GSEB 2024 Chemistry Question Paper with Answer and Solution

(B) The atomic number of the element is $28$, which corresponds to Nickel $(Ni)$.
The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$.
A divalent ion $(Ni^{2+})$ is formed by the loss of two electrons from the $4s$ orbital, resulting in the configuration $[Ar] 3d^8$.
In the $3d^8$ configuration, there are $2$ unpaired electrons $(n=2)$.
The magnetic moment $(\mu)$ is calculated using the spin-only formula: $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n=2$: $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.84 \ BM$.

(D) The reduction reaction for the production of $Al$ from molten $Al_2O_3$ is given by:
$Al^{3+} + 3e^- \rightarrow Al$
From the stoichiometry of the reaction,$1 \ mol$ of $Al$ $(27 \ g)$ requires $3 \ F$ of electricity.
Given mass of $Al = 2.7 \ g$.
Number of moles of $Al = \frac{2.7 \ g}{27 \ g/mol} = 0.1 \ mol$.
Since $1 \ mol$ of $Al$ requires $3 \ F$,then $0.1 \ mol$ of $Al$ requires $0.1 \times 3 = 0.3 \ F$.
Therefore,the required electricity is $0.3 \ F$.

(A) Electronic conductance (metallic conductance) depends on the nature and structure of the metal,the number of valence electrons per atom,and temperature.
It does not depend on the concentration of the electrolyte,as electronic conductance is a property of metals,not electrolytic solutions.
Therefore,the correct option is $A$.

(D) In a dry cell (Leclanché cell),the anode is made of zinc $(Zn)$ and the cathode is a carbon rod surrounded by manganese dioxide $(MnO_2)$.
At the anode,zinc undergoes oxidation: $Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-$.
Since zinc loses electrons,it acts as the reducing agent.
Therefore,the correct option is $D$.

(A) The oxidation reaction for the hydrogen electrode is: $H_2(g) \rightarrow 2H^+(aq) + 2e^-$.
Using the Nernst equation for the oxidation potential: $E_{ox} = E^0_{ox} - \frac{0.0591}{n} \log [H^+]^2$.
Since $E^0_{ox}$ for hydrogen is $0 \ V$ and $n = 2$,we have: $0.177 = 0 - \frac{0.0591}{2} \times 2 \log [H^+]$.
$0.177 = -0.0591 \times \log [H^+]$.
Since $pH = -\log [H^+]$,we get: $0.177 = 0.0591 \times pH$.
$pH = \frac{0.177}{0.0591} = 3$.
Therefore,the correct option is $A$.

(C) metal can displace $H_2$ gas from $HCl$ solution if its standard reduction potential $(E^0_{M^{n+}/M})$ is less than the standard reduction potential of the hydrogen electrode,which is $0.00 \ V$.
Metals with negative reduction potentials act as reducing agents and can reduce $H^+$ ions to $H_2$ gas.
Comparing the given values:
$E^0_{Zn^{2+}/Zn} = -0.76 \ V$ (Negative,can produce $H_2$)
$E^0_{Fe^{2+}/Fe} = -0.44 \ V$ (Negative,can produce $H_2$)
$E^0_{Ni^{2+}/Ni} = -0.25 \ V$ (Negative,can produce $H_2$)
$E^0_{Cu^{2+}/Cu} = +0.34 \ V$ (Positive,cannot produce $H_2$)
Therefore,$Cu$ cannot produce $H_2$ gas.

(A) The cell reaction is: $2Al_{(s)} + 3Zn_{(aq)}^{2+} \rightarrow 2Al_{(aq)}^{3+} + 3Zn_{(s)}$.
Here,the number of electrons involved in the reaction is $n = 6$.
The Nernst equation is given by: $E_{cell} = E_{cell}^0 - \frac{0.059}{n} \log \frac{[Product]}{[Reactant]}$.
Substituting the values: $E_{cell} = E_{cell}^0 - \frac{0.059}{6} \log \frac{[Al^{3+}]^2}{[Zn^{2+}]^3}$.
Thus,option $A$ is correct.

