AIIMS 1991 Chemistry Question Paper with Answer and Solution

(A) According to Planck's quantum theory,energy is always absorbed or emitted in discrete amounts,which are whole number multiples of a quantum $(E = nh\nu)$.
Therefore,the statement that energy is not absorbed or emitted in whole number or multiple of quantum is incorrect.

(B) The spectrum of an atom or ion depends on the number of electrons present in it.
$He$ (Helium) has $2$ electrons.
$Li^{+}$ (Lithium ion) also has $3 - 1 = 2$ electrons.
Since both species are isoelectronic (contain the same number of electrons),their spectra are expected to be similar.

(B) $CO_2$ has a linear geometry with the two $C=O$ bond dipoles pointing in opposite directions,which cancel each other out. Therefore,the net dipole moment of $CO_2$ is $0 \ D$.

(D) The equilibrium constant ($K_c$ or $K_p$) for a given chemical reaction is a function of temperature only.
It is independent of the initial concentrations of reactants,the total pressure of the system,the volume of the reaction vessel,or the presence of a catalyst.
The relationship between the equilibrium constant and temperature is given by the van't Hoff equation: $\ln \frac{K_2}{K_1} = \frac{\Delta H^\circ}{R} (\frac{1}{T_1} - \frac{1}{T_2})$.
Therefore,the value of the equilibrium constant depends on the temperature.

(D) For a reaction to be spontaneous,the Gibbs free energy change $ \Delta G = \Delta H - T \Delta S $ must be negative $( \Delta G < 0 )$.
If $ \Delta H $ is $ +ve $ (endothermic) and $ \Delta S $ is $ -ve $ (decrease in entropy),then $ \Delta G = (+ve) - T(-ve) = (+ve) + (T \times +ve) = +ve $.
Since $ \Delta G $ is always positive regardless of the temperature $ T $,the reaction is impossible under any conditions.

(C) bomb calorimeter is a constant-volume calorimeter used to measure the heat of combustion of a substance.
Since the volume is kept constant $(dV = 0)$,the work done is zero $(w = -P_{ext} \Delta V = 0)$.
According to the first law of thermodynamics,$\Delta U = q + w$.
Since $w = 0$,the heat measured $(q_v)$ is equal to the change in internal energy,$\Delta U$ (or $\Delta E$).

(C) The electronic configuration of the elements are:
$C (Z=6): 1s^2 2s^2 2p^2$
$N (Z=7): 1s^2 2s^2 2p^3$
$O (Z=8): 1s^2 2s^2 2p^4$
$F (Z=9): 1s^2 2s^2 2p^5$
After the removal of the first electron,the configurations for the second ionisation are:
$C^+: 1s^2 2s^2 2p^1$
$N^+: 1s^2 2s^2 2p^2$
$O^+: 1s^2 2s^2 2p^3$
$F^+: 1s^2 2s^2 2p^4$
The second ionisation potential involves removing an electron from the $2p$ orbital. Among these,$O^+$ has a half-filled $2p^3$ configuration,which is exceptionally stable. Therefore,it requires the highest energy to remove the second electron.

(D) The relationship between normality and molarity is given by the formula: $\text{Normality} = \text{Molarity} \times \text{Basicity}$.
For phosphoric acid $(H_3PO_4)$,the basicity is $3$ because it can donate $3 \ H^+$ ions per molecule.
Given that the molarity is $1 \ M$,the normality is calculated as: $1 \ M \times 3 = 3 \ N$.

(C) The hydrogenation of olefins (alkenes) is a catalytic process.
$(C)$: $CH_2=CH_2 + H_2 \xrightarrow{Raney \ Ni, \Delta} CH_3-CH_3$.
Raney $Ni$ acts as a catalyst in the presence of hydrogen gas to facilitate the addition of hydrogen across the double bond.

(B) The relationship between Normality $(N)$ and Molarity $(M)$ is given by the formula: $N = M \times \text{basicity}$.
For sulphuric acid $(H_2SO_4)$,the basicity is $2$ because it can provide $2$ replaceable $H^+$ ions.
Given $M = 2 \, M$.
Therefore,$N = 2 \times 2 = 4 \, N$.

(A) is the correct statement.
$1$. Both gases and liquids exhibit fluidity,which is characterized by viscosity.
$2$. Molecules in the solid state are fixed in position and do not possess translational motion; they only exhibit vibrational motion.
$3$. Gases can be converted directly into solids through the process of deposition.
$4$. While liquids have significant vapour pressure,solids generally have negligible vapour pressure at room temperature.

(D) $Triticale$ is the first man-made cereal crop.
It is an intergeneric hybrid produced by crossing wheat ($Triticum$ species) and rye ($Secale$ species).
The primary goal of this cross was to combine the high yield and grain quality of wheat with the hardiness and disease resistance of rye.

(A) The relationship between the standard Gibbs free energy change and the equilibrium constant is given by the equation: $\Delta G^o = -RT \ln K$ or $\Delta G^o = -2.303 \, RT \log K$.
Therefore,the equilibrium constant $K$ is directly related to the standard free energy change $\Delta G^o$.

(C) Olefins (alkenes) undergo catalytic hydrogenation in the presence of metal catalysts like $Ni$,$Pd$,or $Pt$ and hydrogen gas $(H_2)$.
Raney $Ni$ is a highly active form of nickel catalyst commonly used for the hydrogenation of alkenes to alkanes.
The reaction is represented as:
$-C=C- + H_2 \xrightarrow{\text{Raney } Ni} -CH-CH-$ (hydrogenation)
Therefore,the correct option is $(C)$.

