). The rate remains $3.2 \times 10^{-4}$ (no change).
Therefore,$2^x = 1$,which implies $x = 0$.
Substituting the values of $x$ and $y$ into the rate law,we get $Rate = k [A]^0 [B_2]^1 = k [B_2]$.

(A) For a general reaction $aA + bB \rightarrow cC + dD$,the rate of reaction is given by:
Rate $= -\frac{1}{a} \frac{\Delta [A]}{\Delta t} = -\frac{1}{b} \frac{\Delta [B]}{\Delta t} = \frac{1}{c} \frac{\Delta [C]}{\Delta t} = \frac{1}{d} \frac{\Delta [D]}{\Delta t}$.
For the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,the stoichiometric coefficients are $1, 1, 2$ respectively.
Thus,the rate of reaction is expressed as:
Rate $= -\frac{\Delta [H_2]}{\Delta t} = -\frac{\Delta [I_2]}{\Delta t} = \frac{1}{2} \frac{\Delta [HI]}{\Delta t}$.

(B) According to Kohlrausch's law of independent migration of ions:
$\Lambda_m^o(CH_3COOH) = \Lambda_m^o(CH_3COONa) + \Lambda_m^o(HCl) - \Lambda_m^o(NaCl)$
Substituting the given values:
$\Lambda_m^o(CH_3COOH) = 91 + 426.16 - 126.45$
$\Lambda_m^o(CH_3COOH) = 390.71 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.

(B) For the given cell reaction,the number of electrons transferred is $n = 2$.
Using the Nernst equation at equilibrium: $E_{cell}^o = \frac{0.0591}{n} \log K_c$.
Substituting the values: $1.10 = \frac{0.0591}{2} \log K_c$.
$\log K_c = \frac{1.10 \times 2}{0.0591} \approx 37.22$.
Therefore,$K_c = 10^{37.22} \approx 10^{37}$.

(D) The ability of an ion to bring about coagulation of a given colloid depends upon both the magnitude and sign of its charge. According to the $Hardy-Schulze$ rule,greater the valence of the flocculating ion added,the greater is its power to cause precipitation.

(C) Due to lanthanide contraction,the atomic radii of $Zr$ and $Hf$ are nearly identical.
Lanthanide contraction is explained by the poor shielding effect of $4f$ electrons.
In multi-electron atoms,inner electrons shield outer electrons from the nuclear charge.
The shielding efficiency follows the order $s > p > d > f$.
Because the $4f$ subshell has a very poor shielding effect,the outer electrons experience a higher effective nuclear charge,leading to a decrease in atomic size.
This effect causes elements of the $5d$ series (like $Hf$) to have radii very similar to their counterparts in the $4d$ series (like $Zr$).

(A) When potassium dichromate $(K_2Cr_2O_7)$ reacts with an aqueous solution of sodium hydroxide $(NaOH)$,it forms potassium chromate and sodium chromate.
The chemical equation is: $K_2Cr_2O_7 + 2NaOH \rightarrow K_2CrO_4 + Na_2CrO_4 + H_2O$.
In this reaction,the orange dichromate ion $(Cr_2O_7^{2-})$ is converted into the yellow chromate ion $(CrO_4^{2-})$.

(B) In this coordination complex,the central metal ion is $Cu^{2+}$.
There are $2$ chloro ligands $(Cl^-)$ and $2$ urea ligands $(O=C(NH_2)_2)$.
According to $IUPAC$ nomenclature rules for coordination compounds,ligands are named in alphabetical order.
'Chloro' comes before 'urea'.
Since there are $2$ urea ligands,the prefix 'bis' is used for the ligand name 'urea'.
Thus,the formula is written as $[CuCl_2\{O=C(NH_2)_2\}_2]$.
Therefore,the correct option is $B$.

