GSEB 2024 Chemistry Question Paper with Answer and Solution

(C) The nematode $Meloidogyne$ $incognita$ infects the roots of tobacco plants.
This infection causes a significant reduction in the yield of the crop.
To prevent this,$RNA$ interference $(RNAi)$ technology is used to silence the specific mRNA of the nematode,thereby protecting the plant roots from infestation.

(B) Given differential equation is $\frac{y dx - x dy}{y} = 0$.
Divide both sides by $xy$ (assuming $x \neq 0, y \neq 0$):
$\frac{y dx - x dy}{xy} = 0$
This can be rewritten as:
$\frac{dx}{x} - \frac{dy}{y} = 0$
Integrating both sides with respect to their variables:
$\int \frac{1}{x} dx - \int \frac{1}{y} dy = \int 0 dx$
$\log |x| - \log |y| = \log |C|$
Using the logarithmic property $\log a - \log b = \log(\frac{a}{b})$:
$\log \left| \frac{x}{y} \right| = \log |C|$
Taking the exponential of both sides:
$\frac{x}{y} = C$
Rearranging for $y$:
$y = \frac{1}{C} x$
Since $\frac{1}{C}$ is an arbitrary constant,we can write it as $C_1$:
$y = C_1 x$
Thus,the correct option is $B$.

(C) In an $n-$type semiconductor,the majority charge carriers are electrons and the minority charge carriers are holes.
To create an $n-$type semiconductor,silicon (a tetravalent element) is doped with pentavalent impurity atoms (such as phosphorus,arsenic,or antimony).
These pentavalent atoms provide extra electrons to the conduction band,making electrons the majority carriers.
Therefore,holes are the minority carriers and pentavalent atoms are the dopants.
Thus,the correct statement is $(c)$.

(A) The energy band gap $(E_{g})$ is the energy difference between the top of the valence band and the bottom of the conduction band.
For elements in Group $14$ of the periodic table,as we move down the group,the atomic size increases and the strength of the covalent bonds decreases.
Consequently,the energy required to excite an electron from the valence band to the conduction band decreases.
The energy band gaps for these elements are approximately:
Carbon (Diamond): $(E_{g}) \approx 5.4 \ eV$
Silicon: $(E_{g}) \approx 1.1 \ eV$
Germanium: $(E_{g}) \approx 0.7 \ eV$
Therefore,the correct relationship is $(E_{g})_{C} > (E_{g})_{Si} > (E_{g})_{Ge}$.

(D) $p$-nitrophenol exhibits intermolecular hydrogen bonding,which leads to the association of molecules,thereby increasing the boiling point.
In contrast,$o$-nitrophenol exhibits intramolecular hydrogen bonding,which prevents the association of molecules,resulting in a lower boiling point.
Therefore,both the assertion and the reason are correct,and the reason is the correct explanation for the assertion.
The correct option is $D$.

(D) Aspirin is acetylsalicylic acid with the chemical formula $C_9H_8O_4$.
Its structure consists of a benzene ring,a carboxylic acid group $(-COOH)$,and an ester group $(-OCOCH_3)$.
Counting the bonds:
$1$. The benzene ring has $6$ $C-C$ $\sigma$ bonds,$3$ $C-C$ $\pi$ bonds,and $4$ $C-H$ $\sigma$ bonds.
$2$. The carboxylic acid group has $1$ $C-C$ $\sigma$ bond,$1$ $C=O$ $\sigma$ bond,$1$ $C=O$ $\pi$ bond,$1$ $C-O$ $\sigma$ bond,and $1$ $O-H$ $\sigma$ bond.
$3$. The ester group has $1$ $C-O$ $\sigma$ bond,$1$ $C=O$ $\sigma$ bond,$1$ $C=O$ $\pi$ bond,$1$ $C-C$ $\sigma$ bond,and $3$ $C-H$ $\sigma$ bonds.
Total $\sigma$ bonds = $6 (ring) + 4 (C-H) + 1 (C-C) + 1 (C=O) + 1 (C-O) + 1 (O-H) + 1 (C-O) + 1 (C=O) + 1 (C-C) + 3 (C-H) = 21$.
Total $\pi$ bonds = $3 (ring) + 1 (C=O) + 1 (C=O) = 5$.
Therefore,the number of $\sigma$ and $\pi$ bonds are $21$ and $5$ respectively.

