IIT JEE 1992 Chemistry Question Paper with Answer and Solution

(B) The de-Broglie equation,$\lambda = \frac{h}{p} = \frac{h}{mv}$,relates the wavelength (wave property) to the momentum (particle property) of a moving object,including photons. This equation provides the mathematical basis for the wave-particle duality of light and matter.

(D) An $H_2O$ molecule has two hydrogen atoms attached to an oxygen atom through covalent bonds.
Each hydrogen atom can form one hydrogen bond with the lone pair of an oxygen atom of a neighboring $H_2O$ molecule.
The oxygen atom has two lone pairs,each of which can accept a hydrogen bond from a hydrogen atom of a neighboring $H_2O$ molecule.
Therefore,an $H_2O$ molecule can participate in a total of $4$ hydrogen bonds ($2$ as a donor and $2$ as an acceptor).

(A) The hybridization of the central atom is calculated using the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
For $ClO_2^-$,$V = 7$ (for $Cl$),$M = 0$ (oxygen is divalent),$C = 0$,and $A = 1$.
$H = \frac{1}{2}(7 + 0 - 0 + 1) = \frac{8}{2} = 4$.
$A$ value of $4$ corresponds to $sp^3$ hybridization.
Therefore,the chlorine atom in $ClO_2^-$ is $sp^3$ hybridized,resulting in an angular (bent) shape due to the presence of two lone pairs.

(A) According to the kinetic theory of gases,the average kinetic energy of gas molecules is directly proportional to the absolute temperature $(T)$.
As the temperature increases,the average speed of the molecules increases.
When these faster-moving molecules collide with the walls of the container,they transfer more momentum per collision.
Since pressure is defined as the force exerted per unit area,and force is the rate of change of momentum,the increased momentum transfer results in an increase in the pressure exerted by the gas on the container walls.
Therefore,the correct option is $(A)$.

(D) The Van der Waals equation for $1$ mole of a real gas is given by $\left( P + \frac{a}{V_m^2} \right) (V_m - b) = RT$,where $V_m$ is the molar volume $(V_m = V/n)$.
Substituting $V_m = V/n$ into the equation:
$\left( P + \frac{a}{(V/n)^2} \right) (V/n - b) = RT$
$\left( P + \frac{n^2 a}{V^2} \right) \left( \frac{V - nb}{n} \right) = RT$
Multiplying both sides by $n$ gives the expression for $n$ moles:
$\left( P + \frac{n^2 a}{V^2} \right) (V - nb) = nRT$

(C) The given reaction is $2H_2S_{(g)} \rightleftharpoons 2H_{2(g)} + S_{2(g)}$.
Since the volume of the vessel is $1 L$,the molar concentrations are equal to the number of moles:
$[H_2S] = 0.5 mol L^{-1}$
$[H_2] = 0.10 mol L^{-1}$
$[S_2] = 0.4 mol L^{-1}$
The expression for the equilibrium constant $K_c$ is:
$K_c = \frac{[H_2]^2 [S_2]}{[H_2S]^2}$
Substituting the values:
$K_c = \frac{(0.10)^2 \times (0.4)}{(0.5)^2} = \frac{0.01 \times 0.4}{0.25} = \frac{0.004}{0.25} = 0.016 mol L^{-1}$.

(B) When equal volumes are mixed,the concentration of each ion is halved.
For option $(A)$: $[Ca^{2+}] = 0.5 \times 10^{-4} \ M$,$[F^-] = 0.5 \times 10^{-4} \ M$.
Ionic Product $(IP)$ = $[Ca^{2+}][F^-]^2 = (0.5 \times 10^{-4}) \times (0.5 \times 10^{-4})^2 = 1.25 \times 10^{-13}$.
Since $IP < K_{sp}$,no precipitate is formed.
For option $(B)$: $[Ca^{2+}] = 0.5 \times 10^{-2} \ M$,$[F^-] = 0.5 \times 10^{-3} \ M$.
Ionic Product $(IP)$ = $[Ca^{2+}][F^-]^2 = (0.5 \times 10^{-2}) \times (0.5 \times 10^{-3})^2 = 0.5 \times 10^{-2} \times 0.25 \times 10^{-6} = 1.25 \times 10^{-9}$.
Since $IP > K_{sp}$ $(1.25 \times 10^{-9} > 1.7 \times 10^{-10})$,precipitation occurs.

