IIT JEE 1986 Chemistry Question Paper with Answer and Solution

(C) $CO_2$ has a linear structure with a bond angle of $180^{\circ}$.
$HgCl_2$ also has a linear structure with a bond angle of $180^{\circ}$.
$SnCl_2$ and $SO_2$ are bent (angular) molecules due to the presence of a lone pair on the central atom.
Therefore,$CO_2$ is isostructural with $HgCl_2$.

(B) Rutherford's $\alpha$-particle scattering experiment demonstrated that most of the atom is empty space,with a small,dense,positively charged nucleus at the center. This led to the conclusion that electrons revolve around the nucleus in the extra-nuclear region. Therefore,the correct option is $(B)$.

(C) The correct answer is $(C)$.
For any given value of the azimuthal quantum number $l$,the magnetic quantum number $m$ must range from $-l$ to $+l$ (including zero).
In option $(C)$,$l = 2$,so the possible values for $m$ are $-2, -1, 0, +1, +2$.
Since $m = -3$ is outside this range,this set of quantum numbers is impossible.

(D) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.
This implies that $E \propto \frac{1}{\lambda}$.
Therefore,the ratio of the energies is $\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1}$.
Given $\lambda_1 = 2000 \ \mathring{A}$ and $\lambda_2 = 4000 \ \mathring{A}$.
$\frac{E_1}{E_2} = \frac{4000}{2000} = 2$.

(D) When two similar non-metallic atoms form a bond,they share their valence electrons equally because their electronegativities are identical.
This type of bond is known as a non-polar covalent bond.
Therefore,the electron pair is equally shared between the two atoms.

(A) The correct answer is $A$.
In $N_2O_5$,the structure involves two nitrogen atoms connected by an oxygen atom $(N-O-N)$.
Each nitrogen atom is also bonded to one terminal oxygen atom via a double bond and to another terminal oxygen atom via a coordinate covalent bond (dative bond).
Thus,$N_2O_5$ contains coordinate covalent bonds.

(D) molecule exhibits a dipole moment if it is polar and has a non-zero net dipole moment.
$1, 4-$dichlorobenzene is a symmetric molecule where the dipole moments of the two $C-Cl$ bonds cancel each other out,resulting in a net dipole moment of $0$.
$cis-1, 2-$dichloroethene is polar because the two $Cl$ atoms are on the same side,leading to a non-zero net dipole moment.
$trans-1, 2-$dichloro$-2-$pentene is an unsymmetrical molecule because the groups attached to the double-bonded carbons are different (one side has $H$ and $Cl$,the other has $CH_3$ and $Cl$). Due to this lack of symmetry,the dipole moments do not cancel out,resulting in a non-zero dipole moment.
Therefore,both $(b)$ and $(c)$ exhibit a dipole moment.

(C) The strongest hydrogen bond is found in hydrogen fluoride $(HF)$.
The strength of a hydrogen bond is directly proportional to the electronegativity of the atom involved in the bond.
Since fluorine $(F)$ has the highest electronegativity among the given elements $(F, O, S)$,the $H-F.....F$ bond is the strongest.

(A) coordinate covalent bond (or dative bond) is formed when one atom donates a lone pair of electrons to another atom that is electron-deficient.
In $N_2H_5^+$,the structure is formed by the protonation of hydrazine $(N_2H_4)$.
$N_2H_4$ has a lone pair of electrons on the nitrogen atom. When it reacts with an $H^+$ ion,the nitrogen atom donates its lone pair to the $H^+$ ion to form a coordinate bond.
$N_2H_4 + H^+ \rightarrow [H_2N-NH_3]^+$ or $N_2H_5^+$.
In $BaCl_2$,the bond is ionic. In $HCl$ and $H_2O$,the bonds are purely covalent.

(D) The chemical reaction is: $2NaNO_3(s) \rightleftharpoons 2NaNO_2(s) + O_2(g)$.
According to Le Chatelier's principle:
$1$. Adding more reactant $(NaNO_3)$ shifts the equilibrium in the forward direction to consume the excess reactant.
$2$. Adding more product $(NaNO_2)$ shifts the equilibrium in the reverse direction to consume the excess product.
$3$. Increasing the pressure in a system with gaseous products shifts the equilibrium toward the side with fewer moles of gas. Here,the product side has $1 \text{ mole}$ of $O_2(g)$ and the reactant side has $0 \text{ moles}$ of gas. Thus,increasing pressure shifts the equilibrium to the left (reverse reaction).
Therefore,all the given statements are correct.