(A) $1$. Identify the longest carbon chain containing the principal functional groups (hydroxyl groups). The chain has $6$ carbons,so the parent alkane is hexane.
$2$. Number the chain from the end that gives the lowest locants to the substituents and functional groups. Numbering from right to left gives the hydroxyl groups at positions $1$ and $3$,and the methyl group at position $5$.
$3$. The structure is $(CH_3)_2CH-CH_2-CH(OH)-CH(CH_2OH)-CH_3$. The longest chain is $6$ carbons long. The substituents are a methyl group at $C-5$ and two hydroxyl groups at $C-1$ and $C-3$.
$4$. The correct $IUPAC$ name is $2,5-$dimethylhexane$-1,3-$diol.

(A) The reaction of $2-$methylbut$-2-$ene with ozone $(O_3)$ followed by reductive workup with $Zn / H_2 O$ is known as ozonolysis.
In this reaction,the double bond $(C=C)$ is cleaved to form carbonyl compounds.
The structure of $2-$methylbut$-2-$ene is $CH_3-C(CH_3)=CH-CH_3$.
Upon ozonolysis,the bond breaks at the double bond position:
$CH_3-C(CH_3)=CH-CH_3 \rightarrow CH_3COCH_3 + CH_3CHO$.
The products formed are acetone $(CH_3COCH_3)$ and acetaldehyde $(CH_3CHO)$.

(D) The reaction of sodium acetate $(CH_3COONa)$ with sodalime $(NaOH + CaO)$ is known as decarboxylation.
When sodium acetate is heated with sodalime,it loses a molecule of $CO_2$ to form methane $(CH_4)$ and sodium carbonate $(Na_2CO_3)$.
The chemical equation is: $CH_3COONa + NaOH \xrightarrow{CaO, \Delta} CH_4 + Na_2CO_3$.
Thus,the product obtained is methane.

(C) The nematode $Meloidogyne \ incognita$ infects the roots of tobacco plants. This infection causes a significant reduction in the yield of the crop. To prevent this,$RNA$ interference $(RNAi)$ technology is used to create pest-resistant tobacco plants.

(B) Given the differential equation: $\frac{y dx - x dy}{y} = 0$.
Since $y \neq 0$,we can multiply by $y$ to get $y dx - x dy = 0$.
Rearranging the terms,we have $y dx = x dy$.
Separating the variables,we get $\frac{dx}{x} = \frac{dy}{y}$.
Integrating both sides,we obtain $\int \frac{dx}{x} = \int \frac{dy}{y}$.
This results in $\ln|x| = \ln|y| + \ln|C|$,where $\ln|C|$ is the constant of integration.
Using the property of logarithms,$\ln|x| = \ln|Cy|$.
Taking the exponential of both sides,we get $x = Cy$,which can be rewritten as $y = \frac{1}{C} x$.
Since $\frac{1}{C}$ is also an arbitrary constant,we can write $y = Cx$.

(A) Acidified potassium dichromate $(K_2Cr_2O_7)$ is a strong oxidizing agent.
It oxidizes stannous ions $(Sn^{2+})$ to stannic ions $(Sn^{4+})$ in an acidic medium.
The balanced chemical equation is:
$Cr_2O_7^{2-} + 14H^+ + 3Sn^{2+} \rightarrow 2Cr^{3+} + 3Sn^{4+} + 7H_2O$
Therefore,$Sn^{2+}$ changes to $Sn^{4+}$.

(C) The manganate ion is represented by the chemical formula $MnO_4^{2-}$.
In this ion,manganese is in the $+6$ oxidation state.
In contrast,the permanganate ion is represented by the formula $MnO_4^-$,where manganese is in the $+7$ oxidation state.
Therefore,the correct option is $C$.

(D) molecule is non-polar if its net dipole moment is zero.
$HCl$ is a polar molecule because there is an electronegativity difference between $H$ and $Cl$ atoms.
$NH_3$ has a pyramidal geometry with a net dipole moment due to the lone pair on the nitrogen atom.
$H_2O$ has a bent geometry,which results in a net dipole moment.
$H_2$ is a homonuclear diatomic molecule where both atoms have the same electronegativity,leading to a net dipole moment of $0$. Thus,$H_2$ is non-polar.

