IIT JEE 1981 Chemistry Question Paper with Answer and Solution

(C) Rutherford's experiment on scattering of $\alpha$-particles showed for the first time that the atom has a nucleus.
He observed that the positively charged $\alpha$-particles were repelled and deflected by the concentrated positive charge at the center of the atom.
Rutherford named this positively charged central portion of the atom as the nucleus.

(B) The atomic number of $Ni$ is $28$. The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.
When $Ni$ forms a $Ni^{2+}$ cation,it loses two electrons from the $4s$ orbital.
Thus,the electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
In the $3d^8$ configuration,the electrons are filled in the five $3d$ orbitals according to Hund's rule:
- The first five electrons occupy the five orbitals singly.
- The next three electrons pair up in the first three orbitals.
This leaves two orbitals with single electrons.
Therefore,there are $2$ unpaired electrons in the $Ni^{2+}$ cation.

(D) For a molecule $MX_3$ to have a zero dipole moment,it must possess a trigonal planar geometry.
In a trigonal planar geometry,the central atom $M$ undergoes $sp^2$ hybridization.
These $3$ $sp^2$ hybrid orbitals form $3$ $\sigma$ bonds with $X$ atoms,resulting in a symmetric structure where the individual bond dipoles cancel each other out.

(A) The bond angles for the given molecules are as follows:
$1$. $CO_2$: The molecule is linear with $sp$ hybridization,resulting in a bond angle of $180^\circ$.
$2$. $CH_4$: The molecule has tetrahedral geometry with $sp^3$ hybridization,resulting in a bond angle of $109.5^\circ$.
$3$. $NH_3$: The molecule has trigonal pyramidal geometry with $sp^3$ hybridization and one lone pair,resulting in a bond angle of $107^\circ$.
$4$. $H_2O$: The molecule has bent geometry with $sp^3$ hybridization and two lone pairs,resulting in a bond angle of $104.5^\circ$.
Therefore,the bond angle is greatest in $CO_2$.

(C) The pair of molecules that forms the strongest intermolecular hydrogen bonds is formic acid $(HCOOH)$ and acetic acid $(CH_3COOH)$.
Hydrogen bonding occurs between a hydrogen atom covalently bonded to a highly electronegative atom ($N$,$O$,or $F$) and another electronegative atom with a lone pair.
In the case of carboxylic acids like $HCOOH$ and $CH_3COOH$,they form stable cyclic dimers through two intermolecular hydrogen bonds.
These dimers are exceptionally stable due to the resonance-stabilized structure of the carboxylate group,making the hydrogen bonds in these carboxylic acid pairs stronger than those found in $H_2O$ or $H_2O_2$.

(A) The root mean square velocity $(V_{rms})$ is given by $V_{rms} = \sqrt{\frac{3RT}{M}}$.
The average velocity $(V_{av})$ is given by $V_{av} = \sqrt{\frac{8RT}{\pi M}}$.
The ratio $\frac{V_{rms}}{V_{av}}$ is calculated as $\sqrt{\frac{3RT}{M} \times \frac{\pi M}{8RT}} = \sqrt{\frac{3\pi}{8}}$.
Substituting the value of $\pi \approx 3.14$,we get $\sqrt{\frac{3 \times 3.14}{8}} = \sqrt{1.1775} \approx 1.086$.
Thus,the ratio is $1.086 : 1$.

(B) The temperature at which a real gas behaves like an ideal gas over an appreciable range of pressure is known as the $Boyle$ temperature or $Boyle$ point.
At this temperature,the compressibility factor $Z$ remains close to $1$ for a wide range of pressure.

(D) The equilibrium constant ($K_c$ or $K_p$) is a function of temperature only for a given reaction.
It does not change with the addition of a catalyst,changes in pressure,or changes in the initial concentrations of reactants or products.
Therefore,the correct option is $(D)$.

