GSEB 2022 Chemistry Question Paper with Answer and Solution

(D) The function $y = \tan^{-1} x$ is the inverse of the tangent function restricted to the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$.
By definition,the range of the principal value branch of $y = \tan^{-1} x$ is the open interval $(-\frac{\pi}{2}, \frac{\pi}{2})$.
Therefore,the correct condition is $\frac{-\pi}{2} < y < \frac{\pi}{2}$.
Thus,the correct option is $D$.

(A) The general formula for the unit of the rate constant $(k)$ for a reaction of order $(n)$ is given by: $(Concentration)^{1-n} \times Time^{-1}$.
For a second-order reaction,$n = 2$.
Substituting $n = 2$ into the formula: $(Mol \ L^{-1})^{1-2} \times S^{-1} = (Mol \ L^{-1})^{-1} \times S^{-1}$.
This simplifies to: $Mol^{-1} \ L^1 \ S^{-1}$ or $Mol^{-1} \ L \ S^{-1}$.
Therefore,the correct option is $A$.

(D) bimolecular reaction is one in which two reactant species collide simultaneously to form products.
In the reaction $2NO + O_2 \rightarrow 2NO_2$,the rate-determining step involves the collision of $2$ molecules of $NO$ and $1$ molecule of $O_2$,but specifically,the reaction $2NO + O_2 \rightarrow 2NO_2$ is often cited as a termolecular reaction in elementary steps.
However,looking at the options provided:
Option $A$ involves $4$ molecules ($1$ $N_2$ + $3$ $H_2$).
Option $B$ is a unimolecular decomposition.
Option $C$ is a unimolecular decomposition.
Option $D$ involves $3$ molecules ($2$ $NO$ + $1$ $O_2$).
Actually,in many contexts,the reaction $2NO + O_2 \rightarrow 2NO_2$ is considered termolecular. If we re-evaluate the options for a bimolecular process,none strictly fit perfectly as elementary bimolecular reactions except if we consider the mechanism of $2NO + O_2 \rightarrow 2NO_2$ as $2NO + O_2 \rightarrow 2NO_2$ being a complex reaction where the elementary step $NO + O_2 \rightarrow NO_3$ is bimolecular.
Given the standard textbook examples,$2NO + O_2 \rightarrow 2NO_2$ is the most common example used to discuss molecularity higher than $2$ or complex kinetics. If the question implies which reaction involves two species in its rate-determining step,$D$ is the most appropriate choice.

(C) The unit of the rate constant $K$ for a reaction of order $n$ is given by $(mol \ L^{-1})^{1-n} \ s^{-1}$.
For a second-order reaction $(n = 2)$,the unit is $(mol \ L^{-1})^{1-2} \ s^{-1} = (mol \ L^{-1})^{-1} \ s^{-1} = L \ mol^{-1} \ s^{-1}$.
Since the given unit is $L \ mol^{-1} \ s^{-1}$,the reaction is of the second order.

(D) The complex $[PtCl_2(en)_2]$ contains two monodentate ligands $(Cl^-)$ and two bidentate ligands $(en)$.
It exhibits geometrical isomerism due to the different spatial arrangements of the $Cl^-$ ligands (cis and trans forms).
The cis-isomer of $[PtCl_2(en)_2]$ is optically active because it lacks a plane of symmetry,while the trans-isomer is optically inactive.
Therefore,the complex exhibits both geometrical and optical isomerism.

(C) The primary valency corresponds to the oxidation state of the central metal ion,and the secondary valency corresponds to the coordination number.
For the complex $[Co(C_2O_4)_2(H_2O)_2]^{-}$:
Let the oxidation state of $Co$ be $x$.
$x + 2(-2) + 2(0) = -1$
$x - 4 = -1$
$x = +3$.
Thus,the primary valency is $3$.
The coordination number is the total number of donor atoms attached to the central metal ion.
$C_2O_4^{2-}$ is a bidentate ligand ($2 \times 2 = 4$ donor atoms) and $H_2O$ is a monodentate ligand ($2 \times 1 = 2$ donor atoms).
Total coordination number $= 4 + 2 = 6$.
Thus,the secondary valency is $6$.

(C) The chemical formula for ferric hexacyanoferrate$(III)$ is $Fe[Fe(CN)_6]$.
Upon ionization in an aqueous solution,it dissociates as follows:
$Fe[Fe(CN)_6] \rightarrow Fe^{3+} + [Fe(CN)_6]^{3-}$.
This results in the formation of $1$ ferric ion $(Fe^{3+})$ and $1$ hexacyanoferrate$(III)$ complex ion $([Fe(CN)_6]^{3-})$.
Therefore,the total number of ions obtained is $1 + 1 = 2$.