(A) The reaction of $p$-ethylnitrobenzene with $Br_2$ in the presence of heat or $UV$ light proceeds via a free radical substitution mechanism.
In this mechanism,the bromine radical abstracts a hydrogen atom from the benzylic position because the resulting benzylic radical is stabilized by resonance with the benzene ring.
Specifically,the hydrogen atom is removed from the $\alpha$-carbon (the carbon attached directly to the benzene ring) of the ethyl group.
This leads to the formation of a stable benzylic radical,which then reacts with $Br_2$ to form the major product,$1-nitro-4-(1-bromoethyl)benzene$.
Therefore,the correct structure is $O_2N-C_6H_4-CH(Br)CH_3$.

(A) The reaction of an alkyl aryl ether with $HI$ involves the cleavage of the $C_{alkyl}-O$ bond because the $C_{aryl}-O$ bond has partial double bond character due to resonance and is stronger.
Therefore,the reaction proceeds as follows:
$C_6H_5-O-C_2H_5 + HI \rightarrow C_6H_5OH + C_2H_5I$
Here,$X$ is $C_6H_5OH$ (phenol) and $Y$ is $C_2H_5I$ (ethyl iodide).
Thus,the correct option is $A$.

(C) The reaction of phenol with $H_2CrO_4$ (chromic acid) is an oxidation reaction,not an electrophilic substitution reaction.
In this reaction,phenol is oxidized to $p$-benzoquinone.
Other reactions listed:
$(A)$ Reimer-Tiemann reaction (electrophilic substitution),
$(B)$ Nitration of phenol (electrophilic substitution),
$(D)$ Bromination of phenol (electrophilic substitution).

(B) The reduction of $Butanone$ $(CH_3COCH_2CH_3)$ using a reducing agent like $NaBH_4$ or $LiAlH_4$ yields $Butan-2-ol$ $(CH_3CH(OH)CH_2CH_3)$.
In $Butan-2-ol$,the carbon atom attached to the $-OH$ group is bonded to four different groups: $-H$,$-OH$,$-CH_3$,and $-CH_2CH_3$.
Since this carbon is a chiral center,$Butan-2-ol$ exists as a pair of enantiomers and is optically active.
The other options yield primary alcohols $(Butan-1-ol)$ which are achiral.

(C) The reactivity of alcohols with Lucas's reagent $(conc. \ HCl + anhydrous \ ZnCl_2)$ follows the order: $3^\circ > 2^\circ > 1^\circ$.
This is because the reaction proceeds via an $S_N1$ mechanism involving the formation of a carbocation intermediate.
$3^\circ$ alcohols form the most stable carbocation and thus react the fastest,typically at room temperature.
$(CH_3)_3COH$ is a tertiary $(3^\circ)$ alcohol,while the others are secondary $(2^\circ)$ or primary $(1^\circ)$ alcohols.
Therefore,$(CH_3)_3COH$ gives the fastest reaction.

(B) The reaction of an ether with $HI$ proceeds via an $S_N1$ mechanism when one of the alkyl groups is tertiary.
In the given ether,$(CH_3)_3C-O-C_2H_5$,the tert-butyl group $(CH_3)_3C-$ is capable of forming a stable tertiary carbocation.
Therefore,the $C-O$ bond between the tert-butyl group and the oxygen atom breaks,leading to the formation of tert-butyl iodide $(CH_3)_3C-I$ and ethanol $C_2H_5OH$.
Thus,$X = (CH_3)_3C-I$ and $Y = C_2H_5OH$.

(A) The Reimer-Tiemann reaction involves the treatment of phenol with chloroform $(CHCl_3)$ in the presence of an aqueous base like sodium hydroxide $(NaOH)$.
This reaction leads to the introduction of a formyl group $(-CHO)$ at the ortho position of the phenol ring.
The final product formed is $2$-hydroxybenzaldehyde,commonly known as Salicylaldehyde.
Therefore,the correct option is $A$.