(C) The given reaction is $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) + Q \ \text{cal}$.
Since the reaction is exothermic (indicated by $+ Q \ \text{cal}$),according to Le Chatelier's principle,a low temperature will shift the equilibrium in the forward direction to produce more $SO_3$.
Also,the number of moles of gaseous reactants is $3$ ($2$ moles of $SO_2$ and $1$ mole of $O_2$) and the number of moles of gaseous products is $2$ $(SO_3)$.
Since the volume decreases from left to right,a high pressure will shift the equilibrium in the forward direction to produce more $SO_3$.
Therefore,the most suitable conditions are low temperature and high pressure.

(B) The reaction is $2NH_3 \rightleftharpoons N_2 + 3H_2$.
Initial moles: $a$,$0$,$0$.
Moles at equilibrium: $(a - 2x)$,$x$,$3x$.
Total moles at equilibrium: $(a - 2x) + x + 3x = a + 2x$.
Initial pressure of $a$ moles of $NH_3$ at $27 \ ^{\circ}C$ $(300 \ K)$ is $15 \ atm$.
Let $p$ be the pressure of $a$ moles of $NH_3$ at $347 \ ^{\circ}C$ $(620 \ K)$.
Using $\frac{P_1}{T_1} = \frac{P_2}{T_2}$ (at constant volume),$\frac{15}{300} = \frac{p}{620}$,which gives $p = 31 \ atm$.
Since $P \propto n$ at constant $V$ and $T$,the initial pressure $P_i = 31 \ atm$ corresponds to $a$ moles.
The final pressure $P_f = 50 \ atm$ corresponds to $(a + 2x)$ moles.
Therefore,$\frac{a + 2x}{a} = \frac{50}{31}$.
$1 + \frac{2x}{a} = \frac{50}{31} \implies \frac{2x}{a} = \frac{50}{31} - 1 = \frac{19}{31}$.
Percentage of $NH_3$ decomposed $= \frac{2x}{a} \times 100 = \frac{19}{31} \times 100 \approx 61.29 \% \approx 61.3 \%$.

(B) Molten $NaCl$ exists in the form of $Na^+$ and $Cl^-$ ions.
Since these ions are free to move in the molten state,they act as charge carriers,allowing the substance to conduct electricity.

(B) For pure water,the concentration of hydronium ions is equal to the concentration of hydroxide ions: $[H_3O^{+}] = [OH^{-}]$.
Given $[H_3O^{+}] = 10^{-6}\,M$,therefore $[OH^{-}] = 10^{-6}\,M$.
The ionic product of water is defined as $K_w = [H_3O^{+}][OH^{-}]$.
Substituting the values: $K_w = (10^{-6}) \times (10^{-6}) = 10^{-12}$.

(B) An acidic buffer solution is prepared by mixing a weak acid and its salt with a strong base.
$CH_3COOH$ is a weak acid and $CH_3COONa$ is its salt with a strong base $(NaOH)$.
Therefore,the mixture of $CH_3COOH + CH_3COONa$ forms an acidic buffer.

(C) For a very dilute solution of a strong acid,the contribution of $H^+$ ions from the auto-ionization of water cannot be ignored.
Total $[H^+] = [H^+]_{HCl} + [H^+]_{H_2O} = 10^{-8} + 10^{-7} = 10^{-8} + 10 \times 10^{-8} = 11 \times 10^{-8} \ M$.
$pH = -\log[H^+] = -\log(11 \times 10^{-8}) = 8 - \log(11) \approx 8 - 1.04 = 6.96$.
Since the solution is acidic,the $pH$ must be slightly less than $7$. Thus,$6 < pH < 7$.

(A) The strength of a conjugate base is inversely proportional to the strength of its corresponding acid.
The order of acidic strength of the oxyacids of chlorine is $HClO < HClO_2 < HClO_3 < HClO_4$.
Since $HClO$ is the weakest acid among the given options,its conjugate base,$ClO^-$,is the strongest Bronsted base.

