AIIMS 1982 Chemistry Question Paper with Answer and Solution

(C) The average value of the current over one complete cycle $T$ is given by the formula:
$I_{av} = \frac{1}{T} \int_0^T i(t) dt$
Since the rectifier output is a half-wave rectified signal (as shown in the figure),the current is $I_0 \sin(\omega t)$ for $0 \le t \le T/2$ and $0$ for $T/2 < t < T$.
$I_{av} = \frac{1}{T} \int_0^{T/2} I_0 \sin(\omega t) dt$
Using $\omega = \frac{2\pi}{T}$,we have:
$I_{av} = \frac{I_0}{T} \left[ -\frac{\cos(\omega t)}{\omega} \right]_0^{T/2}$
$I_{av} = \frac{I_0}{T \omega} [-\cos(\omega \cdot T/2) + \cos(0)]$
$I_{av} = \frac{I_0}{T (2\pi/T)} [-\cos(\pi) + 1]$
$I_{av} = \frac{I_0}{2\pi} [1 + 1] = \frac{2I_0}{2\pi} = \frac{I_0}{\pi}$
Wait,looking at the provided figure,it represents a full-wave rectified signal. For a full-wave rectifier,the average value is calculated over the interval $T/2$:
$I_{av} = \frac{1}{T/2} \int_0^{T/2} I_0 \sin(\omega t) dt = \frac{2}{T} \left[ -\frac{I_0 \cos(\omega t)}{\omega} \right]_0^{T/2}$
$I_{av} = \frac{2I_0}{T \omega} [-\cos(\pi) + 1] = \frac{2I_0}{T (2\pi/T)} [1 + 1] = \frac{2I_0}{2\pi} \times 2 = \frac{2I_0}{\pi}$

(B) The molar mass of hydrogen gas $(H_2)$ is $2 \ g/mol$.
$2 \ g$ of $H_2$ contains $6.022 \times 10^{23}$ molecules.
Therefore,$1 \ g$ of $H_2$ contains $\frac{6.022 \times 10^{23}}{2} = 3.011 \times 10^{23}$ molecules.
The value is $3.01 \times 10^{23}$ molecules.

(D) The normality $(N)$ of a solution is given by the formula $N = M \times n$-factor.
For $H_3PO_4$,the basicity is $3$ (it is a tribasic acid),so its $n$-factor is $3$.
Given molarity $(M)$ = $1 \, M$.
Therefore,$N = 1 \times 3 = 3 \, N$.

(B) The given statements describe the conditions for filling degenerate orbitals.
Statement $i$ refers to the preference for single occupancy before pairing.
Statement $ii$ refers to the stability gained by having parallel spins in degenerate orbitals.
These are the fundamental tenets of $Hund's$ rule of maximum multiplicity.

(C) The correct answer is $(C)$.
$BCl_3$ is an electron-deficient compound because the central Boron atom has only $6$ electrons in its valence shell after forming three covalent bonds with Chlorine atoms,which is less than the stable octet of $8$ electrons.

(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) In a water molecule $(H_2O)$,the central oxygen atom undergoes $sp^3$ hybridization.
According to the $VSEPR$ theory,the presence of two lone pairs on the oxygen atom causes repulsion,which compresses the bond angle from the ideal tetrahedral angle of $109^o 28'$ to approximately $104.5^o$ (often represented as $104^o 30'$).
Therefore,the correct option is $(d)$.

(C) $o-$Nitrophenol exhibits intramolecular hydrogen bonding,which reduces its intermolecular attraction.
$p-$Nitrophenol exhibits intermolecular hydrogen bonding,leading to association of molecules and a higher boiling point.
Therefore,$o-$nitrophenol is more volatile than $p-$nitrophenol due to the difference in the type of hydrogen bonding.

(C) The equilibrium constant ($K_c$ or $K_p$) is a constant value for a given reaction at a specific temperature.
It is independent of the initial concentrations of the reactants or products.
It only changes when the temperature of the system is changed.