(C) The standard heat of formation $(\Delta H_f^o)$ is defined as the enthalpy change when $1 \text{ mole}$ of a substance is formed from its constituent elements in their most stable standard states.
For methane $(CH_{4(g)})$,the constituent elements are carbon and hydrogen.
The most stable state of carbon at standard conditions is graphite.
The most stable state of hydrogen is diatomic gas $(H_{2(g)})$.
Therefore,the correct equation is: $C(\text{graphite}) + 2H_{2(g)} \rightarrow CH_{4(g)}$.

(C) In an endothermic reaction,the products have higher enthalpy than the reactants,so $\Delta H > 0$.
From the energy profile diagram,the activation energy $(E_a)$ is the energy difference between the transition state and the reactants.
Since the products are at a higher energy level than the reactants,the energy barrier $(E_a)$ must be greater than the net enthalpy change $(\Delta H)$ of the reaction.
Therefore,$E_a > \Delta H$.

(A) The half-reactions are:
Reduction: $(MnO_4^- + 8H^{+} + 5e^- \to Mn^{2+} + 4H_2O) \times 2$
Oxidation: $(C_2O_4^{2-} \to 2CO_2 + 2e^-) \times 5$
Adding the two half-reactions to balance the electrons:
$2MnO_4^- + 16H^{+} + 10e^- + 5C_2O_4^{2-} \to 2Mn^{2+} + 8H_2O + 10CO_2 + 10e^-$
The balanced equation is:
$2MnO_4^- + 5C_2O_4^{2-} + 16H^{+} \to 2Mn^{2+} + 10CO_2 + 8H_2O$
Thus,the coefficients of $MnO_4^-$,$C_2O_4^{2-}$,and $H^{+}$ are $2, 5$,and $16$ respectively.

(D) The correct answer is $D$.
For transition elements,the filling of $d$-sub-shells is not strictly monotonic due to the stability of half-filled and fully-filled configurations (e.g.,$Cr$ is $3d^5 4s^1$ and $Cu$ is $3d^{10} 4s^1$).
Therefore,the statement that $d$-sub-shells are filled monotonically is incorrect.

(B) Both $CN^{-}$ and $N_2$ are isoelectronic,having $14$ electrons each and a triple bond between the atoms.
However,$N_2$ is a homonuclear diatomic molecule,meaning it has no bond polarity (the electronegativity difference is zero).
In contrast,$CN^{-}$ is a heteronuclear ion with a significant difference in electronegativity between $C$ and $N$,resulting in a polar bond.
The absence of bond polarity in $N_2$ makes it chemically inert compared to $CN^{-}$.

(A) To determine the hybridization of the chlorine atom in $ClO_2^-$,we use the formula: $H = \frac{1}{2} [V + M - C + A]$
Where $V$ is the number of valence electrons of the central atom $(Cl = 7)$,
$M$ is the number of monovalent atoms attached ($0$ in this case,as oxygen is divalent),
$C$ is the cationic charge $(0)$,
$A$ is the anionic charge $(1)$.
$H = \frac{1}{2} [7 + 0 - 0 + 1] = \frac{8}{2} = 4$.
$A$ value of $4$ corresponds to $sp^3$ hybridization.
$ClO_2^-$ has two lone pairs on the chlorine atom,which results in a $V$-shaped or bent structure.

(B) The homolytic fission of the $C-C$ bond in ethane $(CH_3-CH_3)$ results in the formation of two methyl free radicals $(CH_3^{\bullet})$.
$CH_3-CH_3 \xrightarrow{\text{Homolytic fission}} 2CH_3^{\bullet}$
In the methyl free radical,the central carbon atom is bonded to three hydrogen atoms and possesses one unpaired electron in an unhybridized $p$-orbital.
Therefore,the carbon atom is $sp^2$-hybridized with a trigonal planar geometry.

