IIT JEE 1990 Chemistry Question Paper with Answer and Solution

(D) At point $A$,the incident intensity is $I$. Since $25\%$ is reflected,the intensity of ray $AB$ is $I_1 = 0.25I = I/4$. The intensity of the refracted ray is $0.75I = 3I/4$.
At point $C$,the ray is reflected. Again,$25\%$ of the incident energy $(3I/4)$ is reflected. So,the intensity of the ray moving towards $A'$ is $0.25 \times (3I/4) = 3I/16$.
At point $A'$,this ray is refracted. The intensity of the refracted ray $A'B'$ is $0.75 \times (3I/16) = 9I/64$. Thus,$I_2 = 9I/64$.
The ratio of intensities is $I_2/I_1 = (9I/64) / (I/4) = 9/16$.
The ratio of maximum to minimum intensity is given by:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} \right)^2 = \left( \frac{1 + \sqrt{I_2/I_1}}{1 - \sqrt{I_2/I_1}} \right)^2$
Substituting $I_2/I_1 = 9/16$:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{1 + \sqrt{9/16}}{1 - \sqrt{9/16}} \right)^2 = \left( \frac{1 + 3/4}{1 - 3/4} \right)^2 = \left( \frac{7/4}{1/4} \right)^2 = (7)^2 = 49/1$.

(A) The enolic form of acetone is propen$-2-$ol,which has the structure $CH_2=C(OH)-CH_3$.
Counting the bonds: there are $3$ $C-H$ bonds in the methyl group,$2$ $C-H$ bonds in the methylene group,$1$ $O-H$ bond,$1$ $C-O$ bond,$1$ $C-C$ bond,and $1$ $C=C$ bond (which consists of $1$ $\sigma$ and $1$ $\pi$ bond).
Total $\sigma$ bonds = $3 + 2 + 1 + 1 + 1 + 1 = 9$.
Total $\pi$ bonds = $1$.
Oxygen atom in the hydroxyl group has $2$ lone pairs.
Thus,the structure contains $9$ $\sigma$ bonds,$1$ $\pi$ bond,and $2$ lone pairs.

(A) According to Le Chatelier's principle,for an exothermic reaction $(\Delta H < 0)$,a decrease in temperature shifts the equilibrium in the forward direction to release heat.
In the given reaction,the number of moles of gaseous reactants is $1 + 4 = 5$,and the number of moles of gaseous products is $2$.
Since the number of moles decreases in the forward direction $(5 \to 2)$,an increase in pressure shifts the equilibrium towards the side with fewer moles,which is the forward direction.
Therefore,the formation of $AB_{4}$ is favoured at low temperature and high pressure.

(D) The solubility of alkaline earth metal hydroxides increases down the group from $Be$ to $Ba$.
Since $Be(OH)_2$ is the least soluble among the given hydroxides,it has the lowest concentration of ions in a saturated solution.
Therefore,$Be(OH)_2$ has the lowest value of the solubility product $(K_{sp})$ at $25\,^oC$.

(D) The dissociation of the salt $M_2X_3$ is given by:
$M_2X_3(s) \rightleftharpoons 2M^{3+}(aq) + 3X^{2-}(aq)$
If the solubility is $y \ mol \ dm^{-3}$,then the concentration of $M^{3+}$ is $2y \ mol \ dm^{-3}$ and the concentration of $X^{2-}$ is $3y \ mol \ dm^{-3}$.
The solubility product $K_{sp}$ is defined as:
$K_{sp} = [M^{3+}]^2 [X^{2-}]^3$
Substituting the values:
$K_{sp} = (2y)^2 \times (3y)^3$
$K_{sp} = (4y^2) \times (27y^3)$
$K_{sp} = 108y^5 \ mol^5 \ dm^{-15}$
Therefore,the correct option is $(D)$.

(A) The species given are $Mg^{2+}$,$Na^{+}$,$F^{-}$,and $Al$.
$1$. $Al$ is a neutral atom,while $Mg^{2+}$,$Na^{+}$,and $F^{-}$ are ions.
$2$. Comparing the ionic radii of isoelectronic species ($Mg^{2+}$,$Na^{+}$,$F^{-}$): These all have $10$ electrons. The ionic radius decreases as the nuclear charge $(Z)$ increases. The nuclear charges are $F^{-} (Z=9)$,$Na^{+} (Z=11)$,and $Mg^{2+} (Z=12)$. Thus,the order of size is $Mg^{2+} < Na^{+} < F^{-}$.
$3$. The neutral atom $Al$ $(Z=13)$ has a larger atomic radius than these cations. However,comparing $Al$ to the others,the correct order of size is $Mg^{2+} < Na^{+} < F^{-} < Al$.