(D) Calcium oxalate $(CaC_2O_4)$ is a salt of a strong base $(Ca(OH)_2)$ and a weak acid (oxalic acid,$H_2C_2O_4$).
It dissolves in strong mineral acids like $HCl$ and $HNO_3$ because the oxalate ions $(C_2O_4^{2-})$ react with $H^+$ ions to form the weak acid $H_2C_2O_4$,shifting the equilibrium to the right.
However,acetic acid $(CH_3COOH)$ is a weak acid and does not provide a sufficiently high concentration of $H^+$ ions to significantly shift the equilibrium and dissolve the precipitate.

(C) The compound insoluble in acetic acid is $CaC_2O_4$ (Calcium oxalate).
Acetic acid is a weak acid. When $CaCO_3$,$CaO$,or $Ca(OH)_2$ are added to acetic acid,they react to form soluble calcium acetate.
However,$CaC_2O_4$ has a very low solubility product $(K_{sp})$ and is not soluble in weak acids like acetic acid because the concentration of oxalate ions $(C_2O_4^{2-})$ is not significantly reduced by protonation to a level that would dissolve the precipitate.

(D) $CH_3COONa$ is a salt of a weak acid,$CH_3COOH$,and a strong base,$NaOH$.
Since it is a salt of a weak acid and a strong base,it undergoes anionic hydrolysis to produce $OH^-$ ions,making the solution basic.
$CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$.

(A) The combustion reaction of ethane is: $C_2H_6(g) + \frac{7}{2}O_2(g) \to 2CO_2(g) + 3H_2O(l)$.
The standard heat of combustion $\Delta H_c^{\circ}$ is given by: $\Delta H_c^{\circ} = [2 \times \Delta H_f^{\circ}(CO_2) + 3 \times \Delta H_f^{\circ}(H_2O)] - [\Delta H_f^{\circ}(C_2H_6) + \frac{7}{2} \times \Delta H_f^{\circ}(O_2)]$.
Given $\Delta H_f^{\circ}(CO_2) = -94.1 \ kcal$,$\Delta H_f^{\circ}(H_2O) = -68.3 \ kcal$,$\Delta H_f^{\circ}(C_2H_6) = -21.1 \ kcal$,and $\Delta H_f^{\circ}(O_2) = 0 \ kcal$.
Substituting the values: $\Delta H_c^{\circ} = [2(-94.1) + 3(-68.3)] - [-21.1 + 0]$.
$\Delta H_c^{\circ} = [-188.2 - 204.9] + 21.1$.
$\Delta H_c^{\circ} = -393.1 + 21.1 = -372 \ kcal$.

(A) $NaHCO_3$ is an amphoteric substance (acidic salt) and $NaOH$ is a strong base. \\ When they are present in the same solution,they undergo an acid-base neutralization reaction: \\ $NaHCO_3 + NaOH \rightarrow Na_2CO_3 + H_2O$. \\ Therefore,they cannot coexist in a solution as they react to form sodium carbonate and water.

(B) Among the given oxides,$HgO$ (Mercuric oxide) is thermally unstable at moderate temperatures and decomposes to release oxygen gas.
The reaction is: $2HgO_{(s)} \xrightarrow{\Delta} 2Hg_{(l)} + O_{2(g)}$.

(C) $ (C) $ Calcium oxalate $(CaC_2O_4)$ is a salt of a weak acid (oxalic acid) and a strong base,but it is highly insoluble in water and weak acids like acetic acid $(CH_3COOH)$.
It only dissolves in strong mineral acids like hydrochloric acid $(HCl)$ because the oxalate ion $(C_2O_4^{2-})$ reacts with $H^+$ ions to form undissociated oxalic acid,shifting the equilibrium forward.

