IIT JEE 1993 Chemistry Question Paper with Answer and Solution

(D) Isoelectronic species are those that have the same number of electrons.
$I: CH_3^+ = 6 + 3 - 1 = 8 \ e^-$
$II: H_3O^+ = 3(1) + 8 - 1 = 10 \ e^-$
$III: NH_3 = 7 + 3(1) = 10 \ e^-$
$IV: CH_3^- = 6 + 3(1) + 1 = 10 \ e^-$
Thus,$II, III,$ and $IV$ are isoelectronic as each contains $10 \ e^-$. The correct option is $D$.

(D) According to Graham's Law of diffusion,the rate of diffusion $D$ of a gas is inversely proportional to the square root of its density $\rho$.
$\frac{D_A}{D_B} = \sqrt{\frac{\rho_B}{\rho_A}} = (\frac{\rho_B}{\rho_A})^{1/2}$
Therefore,$D_A = D_B (\frac{\rho_B}{\rho_A})^{1/2}$.

(A) The root mean square speed $(u_{rms})$ is defined as the square root of the mean of the squares of the speeds of all molecules.
Mathematically,it is given by the formula:
$u_{rms} = \sqrt{\frac{\sum n_i C_i^2}{\sum n_i}} = \left[ \frac{n_1 C_1^2 + n_2 C_2^2 + n_3 C_3^2 + .....}{n_1 + n_2 + n_3 + .....} \right]^{1/2}$.

(D) . Intensive properties are those that do not depend on the quantity or size of matter present in the system.
$Temperature$ and $refractive \ index$ are independent of the amount of substance,hence they are intensive properties.
$Enthalpy$ and $volume$ are extensive properties because they depend on the amount of matter present.

(C) Noble gases have a stable valence shell configuration,typically $ns^2 np^6$ (except for Helium,which is $1s^2$).
Option $C$ represents the configuration $1s^2, 2s^2 2p^6$,which corresponds to the element Neon ($Ne$,atomic number $10$).
This configuration has a completely filled valence shell,making it a noble gas.

(B) The longest carbon chain containing the aldehyde group $(-CHO)$ has $4$ carbon atoms.
The numbering starts from the aldehyde carbon as $C-1$.
$A$ methyl group is present at the $C-2$ position.
Hence,the $IUPAC$ name is $2-methylbutanal$.

(D) The correct order of stability is $(ii) > (i) > (iii)$.
$(ii)$ $CH_3-CH^{+}-OCH_3$ is the most stable because the lone pair on the oxygen atom provides strong resonance stabilization ($+M$ effect) to the carbocation.
$(i)$ $CH_3-CH^{+}-CH_3$ is a secondary carbocation stabilized by hyperconjugation and the inductive effect $(+I)$ of two methyl groups.
$(iii)$ $CH_3-CH^{+}-COCH_3$ is the least stable because the carbonyl group $(-COCH_3)$ exerts a strong electron-withdrawing effect ($-I$ and $-M$ effects),which destabilizes the positive charge on the carbon atom.

(A) Let the determinant be $D = \left| \begin{array}{ccc} 1 & \log_x y & \log_x z \\ \log_y x & 1 & \log_y z \\ \log_z x & \log_z y & 1 \end{array} \right|$.
Using the change of base formula $\log_a b = \frac{\log b}{\log a}$,we can rewrite the determinant as:
$D = \left| \begin{array}{ccc} 1 & \frac{\log y}{\log x} & \frac{\log z}{\log x} \\ \frac{\log x}{\log y} & 1 & \frac{\log z}{\log y} \\ \frac{\log x}{\log z} & \frac{\log y}{\log z} & 1 \end{array} \right|$.
Now,multiply $R_1$ by $\log x$,$R_2$ by $\log y$,and $R_3$ by $\log z$:
$D = \frac{1}{(\log x)(\log y)(\log z)} \left| \begin{array}{ccc} \log x & \log y & \log z \\ \log x & \log y & \log z \\ \log x & \log y & \log z \end{array} \right|$.
Since all three rows are identical,the value of the determinant is $0$.

(D) Let the center of the circle be $(h, k)$. Since it touches the $y$-axis,its radius $r = |h|$.
The given circle is $x^2 + y^2 - 6x - 6y + 14 = 0$. Its center $C_1 = (3, 3)$ and radius $r_1 = \sqrt{3^2 + 3^2 - 14} = \sqrt{18 - 14} = 2$.
Since the circles touch externally,the distance between centers $C_1(3, 3)$ and $C_2(h, k)$ is equal to the sum of their radii:
$C_1C_2 = r_1 + r_2$
$\sqrt{(h-3)^2 + (k-3)^2} = 2 + |h|$
Squaring both sides:
$(h-3)^2 + (k-3)^2 = (h+2)^2$
$h^2 - 6h + 9 + k^2 - 6k + 9 = h^2 + 4h + 4$
$k^2 - 6k - 10h + 14 = 0$
Replacing $(h, k)$ with $(x, y)$,the locus is:
$y^2 - 10x - 6y + 14 = 0$