(C) The electronic configuration of the given elements corresponds to $Al$ $(3p^1)$,$Si$ $(3p^2)$,$P$ $(3p^3)$,and $Ge$ $(4p^2)$.
Ionization energy generally increases across a period.
However,the $p^3$ configuration represents a half-filled stable subshell.
Due to the extra stability of the half-filled $p$-orbital,the energy required to remove an electron from $[Ne] \, 3s^2 \, 3p^3$ is the highest among the given options.

(D) Zeolite,which is hydrated sodium aluminium silicate $(Na_2Al_2Si_2O_8 \cdot xH_2O)$,acts as an ion exchanger.
When it is treated with hard water,the $Na^+$ ions present in the zeolite structure are exchanged with the $Ca^{2+}$ and $Mg^{2+}$ ions that cause hardness in water.

(B) The term 'lead pencil' is a misnomer. It does not contain lead $(Pb)$.
Instead,it contains a mixture of $Graphite$ and clay,which is used as the core of the pencil.

(C) Aluminium carbide $(Al_4C_3)$ reacts with water to produce methane $(CH_4)$ gas.
The chemical reaction is:
$Al_4C_3 + 12H_2O \longrightarrow 3CH_4 + 4Al(OH)_3$

(B) The Wurtz reaction involves the coupling of alkyl halides in the presence of sodium metal to form higher alkanes.
When a mixture of ethyl iodide $(C_2H_5I)$ and $n-$propyl iodide $(C_3H_7I)$ is used,the following products are formed:
$1$. Self-coupling of ethyl iodide: $C_2H_5-C_2H_5$ ($n-$butane).
$2$. Self-coupling of $n-$propyl iodide: $C_3H_7-C_3H_7$ ($n-$hexane).
$3$. Cross-coupling of ethyl iodide and $n-$propyl iodide: $C_2H_5-C_3H_7$ ($n-$pentane).
Therefore,$n-$propane is not formed in this reaction.

(C) Given equation: $(\cos p - 1){x^2} + (\cos p)x + \sin p = 0$.
Since the roots are real,the discriminant $D \ge 0$.
$D = b^2 - 4ac = (\cos p)^2 - 4(\cos p - 1)(\sin p) \ge 0$.
$\cos^2 p - 4\sin p \cos p + 4\sin p \ge 0$.
For the quadratic equation to be valid,the coefficient of $x^2$ must not be zero,so $\cos p - 1 \neq 0$,which implies $p \neq 2n\pi$.
Analyzing the inequality,for $p \in (0, \pi)$,$\sin p > 0$ and $(1 - \sin p) \ge 0$.
Thus,the condition $D \ge 0$ is satisfied for $p \in (0, \pi)$.

(A) The keto form of acetone is $CH_3-CO-CH_3$. The enolic form is $CH_2=C(OH)-CH_3$.
Counting the bonds in the enolic form $(CH_2=C(OH)-CH_3)$:
- Sigma $(\sigma)$ bonds: $3$ (in $CH_3$) + $1$ ($C$-$C$) + $1$ ($C$=$C$) + $1$ ($C$-$O$) + $1$ ($O$-$H$) + $2$ (in $CH_2$) = $9$ $\sigma$ bonds.
- Pi $(pi)$ bonds: $1$ (in $C$=$C$).
- Lone pairs: Oxygen atom in the hydroxyl group $(-OH)$ has $2$ lone pairs.
Thus,the enolic form contains $9$ $\sigma$ bonds,$1$ $\pi$ bond,and $2$ lone pairs.

(C) Assume (if possible) that $\log_{2} 7 = \frac{p}{q}$ is a rational number,where $p$ and $q$ are coprime integers.
Then $\frac{p}{q} = \log_{2} 7 \implies 7 = 2^{p/q} \implies 2^{p} = 7^{q}$.
This is a contradiction because the $L.H.S.$ is even and the $R.H.S.$ is odd.
Clearly,$\log_{2} 7$ is neither an integer nor a prime number. Therefore,it is an irrational number.

(D) The fraction of supplied heat energy that increases the internal energy of the gas is given by the ratio of the change in internal energy to the heat supplied at constant pressure.
$f = \frac{\Delta U}{(\Delta Q)_{P}}$
Since the change in internal energy $\Delta U$ is always equal to the heat supplied at constant volume $(\Delta Q)_{V} = \mu C_{V} \Delta T$,we have:
$f = \frac{\mu C_{V} \Delta T}{\mu C_{P} \Delta T} = \frac{C_{V}}{C_{P}} = \frac{1}{\gamma}$
For an ideal diatomic gas,the adiabatic index $\gamma = \frac{C_{P}}{C_{V}} = \frac{7}{5}$.
Therefore,the fraction $f = \frac{1}{7/5} = \frac{5}{7}$.