(D) According to Kohlrausch's law of independent migration of ions,the molar conductivity of an electrolyte at infinite dilution is the sum of the molar conductivities of its constituent ions.
We are given:
$\Lambda_{m}^0(NaCl) = 126.4 \ S \ cm^2 \ mol^{-1}$
$\Lambda_{m}^0(HCl) = 425.9 \ S \ cm^2 \ mol^{-1}$
$\Lambda_{m}^0(NaAc) = 91.0 \ S \ cm^2 \ mol^{-1}$
We need to find $\Lambda_{m}^0(HAc)$,which is given by:
$\Lambda_{m}^0(HAc) = \Lambda_{m}^0(H^+) + \Lambda_{m}^0(Ac^-)$
Using the given values:
$\Lambda_{m}^0(HAc) = \Lambda_{m}^0(HCl) + \Lambda_{m}^0(NaAc) - \Lambda_{m}^0(NaCl)$
$\Lambda_{m}^0(HAc) = 425.9 + 91.0 - 126.4$
$\Lambda_{m}^0(HAc) = 516.9 - 126.4 = 390.5 \ S \ cm^2 \ mol^{-1}$
Therefore,the correct option is $D$.

(C) In a Mercury cell,the anode is a zinc-mercury amalgam $(Zn(Hg))$ and the cathode is a paste of mercuric oxide $(HgO)$ and carbon.
The electrolyte is a moist paste of potassium hydroxide $(KOH)$ and zinc oxide $(ZnO)$.
Therefore,the correct option is $C$.

(B) The reduction reaction for the extraction of aluminium from $Al_2O_3$ is given by:
$Al^{3+} + 3e^- \rightarrow Al$
To obtain $1$ mole of $Al$,$3$ moles of electrons are required,which corresponds to $3 \ F$ of charge.
Therefore,to obtain $2$ moles of $Al$,the charge required is:
$2 \times 3 \ F = 6 \ F$.

(C) The standard Gibbs energy change $\Delta G^{\circ}$ is given by the formula: $\Delta G^{\circ} = -nFE^{\circ}_{cell}$.
Here,the number of electrons transferred in the reaction is $n = 2$.
The standard electrode potential is $E^{\circ}_{cell} = 1.1 \ V$.
The Faraday constant is $F = 96487 \ C \ mol^{-1}$.
Substituting these values: $\Delta G^{\circ} = -(2) \times (96487 \ C \ mol^{-1}) \times (1.1 \ V)$.
$\Delta G^{\circ} = -212271.4 \ J \ mol^{-1}$.
Converting to $kJ \ mol^{-1}$: $\Delta G^{\circ} = -212.27 \ kJ \ mol^{-1}$.

(A) An electrochemical cell functions as a galvanic cell when the external potential $(E_{ext})$ is less than the cell potential $(E_{cell})$,allowing spontaneous chemical reactions to produce electricity.
However,when an external potential $(E_{ext})$ greater than the cell potential $(E_{cell})$ is applied,the direction of the current is reversed,and the cell behaves as an electrolytic cell,where non-spontaneous reactions are driven by the external power source.
Therefore,the correct condition is $E_{ext} > E_{cell}$.

(C) Dichloromethane $(CH_2Cl_2)$ is a volatile liquid that is used as a solvent and paint remover. When it comes in direct contact with the eyes,it can cause mild to moderate irritation and can burn the cornea. Therefore,the correct compound is $CH_2Cl_2$.

(D) The reactivity of alkyl halides towards $S_{N}1$ reaction depends on the stability of the carbocation intermediate formed after the loss of the leaving group $(Br^-)$.
$1$. $C_{6}H_{5}CH_{2}^+$ (Primary benzylic carbocation)
$2$. $C_{6}H_{5}CH^+(CH_{3})$ (Secondary benzylic carbocation)
$3$. $C_{6}H_{5}CH^+(C_{6}H_{5})$ (Secondary benzylic carbocation with two phenyl rings)
$4$. $C_{6}H_{5}C^+(CH_{3})(C_{6}H_{5})$ (Tertiary benzylic carbocation with two phenyl rings)
The stability of carbocations follows the order: $3^\circ > 2^\circ > 1^\circ$. Furthermore,the presence of two phenyl groups provides extensive resonance stabilization to the carbocation.
Compound $D$ forms a tertiary benzylic carbocation stabilized by two phenyl rings,making it the most stable carbocation and thus the most reactive towards $S_{N}1$.