(A) In $N_2H_4$,the oxidation state of $H$ is $+1$. Let the oxidation state of $N$ be $x$.
$2x + 4(+1) = 0 \implies 2x = -4 \implies x = -2$.
When $1 \ mol$ of $N_2H_4$ loses $10 \ mol$ of electrons,the total oxidation state increases by $10$.
Let the oxidation state of $N$ in $Y$ be $a$.
Since there are $2$ atoms of $N$,the total change is $2(a - (-2)) = 10$.
$2(a + 2) = 10
$a + $2$ = $5$
$a = +3$.

(C) Anhydrous aluminium chloride exists as a dimer $(Al_2Cl_6)$ in the vapour state and in non-polar solvents.
It is a strong Lewis acid due to the incomplete octet of the $Al$ atom.
Because of its Lewis acidic nature,it undergoes hydrolysis easily in the presence of moisture.
It sublimes at $178\,^oC$ under atmospheric pressure,but it can sublime at lower temperatures under vacuum.

(B) The stability order of carbocations is $3^{\circ} > 2^{\circ} > 1^{\circ} > \text{methyl}$.
$(CH_3)_3C^{+}$ is a $3^{\circ}$ (tert-butyl) carbocation,$(CH_3)_2CH^{+}$ is a $2^{\circ}$ (sec-butyl) carbocation,and $CH_3CH_2^+$ is a $1^{\circ}$ (n-butyl) carbocation.
Since the stability of carbocations increases with the number of alkyl groups attached to the positively charged carbon due to the inductive effect and hyperconjugation,the $3^{\circ}$ carbocation is the most stable.
Therefore,the correct option is $(b)$.

(D) The molecular formula $C_{4}H_{10}O$ represents both alcohols and ethers.
$1$. Positional isomerism: $CH_{3}CH_{2}CH_{2}CH_{2}OH$ (butan-$1$-ol) and $CH_{3}CH(OH)CH_{2}CH_{3}$ (butan-$2$-ol) are position isomers.
$2$. Functional isomerism: $CH_{3}CH_{2}CH_{2}CH_{2}OH$ (an alcohol) and $CH_{3}CH_{2}OCH_{2}CH_{3}$ (an ether) are functional isomers.
$3$. Metamerism: $CH_{3}OCH_{2}CH_{2}CH_{3}$ (methyl propyl ether) and $CH_{3}CH_{2}OCH_{2}CH_{3}$ (diethyl ether) are metamers due to the different distribution of alkyl groups around the oxygen atom.
Therefore,the compound can show all these types of isomerism.

(D) Diethyl ether has the molecular formula $C_4H_{10}O$.
$n$-propyl methyl ether $(CH_3-O-CH_2-CH_2-CH_3)$,butan-$1$-ol $(CH_3-CH_2-CH_2-CH_2-OH)$,and $2$-methylpropan-$2$-ol $((CH_3)_3C-OH)$ all have the molecular formula $C_4H_{10}O$,making them isomers of diethyl ether.
Butanone $(CH_3-CO-CH_2-CH_3)$ has the molecular formula $C_4H_8O$.
Since the molecular formulas are different,butanone is not an isomer of diethyl ether.
Therefore,the correct option is $(D)$.

(C) The carbon-carbon bond distance depends on the hybridization of the carbon atoms.
In $Ethyne$ $(C_2H_2)$,the $C-C$ bond is a triple bond with $sp$ hybridization,having a bond length of $1.20 \ \mathring{A}$.
In $Ethene$ $(C_2H_4)$,the $C-C$ bond is a double bond with $sp^2$ hybridization,having a bond length of $1.34 \ \mathring{A}$.
In $Benzene$ $(C_6H_6)$,the $C-C$ bond length is $1.39 \ \mathring{A}$ due to resonance.
In $Ethane$ $(C_2H_6)$,the $C-C$ bond is a single bond with $sp^3$ hybridization,having a bond length of $1.54 \ \mathring{A}$.
Therefore,the maximum carbon-carbon bond distance is found in $Ethane$.