(B) The stopping potential $(V_S)$ depends only on the frequency of the incident light and the work function of the metal surface. It is independent of the intensity of light or the distance of the source from the cell.
Therefore,the stopping potential remains $0.6 \ V$.
The saturation current $(I_S)$ is directly proportional to the intensity of the incident light. Since the source is a point source,the intensity $I$ follows the inverse square law: $I \propto \frac{1}{r^2}$.
Given $r_1 = 0.2 \ m$ and $r_2 = 0.6 \ m$,the ratio of intensities is $\frac{I_2}{I_1} = \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{0.2}{0.6}\right)^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}$.
Since $I_S \propto I$,the new saturation current $I_{S2} = I_{S1} \times \frac{1}{9} = 18 \ mA \times \frac{1}{9} = 2 \ mA$.
Comparing the options,the correct statements are that the stopping potential remains $0.6 \ V$ (Option $B$) and the saturation current becomes $2 \ mA$ (which is not listed,but $B$ is a correct statement).

(B) The law of multiple proportions was proposed by John Dalton in $1803$ and was later verified by Jons Jacob Berzelius.

(D) For $pH = 1.0$,the concentration of $[H^{+}]$ must be $10^{-1} \ M = 0.1 \ M$.
Option $(d)$:
$M.eq. \ of \ HCl = 75 \ mL \times \frac{1}{5} \ M = 15 \ mmol$.
$M.eq. \ of \ NaOH = 25 \ mL \times \frac{1}{5} \ M = 5 \ mmol$.
Remaining $H^{+}$ ions $= 15 - 5 = 10 \ mmol$.
Total volume $= 75 + 25 = 100 \ mL$.
$[H^{+}] = \frac{10 \ mmol}{100 \ mL} = 0.1 \ M = 10^{-1} \ M$.
Therefore,$pH = -\log(10^{-1}) = 1.0$.

(B) The stopping potential (cut-off voltage) depends only on the frequency of the incident light and the work function of the metal surface. It is independent of the intensity of the light.
Since the distance of the source is increased,the intensity of light decreases,but the frequency remains constant. Therefore,the stopping potential remains unchanged at $0.6\, V$.
The saturation current is directly proportional to the intensity of the incident light. Intensity $I$ follows the inverse square law,$I \propto 1/d^2$.
When the distance increases from $0.2\, m$ to $0.6\, m$ (a factor of $3$),the intensity becomes $(1/3)^2 = 1/9$ of the original value.
Thus,the new saturation current will be $18\, mA / 9 = 2\, mA$.
Therefore,both options $B$ and $D$ (if corrected) are physically correct,but typically in such problems,the focus is on the invariance of the stopping potential. Given the standard format,$B$ is the primary conceptual answer.

(B) Paramagnetism is exhibited by species that possess one or more unpaired electrons.
$1$. In $[Cu(NH_3)_4]Cl_2$,$Cu$ is in the $+2$ oxidation state ($3d^9$ configuration),which has one unpaired electron.
$2$. In $[Ag(NH_3)_2]Cl$,$Ag$ is in the $+1$ oxidation state ($4d^{10}$ configuration),which has no unpaired electrons,making it diamagnetic.
$3$. $NO$ has $15$ valence electrons (odd number),so it has an unpaired electron.
$4$. $NO_2$ has $17$ valence electrons (odd number),so it has an unpaired electron.
Therefore,$[Ag(NH_3)_2]Cl$ is the compound that does not show paramagnetism.

(D) Given equation: $\cos^2 \theta + \sin \theta + 1 = 0$
Using the identity $\cos^2 \theta = 1 - \sin^2 \theta$,we get:
$1 - \sin^2 \theta + \sin \theta + 1 = 0$
Rearranging the terms:
$-\sin^2 \theta + \sin \theta + 2 = 0$
Multiplying by $-1$:
$\sin^2 \theta - \sin \theta - 2 = 0$
Factoring the quadratic equation:
$(\sin \theta - 2)(\sin \theta + 1) = 0$
Since $\sin \theta = 2$ is impossible (as $-1 \le \sin \theta \le 1$),we must have:
$\sin \theta = -1$
For $\sin \theta = -1$,the general solution is $\theta = 2n\pi + \frac{3\pi}{2}$.
For $n=0$,$\theta = \frac{3\pi}{2}$.
Since $\frac{5\pi}{4} < \frac{3\pi}{2} < \frac{7\pi}{4}$,the solution lies in the interval $\left( \frac{5\pi}{4}, \frac{7\pi}{4} \right)$.