(A) An allylic halide is a compound in which the halogen atom is bonded to an $sp^3$ hybridized carbon atom next to a carbon-carbon double bond $(C=C)$.
In $2-$bromocyclohexene,the bromine atom is directly attached to one of the carbon atoms of the double bond,which is $sp^2$ hybridized. This makes it a vinylic halide,not an allylic halide.
In $3-$chloroprop$-1-$ene,$3-$chlorocyclohexene,and $1-$chlorobut$-2-$ene,the halogen is attached to an $sp^3$ carbon adjacent to the double bond,satisfying the definition of an allylic halide.
Therefore,the correct answer is $A$.

(A) The Swarts reaction is a method used to prepare alkyl fluorides by heating alkyl chlorides or alkyl bromides in the presence of metallic fluorides.
Commonly used metallic fluorides include $AgF$,$Hg_2F_2$,$CoF_2$,and $SbF_3$.
$HF$ (hydrogen fluoride) is not a metallic fluoride and is not used as a reagent in the Swarts reaction.
Therefore,the correct option is $A$.

(B) The reactivity of alkyl halides in an $S_N1$ reaction depends on the stability of the carbocation intermediate formed during the rate-determining step.
The stability order of carbocations is: $3^{\circ} > 2^{\circ} > 1^{\circ}$.
Let us analyze the carbocations formed from the given compounds:
$(I)$ $CH_3-CH_2-CH^+(CH_3)$ is a $2^{\circ}$ carbocation.
$(II)$ $(CH_3)_2CH-CH_2^+$ is a $1^{\circ}$ carbocation.
$(III)$ $(CH_3)_3C^+$ is a $3^{\circ}$ carbocation.
Since the stability order is $(II) < (I) < (III)$,the reactivity order in $S_N1$ reaction is $(II) < (I) < (III)$.

(A) The elevation in boiling point is a colligative property,which depends on the van't Hoff factor $(i)$.
$\Delta T_b = i \times K_b \times m$.
For a given molality $(m)$ and solvent,$\Delta T_b$ is directly proportional to $i$.
- For $1 \ m \ FeCl_3$,$i = 4$ $(Fe^{3+} + 3Cl^-)$.
- For $1 \ m \ NaCl$,$i = 2$ $(Na^+ + Cl^-)$.
- For $1 \ m \ BaCl_2$,$i = 3$ $(Ba^{2+} + 2Cl^-)$.
- For $1 \ m \ \text{urea}$,$i = 1$ (non-electrolyte).
Since $FeCl_3$ has the highest van't Hoff factor $(i = 4)$,it will show the highest elevation in boiling point,and thus the highest boiling point. The correct option is $A$.

(D) The solubility of gases in liquids generally decreases with an increase in temperature because the dissolution of gas in a liquid is an exothermic process. According to Le Chatelier's principle,an increase in temperature shifts the equilibrium in the backward direction,thereby decreasing the solubility. Among the given options,$Chlorine$ is a gas,so its solubility in water decreases as the temperature increases.

(C) An ideal solution is one that obeys Raoult's Law over the entire range of concentration and shows no change in volume or enthalpy upon mixing ($\Delta V_{mix} = 0$ and $\Delta H_{mix} = 0$).
$n-hexane$ and $n-heptane$ are structurally similar and have similar intermolecular forces,forming an ideal solution.
$Acetone + Chloroform$ shows negative deviation due to hydrogen bonding.
$Acetone + Ethanol$ and $Phenol + Aniline$ show positive and negative deviations respectively due to changes in intermolecular interactions.
Therefore,the correct option is $C$.