(C) The complexes $[Co(NH_3)_5SO_4]Br$ and $[Co(NH_3)_5Br]SO_4$ exhibit ionisation isomerism.
Ionisation isomerism occurs when the counter ion in a coordination compound is a potential ligand and can displace a ligand which can then become the counter ion.
In this case,the $SO_4^{2-}$ ion and the $Br^-$ ion exchange positions between the coordination sphere and the counter ion position.

(C) The wavelength of absorbed light $(\lambda)$ is inversely proportional to the crystal field splitting energy $(\Delta_o)$.
$\lambda \propto \frac{1}{\Delta_o}$.
Stronger ligands cause greater crystal field splitting,resulting in a higher $\Delta_o$ and a lower wavelength of absorbed light.
The spectrochemical series for the ligands is: $Cl^{-} < H_2O < NH_3 < CN^{-}$.
Among the given complexes,$CN^{-}$ is the strongest ligand,leading to the largest $\Delta_o$ and consequently the lowest wavelength of absorbed light.
Therefore,the correct complex is $[Co(CN)_6]^{3-}$.

(A) The compound $K_2MnO_4$ is potassium manganate.
In this compound,manganese is in the $+6$ oxidation state.
Potassium manganate $(K_2MnO_4)$ is known for its characteristic green color.

(D) transition element is defined as an element which has an incompletely filled $d$-orbital in its ground state or in any of its oxidation states.
$Zn$ $([Ar] 3d^{10} 4s^2)$,$Cd$ $([Kr] 4d^{10} 5s^2)$,and $Hg$ $([Xe] 4f^{14} 5d^{10} 6s^2)$ have completely filled $d$-orbitals in their ground state as well as in their common oxidation states $(+2)$.
$Cu$ $([Ar] 3d^{10} 4s^1)$ has a completely filled $d$-orbital in its ground state,but in its $+2$ oxidation state $(Cu^{2+})$,it has a $3d^9$ configuration,which is incompletely filled.
Therefore,$Cu$ is considered a transition element.

(B) To determine the magnetic moment,we calculate the number of unpaired electrons $(n)$ in each metal ion:
$1$. In $Ni(NO_3)_2$,$Ni$ is in the $+2$ oxidation state $(3d^8)$. It has $n = 2$ unpaired electrons.
$2$. In $MnSO_4$,$Mn$ is in the $+2$ oxidation state $(3d^5)$. It has $n = 5$ unpaired electrons.
$3$. In $CrCl_3$,$Cr$ is in the $+3$ oxidation state $(3d^3)$. It has $n = 3$ unpaired electrons.
$4$. In $FeSO_4$,$Fe$ is in the $+2$ oxidation state $(3d^6)$. It has $n = 4$ unpaired electrons.
The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$. Since $Mn^{2+}$ has the highest number of unpaired electrons $(n=5)$,it will have the highest magnetic moment.

(C) The electronic configurations of the given ions are:
$Ti^{+} (Z=22): [Ar] 3d^2 4s^1$
$V^{+} (Z=23): [Ar] 3d^3 4s^1$
$Cr^{+} (Z=24): [Ar] 3d^5 4s^0$
$Mn^{+} (Z=25): [Ar] 3d^5 4s^1$
Ionisation enthalpy is the energy required to remove an electron from the outermost shell.
$Cr^{+}$ has a stable half-filled $3d^5$ configuration with no electrons in the $4s$ orbital.
Removing an electron from a stable,half-filled $d$-subshell requires significantly more energy compared to removing an electron from the $4s$ orbital of the other ions.
Therefore,$Cr^{+}$ has the highest ionisation enthalpy.

(D) The colour of aqueous solutions of transition metal ions depends on the presence of unpaired $d$-electrons,which allow for $d-d$ transitions.
$1$. $Cu^{2+}$ has $3d^9$ configuration ($1$ unpaired electron),so it is blue.
$2$. $Mn^{2+}$ has $3d^5$ configuration ($5$ unpaired electrons),so it is pink.
$3$. $Ni^{2+}$ has $3d^8$ configuration ($2$ unpaired electrons),so it is green.
$4$. $Zn^{2+}$ has $3d^{10}$ configuration ($0$ unpaired electrons). Since there are no unpaired electrons,$d-d$ transitions are not possible,making the solution colourless.
Therefore,the correct option is $D$.

(A) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
$1$. For $Ti^{3+}$ $(3d^1)$: $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.
$2$. For $Co^{3+}$ $(3d^6)$: In the presence of weak field ligands,$n = 4$,$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.
$3$. For $Cr^{3+}$ $(3d^3)$: $n = 3$,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.
$4$. For $Fe^{3+}$ $(3d^5)$: $n = 5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
Comparing the values,$Ti^{3+}$ has the lowest number of unpaired electrons $(n=1)$,therefore it has the lowest magnetic moment.