(C) $NaBH_4$ (Sodium borohydride) is a selective reducing agent that reduces aldehydes and ketones to their corresponding alcohols,but it does not reduce esters under normal conditions.
In the given molecule,there is a ketone group in the cyclohexane ring and an ester group $(-COOCH_3)$.
$NaBH_4$ will selectively reduce the ketone group to a secondary alcohol $(-CH(OH)-)$ while leaving the ester group intact.
Therefore,the product is a molecule with a cyclohexanol ring attached to a $-CH_2-COOCH_3$ group.
This corresponds to the structure shown in option $C$.

(B) The reaction between benzaldehyde $(C_6H_5CHO)$ and acetophenone $(C_6H_5COCH_3)$ in the presence of a dilute base $(OH^-)$ is a Claisen-Schmidt condensation reaction.
In this reaction,the enolate ion formed from acetophenone attacks the carbonyl carbon of benzaldehyde.
The initial product is a $\beta$-hydroxy ketone,which subsequently undergoes dehydration to form an $\alpha,\beta$-unsaturated ketone.
The reaction is:
$C_6H_5CHO + CH_3COC_6H_5$ $\xrightarrow{OH^-} C_6H_5CH(OH)CH_2COC_6H_5$ $\xrightarrow{-H_2O} C_6H_5CH=CHCOC_6H_5$
The final product is benzalacetophenone (also known as chalcone),which corresponds to option $B$.

(D) Sodium benzoate $(C_6H_5COONa)$ is widely used as a food preservative because it inhibits the growth of bacteria,yeast,and mold in acidic conditions.
Therefore,the correct option is $D$.

(C) The $pK_a$ value is inversely proportional to the acidity of the compound. $A$ stronger acid has a lower $pK_a$ value,while a weaker acid has a higher $pK_a$ value.
Acidity depends on the stability of the conjugate base (carboxylate ion). Electron-withdrawing groups $(EWG)$ increase acidity,while electron-donating groups $(EDG)$ decrease acidity.
Comparing the substituents on the benzene ring:
$1$. $-NO_2$ is a strong $EWG$ (increases acidity).
$2$. $-Cl$ is a weak $EWG$ (increases acidity).
$3$. $-H$ is the reference.
$4$. $-CH_3$ is an $EDG$ (decreases acidity).
Since $-CH_3$ is an electron-donating group,it destabilizes the carboxylate ion,making $p$-methylbenzoic acid the weakest acid among the given options.
Therefore,$p$-methylbenzoic acid has the highest $pK_a$ value.

(B) Fehling reagent is a mild oxidizing agent that is used to distinguish between aliphatic aldehydes and ketones.
Aliphatic aldehydes,such as $CH_3CHO$ (Acetaldehyde),are easily oxidized by Fehling reagent to their corresponding carboxylate ions,resulting in a reddish-brown precipitate of $Cu_2O$.
Ketones (like Acetophenone and Acetone) and aromatic aldehydes (like Benzaldehyde) do not reduce Fehling reagent.
Therefore,the correct answer is $B$ (Acetaldehyde).

(C) . Gattermann-Koch reaction: $C_6H_6 + CO + HCl \xrightarrow{\text{Anhy. } AlCl_3} C_6H_5CHO$ (Produces benzaldehyde).
$B$. Etard reaction: $C_6H_5CH_3 + CrO_2Cl_2 \xrightarrow{CS_2} C_6H_5CHO$ (Produces benzaldehyde).
$C$. Oxidation of toluene with $KMnO_4 + KOH$ yields potassium benzoate $(C_6H_5COOK)$,which upon acidification gives benzoic acid $(C_6H_5COOH)$,not benzaldehyde.
$D$. Rosenmund reduction: $C_6H_5COCl + H_2 \xrightarrow{Pd-BaSO_4} C_6H_5CHO$ (Produces benzaldehyde).
Therefore,the reaction that does not produce benzaldehyde is $C$.

(C) The acidity of carboxylic acids is determined by the stability of the conjugate base (carboxylate ion).
Electron-withdrawing groups (EWGs) stabilize the carboxylate ion through the inductive effect ($-I$ effect),thereby increasing the acidity and the $K_a$ value.
Comparing the substituents:
$1$. $NO_2$ group has a strong $-I$ effect.
$2$. $Br$ group has a moderate $-I$ effect.
$3$. $CCl_3$ group has a very strong $-I$ effect due to three highly electronegative chlorine atoms.
$4$. $CH_3$ group is an electron-donating group ($+I$ effect),which decreases acidity.
Among the given options,$CCl_3COOH$ has the strongest electron-withdrawing effect,making it the strongest acid with the highest $K_a$ value.