(D) When a gas is heated at constant pressure,the heat supplied $(\Delta Q)_p$ is used to increase the internal energy $(\Delta U)$ and to perform work $(\Delta W)$ against external pressure.
$(\Delta Q)_p = \Delta U + \Delta W$
Using the molar heat capacities,we have $\mu C_p \Delta T = \mu C_v \Delta T + P \Delta V$.
The fraction of heat energy used to increase the internal energy is given by the ratio $\frac{\Delta U}{(\Delta Q)_p} = \frac{\mu C_v \Delta T}{\mu C_p \Delta T} = \frac{C_v}{C_p} = \frac{1}{\gamma}$.
For a diatomic gas,the adiabatic index is $\gamma = 7/5$.
Therefore,the fraction is $\frac{1}{7/5} = 5/7$.

(C) For a combination of two thin prisms to produce dispersion without deviation,the net deviation produced by the system must be zero.
The deviation produced by a thin prism is given by $\delta = (\mu - 1)A$.
For the combination,the net deviation $\delta_{net} = \delta_1 + \delta_2 = 0$.
Since the prisms are combined to produce dispersion without deviation,they must be placed in opposite orientations,so $\delta_1 = -\delta_2$.
Thus,$(\mu_1 - 1)A_1 = (\mu_2 - 1)A_2$.
Given: $\mu_1 = 1.54$,$A_1 = 4^o$,$\mu_2 = 1.72$.
Substituting the values: $(1.54 - 1) \times 4^o = (1.72 - 1) \times A_2$.
$0.54 \times 4^o = 0.72 \times A_2$.
$2.16 = 0.72 \times A_2$.
$A_2 = \frac{2.16}{0.72} = 3^o$.

(A) The intensity of the first reflected ray $AB$ is $I_1 = 25\% \text{ of } I = \frac{I}{4}$.
The intensity of the second ray $A'B'$ is determined by the transmission and reflection processes:
$1$. At point $A$, $75\%$ of $I$ is transmitted into the slab: $I_{trans} = \frac{3}{4}I$.
$2$. At point $C$, $25\%$ of $I_{trans}$ is reflected: $I_{refl} = \frac{1}{4} \times \frac{3}{4}I = \frac{3}{16}I$.
$3$. At point $A'$, $75\%$ of $I_{refl}$ is transmitted out of the slab: $I_2 = \frac{3}{4} \times \frac{3}{16}I = \frac{9}{64}I$.
The ratio of maximum to minimum intensity is given by:
$\frac{I_{max}}{I_{min}} = \left( \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} \right)^2$
Substituting the values:
$\sqrt{I_1} = \sqrt{\frac{I}{4}} = \frac{1}{2}\sqrt{I}$
$\sqrt{I_2} = \sqrt{\frac{9}{64}I} = \frac{3}{8}\sqrt{I}$
$\frac{I_{max}}{I_{min}} = \left( \frac{\frac{1}{2} + \frac{3}{8}}{\frac{1}{2} - \frac{3}{8}} \right)^2 = \left( \frac{\frac{4+3}{8}}{\frac{4-3}{8}} \right)^2 = \left( \frac{7}{1} \right)^2 = \frac{49}{1}$

(B) For dispersion without deviation,the net deviation produced by the combination must be zero.
The condition for dispersion without deviation is given by the formula:
$(\mu - 1)A + (\mu' - 1)A' = 0$
Here,the prisms are combined to produce dispersion without deviation,which implies the deviations are in opposite directions:
$(\mu - 1)A = -(\mu' - 1)A'$
Taking the magnitudes:
$\frac{A}{A'} = \frac{(\mu' - 1)}{(\mu - 1)}$
Given:
$A = 4^o$,$\mu = 1.54$,$\mu' = 1.72$
Substituting the values:
$\frac{4}{A'} = \frac{(1.72 - 1)}{(1.54 - 1)}$
$\frac{4}{A'} = \frac{0.72}{0.54}$
Solving for $A'$:
$A' = \frac{4 \times 0.54}{0.72}$
$A' = \frac{2.16}{0.72} = 3^o$

(C) The deviation produced by a thin prism of prism angle $A$ is given by $\delta = (\mu - 1) A$.
Since the combined prism produces dispersion without deviation,the total deviation is zero.
$\delta = \delta_1 + \delta_2 = 0$
$(\mu_1 - 1) A_1 + (\mu_2 - 1) A_2 = 0$
$(\mu_1 - 1) A_1 = -(\mu_2 - 1) A_2$
The negative sign indicates that the refracting angles of the two prisms are in opposite directions.
$A_2 = \frac{(\mu_1 - 1) A_1}{(\mu_2 - 1)} = \frac{(1.54 - 1) \times 4^o}{(1.72 - 1)}$
$A_2 = \frac{0.54 \times 4^o}{0.72} = \frac{3}{4} \times 4^o = 3^o$

(C) The correct answer is $(C)$.
In $CrCl_3$,the chromium ion is $Cr^{3+}$.
The electronic configuration of $Cr^{3+}$ is $[Ar] 3d^3$.
Since it contains $3$ unpaired electrons in the $d$-orbital,it exhibits $d-d$ transitions,which results in the formation of coloured solutions.

