GSEB 2019 Chemistry Question Paper with Answer and Solution

(B) We know that $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$.
Given integral is $I = \int e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) dx$.
Using trigonometric identities: $1 + \sin x = 1 + 2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $1 + \cos x = 2 \cos^2 \frac{x}{2}$.
So,$\frac{1 + \sin x}{1 + \cos x} = \frac{1 + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} = \frac{1}{2} \sec^2 \frac{x}{2} + \tan \frac{x}{2}$.
Let $f(x) = \tan \frac{x}{2}$. Then $f'(x) = \sec^2 \frac{x}{2} \cdot \frac{1}{2} = \frac{1}{2} \sec^2 \frac{x}{2}$.
Thus,the integral becomes $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C = e^x \tan \frac{x}{2} + C$.
Therefore,the correct option is $B$.

(B) According to the Arrhenius equation,$K = A e^{-E_a / RT}$.
Taking the logarithm on both sides to the base $10$,we get $\log_{10} K = \log_{10} A - \frac{E_a}{2.303 RT}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log_{10} K$ and $x = \frac{1}{T}$,the slope $m$ is equal to $-\frac{E_a}{2.303 R}$.

(B) The balanced chemical equation for the decomposition of ammonia is: $2NH_{3(g)} \rightarrow N_{2(g)} + 3H_{2(g)}$
For a zero-order reaction,the rate of reaction is equal to the rate constant $K$.
Rate $= K = 2.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$
From the stoichiometry of the reaction,the rate of reaction is related to the rate of production of $H_2$ as: $\text{Rate} = \frac{1}{3} \frac{d[H_2]}{dt}$
Therefore,$\frac{d[H_2]}{dt} = 3 \times \text{Rate} = 3 \times (2.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}) = 7.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$

(D) For a reaction of order $n$,the half-life time $(t_{1/2})$ is related to the initial concentration $[R]_0$ by the expression:
$t_{1/2} \propto [R]_0^{1-n}$.
Comparing this with the given relation $t_{1/2} \propto \frac{1}{[R]_0^{n-1}}$,we can rewrite the given relation as:
$t_{1/2} \propto [R]_0^{-(n-1)} = [R]_0^{1-n}$.
Since the given expression matches the standard formula for a reaction of order $n$,the order of the reaction is $n$.

(A) An elementary reaction is a single-step reaction where the rate of reaction is directly proportional to the product of the concentrations of the reactants raised to the power of their stoichiometric coefficients.
Therefore,for an elementary reaction,the order of the reaction is equal to its molecularity.

(C) The rate law for the reaction is given by: $\text{Rate} = k[A]^x$ ...$(1)$
When the concentration of $A$ is doubled,the rate increases by $1.59$ times:
$1.59 \times \text{Rate} = k[2A]^x$ ...$(2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{1.59 \times \text{Rate}}{\text{Rate}} = \frac{k[2A]^x}{k[A]^x}$
$1.59 = (2)^x$
Taking $\log$ on both sides:
$\log(1.59) = x \log(2)$
$0.2014 = x \times 0.3010$
$x = \frac{0.2014}{0.3010} \approx 0.669$
Since $0.669 \approx \frac{2}{3}$,the order of the reaction is $\frac{2}{3}$.

(D) Linkage isomerism occurs in coordination compounds containing ambidentate ligands,which can coordinate to the central metal atom through two different donor atoms.
In the pair $[Co(NO_2)(NH_3)_5]Cl_2$ and $[Co(ONO)(NH_3)_5]Cl_2$,the ligand $NO_2^-$ is an ambidentate ligand.
In the first complex,the nitrogen atom $(N)$ is the donor atom $(Co-NO_2)$,
while in the second complex,the oxygen atom $(O)$ is the donor atom $(Co-ONO)$.
Therefore,this pair represents linkage isomerism.

(B) The chemical formula for ammonium diammine dioxalato cobaltate$(III)$ is $(NH_4)[Co(NH_3)_2(C_2O_4)_2]$.
In this complex,the central metal ion is $Co^{3+}$.
Primary valency corresponds to the oxidation state of the metal ion,which is $3$.
Secondary valency corresponds to the coordination number of the metal ion.
The ligands are $2$ $NH_3$ (monodentate) and $2$ $C_2O_4^{2-}$ (bidentate).
Coordination number = $(2 \times 1) + (2 \times 2) = 2 + 4 = 6$.
Thus,the primary valency is $3$ and the secondary valency is $6$.

(C) The magnitude of crystal field splitting energy $(\Delta_0)$ depends on the strength of the ligand according to the spectrochemical series.
Stronger ligands cause larger splitting,while weaker ligands cause smaller splitting.
The spectrochemical series for the given ligands is: $CN^- > NH_3 > C_2O_4^{2-} > H_2O$.
However,comparing the specific complexes provided,$H_2O$ is a weak field ligand compared to $CN^-$,$NH_3$,and $C_2O_4^{2-}$.
Among the given options,$[Co(H_2O)_6]^{3+}$ contains the weakest ligand,resulting in the lowest value of $\Delta_0$.