(A) According to Baeyer's strain theory,the angle strain in cycloalkanes is related to the deviation of the bond angle from the ideal tetrahedral angle of $109.5^o$.
$Cyclopropane$ has a bond angle of $60^o$,which is the largest deviation from the ideal tetrahedral angle,making it the most strained and most reactive cycloalkane.
Therefore,the correct option is $(A)$.

(C) The correct answer is $(C)$.
In $CH_{2}=C=CH_{2}$ (Propadiene),the terminal carbon atoms are $sp^{2}$ hybridized because they are bonded to two hydrogen atoms and one carbon atom via a double bond.
The central carbon atom is $sp$ hybridized because it is bonded to two carbon atoms via two double bonds.

(A) The reactivity order of halogens in free radical halogenation is $F_2 > Cl_2 > Br_2 > I_2$.
Fluorine is the most reactive halogen due to its small size and high electronegativity,making the reaction highly exothermic and often explosive.

(C) The bond order of individual carbon-carbon bonds in benzene is $1.5$,which is between $1$ and $2$.
This is due to the resonance structure of $C_6H_6$.
In the benzene molecule,the $\pi$-electrons are delocalized over the entire ring.
Each carbon-carbon bond in benzene is equivalent,and the bond order is calculated as the average of the single and double bonds in the resonance structures,which is $(1+2)/2 = 1.5$.

(D) The molecular formula of diethyl ether $(C_2H_5-O-C_2H_5)$ is $C_4H_{10}O$.
Isomers are compounds with the same molecular formula but different structural arrangements.
$n-$propyl methyl ether $(CH_3-O-CH_2-CH_2-CH_3)$ has the formula $C_4H_{10}O$.
Butan$-1-$ol $(CH_3-CH_2-CH_2-CH_2-OH)$ has the formula $C_4H_{10}O$.
$2-$methylpropan$-2-$ol $(CH_3-C(OH)(CH_3)-CH_3)$ has the formula $C_4H_{10}O$.
Butanone $(CH_3-CO-CH_2-CH_3)$ has the molecular formula $C_4H_8O$.
Therefore,butanone is not an isomer of diethyl ether.

(D) The reaction of ethanol with concentrated $H_2SO_4$ at $170\,^{\circ}C$ is a dehydration reaction.
In this process,ethanol undergoes elimination of a water molecule to form ethene.
The chemical equation is:
$C_2H_5OH \xrightarrow{\text{Conc. } H_2SO_4, 170\,^{\circ}C} C_2H_4 + H_2O$
Therefore,the correct option is $D$.

(C) The electronic configurations of the elements are: $C (2s^2 2p^2)$,$N (2s^2 2p^3)$,$O (2s^2 2p^4)$,$F (2s^2 2p^5)$.
After the removal of the first electron,the configurations become: $C^+ (2s^2 2p^1)$,$N^+ (2s^2 2p^2)$,$O^+ (2s^2 2p^3)$,$F^+ (2s^2 2p^4)$.
The second ionization potential involves removing an electron from these cations.
$O^+$ has a stable half-filled $2p^3$ configuration,making it the most difficult to remove an electron from.
Following the periodic trend and stability,the order of second ionization potential is $O > F > N > C$.