(C) According to $Le \ Chatelier's \ principle$,if the concentration of one or more products is increased in a system at equilibrium,the system will shift in the reverse or backward direction to counteract the change and re-establish equilibrium.
Therefore,option $(C)$ is correct.

(D) The solubility product $(K_{sp})$ represents the maximum concentration of ions that can exist in a saturated solution at a given temperature.
When the ionic product $(Q_{ip})$ of a solution is less than $K_{sp}$,the solution is unsaturated.
When $Q_{ip} = K_{sp}$,the solution is saturated.
When $Q_{ip} > K_{sp}$,the solution becomes supersaturated,and the excess solute precipitates out of the solution.

(D) buffer solution consists of a weak acid and its conjugate base (salt) or a weak base and its conjugate acid (salt).
$NH_4OH$ is a weak base and $NH_4Cl$ is its salt with a strong acid $(HCl)$.
Therefore,the pair ($NH_4OH$ and $NH_4Cl$) forms a basic buffer solution.
The correct option is $(D)$.

(C) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by the equation: $\Delta H = \Delta E + \Delta n_g RT$.
For the reaction $CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)}$,the change in the number of moles of gaseous species is calculated as: $\Delta n_g = n_{p(g)} - n_{r(g)} = 1 - (1 + 0.5) = 1 - 1.5 = -0.5$.
Since $\Delta n_g = -0.5$ (which is negative),the term $\Delta n_g RT$ is negative.
Therefore,$\Delta H = \Delta E - 0.5 RT$,which implies that $\Delta H < \Delta E$.

(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.

(D) The correct answer is $(D)$.
In the periodic table,as we move from left to right across a period,the atomic radius decreases due to an increase in effective nuclear charge.
All the given elements ($Mg$,$Al$,$Si$,and $P$) belong to the $3^{rd}$ period.
Their positions are: $Mg$ (Group $2$),$Al$ (Group $13$),$Si$ (Group $14$),and $P$ (Group $15$).
Since $Mg$ is the leftmost element among these in the $3^{rd}$ period,it has the largest atomic radius.

(B) The electropositive character (metallic character) of an element increases down a group and decreases across a period from left to right.
Among the given options,$[He] \, 2s^1$ is Lithium $(Li)$,$[Xe] \, 6s^1$ is Cesium $(Cs)$,$[He] \, 2s^2$ is Beryllium $(Be)$,and $[Xe] \, 6s^2$ is Barium $(Ba)$.
Cesium $(Cs)$ is an alkali metal located at the bottom of the alkali metal group (Group $1$),making it the most electropositive element among the choices provided.
Therefore,the correct configuration is $[Xe] \, 6s^1$.

(D) $CCl_4 + H_2O \to$ No reaction.
Carbon atom does not have vacant $d$-orbitals to accept the lone pair of electrons from water molecules,hence it cannot undergo hydrolysis.

(C) The correct answer is $(C)$.
$NaF$ has the highest melting point among the given halides.
According to Fajan's rule,as the size of the anion increases,the covalent character increases,and the ionic character decreases.
Since the fluoride ion $(F^-)$ is the smallest among the halides,$NaF$ has the maximum ionic character and the strongest electrostatic forces of attraction,leading to the highest melting point.

(B) The melting point of an ionic compound is directly proportional to its ionic character. According to Fajan's rule,the polarizing power of a cation is small and the polarizability of the anion is small when the electronegativity difference between the bonded atoms is large. In $NaF$,the electronegativity difference between $Na$ $(0.9)$ and $F$ $(4.0)$ is the largest among alkali metal halides,resulting in the maximum ionic character. Therefore,$NaF$ has the highest melting point.

(C) . Metamerism is a special type of isomerism shown by compounds containing polyvalent functional groups like ethers $(-O-)$,ketones $(-CO-)$,and secondary amines $(-NH-)$. Alkenes do not contain such polyvalent functional groups,hence they do not exhibit metamerism.