(A) In a dry cell (Leclanché cell),the anode is made of zinc $(Zn)$ and the cathode is a carbon rod surrounded by powdered manganese dioxide $(MnO_2)$ and carbon.
During the discharge process,the $MnO_2$ acts as an oxidising agent where $Mn^{4+}$ is reduced to $Mn^{3+}$.
The reaction at the cathode is: $MnO_2 + NH_4^+ + e^- \rightarrow MnO(OH) + NH_3$.
Therefore,$MnO_2$ is the oxidising agent.

(A) The reduction reaction is $Al^{3+} + 3e^- \rightarrow Al$.
To produce $1 \ mol$ $(27 \ g)$ of $Al$,$3 \ F$ of electricity is required.
For $40.0 \ g$ of $Al$,the electricity required is calculated as:
$\text{Electricity} = \frac{3 \times 40.0}{27} = 4.44 \ F$.

(B) The molar conductivity at infinite dilution for methanoic acid $(HCOOH)$ is calculated using Kohlrausch's law: $\Lambda_m^0(HCOOH) = \lambda^0(H^+) + \lambda^0(HCOO^-)$.
Substituting the given values: $\Lambda_m^0(HCOOH) = 349.6 + 54.6 = 404.2 \ S \ cm^2 \ mol^{-1}$.
The degree of dissociation $(\alpha)$ is given by the ratio of molar conductivity at a given concentration $(\Lambda_m)$ to the molar conductivity at infinite dilution $(\Lambda_m^0)$: $\alpha = \frac{\Lambda_m}{\Lambda_m^0}$.
$\alpha = \frac{46.1}{404.2} \approx 0.114$.

(D) Electronic conductance in metals is primarily determined by the number of valence electrons per atom,the nature and structure of the metal,and temperature (as temperature increases,the vibration of metal ions increases,causing more scattering of electrons and thus decreasing conductance). Pressure has a negligible effect on the electronic conductance of metals. Therefore,the correct option is $D$.

(A) The reduction half-reaction for $MnO_4^{-}$ to $Mn^{2+}$ is given by:
$MnO_4^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_2O$
From the balanced equation,$1 \ mol$ of $MnO_4^{-}$ requires $5 \ mol$ of electrons,which is equal to $5 \ F$ of electricity.
Therefore,for $2 \ mol$ of $MnO_4^{-}$,the electricity required is $2 \times 5 \ F = 10 \ F$.

(B) The potential of a hydrogen electrode is given by the Nernst equation: $E_{H^{+}/H_2} = E^{\circ}_{H^{+}/H_2} - \frac{0.059}{n} \log \frac{1}{[H^{+}]}$.
Given $pH = 1$,we know that $-\log [H^{+}] = 1$,which implies $[H^{+}] = 10^{-1} \ M$.
For a standard hydrogen electrode,$E^{\circ}_{H^{+}/H_2} = 0.0 \ V$ and $n = 1$.
Substituting the values: $E = 0.0 - \frac{0.059}{1} \log \frac{1}{10^{-1}}$.
$E = -0.059 \times \log(10) = -0.059 \times 1 = -0.059 \ V$.

(B) The chemical structure of $DDT$ is $1,1,1-\text{trichloro}-2,2-\text{bis}(p-\text{chlorophenyl})\text{ethane}$.
Its molecular formula is $C_{14}H_9Cl_5$.
In the structure,there are two benzene rings,each containing $3$ $(\pi)$ bonds,totaling $6$ $(\pi)$ bonds.
Counting the $(\sigma)$ bonds: There are $29$ $(\sigma)$ bonds in the entire molecule,including the $C-C$,$C-H$,and $C-Cl$ bonds.
Thus,the correct answer is $29$ $(\sigma)$ and $6$ $(\pi)$ bonds.

(C) Osmotic pressure $(\pi)$ is a colligative property given by the formula $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature.
Since $C$,$R$,and $T$ are identical for all solutions,$\pi$ depends directly on the van't Hoff factor $(i)$.
For $1 \ M \ BaCl_2$,$i = 3$ (dissociates into $Ba^{2+} + 2Cl^-$).
For $1 \ M \ NaCl$,$i = 2$ (dissociates into $Na^+ + Cl^-$).
For $1 \ M \ FeCl_3$,$i = 4$ (dissociates into $Fe^{3+} + 3Cl^-$).
For $1 \ M \ glucose$,$i = 1$ (does not dissociate).
Since $FeCl_3$ has the highest van't Hoff factor $(i = 4)$,it will have the highest osmotic pressure.