(B) The Iodoform reaction is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$1$. Propanal $(CH_3CH_2CHO)$ does not contain the $CH_3CO-$ group.
$2$. Acetone $(CH_3COCH_3)$ contains the $CH_3CO-$ group and thus gives a positive Iodoform test.
$3$. Pent-$3$-one $(CH_3CH_2COCH_2CH_3)$ does not contain a methyl group attached to the carbonyl carbon.
$4$. Benzophenone $(C_6H_5COC_6H_5)$ does not contain a methyl group attached to the carbonyl carbon.
Therefore,the correct answer is $B$.

(A) The reaction of a ketone with hydroxylamine $(NH_2OH)$ in an acidic medium is a nucleophilic addition-elimination reaction (condensation reaction).
Cyclopentanone reacts with hydroxylamine to form cyclopentanone oxime and water.
The reaction is represented as:
$C_5H_8O + NH_2OH \xrightarrow{H^+} C_5H_8NOH + H_2O$.
Therefore,the correct product is cyclopentanone oxime.

(D) The Cannizzaro reaction is a disproportionation reaction given by aldehydes that do not possess an $\alpha$-hydrogen atom.
$HCHO$ (formaldehyde) has no $\alpha$-hydrogen.
$C_6H_5CHO$ (benzaldehyde) has no $\alpha$-hydrogen.
$1\text{-methylcyclohexanecarbaldehyde}$ has no $\alpha$-hydrogen at the carbonyl carbon position.
$CH_3CHO$ (acetaldehyde) possesses three $\alpha$-hydrogen atoms,so it undergoes aldol condensation instead of the Cannizzaro reaction.

(C) The Gatterman-Koch reaction is a specific form of Friedel-Crafts acylation used to introduce a formyl group $(-CHO)$ into an aromatic ring.
In this reaction,benzene is treated with carbon monoxide $(CO)$ and hydrogen chloride $(HCl)$ in the presence of anhydrous aluminum chloride $(AlCl_3)$ or cuprous chloride $(CuCl)$ to produce benzaldehyde.
The reaction is represented as: $C_6H_6 + CO + HCl \xrightarrow{anhydrous \ AlCl_3 / CuCl} C_6H_5CHO$.

(C) The chemical names of the vitamins are as follows:
$(i)$ Vitamin $B_6$ is Pyridoxine $(S)$.
$(ii)$ Vitamin $B_1$ is Thiamine $(R)$.
$(iii)$ Vitamin $B_2$ is Riboflavin $(Q)$.
$(iv)$ Vitamin $C$ is Ascorbic acid $(P)$.
Therefore,the correct matching is $(i)$ $\rightarrow S, (ii)$ $\rightarrow R, (iii)$ $\rightarrow Q, (iv)$ $\rightarrow P$.

(A) The carbohydrate stored in the liver of animals is $Glycogen$.
It is often referred to as animal starch because its structure is similar to $Amylopectin$ and it serves as the primary storage form of glucose in animals.

(D) $Testosterone$ is the primary male sex hormone and an anabolic steroid.
It plays a key role in the development of male reproductive tissues such as the testes and prostate,as well as promoting secondary sexual characteristics such as increased muscle and bone mass,and the growth of body hair and deepening of the voice.
Therefore,the correct option is $D$.

(D) In nucleic acids,the nitrogenous bases are classified into two types based on their structure:
$1$. $\text{Purines}$ (bicyclic bases with two fused rings): Adenine $(A)$ and Guanine $(G)$.
$2$. $\text{Pyrimidines}$ (monocyclic bases with one ring): Cytosine $(C)$,Thymine $(T)$,and Uracil $(U)$.
Since Guanine $(G)$ is a purine,it possesses a bicyclic (double-ring) structure. Therefore,the correct answer is $G$.

(C) An amino acid is optically active if it contains a chiral carbon atom (a carbon atom bonded to four different groups).
Glycine has the structure $NH_2-CH_2-COOH$.
In glycine,the central carbon atom is bonded to two identical hydrogen atoms,meaning it is not a chiral center.
Therefore,glycine is the only amino acid among the options that is not optically active.