(B) To determine the highest paramagnetism,we calculate the number of unpaired electrons in each complex:
$1$. $[Cr(H_2O)_6]^{3+}$: $Cr$ is in $+3$ oxidation state. $Cr(Z=24)$ configuration is $[Ar] 3d^5 4s^1$. $Cr^{3+}$ is $[Ar] 3d^3$. It has $3$ unpaired electrons.
$2$. $[Fe(H_2O)_6]^{2+}$: $Fe$ is in $+2$ oxidation state. $Fe(Z=26)$ configuration is $[Ar] 3d^6 4s^2$. $Fe^{2+}$ is $[Ar] 3d^6$. Since $H_2O$ is a weak field ligand,the electrons are arranged as $t_{2g}^4 e_g^2$,resulting in $4$ unpaired electrons.
$3$. $[Cu(H_2O)_6]^{2+}$: $Cu$ is in $+2$ oxidation state. $Cu(Z=29)$ configuration is $[Ar] 3d^{10} 4s^1$. $Cu^{2+}$ is $[Ar] 3d^9$. It has $1$ unpaired electron.
$4$. $[Zn(H_2O)_6]^{2+}$: $Zn$ is in $+2$ oxidation state. $Zn(Z=30)$ configuration is $[Ar] 3d^{10} 4s^2$. $Zn^{2+}$ is $[Ar] 3d^{10}$. It has $0$ unpaired electrons.
Since $[Fe(H_2O)_6]^{2+}$ has the maximum number of unpaired electrons $(4)$,it exhibits the highest paramagnetism.

(C) The net reaction is obtained by adding the two given processes:
$3({}_1H^2) \to {}_2He^4 + p + n$
Calculate the mass defect $(\Delta m)$:
$\Delta m = 3 \times M(H^2) - [M(He^4) + M(p) + M(n)]$
$\Delta m = 3(2.014) - [4.001 + 1.007 + 1.008] = 6.042 - 6.016 = 0.026 \, amu$
Calculate the energy released $(Q)$ per net reaction:
$Q = 0.026 \times 931.5 \, MeV \approx 24.22 \, MeV$
$Q = 24.22 \times 1.6 \times 10^{-13} \, J \approx 3.875 \times 10^{-12} \, J$
Total energy produced by $10^{40}$ deuterons:
Since $3$ deuterons are used per reaction,total reactions $N = \frac{10^{40}}{3}$.
Total Energy $E_{total} = N \times Q = \frac{10^{40}}{3} \times 3.875 \times 10^{-12} \approx 1.29 \times 10^{28} \, J$
Time taken to exhaust the supply $(t)$:
$t = \frac{E_{total}}{Power} = \frac{1.29 \times 10^{28}}{10^{16}} = 1.29 \times 10^{12} \, sec$
Thus,the time is of the order of $10^{12} \, sec$.

(C) The reaction of an alcohol $(ROH)$ with methyl magnesium iodide $(CH_3MgI)$ produces methane gas $(CH_4)$:
$ROH + CH_3MgI \to CH_4 \uparrow + Mg(OR)I$
According to the stoichiometry,$1 \, mol$ of alcohol produces $1 \, mol$ of $CH_4$ gas.
At $STP$,$1 \, mol$ of any gas occupies $22400 \, mL$.
Given that $1.12 \, mL$ of $CH_4$ is produced from $4.12 \, mg$ of alcohol.
Therefore,$22400 \, mL$ of $CH_4$ would be produced by:
$\text{Molecular mass} = \frac{4.12 \, mg}{1.12 \, mL} \times 22400 \, mL/mol$
$= 4.12 \times 20000 \, mg/mol = 82400 \, mg/mol = 82.4 \, g/mol$.
Thus,the molecular mass of the alcohol is $82.4 \, g/mol$.

(B) Fluorspar $(CaF_2)$ is added in small quantities during the electrolytic reduction of alumina $(Al_2O_3)$ dissolved in fused cryolite $(Na_3AlF_6)$.
Its primary functions are to lower the melting point of the mixture and to increase the electrical conductivity of the melt,which facilitates the electrolysis process at a lower temperature of approximately $1140 \, K$.

(B) The molarity of a solution is defined as the number of moles of solute per liter of solution.
Since the volume of a solution increases with an increase in temperature due to thermal expansion,the molarity $(M = \frac{n}{V})$ decreases.

(C) According to Raoult's law for dilute solutions,the relative lowering of vapour pressure is equal to the mole fraction of the solute: $\frac{P^o - P_s}{P^o} = \frac{n_2}{n_1} = \frac{W_2 / M_2}{W_1 / M_1}$.
Given: $P^o = 3000 \ N/m^2$,$P_s = 2985 \ N/m^2$,$W_2 = 5 \ g$,$W_1 = 100 \ g$,$M_1 = 18 \ g/mol$ (for water).
Substituting the values: $\frac{3000 - 2985}{3000} = \frac{5 / M_2}{100 / 18}$.
$\frac{15}{3000} = \frac{5 \times 18}{100 \times M_2}$.
$0.005 = \frac{90}{100 \times M_2}$.
$0.005 = \frac{0.9}{M_2}$.
$M_2 = \frac{0.9}{0.005} = 180 \ g/mol$.

(B) To determine the paramagnetism,we calculate the number of unpaired electrons in each complex:
$1$. In $[Cr(H_2O)_6]^{3+}$,$Cr^{3+}$ is $3d^3$,which has $3$ unpaired electrons.
$2$. In $[Fe(H_2O)_6]^{2+}$,$Fe^{2+}$ is $3d^6$. Since $H_2O$ is a weak field ligand,the configuration is $t_{2g}^4 e_g^2$,resulting in $4$ unpaired electrons.
$3$. In $[Cu(H_2O)_6]^{2+}$,$Cu^{2+}$ is $3d^9$,which has $1$ unpaired electron.
$4$. In $[Zn(H_2O)_6]^{2+}$,$Zn^{2+}$ is $3d^{10}$,which has $0$ unpaired electrons.
Since $[Fe(H_2O)_6]^{2+}$ has the highest number of unpaired electrons $(4)$,it exhibits the highest paramagnetism.