(C) The correct answer is $(C)$.
Alcoholic $KOH$ contains alkoxide ions $(RO^-)$,which act as strong bases.
These ions abstract a proton from the $\beta$-carbon of an alkyl halide,leading to an elimination reaction known as dehydrohalogenation.
The reaction is: $CH_3-CH_2-Br + KOH_{(alc)} \xrightarrow{\Delta} CH_2=CH_2 + KBr + H_2O$.
Mechanism:
$ROH + KOH \to ROK + H_2O$
$ROK \to RO^- + K^+$
$RO^- + H-CH_2(\beta)-CH_2(\alpha)-Br \to ROH + CH_2=CH_2 + Br^-$

(B) Neopentyl bromide $(CH_3-C(CH_3)_2-CH_2-Br)$ is a primary alkyl halide with significant steric hindrance at the $\beta-$carbon.
Upon reaction with alcoholic $KOH$,the formation of a primary carbocation is unfavorable.
Instead,it undergoes a concerted rearrangement via a $1, 2-$methyl shift to form a more stable tertiary carbocation $(CH_3-C^{+}(CH_3)-CH_2-CH_3)$.
Elimination of a proton from this intermediate according to Saytzeff's rule yields $2-$methyl$-2-$butene as the major product.

(B) The standard reduction potential of $Mg^{2+}/Mg$ is $-2.37 \ V$,which is much lower than the reduction potential of water $(E^o = -0.83 \ V)$.
In an aqueous solution,water molecules are reduced more easily than $Mg^{2+}$ ions.
Therefore,during electrolysis,$H_2$ gas is evolved at the cathode instead of $Mg$ metal.
Thus,$Mg$ cannot be obtained by the electrolysis of its aqueous salt solution.

(D) The cell reaction is $Fe^{2+} + Sn \to Fe + Sn^{2+}$.
Here,$Fe^{2+}$ is reduced to $Fe$ (cathode) and $Sn$ is oxidized to $Sn^{2+}$ (anode).
The standard cell potential is calculated as $E^o_{cell} = E^o_{cathode} - E^o_{anode}$.
Given $E^o_{Fe^{2+}/Fe} = -0.44 \ V$ and $E^o_{Sn^{2+}/Sn} = -0.14 \ V$.
$E^o_{cell} = (-0.44 \ V) - (-0.14 \ V) = -0.44 + 0.14 = -0.30 \ V$.

(B) $Mg$ and $Al$ cannot be obtained by the electrolysis of aqueous solutions of their salts.
This is because the reduction potential of $H_2O$ is higher than that of $Mg^{2+}$ and $Al^{3+}$ ions.
Consequently,$H_2$ gas is liberated at the cathode instead of the metal.

(D) The reaction proceeds in two steps:
$1$. Chlorination of toluene in the presence of light and heat (free radical substitution) yields benzyl chloride $(C_6H_5CH_2Cl)$.
$2$. Treatment of benzyl chloride with aqueous $NaOH$ (nucleophilic substitution) replaces the chlorine atom with a hydroxyl group,resulting in the formation of benzyl alcohol $(C_6H_5CH_2OH)$.

(C) The reaction of a carbonyl compound $(R_2C=O)$ with hydrogen cyanide $(HCN)$ involves the attack of the nucleophile $(CN^-)$ on the electrophilic carbonyl carbon atom.
This is a characteristic reaction of aldehydes and ketones known as nucleophilic addition reaction.

(D) The depression in freezing point $(\Delta T_f)$ is a colligative property,which is directly proportional to the van't Hoff factor $(i)$,where $\Delta T_f = i \times K_f \times m$.
For equimolal solutions,the freezing point is highest when the depression in freezing point is minimum.
This occurs when the number of particles produced in the solution is minimum.
$C_6H_{12}O_6$ (glucose) is a non-electrolyte,so $i = 1$.
$C_6H_5NH_3^+Cl^-$ dissociates into $2$ ions $(i = 2)$.
$Ca(NO_3)_2$ dissociates into $3$ ions $(i = 3)$.
$La(NO_3)_3$ dissociates into $4$ ions $(i = 4)$.
Since glucose has the lowest $i$ value,it shows the minimum depression in freezing point,resulting in the highest freezing point.