(A) Meso-tartaric acid contains a plane of symmetry which divides the molecule into two identical halves.
The optical rotation caused by one half is cancelled by the other half within the same molecule.
This is known as internal compensation or molecular symmetry.

(C) The correct electronic configuration of $Cu$ $(Z=29)$ is $[Ar] 4s^1 3d^{10}$ due to the extra stability of the fully filled $d$-subshell.
Option $C$ shows the configuration as $4s^2 3d^9$,which is incorrect for the ground state of $Cu$.

(D) The motional electromotive force $(EMF)$ induced in a conductor moving in a magnetic field is given by the formula $\varepsilon = (\vec{v} \times \vec{B}) \cdot \vec{l}$.
For an induced potential difference to exist,the velocity of the conductor $\vec{v}$,the magnetic field $\vec{B}$,and the length of the conductor $\vec{l}$ must be mutually perpendicular.
In the given diagram,the magnetic field $\vec{B}$ is directed from the North $(N)$ pole to the South $(S)$ pole (along the $L-Q$ axis).
The length of the conductor is along the axis perpendicular to the page.
To induce an $EMF$,the velocity $\vec{v}$ must be perpendicular to both $\vec{B}$ and the length of the conductor.
Moving the conductor in direction $M$ (upward) or $P$ (downward) satisfies the condition of being perpendicular to the magnetic field lines ($L-Q$ direction) and the length of the conductor.
However,looking at the standard orientation of such problems,direction $M$ is the most appropriate choice to ensure the motional $EMF$ is generated across the ends of the conductor.

(A) The atomic number of Radon $(Rn)$ is $86$.
For an element with atomic number $Z = 106$,the electronic configuration is filled as follows:
$1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^{10}, 4p^6, 5s^2, 4d^{10}, 5p^6, 6s^2, 4f^{14}, 5d^{10}, 6p^6, 7s^2, 5f^{14}, 6d^4$.
However,according to the stability rules for $d$-block elements,the configuration is $[Rn] \, 5f^{14} \, 6d^4 \, 7s^2$.
Wait,checking the specific configuration for Seaborgium $(Z=106)$: The ground state configuration is $[Rn] \, 5f^{14} \, 6d^4 \, 7s^2$.

(B) Thallium is a $p$-block element belonging to group $13$ and period $6$.
The electronic configuration of Thallium is $[Xe] \, 4f^{14} \, 5d^{10} \, 6s^2 \, 6p^1$.
The $d$ and $f$ electrons of inner shells provide poor shielding,which increases the effective nuclear charge on the $6s^2$ electrons,making them reluctant to participate in bonding. This phenomenon is known as the inert pair effect.
Due to this effect,the $6s^2$ electrons remain paired,while the $6p^1$ electron is easily lost,leading to the $+1$ oxidation state. The $+3$ state is also possible but less stable than the $+1$ state for Thallium.

(B) The brown ring test is used to detect the presence of nitrate ions $(NO_3^-)$ in a solution.
In this test,$Fe(II)$ ions react with $NO$ (nitric oxide) to form a brown-colored coordination complex.
The reaction is: $[Fe(H_2O)_6]^{2+} + NO \to [Fe(H_2O)_5(NO)]^{2+} + H_2O$.
This complex,$[Fe(H_2O)_5(NO)]^{2+}$,is responsible for the brown ring observed at the junction of the two layers.

(A) Pseudohalide ions are monovalent negative ions composed of two or more electronegative atoms that exhibit chemical properties similar to those of halide ions $(X^-)$.
These ions can form dimeric molecules known as pseudohalogens,which behave similarly to halogens $(X_2)$.
Among the given options,the cyanide ion $(CN^-)$ is a well-known pseudohalide,and its corresponding pseudohalogen is cyanogen $(CN)_2$.

