AIIMS 1996 Chemistry Question Paper with Answer and Solution

(C) The charge on an atom or ion is calculated by the difference between the number of protons and the number of electrons.
Charge = (Number of protons) - (Number of electrons)
Charge = $17 - 18 = -1$
Therefore,the species is a chloride ion $(Cl^-)$ which has $17$ protons,$18$ neutrons,and $18$ electrons.

(C) The correct answer is $(C)$.
$C_2H_2$ (acetylene) has a linear structure because both carbon atoms are $sp$-hybridized.
In $sp$-hybridization,the bond angle is $180^\circ$,resulting in a linear geometry.

(D) The exceptionally high boiling point of water is due to the presence of strong intermolecular $ \text{Hydrogen bonding} $ between $ H_2O $ molecules. This requires more energy to break the bonds,thereby increasing the boiling point.

(C) The relationship between $K_p$ and $K_c$ is given by the equation: $K_p = K_c(RT)^{\Delta n}$.
For $K_p = K_c$,the value of $\Delta n$ must be $0$.
$\Delta n$ is defined as the difference between the sum of the stoichiometric coefficients of gaseous products and gaseous reactants.
For option $(C)$: $H_{2_{(g)}} + Cl_{2_{(g)}} \rightleftharpoons 2HCl_{(g)}$,$\Delta n = 2 - (1 + 1) = 0$.
Therefore,for this reaction,$K_p = K_c$.

(D) According to Le Chatelier's principle,increasing the volume of the container decreases the total pressure of the system.
To counteract this,the equilibrium shifts in the direction where the total number of gaseous moles increases (i.e.,$\Delta n_g > 0$).
For option $(A)$: $\Delta n_g = 2 - (1+1) = 0$.
For option $(B)$: $\Delta n_g = 2 - (2+1) = -1$.
For option $(C)$: $\Delta n_g = 2 - (1+3) = -2$.
For option $(D)$: $\Delta n_g = (1+1) - 1 = +1$.
Since $\Delta n_g$ is positive only for option $(D)$,the equilibrium will shift to the right when the volume is doubled.

(A) Given that $[OH^{-}] = 10^{-7} \ M$.
$pOH = -\log[OH^{-}] = -\log(10^{-7}) = 7$.
We know that at $25^{\circ}C$,$pH + pOH = 14$.
Therefore,$pH = 14 - pOH = 14 - 7 = 7$.

(D) At $100 ^oC$ and $1 \text{atm}$ pressure,water is in equilibrium with its vapor,i.e.,$H_2O_{(l)} \rightleftharpoons H_2O_{(g)}$.
For any process at equilibrium,the change in Gibbs free energy is $\Delta G = 0$.
Since $\Delta G = \Delta H - T\Delta S$,at equilibrium,we have $0 = \Delta H - T\Delta S$.
Therefore,$\Delta H = T\Delta S$.

(B) $H_2O_2$ can act as an oxidizing agent by gaining electrons (reducing to $H_2O$) and as a reducing agent by losing electrons (oxidizing to $O_2$).
Therefore,it acts as both an oxidizing and a reducing agent.

(C) The oxidation states of the given elements are as follows:
$N$: $-3$ to $+5$
$O$: $-2$ to $+2$
$Cl$: $-1$ to $+7$
$C$: $-4$ to $+4$
Comparing these,$Cl$ exhibits the highest oxidation number of $+7$.

(B) Baking soda is used for baking purposes.
The chemical formula of baking soda is $NaHCO_3$.
Baking soda is chemically known as sodium bicarbonate or sodium hydrogen carbonate.

(A) To find the percentage of nitrogen,we calculate the mass percentage for each compound:
$1$. Urea $(NH_2CONH_2)$: Molar mass = $60 \ g/mol$. Nitrogen mass = $28 \ g$. Percentage = $(28/60) \times 100 = 46.66 \%$.
$2$. Ammonium sulphate $(NH_4)_2SO_4$: Molar mass = $132 \ g/mol$. Nitrogen mass = $28 \ g$. Percentage = $(28/132) \times 100 = 21.21 \%$.
$3$. Ammonium nitrate $(NH_4NO_3)$: Molar mass = $80 \ g/mol$. Nitrogen mass = $28 \ g$. Percentage = $(28/80) \times 100 = 35 \%$.
$4$. Calcium nitrate $Ca(NO_3)_2$: Molar mass = $164 \ g/mol$. Nitrogen mass = $28 \ g$. Percentage = $(28/164) \times 100 = 17.07 \%$.
Thus,Urea has the highest percentage of nitrogen.