(B) Let $f(x) = ax^2 + bx + c$ and $g(x) = 1 + \cos^8 x$. The given equation is $\int_{0}^{1} g(x)f(x) dx = \int_{0}^{2} g(x)f(x) dx$.
This can be rewritten as $\int_{0}^{1} g(x)f(x) dx = \int_{0}^{1} g(x)f(x) dx + \int_{1}^{2} g(x)f(x) dx$.
Subtracting $\int_{0}^{1} g(x)f(x) dx$ from both sides,we get $\int_{1}^{2} g(x)f(x) dx = 0$.
Since $g(x) = 1 + \cos^8 x$ is always positive for all real $x$,the integral of $g(x)f(x)$ over the interval $(1, 2)$ can only be zero if $f(x)$ changes sign in the interval $(1, 2)$.
By the Intermediate Value Theorem,if a continuous function $f(x)$ changes sign in an interval $(1, 2)$,there must exist at least one point $c \in (1, 2)$ such that $f(c) = 0$.
Since $(1, 2) \subset (0, 2)$,the quadratic equation $ax^2 + bx + c = 0$ must have at least one root in $(0, 2)$.

(C) Rutherford's experiment on the scattering of $\alpha-$particles showed for the first time that the atom has a nucleus.
He observed that the positively charged $\alpha-$particles were repelled and deflected by the concentrated positive charge at the center of the atom.
Rutherford named this positively charged central portion of the atom the nucleus.

(B) In the molten state,the ionic lattice of $NaCl$ breaks down,allowing the $Na^+$ and $Cl^-$ ions to move freely.
Since electricity is conducted by the movement of charged particles,the presence of these free ions enables molten $NaCl$ to conduct electricity.

(B) The radioactive decay law is given by $N = N_0 e^{-\lambda t}$,where $N/N_0 = 1/20$.
Taking the natural logarithm on both sides: $\ln(1/20) = -\lambda t$,which implies $\ln(20) = \lambda t$.
We know that $\lambda = \frac{\ln 2}{t_{1/2}}$,where $t_{1/2} = 3.8 \ days$.
Substituting $\lambda$: $\ln(20) = \frac{\ln 2}{3.8} \times t$.
$t = 3.8 \times \frac{\ln 20}{\ln 2} = 3.8 \times \log_2(20)$.
Using the change of base formula: $t = 3.8 \times \frac{\log_{10} 20}{\log_{10} 2} = 3.8 \times \frac{\log_{10} (2 \times 10)}{\log_{10} 2} = 3.8 \times \frac{1 + \log_{10} 2}{\log_{10} 2}$.
Given $\log_{10} 2 \approx 0.3010$,then $t = 3.8 \times \frac{1 + 0.3010}{0.3010} = 3.8 \times \frac{1.3010}{0.3010} \approx 3.8 \times 4.322 = 16.42 \ days$.
Rounding to the nearest option,the answer is $16.5 \ days$.

(C) The photographic film is coated with silver halide $(AgX)$ crystals.
When a picture is captured,the film is exposed to light,causing $Ag^+$ ions in $AgX$ to be reduced to opaque silver metal.
This opaque metal forms the image. To obtain a clear image,the unreacted $AgX$ must be removed.
This is achieved using a hypo-solution,which is a solution of sodium thiosulphate $(Na_2S_2O_3)$.
Sodium thiosulphate reacts with unreacted silver bromide $(AgBr)$ to form a water-soluble complex,$Na_3[Ag(S_2O_3)_2]$,which is then washed away.
The reaction is: $2Na_2S_2O_3 + AgBr \rightarrow Na_3[Ag(S_2O_3)_2] + NaBr$.
Thus,hypo is used due to its complex-forming behaviour.

(B) The stability of carbonium ions (carbocations) follows the order: $3^\circ > 2^\circ > 1^\circ > \text{methyl}$.
$tert$-butyl carbocation is a $3^\circ$ carbocation,which is stabilized by the inductive effect $(+I)$ of three methyl groups and hyperconjugation.
Therefore,the $tert$-butyl carbocation is the most stable among the given options.