(C) $1$. Coordination number: $NH_3$ is a monodentate ligand $(4 \times 1 = 4)$ and $CO_3^{2-}$ is a bidentate ligand $(1 \times 2 = 2)$. Total coordination number = $4 + 2 = 6$.
$2$. Oxidation number: Let the oxidation state of $Co$ be $x$. The complex is $[Co(NH_3)_4CO_3]ClO_4$. The charge on $ClO_4$ is $-1$,so the complex ion $[Co(NH_3)_4CO_3]^+$ has a charge of $+1$. Thus,$x + 4(0) + 1(-2) = +1$,which gives $x = +3$.
$3$. Number of electrons in $d$-orbital: $Co$ $(Z=27)$ has the configuration $[Ar] 3d^7 4s^2$. $Co^{3+}$ has the configuration $[Ar] 3d^6$. Thus,there are $6$ electrons in the $d$-orbital.
$4$. Number of unpaired electrons: In the presence of strong field ligands like $NH_3$ and $CO_3^{2-}$,the $6$ electrons in the $3d$ orbitals of $Co^{3+}$ pair up in the $t_{2g}$ set. Therefore,the number of unpaired electrons is $0$.

(A) Ionisation isomerism occurs when the counter ion in a coordination compound is a potential ligand and can displace a ligand that is currently bonded to the central metal atom.
In the complex $[Pt(NH_3)_4 Cl_2] Br_2$,the $Br^-$ ions are outside the coordination sphere and can exchange places with the $Cl^-$ ligands inside the coordination sphere to form $[Pt(NH_3)_4 Cl Br] Cl Br$ or $[Pt(NH_3)_4 Br_2] Cl_2$.
Therefore,the correct option is $A$.

(D) The electronic configuration of $Ti^{3+}$ $(Z=22)$ is $[Ar] 3d^1$.
In an octahedral crystal field,the $d$-orbitals split into $t_{2g}$ and $e_g$ sets.
The single electron occupies the lower energy $t_{2g}$ orbital,giving the configuration $t_{2g}^1 e_g^0$.
Upon absorbing visible light,the electron is excited to the higher energy $e_g$ orbital.
Therefore,the electronic transition is $t_{2g}^1 e_g^0 \rightarrow t_{2g}^0 e_g^1$.

(C) The colour of transition metal ions is due to $d-d$ transitions,which require unpaired electrons in the $d$-orbitals.
$Zn^{2+}$ has the electronic configuration $[Ar] 3d^{10}$,meaning all its $d$-orbitals are completely filled.
Since there are no unpaired electrons,no $d-d$ transitions occur,making its aqueous solution colourless.

(A) The electronic configuration of $Co$ $(Z=27)$ is $[Ar] 3d^7 4s^2$.
For $Co^{2+}$,the configuration is $[Ar] 3d^7$.
In the $3d$ subshell,there are $3$ unpaired electrons $(n=3)$.
The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n=3$: $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.

(D) The actinide series consists of elements from atomic number $89$ to $103$.
These elements involve the filling of the $5f$ orbitals.
The general electronic configuration for the actinide series is given by $[Rn] 5f^{0-14} 6d^{0-2} 7s^2$.
Here,$[Rn]$ represents the noble gas core of Radon $(Z=86)$.

(A) The correct answer is $A$.
An electrolytic cell is a device that converts electrical energy into chemical energy,which is a non-spontaneous process.
In contrast,Leclanche cells,storage cells,and fuel cells are all types of electrochemical (galvanic/voltaic) cells that convert chemical energy into electrical energy through spontaneous redox reactions.

(C) The resistance $R$ of a conductor is directly proportional to its length $l$ and inversely proportional to its area of cross-section $A$.
Mathematically,this is expressed as $R \propto \frac{l}{A}$.
Therefore,the correct relation is $R \propto \frac{l}{A}$.

(B) In a mercury cell,the electrolyte is a paste of potassium hydroxide $(KOH)$ and zinc oxide $(ZnO)$.
Therefore,the correct mixture used as an electrolyte paste is $KOH + ZnO$.

(B) For a reaction to be spontaneous in an electrochemical cell,the cell potential $E_{cell}^0$ must be positive.
$E_{cell}^0 = E_{cathode}^0 - E_{anode}^0$.
Given $E_{Cl_2 \mid 2Cl^-}^0 = 1.36 \ V$ and $E_{Br_2 \mid 2Br^-}^0 = 1.09 \ V$.
For option $B$: $2Br^- + Cl_2 \rightarrow Br_2 + 2Cl^-$.
Here,$Cl_2$ is reduced to $Cl^-$ (cathode) and $Br^-$ is oxidized to $Br_2$ (anode).
$E_{cell}^0 = E_{Cl_2 \mid 2Cl^-}^0 - E_{Br_2 \mid 2Br^-}^0 = 1.36 \ V - 1.09 \ V = 0.27 \ V$.
Since $E_{cell}^0 > 0$,the reaction is spontaneous.

