IIT JEE 1985 Chemistry Question Paper with Answer and Solution

(B) For a parallel beam of light to emerge as a parallel beam from a combination of a convex and a concave lens,the second focal point of the convex lens must coincide with the first focal point of the concave lens.
Let $f_1 = 20 \ cm$ be the focal length of the convex lens and $f_2 = 5 \ cm$ be the focal length of the concave lens.
The distance between the lenses is given by $d = f_1 - f_2$.
Substituting the values,we get $d = 20 \ cm - 5 \ cm = 15 \ cm$.

(A) The radius of an atom is typically in the range of $10^{-10} \ m$ or $1 \ \mathring{A}$.
Converting this to centimeters,$10^{-10} \ m = 10^{-8} \ cm$.
Therefore,the order of the radius of an atom is $10^{-8} \ cm$.

(B) Bohr's model is applicable only to hydrogen-like species,which are atoms or ions containing only one electron,such as $H$,$He^+$,$Li^{2+}$,etc.

(A) According to Molecular Orbital Theory $(MOT)$,the electronic configurations are as follows:
$A].$ For $NO$ ($15$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^1$. It has $1$ unpaired electron.
$B].$ For $CO$ ($14$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2$. All electrons are paired.
$C].$ For $CN^{-}$ ($14$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2$. All electrons are paired.
$D].$ For $O_2$ ($16$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^1, \pi^* 2p_y^1$. It has $2$ unpaired electrons.
Thus,$NO$ is the correct molecule.

(D) According to Graham's law of diffusion,the rate of diffusion $(r)$ of a gas is inversely proportional to the square root of its density $(d)$ or its molar mass $(M)$.
Mathematically,$r \propto \frac{1}{\sqrt{d}}$ or $r \propto \frac{1}{\sqrt{M}}$.
Therefore,the rate of diffusion is inversely proportional to the square root of its molecular mass.

(D) Lewis acid is defined as an electron-pair acceptor.
$AlCl_3$ and $BeCl_2$ are electron-deficient compounds (incomplete octet),so they act as Lewis acids.
$HSO_4^-$ can accept a proton or an electron pair in certain conditions.
$NH_3$ has a lone pair of electrons on the nitrogen atom,which it can donate.
Therefore,$NH_3$ acts as a Lewis base,not a Lewis acid.

(A) The conjugate acid of a base is formed by the addition of a proton $(H^+)$ to the base.
For the base $NH_2^-$,the conjugate acid is formed as follows:
$NH_2^- + H^+ \rightarrow NH_3$
Therefore,the correct option is $(A)$.

(D) The reactivity series of metals indicates that hydrogen can reduce oxides of metals that are less reactive than hydrogen.
$Cu$,$Fe$,and $Sn$ are less reactive than hydrogen,so their oxides ($CuO$,$Fe_2O_3$,$SnO_2$) can be reduced by $H_2$.
However,$Al$ is more reactive than hydrogen,so $H_2$ cannot reduce $Al_2O_3$ to $Al$.
Therefore,the correct option is $(D)$.

(D) Glauber's salt is the common name for sodium sulphate decahydrate,which has the chemical formula $Na_2SO_4 \cdot 10H_2O$.
It exists as white or colourless monoclinic crystals.
When exposed to dry air,it undergoes efflorescence,losing its water of crystallization to form powdery anhydrous sodium sulphate.

(D) Phenolphthalein is the correct indicator for the titration of a weak acid with a strong base.
This is because the equivalence point of such a titration lies in the basic range ($pH$ range $8-10$).
Phenolphthalein changes its color from colourless to pink in this $pH$ range.

(C) The Anti-Markownikoff addition of $HBr$ (also known as the peroxide effect or Kharasch effect) is only observed in unsymmetrical alkenes.
$But-2-ene$ $(CH_3-CH=CH-CH_3)$ is a symmetrical alkene.
In a symmetrical alkene,the addition of $HBr$ results in the same product regardless of whether it follows Markownikoff's rule or the Anti-Markownikoff mechanism.
Therefore,the Anti-Markownikoff addition is not observed for $But-2-ene$.

(A) The acidic nature of hydrogen in hydrocarbons depends on the hybridization of the carbon atom to which it is attached.
In $CH \equiv CH$ (Ethyne),the carbon is $sp$ hybridized,which has $50\%$ $s$-character,making it more electronegative and thus the $C-H$ bond is polar,allowing the hydrogen to be acidic.
Reaction with sodium metal confirms this: $2CH \equiv CH + 2Na \to 2CH \equiv C^{-}Na^{+} + H_2 \uparrow$.
Therefore,ethyne has acidic hydrogen.