(D) The standard reduction potential of halogens is positive and decreases from $F_2$ to $I_2$.
Therefore,halogens act as oxidizing agents,and their oxidizing power decreases in the order: $F_2 > Cl_2 > Br_2 > I_2$.
Thus,$F_2$ is the strongest oxidizing agent.

(A) $F_2$ is the strongest oxidizing agent in the periodic table.
The oxidizing power of halogens is determined by their standard reduction potential $(E^\circ)$,which is influenced by factors such as enthalpy of atomization,electron gain enthalpy,and hydration enthalpy.
The standard reduction potentials ($E^\circ$ in $V$) for halogens are:
$F_2: +2.87 \ V$
$Cl_2: +1.36 \ V$
$Br_2: +1.07 \ V$
$I_2: +0.54 \ V$
Due to the low bond dissociation enthalpy of $F-F$ and high hydration enthalpy of the $F^-$ ion,$F_2$ exhibits the highest reduction potential,making it the strongest oxidizing agent.

(C) The gas liberated is $SO_2$ (sulphur dioxide).
$(i)$ $SO_2$ reacts with baryta water $(Ba(OH)_2)$ to form $BaSO_3$,which causes turbidity: $Ba(OH)_2 + SO_2 \to BaSO_3 \downarrow + H_2O$.
$(ii)$ $SO_2$ reduces acidified potassium dichromate $(K_2Cr_2O_7)$ to chromium$(III)$ sulphate,which is green in colour: $K_2Cr_2O_7 + 3SO_2 + H_2SO_4 \to K_2SO_4 + Cr_2(SO_4)_3 + H_2O$.
These tests are characteristic of the sulphite ion $(SO_3^{2-})$.

(C) For an endothermic reaction,the enthalpy change $\Delta H$ is positive.
The relationship between the activation energy of the forward reaction $(E_{a(f)})$,the activation energy of the backward reaction $(E_{a(b)})$,and the enthalpy of the reaction is given by the equation: $E_{a(f)} = E_{a(b)} + \Delta H$.
Since $E_{a(b)}$ must always be a positive value for any reaction to occur,it follows that $E_{a(f)} > \Delta H$.
Therefore,the minimum value for the energy of activation is more than $\Delta H$.

(D) The chemical formulas for the given minerals are:
$1$. Cryolite: $Na_{3}AlF_{6}$ (contains $Al$)
$2$. Mica: $K_{2}O \cdot 3Al_{2}O_{3} \cdot 6SiO_{2} \cdot 2H_{2}O$ (contains $Al$)
$3$. Feldspar: $KAlSi_{3}O_{8}$ (contains $Al$)
$4$. Fluorspar: $CaF_{2}$ (does not contain $Al$)
Therefore,Fluorspar is the correct answer.

(B) . In $[Ag(NH_3)_2]Cl$,the central metal ion is $Ag^+$,which has a $d^{10}$ electronic configuration,meaning all electrons are paired,making it diamagnetic.
In $[Cu(NH_3)_4]Cl_2$,$Cu^{2+}$ has a $d^9$ configuration,which has one unpaired electron.
$NO$ and $NO_2$ are odd-electron molecules,which are paramagnetic.

(C) The reaction of an acid chloride with $H_2$ in the presence of $Pd-BaSO_4$ is known as the Rosenmund reduction.
This reaction specifically reduces the acid chloride group $(-COCl)$ to an aldehyde group $(-CHO)$.
Therefore,the product $P$ is $RCHO$.

(D) The reaction of ethyl formate $(HCOOC_2H_5)$ with excess $CH_3MgI$ proceeds in two steps.
First,the nucleophilic attack of $CH_3^-$ on the carbonyl carbon of ethyl formate leads to the formation of an intermediate,which eliminates an ethoxide ion to form acetaldehyde $(CH_3CHO)$.
Since $CH_3MgI$ is in excess,it further reacts with the formed acetaldehyde to produce an alkoxide intermediate.
Finally,upon hydrolysis,this intermediate yields a secondary alcohol.
The overall reaction is: $HCOOC_2H_5 + 2CH_3MgI \xrightarrow{H_3O^+} (CH_3)_2CHOH + Mg(OH)I + C_2H_5OH$.
The product formed is isopropyl alcohol $(CH_3CH(OH)CH_3)$.