(B) In the double helix structure of $DNA$,nitrogenous bases form specific base pairs through hydrogen bonding.
$Adenine$ $(A)$ pairs with $Thymine$ $(T)$ via two hydrogen bonds.
$Guanine$ $(G)$ pairs with $Cytosine$ $(C)$ via three hydrogen bonds.
Therefore,the correct pair among the options is $Adenine$ and $Thymine$.

(A) The correct answer is $A$.
Ascorbic acid is chemically known as $Vitamin \ C$,which is an essential water-soluble vitamin required for the synthesis of collagen and as an antioxidant.

(C) $Lactose$ is a disaccharide composed of two monosaccharide units.
Upon hydrolysis,$Lactose$ breaks down into one molecule of $D-(+)-Glucose$ and one molecule of $D-(+)-Galactose$.
The reaction is: $C_{12}H_{22}O_{11} + H_2O \rightarrow C_6H_{12}O_6 (Glucose) + C_6H_{12}O_6 (Galactose)$.

(B) According to the Arrhenius equation,$\ln K = \ln A - \frac{E_{a}}{RT}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \ln K$,$x = \frac{1}{T}$,and $m$ is the slope.
The slope $m = -\frac{E_{a}}{R}$.
Given $E_{a} = 33.256 \ J \ mol^{-1}$ and the gas constant $R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Slope $m = -\frac{33.256}{8.314} = -4$.

(C) catalyst provides an alternative reaction pathway with a lower activation energy $(E_a)$.
It increases the rate of both the forward and backward reactions equally.
However,it does not affect the thermodynamic properties of the reaction,such as the enthalpy change $(\Delta H)$.
Therefore,the heat evolved or absorbed during the reaction remains unchanged.

(D) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 5 \ min$,so $k = \frac{0.693}{5} \ min^{-1}$.
The time $t$ required for $99.9 \%$ completion is given by the formula $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.
For $99.9 \%$ completion,$[A]_t = [A]_0 - 0.999[A]_0 = 0.001[A]_0$.
Substituting the values: $t = \frac{2.303}{0.693/5} \log \frac{[A]_0}{0.001[A]_0}$.
$t = \frac{2.303 \times 5}{0.693} \log(1000)$.
Since $\log(1000) = 3$ and $\frac{2.303}{0.693} \approx 3.32$,we have $t \approx 3.32 \times 5 \times 3 = 49.8 \approx 50 \ min$.
Thus,the correct option is $D$.

(D) The rate law for the reaction is given by: $Rate = k[A]^2[B]^0 = k[A]^2$.
When the concentration of $A$ is doubled $([A]' = 2[A])$ and the concentration of $B$ is halved $([B]' = 0.5[B])$,the new rate $(Rate')$ is:
$Rate' = k(2[A])^2(0.5[B])^0 = k(4[A]^2)(1) = 4k[A]^2$.
Comparing the new rate with the initial rate,we get $Rate' = 4 \times Rate$.
Therefore,the rate becomes $4$ times the original rate.

(A) The unit of the rate constant $K$ for a reaction of order $n$ is given by the formula: $(mol \ L^{-1})^{1-n} \ s^{-1}$.
Given the unit is $L^2 \ mol^{-2} \ s^{-1}$,which can be rewritten as $mol^{-2} \ L^2 \ s^{-1}$.
Comparing this with $(mol \ L^{-1})^{1-n} \ s^{-1}$,we have:
$(mol \ L^{-1})^{1-n} = mol^{-2} \ L^2$.
This implies $1-n = -2$.
Solving for $n$,we get $n = 1 + 2 = 3$.
Therefore,the order of the reaction is $3$.

(D) For a zero order reaction,the rate law is given by $Rate = k[R]^0 = k$.
Integrating this,we get $[R]_t = [R]_0 - kt$.
At half-life $(t = t_{1/2})$,the concentration $[R]_t = \frac{[R]_0}{2}$.
Substituting this into the integrated rate equation: $\frac{[R]_0}{2} = [R]_0 - kt_{1/2}$.
Rearranging for $t_{1/2}$,we get $kt_{1/2} = \frac{[R]_0}{2}$,which simplifies to $t_{1/2} = \frac{[R]_0}{2k}$.
Since $k$ is a constant,$t_{1/2} \propto [R]_0$.
Therefore,the correct relation is $t_{1/2} \propto [R]_0$.