(A) Electrophilic aromatic substitution is facilitated by electron-donating groups on the benzene ring,which increase the electron density of the ring.
$1$. The $-CH_3$ group in Toluene is an electron-donating group due to the $+I$ effect and hyperconjugation.
$2$. Benzene has no substituents.
$3$. The $-COOH$ group in Benzoic acid and the $-NO_2$ group in Nitrobenzene are electron-withdrawing groups,which decrease the electron density of the ring.
Therefore,Toluene is the most reactive towards electrophilic nitration among the given options.

(A) The conjugate acid of a base is formed by adding a proton $(H^+)$ to the base.
For the base $NH_2^-$,the conjugate acid is formed as follows:
$NH_2^- + H^+ \rightarrow NH_3$
Therefore,the conjugate acid of $NH_2^-$ is $NH_3$.

(A) According to Molecular Orbital Theory $(MOT)$:
$NO$ has a total of $15$ electrons ($7$ from $N$ and $8$ from $O$).
The electronic configuration is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$.
Since there is one electron in the antibonding $\pi^* 2p_x$ orbital,$NO$ is paramagnetic and contains one unpaired electron.
$CO$ ($14$ electrons) and $CN^{-}$ ($14$ electrons) are diamagnetic (zero unpaired electrons).
$O_2$ ($16$ electrons) has two unpaired electrons in the $\pi^* 2p$ orbitals.

(C) When the block $m$ is suspended in air,the reading of spring balance $A$ is equal to the weight of the block,$T = mg = 2 \ kgf$.
When the block is immersed in the liquid,an upward buoyant force $F_B$ acts on the block. The new tension in the spring $A$ becomes $T' = mg - F_B$. Since $F_B > 0$,$T' < mg$,so the reading of balance $A$ will be less than $2 \ kg$.
According to Newton's third law,the block exerts an equal and opposite downward force $F_B$ on the liquid. This force is transmitted to the beaker and consequently to the pan of balance $B$. Thus,the new reading of balance $B$ becomes $W_{beaker+liquid} + F_B$. Since $F_B > 0$,the reading of balance $B$ will be more than $5 \ kg$.

(C) The equation of the circle is $(x - 1)^2 + y^2 = 1$,which simplifies to $x^2 - 2x + 1 + y^2 = 1$,or $x^2 + y^2 - 2x = 0$.
Let $(h, k)$ be the midpoint of a chord passing through the origin $(0, 0)$.
The equation of a chord with midpoint $(h, k)$ is given by $T = S_1$,where $T = xh + yk - (x + h)$ and $S_1 = h^2 + k^2 - 2h$.
So,$xh + yk - x - h = h^2 + k^2 - 2h$.
Since this chord passes through the origin $(0, 0)$,we substitute $x = 0$ and $y = 0$ into the equation:
$0(h) + 0(k) - 0 - h = h^2 + k^2 - 2h$.
$-h = h^2 + k^2 - 2h$.
$h^2 + k^2 - h = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 - x = 0$.

(D) Given $|z_1| = |z_2| = 1$,we can write $z_1 = \cos \theta_1 + i \sin \theta_1$ and $z_2 = \cos \theta_2 + i \sin \theta_2$,where $\theta_1 = \arg(z_1)$ and $\theta_2 = \arg(z_2)$.
Since $z_1 = a + ib$ and $z_2 = c + id$,we have $a = \cos \theta_1, b = \sin \theta_1, c = \cos \theta_2, d = \sin \theta_2$.
Given $R(z_1\overline{z_2}) = 0$,we have $R[(\cos \theta_1 + i \sin \theta_1)(\cos \theta_2 - i \sin \theta_2)] = 0$.
This simplifies to $R[\cos(\theta_1 - \theta_2) + i \sin(\theta_1 - \theta_2)] = \cos(\theta_1 - \theta_2) = 0$.
Thus,$\theta_1 - \theta_2 = \pm \frac{\pi}{2}$,which implies $\theta_1 = \theta_2 \pm \frac{\pi}{2}$.
Now,$w_1 = a + ic = \cos \theta_1 + i \cos \theta_2$. Since $\cos \theta_2 = \sin \theta_1$ (or $-\sin \theta_1$),we have $|w_1| = \sqrt{\cos^2 \theta_1 + \sin^2 \theta_1} = 1$.
Similarly,$w_2 = b + id = \sin \theta_1 + i \sin \theta_2$. Since $\sin \theta_2 = \cos \theta_1$ (or $-\cos \theta_1$),we have $|w_2| = \sqrt{\sin^2 \theta_1 + \cos^2 \theta_1} = 1$.
Finally,$w_1\overline{w_2} = (a + ic)(b - id) = (ab + cd) + i(bc - ad)$.
$R(w_1\overline{w_2}) = ab + cd = \cos \theta_1 \sin \theta_1 + \cos \theta_2 \sin \theta_2 = \frac{1}{2}(\sin 2\theta_1 + \sin 2\theta_2)$.
Using $\theta_1 = \theta_2 + \frac{\pi}{2}$,we get $\sin 2\theta_1 = \sin(2\theta_2 + \pi) = -\sin 2\theta_2$,so $ab + cd = 0$.
Thus,all conditions are satisfied.