(A) Let the mass of the bullet be $m\,g$.
The heat required for the bullet to reach its melting point and melt is given by $Q_1 = mc\Delta\theta + mL$.
Substituting the values: $Q_1 = m \times 0.03 \times (327 - 27) + m \times 6 = m \times 0.03 \times 300 + 6m = 9m + 6m = 15m\,cal$.
Converting this to Joules: $Q_1 = 15m \times 4.2 = 63m\,J$.
The kinetic energy of the bullet is $K.E. = \frac{1}{2}mv^2$,where $m$ is in $kg$. Since $m$ is in $g$,$m_{kg} = m \times 10^{-3}$.
So,$K.E. = \frac{1}{2} \times (m \times 10^{-3}) \times v^2 = 0.5 \times 10^{-3} mv^2\,J$.
Since $25\%$ of the heat is absorbed by the obstacle,$75\%$ of the kinetic energy is absorbed by the bullet to melt it.
Therefore,$0.75 \times (0.5 \times 10^{-3} mv^2) = 63m$.
$0.375 \times 10^{-3} v^2 = 63$.
$v^2 = \frac{63}{0.375 \times 10^{-3}} = \frac{63}{0.000375} = 168000$.
$v = \sqrt{168000} \approx 409.88\,m/s \approx 410\,m/s$.

(A) The kinetic energy of the bullet is $K.E. = \frac{1}{2}mv^2$.
Since $25\%$ of the heat is absorbed by the obstacle,$75\%$ of the kinetic energy is used to heat and melt the bullet.
Heat required to raise the temperature of the bullet to its melting point: $Q_1 = mc\Delta T = m \times 0.03 \times (327 - 27) = m \times 0.03 \times 300 = 9m\,cal$.
Heat required to melt the bullet: $Q_2 = mL = m \times 6 = 6m\,cal$.
Total heat required: $Q = Q_1 + Q_2 = 15m\,cal$.
Converting to Joules: $Q = 15m \times 4.2 = 63m\,J$.
Equating $75\%$ of $K.E.$ to $Q$: $0.75 \times \frac{1}{2}mv^2 = 63m$.
$0.375v^2 = 63$.
$v^2 = \frac{63}{0.375} = 168$.
Wait,recalculating with units: $Q = 15m\,cal = 15m \times 4.2 = 63m\,J$. $0.75 \times 0.5 \times v^2 = 63 \implies 0.375v^2 = 63 \implies v^2 = 168$. This suggests a unit mismatch in the provided solution. Re-evaluating: $15m\,cal = 15 \times 4.2 = 63\,J$. If $m$ is in grams,$K.E. = \frac{1}{2} \times (m \times 10^{-3}) \times v^2$. Thus $0.75 \times 0.5 \times m \times 10^{-3} \times v^2 = 15m \times 4.2$.
$0.375 \times 10^{-3} \times v^2 = 63$.
$v^2 = \frac{63}{0.375} \times 1000 = 168000$.
$v = \sqrt{168000} \approx 409.87 \approx 410\,m/s$.

(C) When a solution shows positive deviation from Raoult's law,the intermolecular forces present are weaker than those present in the pure components.
This leads to an increase in the total vapour pressure of the solution compared to the ideal case.
Because the vapour pressure is higher,the solution reaches the atmospheric pressure at a lower temperature.
Therefore,the azeotropic mixture boils at a temperature lower than either of its pure components.
Hence,the correct option is $C$.

(B) Molten sodium chloride $(NaCl)$ exists in a liquid state where the ionic lattice breaks down,resulting in the presence of free $Na^+$ and $Cl^-$ ions. These free ions are responsible for the conduction of electricity.

(C) In photography,sodium thiosulphate $(Na_2S_2O_3 \cdot 5H_2O)$,commonly known as 'hypo',is used as a fixing agent.
It dissolves the unexposed silver bromide $(AgBr)$ from the photographic film by forming a soluble complex,sodium dithiosulphatoargentate$(I)$,according to the reaction:
$AgBr(s) + 2Na_2S_2O_3(aq) \rightarrow Na_3[Ag(S_2O_3)_2](aq) + NaBr(aq)$.
Therefore,its utility is due to its complex-forming behaviour.