(A) catalyst provides an alternative reaction pathway with a lower activation energy.
It does not change the Gibbs energy $( \Delta G )$ of the reaction,nor does it alter the equilibrium constant $( K_{eq} )$.
It simply speeds up the attainment of equilibrium by lowering the activation energy barrier.
Therefore,the statement that it does not alter Gibbs energy is true.

(B) The rate law for the reaction is given by $Rate = k[A]^1[B]^2$.
If the concentration of $B$ is increased three times,the new concentration becomes $[B]' = 3[B]$.
The new rate of reaction $Rate'$ will be $Rate' = k[A]^1(3[B])^2$.
$Rate' = k[A]^1(9[B]^2) = 9 \times k[A]^1[B]^2$.
$Rate' = 9 \times Rate$.
Therefore,the rate of reaction increases by $9$ times.
Note: The rate constant $k$ remains unaffected by changes in concentration.

(C) The Arrhenius equation is given by $K = A e^{-E_a / RT}$.
Taking the natural logarithm on both sides,we get $\ln K = \ln A - \frac{E_a}{RT}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \ln K$,$x = \frac{1}{T}$,$m$ is the slope,and $c$ is the intercept.
Here,the slope $m = -\frac{E_a}{R}$.
Therefore,the correct option is $C$.

(A) 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.

(C) The complex $[Ni(CO)_4]$ involves $Ni$ in the $0$ oxidation state. The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$. In the presence of strong field ligand $CO$,the $4s$ electrons pair up into the $3d$ orbitals,resulting in a $3d^{10}$ configuration. This leads to $sp^3$ hybridization,which corresponds to a tetrahedral geometry. Since all electrons are paired,it is diamagnetic. The other options like $[NiCl_4]^{2-}$ are tetrahedral but paramagnetic due to the presence of unpaired electrons in the $3d$ orbitals.

(B) In a coordination complex,the crystal field splitting energy $(\Delta_0)$ and the pairing energy $(P)$ determine the electronic configuration of the metal ion.
For a high spin complex,the crystal field splitting energy is less than the pairing energy $(\Delta_0 < P)$.
This means that electrons prefer to occupy higher energy orbitals rather than pairing up in the lower energy orbitals,resulting in a maximum number of unpaired electrons.
Therefore,the correct condition is $\Delta_0 < P$.

(D) Chlorophyll is a green pigment found in plants that plays a vital role in photosynthesis.
It is a porphyrin derivative containing a central metal ion.
The central metal ion in the chlorophyll molecule is $Mg^{2+}$ (Magnesium ion).
Therefore,chlorophyll is a coordination compound of Magnesium.

(C) Optical isomerism is shown by complexes that lack a plane of symmetry and a center of inversion.
$Cis-[CrCl_2(ox)_2]^{3-}$ contains two bidentate oxalate $(ox^{2-})$ ligands and two chloride ligands in a cis configuration.
This complex lacks a plane of symmetry and is chiral,meaning it exists as a pair of non-superimposable mirror images (enantiomers).
In contrast,$Trans-[PtCl_2(en)_2]^{2+}$ has a plane of symmetry,and the other options are either achiral or do not exhibit optical isomerism due to their specific geometry.

(D) The chemical formula of Iron $(III)$ hexacyanoferrate $(II)$ is $Fe_4[Fe(CN)_6]_3$.
On ionization,it dissociates as follows:
$Fe_4[Fe(CN)_6]_3 \rightarrow 4Fe^{3+} + 3[Fe(CN)_6]^{4-}$
Total number of ions = $4 + 3 = 7$.

(A) The stability of a coordination complex is influenced by the chelate effect.
$C_2O_4^{2-}$ (oxalate) is a bidentate ligand,which forms a five-membered chelate ring with the central metal ion.
Complexes containing chelate rings are significantly more stable than those containing only monodentate ligands like $NH_3$,$H_2O$,or $Cl^-$.
Therefore,$[Fe(C_2O_4)_3]^{3-}$ is the most stable complex among the given options due to the chelate effect.