(B) desiccator is a container used to keep substances dry or to remove moisture from them. $A$ substance used in a desiccator must be a strong dehydrating agent or a desiccant.
$Conc. H_2SO_4$,$CaCl_2$,and $P_4O_{10}$ are all well-known dehydrating agents and are commonly used as desiccants.
$Na_2SO_4$ (anhydrous sodium sulfate) is not a strong dehydrating agent; it is often used as a drying agent for organic solvents,but it is not suitable for use in a desiccator to maintain a dry atmosphere for other substances.

(A) The correct answer is $A$.
$AgI$ (Silver iodide) is insoluble in water due to its high covalent character,which arises from the large size of the iodide ion $(I^-)$ and the polarizing power of the silver ion $(Ag^+)$ according to Fajan's rule.
In contrast,$KBr$,$CaCl_2$,and $AgF$ are ionic compounds and are generally soluble in water.

(D) In the titration of $KMnO_4$ with $Fe^{2+}$ ions,$KMnO_4$ acts as a self-indicator.
As soon as the equivalence point is reached,the addition of one extra drop of $KMnO_4$ imparts a permanent pink color to the solution.
Therefore,no external indicator is required.

(B) Ethane $(C_2H_6)$ is a saturated hydrocarbon (alkane).
$(A)$ It undergoes free radical substitution with chlorine to form chloroethane.
$(B)$ Catalytic hydrogenation is a process used for unsaturated hydrocarbons (alkenes or alkynes) to convert them into alkanes. Since ethane is already saturated,it cannot be catalytically hydrogenated.
$(C)$ Complete combustion of alkanes produces $CO_2$ and $H_2O$.
$(D)$ Ethane $(C_2H_6)$ and iso-butane $(C_4H_{10})$ belong to the same homologous series (alkanes),differing by two $CH_2$ units,so they are homologues.
Therefore,the statement that is not true is $(B)$.

(A) Petroleum refining is the process of separating crude oil into various useful products such as gasoline,diesel,kerosene,and lubricating oils. This is primarily achieved through fractional distillation,where the components are separated based on their different boiling points.

(C) The reaction of sodium benzoate $(C_6H_5COONa)$ with sodalime $(NaOH + CaO)$ is a decarboxylation reaction.
When sodium benzoate is heated with sodalime,it undergoes decarboxylation to produce benzene $(C_6H_6)$ and sodium carbonate $(Na_2CO_3)$.
The chemical equation is:
$C_6H_5COONa + NaOH \xrightarrow{CaO} C_6H_6 + Na_2CO_3$
Therefore,the correct option is $(C)$.

(B) The catalytic hydrogenation of benzene $(C_6H_6)$ involves the addition of $3$ moles of hydrogen $(H_2)$ in the presence of a catalyst like nickel $(Ni)$,palladium $(Pd)$,or platinum $(Pt)$ at high temperature and pressure.
This reaction results in the saturation of the aromatic ring to form cyclohexane $(C_6H_{12})$.
The chemical equation is: $C_6H_6 + 3H_2 \xrightarrow{Ni} C_6H_{12}$ (Cyclohexane).

(B) catalyst is a substance that increases the rate of a chemical reaction by providing an alternative pathway with lower activation energy.
$1.$ Haber$'$s process uses $Fe$ as a catalyst for the synthesis of $NH_3$.
$2.$ The Thermite process involves the reduction of metal oxides (like $Cr_2O_3$ or $Fe_2O_3$) by aluminum $(Al)$. The reaction $2Al + Cr_2O_3 \rightarrow Al_2O_3 + 2Cr$ is a highly exothermic,self-sustaining redox reaction that does not require a catalyst.
$3.$ The Ostwald process uses $Pt/Rh$ gauze as a catalyst for the oxidation of $NH_3$.
$4.$ The Contact process uses $V_2O_5$ as a catalyst for the oxidation of $SO_2$ to $SO_3$.
Therefore,the Thermite process does not involve a catalyst.

(D) The catalytic oxidation of ammonia (Ostwald process) is represented by the reaction:
$4NH_3 + 5O_2 \xrightarrow{Pt, 1100 \ K} 4NO + 6H_2O$
The oxide formed is $NO$ (Nitric oxide).
This $NO$ is further oxidized to $NO_2$ and then absorbed in water to produce $HNO_3$:
$2NO + O_2 \to 2NO_2$
$4NO_2 + 2H_2O + O_2 \to 4HNO_3$
Thus,the oxide used in the preparation of $HNO_3$ is $NO$.