(D) The electronic configuration of nitrogen ($N$,$Z=7$) is $1s^2, 2s^2, 2p^3$.
Nitrogen belongs to the second period and lacks $d$-orbitals in its valence shell.
Therefore,the statement that $d$-orbitals are available for bonding is incorrect.

(B) The anhydride of an acid is formed by the removal of water from the acid. For nitrous acid $(HNO_2)$,the reaction is:
$2HNO_2 \to H_2O + N_2O_3$.
Thus,$N_2O_3$ is the anhydride of nitrous acid.

(C) The $-COOC_2H_5$ group present in ethyl benzoate is an electron-withdrawing group (deactivating group) that directs incoming electrophiles to the meta position.
Therefore,ethyl benzoate undergoes meta substitution upon monochlorination.
The reaction is:
$C_6H_5COOC_2H_5 + Cl_2 \xrightarrow{FeCl_3} m-Cl-C_6H_4COOC_2H_5 + HCl$

(C) The molality $(m)$ of the solution is calculated as: $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.1 \, mol}{0.2 \, kg} = 0.5 \, mol \, kg^{-1}$.
Using the formula for elevation in boiling point: $\Delta T_b = K_b \times m$.
$\Delta T_b = 0.513 \, ^oC \, kg \, mol^{-1} \times 0.5 \, mol \, kg^{-1} = 0.2565 \, ^oC$.
The boiling point of the solution $(T_b)$ is: $T_b = T_b^0 + \Delta T_b = 100 \, ^oC + 0.2565 \, ^oC = 100.2565 \, ^oC$.
Rounding to two decimal places,the value is approximately $100.25 \, ^oC$.

(D) For a $bcc$ structure,the nearest neighbour distance $d$ is related to the edge length $a$ by the formula $d = \frac{\sqrt{3}}{2} a$.
Therefore,$a = \frac{2d}{\sqrt{3}} = \frac{2 \times 4.52 \ \mathring{A}}{1.732} \approx 5.219 \ \mathring{A} = 5.219 \times 10^{-10} \ m$.
For a $bcc$ unit cell,the number of atoms per unit cell $Z = 2$.
The density $\rho$ is given by $\rho = \frac{Z \times M}{N_A \times a^3}$.
Substituting the values: $\rho = \frac{2 \times 39 \ g/mol}{(6.022 \times 10^{23} \ mol^{-1}) \times (5.219 \times 10^{-10} \ m)^3}$.
$\rho = \frac{78}{6.022 \times 10^{23} \times 1.421 \times 10^{-28}} \approx 911.5 \ kg \ m^{-3}$.
Rounding to the nearest given option,the density is $910 \ kg \ m^{-3}$.

(C) When a cation shifts from its original lattice site to an interstitial position,the defect is known as a $Frenkel$ defect. This defect is common in ionic crystals where the cation is smaller than the anion.

(B) Let the rate law be $Rate = k[A]^n$.
When concentration is doubled,$Rate' = k[2A]^n = 4 \times Rate$.
So,$2^n = 4$,which implies $n = 2$.
When concentration is tripled,$Rate'' = k[3A]^n = 9 \times Rate$.
So,$3^n = 9$,which implies $n = 2$.
Thus,the rate is proportional to the square of the concentration of $A$ $([A]^2)$.

(D) In electrolytic conductors,current is carried by the movement of ions,not by a stream of electrons. Electrons flow through metallic conductors,whereas ions move towards electrodes in electrolytic solutions to form new products. Therefore,the statement that a single stream of electrons flows from cathode to anode is not applicable to electrolytic conductors.

(A) The reaction of ethyl alcohol $(CH_3CH_2OH)$ with bleaching powder $(CaOCl_2)$ involves the following steps:
$1$. Bleaching powder reacts with water to produce chlorine: $CaOCl_2 + H_2O \to Ca(OH)_2 + Cl_2$.
$2$. Chlorine oxidizes ethyl alcohol to acetaldehyde: $CH_3CH_2OH + Cl_2 \to CH_3CHO + 2HCl$.
$3$. Acetaldehyde is then chlorinated to form chloral $(CCl_3CHO)$: $CH_3CHO + 3Cl_2 \to CCl_3CHO + 3HCl$.
$4$. Finally,chloral reacts with calcium hydroxide to form chloroform: $2CCl_3CHO + Ca(OH)_2 \to 2CHCl_3 + (HCOO)_2Ca$.

(B) In non-polar solvents like benzene,the solvation effect (which stabilizes the conjugate acid in water) is absent.
Therefore,the basic strength is primarily determined by the inductive effect ($+I$ effect) of the alkyl groups.
As the number of methyl groups increases,the electron density on the nitrogen atom increases,making the lone pair more available for donation.
The order of basicity based on the $+I$ effect is: $(CH_3)_3N > (CH_3)_2NH > CH_3NH_2$.
However,the provided option $(B)$ represents the order in aqueous solution,which is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N$ due to a combination of inductive effect,solvation,and steric hindrance.
Given the standard textbook context for this specific question,option $(B)$ is the accepted answer.

(B) The dissolution of $AgCl$ in aqueous ammonia occurs because of the formation of a soluble complex ion,diamminesilver$(I)$ chloride.
The chemical equation for the reaction is:
$AgCl(s) + 2NH_3(aq) \to [Ag(NH_3)_2]Cl(aq)$
Thus,the correct product formed is $[Ag(NH_3)_2]Cl$.