(C) An ambidentate ligand is a ligand that can coordinate to the central metal atom through two different donor atoms.
$SCN^{-}$ is an ambidentate ligand because it can coordinate through the sulfur atom (thiocyanato-$S$) or the nitrogen atom (isothiocyanato-$N$).
$H_2O$ is a monodentate ligand,$C_2O_4^{2-}$ is a bidentate ligand,and $[EDTA]^{4-}$ is a hexadentate ligand.

(D) The central metal atom is $Ni$ with atomic number $28$. The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.
In the complex ion $[Ni(CN)_4]^{2-}$,the oxidation state of $Ni$ is $x + 4(-1) = -2$,which gives $x = +2$.
Thus,$Ni^{2+}$ has the configuration $[Ar] 3d^8$.
$CN^-$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.
After pairing,all $8$ electrons occupy the $d_{xy}, d_{yz}, d_{xz},$ and $d_{z^2}$ orbitals in pairs,leaving no unpaired electrons.
Therefore,the number of unpaired electrons is $0$.

(A) The given complexes $[Co(NH_3)_5Cl]SO_4$ and $[Co(NH_3)_5SO_4]Cl$ differ in the counter ion present outside the coordination sphere.
In the first complex,the $SO_4^{2-}$ ion is the counter ion,while in the second complex,the $Cl^-$ ion is the counter ion.
When these complexes are dissolved in water,they give different ions in the solution.
This type of isomerism,where the counter ion is exchanged with a ligand from the coordination sphere,is known as ionisation isomerism.
Therefore,the correct option is $A$.

(B) In the coordination compound $[Co(en)_3]Cl_3$,the central metal atom is $Co$.
Primary valency corresponds to the oxidation state of the central metal atom. Let the oxidation state of $Co$ be $x$. Since $en$ (ethylenediamine) is a neutral ligand and $Cl$ has a charge of $-1$,we have $x + 3(0) + 3(-1) = 0$,which gives $x = +3$. Thus,the primary valency is $3$.
Secondary valency corresponds to the coordination number of the central metal atom. $en$ is a bidentate ligand,so $3$ molecules of $en$ provide $3 \times 2 = 6$ donor atoms. Thus,the coordination number is $6$.
Therefore,the primary and secondary valencies are $3$ and $6$ respectively.

(A) The ion $MnO_4^{2-}$ (manganate ion) is green in colour.
$MnO_4^{-}$ (permanganate ion) is purple.
$Cr_2O_7^{2-}$ (dichromate ion) is orange.
$CrO_4^{2-}$ (chromate ion) is yellow.
Therefore,the correct option is $A$.

(B) Bronze is an alloy primarily composed of copper $(Cu)$ and tin $(Sn)$. Therefore,the correct option is $B$.

(D) The Wacker process is an industrial method for the oxidation of ethylene to acetaldehyde.
In this process,$PdCl_2$ (palladium$(II)$ chloride) is used as the primary catalyst,typically in the presence of $CuCl_2$ as a co-catalyst.
Therefore,the correct option is $D$.

(A) The magnetic moment (spin-only) is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
$1$. For $Fe^{3+}$ $(Z=26)$: Electronic configuration is $[Ar] 3d^5$. Number of unpaired electrons $(n)$ = $5$.
$2$. For $Cr^{3+}$ $(Z=24)$: Electronic configuration is $[Ar] 3d^3$. Number of unpaired electrons $(n)$ = $3$.
$3$. For $Co^{3+}$ $(Z=27)$: Electronic configuration is $[Ar] 3d^6$. Number of unpaired electrons $(n)$ = $4$.
$4$. For $Ti^{3+}$ $(Z=22)$: Electronic configuration is $[Ar] 3d^1$. Number of unpaired electrons $(n)$ = $1$.
Since $Fe^{3+}$ has the highest number of unpaired electrons $(n=5)$,it has the highest magnetic moment.

(D) Transition elements are characterized by the presence of partially filled $d$-orbitals,which allow them to exhibit variable oxidation states.
However,$Scandium$ ($Sc$,atomic number $21$) has an electronic configuration of $[Ar] 3d^1 4s^2$.
In its compounds,it loses all three valence electrons to form the $Sc^{3+}$ ion,which has a stable noble gas configuration $([Ar])$.
Because it does not have other stable oxidation states,$Scandium$ is the only transition element among the choices that does not exhibit variable oxidation states.
Therefore,the correct option is $D$.