(C) The $IUPAC$ name $2, 2, 3-$trimethylhexane indicates a hexane parent chain (six carbon atoms).
At position $2$,there are two methyl groups,and at position $3$,there is one methyl group.
The structure is $CH_3-C(CH_3)_2-CH(CH_3)-CH_2-CH_2-CH_3$ is not correct as per the name.
Let us re-evaluate: $2, 2, 3-$trimethylhexane corresponds to the structure $CH_3-C(CH_3)_2-CH(CH_3)-CH_2-CH_2-CH_3$ is actually $2, 2, 3-$trimethylhexane.
Wait,let us count the carbons: $CH_3(1)-C(2)(CH_3)_2-CH(3)(CH_3)-CH_2(4)-CH_2(5)-CH_3(6)$. This is $2, 2, 3-$trimethylhexane.
Option $A$ is $2, 2, 3-$trimethylpentane.
Option $B$ is $2, 2, 4-$trimethylpentane.
Option $C$ is $2, 2, 3-$trimethylhexane.

(A) To determine the number of monochloro isomers,we count the number of non-equivalent hydrogen atoms in the molecule.
$1$. For $n-$butane $(CH_3-CH_2-CH_2-CH_3)$,there are two sets of equivalent hydrogens: the terminal $CH_3$ groups and the internal $CH_2$ groups. Thus,it gives two monochloro products: $1-$chlorobutane and $2-$chlorobutane.
$2$. For $2,4-$dimethyl pentane,there are three sets of equivalent hydrogens.
$3$. For Benzene,all six hydrogen atoms are equivalent,giving only one monochloro product (chlorobenzene).
$4$. For $1-$methyl propane (isobutane),there are two sets of equivalent hydrogens,but the terminal $CH_3$ groups are equivalent to each other,and the central $CH$ is unique. However,the question asks for the specific case of $n-$butane as the standard answer for two isomers.

(B) The correct answer is $(B)$ $n-$octane.
$1.$ Boiling point depends on molecular mass. $A$ higher molecular mass leads to a higher boiling point.
$2.$ For isomers with the same molecular mass,the boiling point depends on the surface area. Straight-chain (linear) alkanes have a larger surface area compared to branched alkanes,leading to stronger van der Waals forces and thus a higher boiling point.
$3.$ Among the given options,$n-$octane $(C_8H_{18})$ has the highest molecular mass compared to $n-$butane $(C_4H_{10})$ and is a straight-chain isomer compared to iso-octane and $2, 2, 3, 3-$tetramethylbutane,which are branched.

(A) The free radical chlorination of alkanes is a chain reaction that continues until all hydrogen atoms are replaced by chlorine atoms if $Cl_2$ is in excess.
To obtain the mono-substituted product $(C_2H_5Cl)$ as the major product,the alkane $(C_2H_6)$ must be taken in excess.
Therefore,the reaction $C_2H_6$ (excess) $+ Cl_2 \xrightarrow{UV \text{ Light}} C_2H_5Cl + HCl$ provides the maximum yield of the mono-chloro product.

(C) $CaC_2O_4$ (calcium oxalate) is the salt of oxalic acid.
Since oxalic acid is a stronger acid than acetic acid,$CaC_2O_4$ does not dissolve in acetic acid.
In contrast,$CaCO_3$,$CaO$,and $Ca(OH)_2$ react with acetic acid to form soluble calcium acetate.

(C) The central atom in $SO_2$ is sulphur $(S)$.
Sulphur has $6$ valence electrons.
In $SO_2$,sulphur forms two double bonds with two oxygen atoms.
Number of electron pairs around sulphur = $2$ (bonding pairs) + $1$ (lone pair) = $3$.
Since the steric number is $3$,the hybridization is $sp^2$.

(B) The average kinetic energy of gas molecules is directly proportional to the absolute temperature $(KE \propto T)$.
Since the process occurs at a constant temperature,the kinetic energy of the molecules remains constant.

(B) Ethanol $(C_2H_5OH)$ and dimethyl ether $(CH_3OCH_3)$ both have the same molecular formula,$C_2H_6O$.
Since they have the same molecular formula but different structural arrangements (functional groups),they are functional isomers of each other.
Therefore,the correct option is $(B)$.

(D) The radioisotope of hydrogen is Tritium,represented as $_1H^3$ or $_1^3H$.
It consists of $1$ proton and $2$ neutrons.
The sum of the number of neutrons and protons is $1 + 2 = 3$.