(B) In a positive deviation from Raoult's law,the solute-solvent interactions are weaker than the solute-solute and solvent-solvent interactions.
This results in an increase in total vapor pressure and a positive enthalpy of mixing $(\Delta H_{mix} > 0)$.
Among the given options,the mixture of $Ethanol + Acetone$ exhibits positive deviation because the hydrogen bonding between ethanol molecules is disrupted by the addition of acetone.
Conversely,$Phenol + Aniline$ and $Chloroform + Acetone$ show negative deviation due to strong intermolecular hydrogen bonding between the components.
$Nitric \ acid + Water$ also shows negative deviation due to the formation of strong hydrogen bonds.

(B) The elevation in boiling point is given by the formula $\Delta T_b = K_b \times m$.
Given $K_b = 0.52 \ K \ kg \ mol^{-1}$ and molality $m = 1 \ m$,we have $\Delta T_b = 0.52 \times 1 = 0.52 \ K$.
The boiling point of pure water is $373.15 \ K$.
Therefore,the boiling point of the solution is $T_b = T_b^0 + \Delta T_b = 373.15 \ K + 0.52 \ K = 373.67 \ K$.
Thus,the correct option is $B$.

(B) Given: $10\%$ w/w aqueous solution of $NaOH$ means $10 \ g$ of $NaOH$ is present in $100 \ g$ of solution.
Mass of solute $(NaOH)$ = $10 \ g$.
Mass of solvent $(H_2O)$ = $100 \ g - 10 \ g = 90 \ g = 0.09 \ kg$.
Molar mass of $NaOH$ = $40 \ g \ mol^{-1}$.
Number of moles of $NaOH$ = $\frac{10 \ g}{40 \ g \ mol^{-1}} = 0.25 \ mol$.
Molality $(m)$ = $\frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.25 \ mol}{0.09 \ kg} \approx 2.78 \ m$.

(D) solution where the solute is a gas and the solvent is a solid is known as a gas-in-solid solution.
In the case of $H_2$ gas adsorbed on a palladium $(Pd)$ metal surface,hydrogen acts as the solute (gas) and palladium acts as the solvent (solid).
Therefore,the correct option is $D$.

(B) Two solutions are isotonic if they have the same osmotic pressure,which implies they have the same molar concentration of particles (osmolarity).
For $H_2SO_4$,it dissociates as: $H_2SO_4 \rightarrow 2H^+ + SO_4^{2-}$.
The van't Hoff factor $i = 3$.
Osmolarity of $0.2 \ M \ H_2SO_4 = i \times M = 3 \times 0.2 = 0.6 \ M$.
Now,check the options:
$A$: $0.4 \ M \ HCl \rightarrow i = 2, \text{osmolarity} = 2 \times 0.4 = 0.8 \ M$.
$B$: $0.3 \ M \ HCl \rightarrow i = 2, \text{osmolarity} = 2 \times 0.3 = 0.6 \ M$.
$C$: $0.1 \ M \ HNO_3 \rightarrow i = 2, \text{osmolarity} = 2 \times 0.1 = 0.2 \ M$.
$D$: $0.2 \ M \ HNO_3 \rightarrow i = 2, \text{osmolarity} = 2 \times 0.2 = 0.4 \ M$.
Since $0.3 \ M \ HCl$ has the same osmolarity $(0.6 \ M)$ as $0.2 \ M \ H_2SO_4$,it is isotonic. The correct option is $B$.

(A) An ideal solution is one that obeys Raoult's law over the entire range of concentration and shows no change in enthalpy or volume upon mixing.
Pairs like $Benzene + Toluene$,$n-hexane + n-heptane$,and $Bromoethane + Chloroethane$ form ideal solutions because the intermolecular forces between the components are similar to those in the pure liquids.
$Phenol + Aniline$ is a non-ideal solution because it exhibits negative deviation from Raoult's law due to the formation of strong intermolecular hydrogen bonding between the phenol and aniline molecules.
Therefore,the correct option is $A$.

(C) To find the molarity of the diluted solution,we use the dilution formula: $M_1 \cdot V_1 = M_2 \cdot V_2$
Given:
$M_1 = 0.5 \ M$
$V_1 = 30 \ mL$
$V_2 = 500 \ mL$
Substituting the values into the formula:
$(0.5 \ M) \cdot (30 \ mL) = M_2 \cdot (500 \ mL)$
$M_2 = \frac{0.5 \times 30}{500}$
$M_2 = \frac{15}{500} = 0.03 \ M$
Therefore,the molarity of the diluted solution is $0.03 \ M$.