Pseudohalide	Pseudohalogen
$CN^-$ (Cyanide)	$(CN)_2$ (Cyanogen)
$SCN^-$ (Thiocyanate)	$(SCN)_2$ (Thiocyanogen)

(B) The brown ring test is used to detect the presence of nitrate ions $(NO_3^-)$ in a solution.
When a freshly prepared ferrous sulfate solution is added to the nitrate solution followed by the careful addition of concentrated sulfuric acid along the sides of the test tube,a brown ring is formed at the junction of the two layers.
The reaction involves the reduction of nitrate to nitric oxide $(NO)$,which then reacts with the $Fe(II)$ ions to form the brown complex $[Fe(H_2O)_5(NO)]SO_4$.

(C) $Na_2CO_3$ is a strong electrolyte that provides a high concentration of $CO_3^{2-}$ ions.
This high concentration of carbonate ions causes the precipitation of $Mg^{2+}$ ions as $MgCO_3$ along with the fifth group radicals $(Ba^{2+}, Sr^{2+}, Ca^{2+})$.
$(NH_4)_2CO_3$ is a weaker electrolyte and provides a lower concentration of $CO_3^{2-}$ ions,which is sufficient to exceed the solubility product $(K_{sp})$ of the fifth group carbonates but not $MgCO_3$.

(A) The group $III$ radicals $(Fe^{3+}, Al^{3+}, Cr^{3+})$ are precipitated as hydroxides using $NH_4OH$ in the presence of $NH_4Cl$.
$NH_4Cl$ provides $NH_4^+$ ions,which suppress the ionization of $NH_4OH$ due to the common ion effect,ensuring that the concentration of $OH^-$ ions is low enough to prevent the precipitation of group $IV$ and $V$ radicals.
Any ammonium salt that provides $NH_4^+$ ions without introducing interfering anions can be used.
$NH_4NO_3$ is a suitable substitute because it provides $NH_4^+$ ions and does not interfere with the precipitation process.
$(NH_4)_2SO_4$ cannot be used because $SO_4^{2-}$ ions would precipitate $Ba^{2+}, Sr^{2+},$ and $Ca^{2+}$ as sulfates.

(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.

(C) The number of half-lives $(n)$ is calculated as: $n = \frac{\text{Total time}}{t_{1/2}} = \frac{50 \ minutes}{25 \ minutes} = 2$.
The fraction of the substance remaining is given by the formula: $\text{Amount left} = (1/2)^n$.
Substituting the value of $n$: $\text{Amount left} = (1/2)^2 = 1/4$ (One-fourth).

(A) According to Faraday's first law of electrolysis,the mass of a substance deposited is given by $W = Z \times Q$,where $Z$ is the electrochemical equivalent and $Q$ is the charge in Faradays.
For $1 \ F$ of electricity,the mass deposited is equal to the equivalent weight of the metal.
Equivalent weight $= \frac{\text{Atomic weight}}{\text{Valency factor (n-factor)}}$.
For $Ag^{+}$ $(n=1)$: $Eq. wt. = \frac{108}{1} = 108 \ g$.
For $Ni^{2+}$ $(n=2)$: $Eq. wt. = \frac{59}{2} = 29.5 \ g$.
For $Cr^{3+}$ $(n=3)$: $Eq. wt. = \frac{52}{3} \approx 17.33 \ g$.
Thus,the masses deposited are $108 \ g$,$29.5 \ g$,and $17.3 \ g$ respectively.

(C) In electrolytic cells,the electrodes must be chemically inert so that they do not react with the electrolyte or the products of electrolysis.
$Na$ is a highly reactive alkali metal that reacts vigorously with water.
$S$ (sulfur) is a non-metal and a poor conductor of electricity.
$Hg$ (mercury) is a liquid metal and can be used as an electrode in specific cases (like the Castner-Kellner cell),but $Pt$ (platinum) and graphite are the standard inert electrodes used in aqueous electrolytic cells because they are chemically stable and conduct electricity well.
Therefore,$Pt$ and graphite are the most suitable choices for general electrolytic cells.