(C) Let the intensity of each source be $I$. Since amplitude is $A$,$I \propto A^2$.
In the first case (coherent sources),the intensity at the midpoint (where path difference is $0$) is given by $I_1 = I + I + 2\sqrt{I}\sqrt{I} \cos(0) = 4I$.
In the second case (incoherent sources),the intensities simply add up,so $I_2 = I + I = 2I$.
The ratio of the intensities is $\frac{I_1}{I_2} = \frac{4I}{2I} = \frac{2}{1}$.

(C) The correct option is $C$.
By definition,molarity $(M)$ is defined as the number of moles of solute dissolved in one litre of the solution.
Therefore,a molar solution contains $1 \ mole$ of solute in $1 \ L$ of the solution.

(B) In the Hall-Heroult process,pure alumina $(Al_2O_3)$ has a very high melting point (about $2323 \ K$) and is a poor conductor of electricity.
Cryolite $(Na_3AlF_6)$ is added to alumina to lower its melting point to about $1140 \ K$ and to increase its electrical conductivity,making the electrolysis process more efficient.

(A) In the presence of dilute $HCl$,$H_2S$ provides a low concentration of $S^{2-}$ ions due to the common ion effect.
This concentration is sufficient to exceed the solubility product $(K_{sp})$ of group $II$ metal sulphides (like $Bi_2S_3, SnS_2, HgS, CuS$) but not group $IV$ metal sulphides (like $ZnS, NiS$).
$Bi^{3+}$ and $Sn^{4+}$ both belong to group $II$ and will both precipitate as sulphides in dilute $HCl$,making them inseparable by this method.

(D) The reaction of toluene with chlorine in the presence of a Lewis acid like ferric chloride $(FeCl_3)$ is an electrophilic aromatic substitution reaction (chlorination).
The methyl group $(-CH_3)$ attached to the benzene ring is an ortho/para-directing group due to its electron-donating inductive effect and hyperconjugation.
Therefore,the chlorine electrophile $(Cl^+)$ attacks the ortho and para positions of the toluene ring,resulting in the formation of $o-$chlorotoluene and $p-$chlorotoluene as the major products.
The reaction is: $C_6H_5CH_3 + Cl_2 \xrightarrow{FeCl_3} C_6H_4(CH_3)Cl + HCl$.

(A) The molality $(m)$ of a solution is defined as the number of moles of solute dissolved per kilogram $(1000 \ g)$ of the solvent.
Therefore,a molal solution contains one mole of solute in $1000 \ g$ of the solvent.

(D) The number of half-lives $(n)$ is calculated as $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{560}{140} = 4$.
The amount of radioactive substance remaining $(N)$ is given by the formula $N = \frac{N_o}{2^n}$,where $N_o$ is the initial amount.
Given $N_o = 1 \ g$ and $n = 4$,we have $N = \frac{1}{2^4} = \frac{1}{16} \ g$.

(D) The acidity of a compound depends on the stability of its conjugate base.
$1$. Acetic acid $(CH_3COOH)$ is a carboxylic acid,and its conjugate base (acetate ion) is stabilized by resonance between two equivalent oxygen atoms,making it more acidic than phenol ($pK_a \approx 4.75$ vs $pK_a \approx 10$).
$2$. $p-$nitrophenol is more acidic than phenol because the nitro group $(-NO_2)$ exerts a strong electron-withdrawing effect ($-I$ and $-M$ effects),which stabilizes the phenoxide ion.
Therefore,phenol is less acidic than both acetic acid and $p-$nitrophenol.

(C) The reaction of alcohols with $ZnCl_2 +$ conc. $HCl$ is known as the $Lucas$ test.
$3^o$ alcohols react immediately to form turbidity.
$2^o$ alcohols react after about $5$ minutes.
$1^o$ alcohols do not react at room temperature.
Among the given options,$(CH_3)_3C-OH$ ($2-$methylpropan$-2-$ol) is a $3^o$ alcohol,which reacts immediately to produce turbidity.

(D) The reaction of alcohols with hydrogen halides $(HX)$ follows the order: $3^{\circ} > 2^{\circ} > 1^{\circ}$ alcohols.
This is because the reaction proceeds via the formation of a carbocation intermediate,and the stability of the carbocation follows the order: $3^{\circ} > 2^{\circ} > 1^{\circ}$.
$2$-methyl propane-$2$-ol is a tertiary $(3^{\circ})$ alcohol,which forms a stable tertiary carbocation,making it the most reactive towards $HBr$.