(D) $HF$ is a very weak reducing agent because the $F^-$ ion has a very high hydration energy and the $F-F$ bond is very strong.
Consequently,$HF$ does not act as a reducing agent and cannot be oxidized by any of the given oxidizing agents.

(D) An azeotropic mixture is a constant boiling mixture that behaves like a pure liquid.
Since it boils at a constant temperature without changing its composition,it is not possible to separate the components of an azeotropic mixture by simple distillation.
Therefore,neither $HCl$ nor $H_2O$ can be obtained in their pure states from this mixture.

(C) When a solution shows a positive deviation from Raoult's law,the intermolecular forces between the solute and solvent molecules are weaker than those present in the pure components.
This results in an increase in the total vapour pressure of the solution compared to the ideal solution.
Because the vapour pressure is higher,the solution reaches its boiling point at a lower temperature than either of the pure components. This is known as a minimum-boiling azeotrope.

(A) An $\alpha$-particle is a helium nucleus,represented as $_{2}He^{4}$.
When a radioactive nucleus emits an $\alpha$-particle,its mass number decreases by $4$ and its atomic number decreases by $2$.
The nuclear reaction is: $_{92}U^{238} \to _{90}Th^{234} + _{2}He^{4}$.
Thus,the product has a mass number of $234$ and an atomic number of $90$.

(D) The specific rate constant $(k)$ of a reaction is independent of the concentration of reactants or products and the time of the reaction.
It is a characteristic constant for a given reaction at a specific temperature.
According to the Arrhenius equation,$k = A e^{-E_a / RT}$,the rate constant depends significantly on the temperature of the reaction.

(A) The strength of a reducing agent is inversely proportional to its standard reduction potential $(E^\circ)$.
Comparing the given values:
$E^\circ (Zn^{2+}/Zn) = -0.762 \ V$
$E^\circ (Cr^{3+}/Cr) = -0.740 \ V$
$E^\circ (H^+/H_2) = 0.00 \ V$
$E^\circ (Fe^{3+}/Fe^{2+}) = 0.770 \ V$
Since $Zn_{(s)}$ has the most negative standard reduction potential $(-0.762 \ V)$,it has the highest tendency to lose electrons and act as the strongest reducing agent.

(B) Chlorine is more reactive than bromine and it displaces bromine from its salt solution.
When chlorine is added to a solution of potassium bromide,the bromine is displaced by the chlorine to form potassium chloride.
The chemical reaction is: $2 KBr + Cl_2 \rightarrow 2 KCl + Br_2$.
Since chlorine has a higher reduction potential than bromine,it acts as a stronger oxidizing agent and effectively replaces bromine.

(C) The reaction of alcohols with $ZnCl_2 +$ conc. $HCl$ is known as the $Lucas$ test.
$3^o$ alcohols react immediately to form turbidity.
$2^o$ alcohols react after about $5$ minutes.
$1^o$ alcohols do not react at room temperature.
Among the given options,$(CH_3)_3C-OH$ ($2-$methylpropan$-2-$ol) is a $3^o$ alcohol,which reacts immediately to produce turbidity.

(A) Secondary amines react with nitrous acid $(HNO_2)$ at low temperature $(273-278 \ K)$ to form $N$-nitrosoamines,which are yellow oily liquids.
$(C_2H_5)_2NH + HONO \to (C_2H_5)_2N-N=O + H_2O$
Diethylamine is a secondary amine,hence it forms diethyl nitrosoamine.

(D) . $F^-$ is the strongest reducing agent among halides,but $F_2$ is the strongest oxidizing agent. The oxidation of $F^-$ to $F_2$ is not possible by any chemical oxidizing agent because the standard oxidation potential of $F^-$ is very low (highly negative). Therefore,$F^-$ can be oxidized to $F_2$ only by electrolysis.