(B) The freezing point depression $\Delta T_f$ is a colligative property,which depends on the molality $(m)$ of the solution. $\Delta T_f = K_f \times m$. For the freezing points to be equal,the molality of the two solutions must be equal.
Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$.
For option $B$:
Moles of urea $= \frac{6 \ g}{60 \ g/mol} = 0.1 \ mol$.
Moles of glucose $= \frac{18 \ g}{180 \ g/mol} = 0.1 \ mol$.
Since the mass of the solvent $(100 \ g \ H_2O)$ is the same for both,and the number of moles of solute is the same $(0.1 \ mol)$,the molality of both solutions is identical.
Therefore,they will have the same freezing point.

(C) According to Henry's law,the partial pressure of a gas $(p)$ is directly proportional to its mass $(m)$ dissolved in a given volume of solvent at a constant temperature: $p \propto m$ or $p = k \times m$.
Given: $p_1 = 1 \ bar$,$m_1 = 7.14 \times 10^{-3} \ g$.
We need to find $p_2$ when $m_2 = 5 \times 10^{-2} \ g$.
Using the ratio: $\frac{p_2}{p_1} = \frac{m_2}{m_1}$.
$\frac{p_2}{1} = \frac{5 \times 10^{-2}}{7.14 \times 10^{-3}} = \frac{50 \times 10^{-3}}{7.14 \times 10^{-3}} \approx 7 \ bar$.
Thus,the partial pressure is $7 \ bar$.

(C) Mass of solute in the first solution $= 400 \ g \times 0.25 = 100 \ g$.
Mass of solute in the second solution $= 600 \ g \times 0.40 = 240 \ g$.
Total mass of solute $= 100 \ g + 240 \ g = 340 \ g$.
Total mass of the resulting solution $= 400 \ g + 600 \ g = 1000 \ g$.
Mass percentage of the resulting solution $= (\text{Total mass of solute} / \text{Total mass of solution}) \times 100 = (340 \ g / 1000 \ g) \times 100 = 34 \%$.
Therefore,the correct option is $C$.

(D) When an unripe mango is placed in a concentrated salt solution,the concentration of salt outside the mango is much higher than the concentration of solutes inside the mango cells.
This creates a concentration gradient where water moves from the region of lower solute concentration (inside the mango) to the region of higher solute concentration (the salt solution) through the semi-permeable cell membranes of the mango.
This process is known as osmosis.
As a result,the mango loses water and shrivels.

(B) An ideal solution is one that obeys Raoult's law over the entire range of concentration and shows no change in enthalpy $(\Delta H_{mix} = 0)$ or volume $(\Delta V_{mix} = 0)$ upon mixing.
Benzene and Toluene form an ideal solution because the intermolecular forces between benzene-benzene,toluene-toluene,and benzene-toluene are nearly identical.
Chloroform and acetone show negative deviation,while ethanol and acetone,and water and ethanol show positive deviation from Raoult's law.

(D) The elevation in boiling point is given by the formula: $\Delta T_{b} = i \times K_{b} \times m$.
For glucose,which is a non-electrolyte,the van't Hoff factor $(i)$ is $1$.
Given the molality $(m)$ is $1 \ m$,we substitute these values into the equation:
$\Delta T_{b} = 1 \times K_{b} \times 1 = K_{b}$.
Therefore,$\Delta T_{b} = K_{b}$.

(C) The mass percentage of a solute is calculated using the formula: $\text{Mass percentage} = \frac{\text{Mass of solute}}{\text{Mass of solute} + \text{Mass of solvent}} \times 100$.
Given: $\text{Mass of benzene (solute)} = 22 \ g$.
$\text{Mass of carbon tetrachloride (solvent)} = 122 \ g$.
$\text{Total mass of solution} = 22 \ g + 122 \ g = 144 \ g$.
$\text{Mass percentage of benzene} = \frac{22}{144} \times 100 = 15.277 \% \approx 15.28 \%$.
Therefore,the correct option is $C$.

(A) solid solution is a solution in which the solute and solvent are both in the solid state.
$A$ Copper dissolved in gold is an example of a solid solution (alloy),where both components are solids.
$B$ Glucose dissolved in water is a solid-in-liquid solution.
$C$ Camphor in nitrogen gas is a solid-in-gas solution.
$D$ Ethanol dissolved in water is a liquid-in-liquid solution.
Therefore,the correct answer is $A$.