(A) The given complex is $[Co(NO_2)_2(NH_3)_2]$. Assuming a square planar geometry (as it is a $4$-coordinate complex),it follows the form $[MA_2B_2]$.
For a square planar complex of the type $[MA_2B_2]$,there are $2$ possible geometrical isomers: $cis$ and $trans$.
In the $cis$ isomer,the identical ligands are adjacent to each other.
In the $trans$ isomer,the identical ligands are opposite to each other.
Therefore,the total number of geometrical isomers is $2$.

(D) The presence of electron-withdrawing groups like $-NO_2$ at the ortho and para positions relative to the halogen atom significantly activates the benzene ring toward nucleophilic aromatic substitution.
This occurs because the $-NO_2$ group stabilizes the carbanion intermediate (Meisenheimer complex) formed during the reaction by withdrawing electron density through both inductive and resonance effects.
In $2,4-$dinitrochlorobenzene,the two $-NO_2$ groups are located at the ortho and para positions,which effectively delocalize the negative charge of the intermediate,thereby lowering the activation energy for the substitution reaction.

(C) Primary alcohols upon oxidation yield aldehydes,whereas secondary alcohols upon oxidation yield ketones.
$R-CH(OH)-R_1 \xrightarrow{[O]} R-C(=O)-R_1 + H_2O$.
Hydrolysis of esters yields carboxylic acids and alcohols.
Reaction of acid halides with alcohols yields esters.

(B) Phenylmethanol (benzyl alcohol) can be prepared from benzaldehyde by reduction. While common laboratory reagents for reducing aldehydes to alcohols include $NaBH_4$ or $LiAlH_4$,the reaction of benzaldehyde with $Zn/HCl$ (Clemmensen reduction conditions) typically reduces the carbonyl group to a methylene group (forming toluene). However,in the context of this specific question,$Zn/HCl$ is provided as the intended reducing agent to produce the alcohol. The reaction is: $C_6H_5CHO + 2[H] \xrightarrow{Zn/HCl} C_6H_5CH_2OH$.

(B) The correct answer is $B$. $N$-acetyl-para-aminophenol (also known as Paracetamol) is a non-narcotic analgesic.
Non-narcotic analgesics are drugs that relieve pain without causing addiction or mood modification.
Morphine is a narcotic analgesic,while Diazepam is a tranquilizer and Tetrahydrocannabinol is a psychoactive substance.

(B) The reaction between a primary amine,chloroform,and alcoholic $KOH$ is known as the carbylamine reaction.
The chemical equation is: $CHCl_3 + C_2H_5NH_2 + 3KOH \to C_2H_5NC + 3KCl + 3H_2O$.
In this reaction,ethyl amine $(C_2H_5NH_2)$ reacts with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ to produce ethyl isocyanide $(C_2H_5NC)$,which has a characteristic foul smell.

(D) Teflon (Polytetrafluoroethylene) has great chemical inertness and high thermal stability,hence it is used for making non-stick utensils.
For this purpose,a thin layer of Teflon is coated on the inner side of the vessel.

(D) The primary function of enzymes in living systems is to catalyse biochemical reactions.
Enzymes are highly substrate-specific and accelerate reactions by providing an alternate pathway with lower activation energy.

(B) Haemoglobin is a globular protein. It is a conjugated protein that transports oxygen in the blood.

(B) The conversion of an alkyl halide $(R-X)$ to an alcohol $(R-OH)$ is a nucleophilic substitution reaction.
When an alkyl halide is treated with an aqueous alkali like $Aq. KOH$ or $NaOH$,the halide ion $(X^-)$ is replaced by the hydroxyl group $(OH^-)$.
Reaction: $R-X + KOH (aq) \rightarrow R-OH + KX$.
This is a nucleophilic substitution reaction ($S_N1$ or $S_N2$ depending on the substrate).
Therefore,the correct option is $B$.

(D) Ionic reactions involve the interaction of oppositely charged ions in solution. Because these ions exert strong electrostatic forces of attraction on each other,they combine almost immediately upon mixing. Therefore,ionic reactions are considered to be instantaneous. The Assertion is incorrect,while the Reason is correct.