(D) We use the relation $\Delta G^o = -nFE^o$.
For reaction $(i): Fe^{2+} + 2e^- \to Fe$,$\Delta G_1^o = -2 \times F \times (-0.440 \ V) = 0.880 \ F$.
For reaction $(ii): Fe^{3+} + 3e^- \to Fe$,$\Delta G_2^o = -3 \times F \times (-0.036 \ V) = 0.108 \ F$.
We want the reaction: $Fe^{3+} + e^- \to Fe^{2+}$.
This can be obtained by $(ii) - (i)$:
$\Delta G_3^o = \Delta G_2^o - \Delta G_1^o = 0.108 \ F - 0.880 \ F = -0.772 \ F$.
Since $\Delta G_3^o = -nFE^o$ where $n=1$:
$-0.772 \ F = -1 \times F \times E^o$.
$E^o = +0.772 \ V \approx +0.771 \ V$.

(A) When an excess of $NH_3$ solution is added to a $CuSO_4$ solution,the pale blue precipitate of $Cu(OH)_2$ initially formed dissolves to form a deep blue solution.
This deep blue colour is due to the formation of the tetraamminecopper$(II)$ complex,$[Cu(NH_3)_4]^{2+}$.
The reaction is: $Cu^{2+}(aq) + 4NH_3(aq) \rightarrow [Cu(NH_3)_4]^{2+}(aq)$.

(A) Lucas reagent is a solution of anhydrous $ZnCl_2$ in concentrated $HCl$.
This reagent is used to classify alcohols of low molecular weight. The reaction involves the substitution of the hydroxyl group $(-OH)$ with a chloride ion $(Cl^-)$.
The reaction proceeds via an $SN1$ mechanism,where the rate of reaction depends on the stability of the carbocation intermediate. Tertiary alcohols react fastest,followed by secondary,while primary alcohols react very slowly or not at all at room temperature.
Therefore,the correct composition is anhydrous $ZnCl_2$ and concentrated $HCl$.

(A) The chemical name $DDT$ stands for dichlorodiphenyl trichloroethane.
Its structure consists of two chlorophenyl groups attached to a central carbon atom,which is also bonded to a hydrogen atom and a trichloromethyl group $(CCl_3)$.
The correct formula is $(Cl-C_6H_4)_2CH-CCl_3$.

(A) The reaction between formaldehyde $(HCHO)$ and ammonia $(NH_3)$ produces hexamethylenetetramine,commonly known as urotropine.
The balanced chemical equation is:
$6HCHO + 4NH_3 \to (CH_2)_6N_4 + 6H_2O$
Thus,the formula of urotropine is $(CH_2)_6N_4$.

(C) The acidity of carboxylic acids is increased by the presence of electron-withdrawing groups $(EWG)$ due to the inductive effect ($-I$ effect).
Greater the number of $EWG$ and closer they are to the carboxyl group,the stronger the acid.
In $CH_3-CH_2-CCl_2-COOH$,there are two chlorine atoms on the $\alpha$-carbon,which exert a strong $-I$ effect,stabilizing the carboxylate anion significantly.
Therefore,$CH_3-CH_2-CCl_2-COOH$ is the strongest acid among the given options and will be most highly ionised in water.

(B) Sugars contain a carbonyl group which can undergo enolization in the presence of the $OH^-$ ions found in an alkaline medium.
This enolization process leads to the rearrangement of the sugar structure (e.g.,Lobry de Bruyn-van Ekenstein transformation).
Therefore,rearrangement is possible in an alkaline medium.

(B) The correct answer is $B$.
Sometimes,yellow turbidity is observed upon passing $H_2S$ gas even when second group radicals are absent.
This occurs because certain fourth group radicals (such as $Zn^{2+}$,$Mn^{2+}$,$Ni^{2+}$,$Co^{2+}$) may precipitate as sulphides if the concentration of $S^{2-}$ ions in the solution is sufficiently high,leading to the formation of a colloidal suspension or turbidity.