(D) The radius of an atom is typically in the range of $10^{-10} \, m$ or $1 \, \mathring{A}$.
Converting this to centimeters,$1 \, \mathring{A} = 10^{-8} \, cm$.
Therefore,the order of the radius of an atom is $10^{-8} \, cm$.

(B) The conjugate acid of a base is formed by adding a proton $(H^+)$ to the base.
For the amide ion $(NH_2^-)$,the reaction is:
$NH_2^- + H^+ \rightarrow NH_3$
Therefore,the conjugate acid of $NH_2^-$ is $NH_3$.

(B) The conjugate acid of a base is formed by the addition of a proton $(H^+)$ to the base.
For the base $NH_2^-$,the conjugate acid is obtained by adding $H^+$:
$NH_2^- + H^+ \rightarrow NH_3$
Therefore,the conjugate acid of $NH_2^-$ is $NH_3$.

(A) Heating $KNO_3$ produces potassium nitrite and oxygen gas: $2KNO_{3(s)} \xrightarrow{\Delta} 2KNO_{2(s)} + O_{2(g)}$.
Heating heavy metal nitrates like $Pb(NO_3)_2$,$Cu(NO_3)_2$,and $AgNO_3$ produces nitrogen dioxide $(NO_2)$ along with oxygen and the metal oxide.

(D) Molality is defined as the number of moles of solute present in $1 \ kg$ of solvent.
The formula is: $\text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent in } kg}$.

(B) According to Raoult's law for a solution containing a non-volatile solute,the relative lowering of vapour pressure is equal to the mole fraction of the solute present in the solution.
Mathematically,it is expressed as: $\frac{P^o - P_s}{P^o} = x_2$,where $P^o$ is the vapour pressure of the pure solvent,$P_s$ is the vapour pressure of the solution,and $x_2$ is the mole fraction of the solute.

(C) In negative $\beta$-decay,a neutron $(n)$ inside the nucleus transforms into a proton $(p)$,an electron $(_{-1}e^0)$,and an antineutrino $(\bar{\nu})$.
The process is represented as: $n \to p + _{-1}e^0 + \bar{\nu}$.
Therefore,the correct statement is that a neutron in the nucleus decays,emitting an electron.

(C) The given cell reaction is: $1/2 H_{2(g)} + AgCl_{(s)} \to H^{+}_{(aq)} + Cl^{-}_{(aq)} + Ag_{(s)}$.
In this reaction,$H_2$ is oxidized to $H^+$ at the anode,and $AgCl$ is reduced to $Ag$ at the cathode.
The anode reaction is: $1/2 H_{2(g)} \to H^{+}_{(aq)} + e^{-}$,which corresponds to a standard hydrogen electrode $(Pt | H_{2(g)}, HCl_{(soln)})$.
The cathode reaction is: $AgCl_{(s)} + e^{-} \to Ag_{(s)} + Cl^{-}_{(aq)}$,which corresponds to a silver-silver chloride electrode $(AgCl_{(s)} | Ag)$.
Therefore,the cell representation is $Pt | H_{2(g)}, HCl_{(soln)} || AgCl_{(s)} | Ag$.

(C) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$Diethyl \ ketone$ $(CH_3CH_2-CO-CH_2CH_3)$ does not contain the $CH_3CO-$ group,therefore it does not form a yellow precipitate of iodoform $(CHI_3)$.

$CH_3COCH_3 \xrightarrow{NaOH/I_2} CHI_3$	$Yellow \ ppt$
$C_2H_5OH \xrightarrow{NaOH/I_2} CHI_3$
$CH_3-CH(OH)-CH_3 \xrightarrow{NaOH/I_2} CHI_3$