(B) According to Kohlrausch's law of independent migration of ions,the molar conductivity at infinite dilution for a weak electrolyte like $NH_4OH$ can be calculated using strong electrolytes.
$\Lambda_{m(NH_4OH)}^0 = \lambda_{NH_4^+}^0 + \lambda_{OH^-}^0$
To obtain this,we combine the molar conductivities of $NH_4Cl$,$NaOH$,and $NaCl$ as follows:
$\Lambda_{m(NH_4Cl)}^0 = \lambda_{NH_4^+}^0 + \lambda_{Cl^-}^0$
$\Lambda_{m(NaOH)}^0 = \lambda_{Na^+}^0 + \lambda_{OH^-}^0$
$\Lambda_{m(NaCl)}^0 = \lambda_{Na^+}^0 + \lambda_{Cl^-}^0$
By performing the operation $\Lambda_{m(NH_4Cl)}^0 + \Lambda_{m(NaOH)}^0 - \Lambda_{m(NaCl)}^0$,we get:
$(\lambda_{NH_4^+}^0 + \lambda_{Cl^-}^0) + (\lambda_{Na^+}^0 + \lambda_{OH^-}^0) - (\lambda_{Na^+}^0 + \lambda_{Cl^-}^0) = \lambda_{NH_4^+}^0 + \lambda_{OH^-}^0 = \Lambda_{m(NH_4OH)}^0$
Therefore,the correct expression is $\Lambda_{m(NH_4Cl)}^0 + \Lambda_{m(NaOH)}^0 - \Lambda_{m(NaCl)}^0$.

(C) The mixture of acetone $(CH_3COCH_3)$ and carbon disulphide $(CS_2)$ forms a non-ideal solution showing positive deviation from Raoult's law.
For solutions showing positive deviation,the intermolecular forces between the components are weaker than the pure components.
This leads to an increase in volume upon mixing,i.e.,$\Delta V_{mixture} > 0$,and an absorption of heat,i.e.,$\Delta H_{mixture} > 0$.

(A) The concentration is given as $0.05 \% w/v$,which means $0.05 \ g$ of $CaCl_2$ is present in $100 \ mL$ of the solution.
To convert $w/v$ percentage to $ppm$ (parts per million),we use the formula: $ppm = (w/v \%) \times 10^4$.
$ppm = 0.05 \times 10^4 = 500 \ ppm$.
Therefore,the correct option is $A$.

(B) The boiling point elevation is given by $\Delta T_b = i \times K_b \times m$. The solution with the highest boiling point will have the highest value of the van't Hoff factor $(i)$ multiplied by molality $(m)$.
For $A$: $i = 2$ $(NaCl \rightarrow Na^+ + Cl^-)$,$m = 0.1$,so $i \times m = 0.2$.
For $B$: $i = 3$ $(Ba(NO_3)_2 \rightarrow Ba^{2+} + 2NO_3^-)$,$m = 0.2$,so $i \times m = 0.6$.
For $C$: $i = 4$ $(Na_3PO_4 \rightarrow 3Na^+ + PO_4^{3-})$,$m = 0.01$,so $i \times m = 0.04$.
For $D$: $i = 2$ $(KNO_3 \rightarrow K^+ + NO_3^-)$,$m = 0.03$,so $i \times m = 0.06$.
Comparing the values,$0.6$ is the highest,therefore $0.2 \ m$ $Ba(NO_3)_2$ has the highest boiling point.

(C) Concentration units that involve volume (such as $Molarity$,$Normality$,and $Formality$) are temperature-dependent because volume changes with temperature.
$Molality$ is defined as the number of moles of solute per kilogram of solvent.
Since mass does not change with temperature,$Molality$ remains independent of temperature.

(A) The boiling point elevation is given by the formula $\Delta T_b = i \times K_b \times m$.
Since the molality $(m)$ and the ebullioscopic constant $(K_b)$ are identical for all solutions,the boiling point depends on the van't Hoff factor $(i)$.
For $K_4[Fe(CN)_6]$,$i = 5$ $(4K^+ + [Fe(CN)_6]^{4-})$.
For $Na_2SO_4$,$i = 3$ $(2Na^+ + SO_4^{2-})$.
For Urea,$i = 1$ (non-electrolyte).
For $NaCl$,$i = 2$ $(Na^+ + Cl^-)$.
Since $K_4[Fe(CN)_6]$ has the highest van't Hoff factor $(i=5)$,it will show the highest boiling point elevation and thus the highest boiling point.

(B) According to Henry's Law,the solubility of a gas in a liquid is given by the relation $P = K_H \times x$,where $P$ is the partial pressure of the gas,$K_H$ is Henry's law constant,and $x$ is the mole fraction of the gas in the solution.
Rearranging for solubility $(x)$,we get $x = P / K_H$.
Since $K_H$ is in the denominator,as the temperature increases,$K_H$ increases,which leads to a decrease in the solubility $(x)$ of the gaseous solute.