(A) $NCl_3$ (Nitrogen trichloride) is highly reactive and thermally unstable due to the weak $N-Cl$ bond and the small size of the nitrogen atom,which leads to significant steric repulsion between the chlorine atoms.
Consequently,it decomposes explosively into nitrogen gas and chlorine gas.
Therefore,the correct option is $A$.

(A) The brown ring test is a common chemical test used to detect the presence of the nitrate ion $(NO_3^-)$ in a solution.
When a freshly prepared ferrous sulfate $(FeSO_4)$ solution is added to the nitrate solution,followed by the careful addition of concentrated sulfuric acid $(H_2SO_4)$ along the sides of the test tube,a brown ring is formed at the junction of the two layers.
The brown ring is due to the formation of the complex $[Fe(H_2O)_5NO]SO_4$.

(A) The body-centered cubic $(BCC)$ structure is characterized by atoms at the corners and one atom at the center of the unit cell.
Among the given options,$Sodium$ $(Na)$ crystallizes in a $BCC$ structure at room temperature.
$Magnesium$ $(Mg)$ and $Zinc$ $(Zn)$ typically exhibit hexagonal close-packed $(HCP)$ structures,while $Copper$ $(Cu)$ exhibits a face-centered cubic $(FCC)$ structure.
Therefore,the correct option is $A$.

(B) In a body-centered cubic $(BCC)$ lattice,each atom at the center of the unit cell is surrounded by $8$ atoms located at the corners of the unit cell.
Therefore,the coordination number of a body-centered cubic lattice is $8$.

(D) catalyst is a substance that increases the rate of a chemical reaction without itself undergoing any permanent chemical change.
$1$. It provides an alternative pathway with lower activation energy $(E_a)$.
$2$. Although it may participate in the reaction mechanism,it is regenerated at the end of the process.
$3$. It speeds up both the forward and backward reactions equally,thus it does not change the equilibrium constant $(K_{eq})$ or the position of equilibrium.
Therefore,all the given statements are correct.

(B) catalyst is a substance that lowers the activation energy of a reaction by providing an alternate pathway. Generally,a catalyst increases the rate of a reaction.
In the Haber's process,$Fe$ is used as a catalyst.
In the Ostwald process,$Pt/Rh$ is used as a catalyst.
In the Contact process,$V_2O_5$ is used as a catalyst.
The Thermite process involves the reduction of metal oxides by aluminum,such as $2 Al + Fe_2O_3 \rightarrow Al_2O_3 + 2 Fe$. This is a highly exothermic,self-sustaining redox reaction that does not require a catalyst.

(C) The oxidation of sulfur dioxide $(SO_2)$ to sulfur trioxide $(SO_3)$ is a key step in the Contact Process for the manufacture of sulfuric acid.
This reaction is catalyzed by vanadium$(V)$ oxide $(V_2O_5)$.
The chemical equation is: $2SO_2(g) + O_2(g) \xrightarrow{V_2O_5} 2SO_3(g)$.

(C) When phenol is treated with bromine water,the $-OH$ group strongly activates the benzene ring towards electrophilic substitution.
Due to this high activation,bromine atoms substitute at all available ortho and para positions,resulting in the formation of $2,4,6-$tribromophenol.
This product is obtained as a white precipitate.
The reaction is: $C_6H_5OH + 3Br_2 (aq) \rightarrow C_6H_2Br_3OH + 3HBr$.

(A) Bakelite is a thermosetting polymer formed by the condensation polymerization of phenol and formaldehyde in the presence of an acid or base catalyst.
Polystyrene is prepared by the polymerization of styrene monomers.
Nylon is a polyamide formed by the condensation polymerization of diamines and dicarboxylic acids.
$PVC$ (polyvinyl chloride) is formed by the addition polymerization of vinyl chloride monomers.
Therefore,phenol is a key constituent in the manufacture of Bakelite.

(D) When calcium acetate $(CH_3COO)_2Ca$ is subjected to dry distillation (dry heating),it undergoes thermal decomposition to produce acetone $(CH_3COCH_3)$ and calcium carbonate $(CaCO_3)$.
The reaction is as follows:
$(CH_3COO)_2Ca \xrightarrow{\Delta} CH_3COCH_3 + CaCO_3$
Therefore,the correct option is $(D)$.

(C) Acetophenone is prepared by the Friedel-Crafts acylation of benzene.
In this reaction,benzene reacts with acetyl chloride $(CH_3COCl)$ in the presence of an anhydrous Lewis acid catalyst,$AlCl_3$.
The reaction is: $C_6H_6 + CH_3COCl \xrightarrow{AlCl_3} C_6H_5COCH_3 + HCl$.
Thus,the correct reactants are benzene and acetyl chloride.

(A) The correct answer is $A$.
Fehling solution is a mild oxidizing agent used to distinguish between aldehydes and ketones.
Aldehydes (like acetaldehyde and formalin) and reducing sugars (like $D^{-}$-glucose) are oxidized by Fehling solution to give a brick red precipitate of $Cu_2O$.
Ketones,such as acetone,are not easily oxidized by Fehling solution because they require a stronger oxidizing agent to break the $C-C$ bond.
Therefore,acetone does not give a brick red precipitate.

(D) $CH_3CHO$ gives a positive Tollen's test (silver mirror test) because it is an aldehyde,forming a silver mirror: $CH_3CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow CH_3COO^- + 2Ag_{\downarrow} + 4NH_3 + 2H_2O$.
Acetone is a ketone and does not react with Tollen's reagent.

(A) Acetaldehyde $(CH_3CHO)$ is an aldehyde.
When treated with a strong oxidizing agent like concentrated $KMnO_4$,it undergoes oxidation.
The aldehyde group $(-CHO)$ is oxidized to a carboxylic acid group $(-COOH)$.
Therefore,the reaction is: $CH_3CHO + [O] \xrightarrow{KMnO_4} CH_3COOH$ (Acetic acid).

(C) The reaction between ethyl alcohol $(C_2H_5OH)$ and acetic acid $(CH_3COOH)$ in the presence of concentrated $H_2SO_4$ is given by:
$C_2H_5OH + CH_3COOH \xrightarrow{Conc. H_2SO_4} CH_3COOC_2H_5 + H_2O$
This reaction produces an ester (ethyl acetate),which has a characteristic fruity smell.
Therefore,this reaction is known as esterification.

(B) The reaction of aniline $(C_6H_5NH_2)$ with nitrous acid $(HNO_2)$,which is generated in situ by the reaction of $NaNO_2$ and dil. $HCl$,at a low temperature of $0^{\circ} - 5^{\circ}C$ is known as the diazotization reaction.
The chemical equation for this reaction is:
$C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{0^{\circ} - 5^{\circ}C} C_6H_5N_2^+Cl^- + NaCl + 2H_2O$
The product formed is benzene diazonium chloride.

(D) Enzymes are biological catalysts that speed up biochemical reactions.
They are primarily proteinaceous in nature,meaning they consist of amino acids.
They exhibit optimum activity at specific temperatures,which for human enzymes is typically around the body temperature of $37.5^{\circ} C$.
Therefore,all the given statements are correct.

(D) Proteins are polymers of amino acids linked by peptide bonds.
On hydrolysis,proteins break down into their constituent amino acids.
The reaction is represented as:
$Dipeptide + H_2O \rightarrow Amino \ Acid_1 + Amino \ Acid_2$

(D) Nucleic acids are polymers of nucleotides. Each nucleotide consists of a nitrogenous base,a pentose sugar,and a phosphate group.
In the structure of a nucleic acid,the backbone is formed by the alternating sequence of sugar and phosphate groups,while the nitrogenous bases are attached to the sugar units.
Therefore,the repeating unit in the backbone is sugar-phosphate,with the base attached to the sugar. Among the given options,the sequence of components in a nucleotide unit is often represented as sugar-phosphate-base (where the base is attached to the sugar). However,if referring to the backbone chain specifically,it is sugar-phosphate. Given the options provided,$D$ is the most appropriate representation of the structural arrangement.

(B) Compounds containing the $CH_3CO-$ group or those that can be oxidized to this group (like $CH_3CH(OH)-$ group) give a positive iodoform test.
Ethyl alcohol $(CH_3CH_2OH)$ is oxidized to acetaldehyde $(CH_3CHO)$ by $I_2/NaOH$,which then reacts to form iodoform $(CHI_3)$.
The reaction is:
$CH_3CH_2OH$ $\xrightarrow{I_2/NaOH} CH_3CHO$ $\xrightarrow{I_2/NaOH} CI_3CHO$ $\xrightarrow{NaOH} CHI_3 + HCOONa$