JEE Main 2025 April 7 Shift 1 Chemistry Question Paper, Exam Analysis, and Answer Keys (Available)

Step 1: Analyze Statement I

Ozonolysis is a reaction where an alkene undergoes cleavage in the presence of ozone (\( O_3 \)), forming an ozonide intermediate. This ozonide is then reduced by zinc (\( Zn \)) and water (\( H_2 O \)) to give two carbonyl compounds.


For cis-2-butene, the reaction proceeds as follows: \[ cis-2-butene \xrightarrow{O_3} ozonide intermediate \xrightarrow{Zn, H_2O} ethanal (acetaldehyde). \]
When cis-2-butene undergoes ozonolysis, the double bond is cleaved symmetrically, producing two molecules of ethanal (acetaldehyde), which is a simple aldehyde.
Therefore, Statement I is true.


Step 2: Analyze Statement II

For \textit{3, 6-dimethyloct-4-ene, when ozonolysis occurs followed by treatment with \( Zn \) and \( H_2 O \), the reaction will lead to cleavage of the double bond, forming two products. The key point here is the symmetry of the molecule.

3, 6-dimethyloct-4-ene is a symmetric compound, and the cleavage of the double bond leads to the formation of two carbonyl groups. The product is symmetrical, and therefore, it will not have any chiral centers because both products are non-chiral and identical.

In other words, the two products obtained are symmetrical and lack any chiral centers, making Statement II false. Quick Tip: For ozonolysis, always check the symmetry of the reactant molecule. Symmetrical cleavage products will not have chiral centers, while asymmetrical products will have chiral centers.

The carbamylamine test is used to detect primary amines. It results in the formation of an isocyanide (carbamylamine).

Option A \( NH_2 \) (Phenylamine) is a primary amine and will give a positive result.

Option B \( (CH_3)_2NH \) (Dimethylamine) is a secondary amine and does not give a positive result.

Option C \( CH_3NH_2 \) (Methylamine) is a primary amine and will give a positive result.

Option D \( (CH_3)_3N \) (Trimethylamine) is a tertiary amine and does not give a positive result.

Option E \( HNCH_3 \) (Methylamine attached to a benzene ring) is a primary amine and will give a positive result.

Thus, the correct options are A and C. Quick Tip: Only primary amines show a positive result in the carbamylamine test, forming isocyanides (carbamylamine).

Step 1: Analyze Statement (1)

The total pressure at \(t = \infty\) is 240 mm Hg, which is the pressure due to the products of the reaction (2B + C). The initial pressure of A was 80 mm Hg, and since 2 moles of B and 1 mole of C are produced per mole of A, the increase in total pressure from the initial 80 mm Hg to the final 240 mm Hg indicates that 160 mm Hg pressure is due to the products.

So, the initial pressure of A is indeed 80 mm Hg. Therefore, statement (1) is correct.

Step 2: Analyze Statement (2)

For a first-order reaction, the reaction does not go to completion. It asymptotically approaches a state where the concentration of the reactant is very low but not zero. As time progresses, the pressure will increase, but it will not go to 240 mm Hg immediately. The reaction approaches this value asymptotically. Therefore, Statement (2) is correct.

Step 3: Analyze Statement (3) - Rate Constant Calculation
For a first-order reaction, the integrated rate law is given by: \[ \ln \left( \frac{[A]_0}{[A]} \right) = kt \]
Where:

\([A]_0\) is the initial pressure of A (80 mm Hg),

\([A]\) is the pressure of A after 10 minutes (the difference between the initial and the total pressure),

\(k\) is the rate constant,

\(t\) is the time in minutes.


At \(t = 10\) minutes, the total pressure is 160 mm Hg, and the pressure of A is: \[ P_A = 80 - (160 - 240) = 40 mm Hg \]
Now, applying the rate law: \[ \ln \left( \frac{80}{40} \right) = k \times 10 \] \[ \ln(2) = k \times 10 \] \[ k = \frac{\ln(2)}{10} = \frac{0.693}{10} = 0.0693 \, min^{-1} \]
The rate constant is approximately \(0.0693 \, min^{-1}\), not 1.693 min\(^{-1}\).
Thus, Statement (3) is incorrect.

Step 4: Analyze Statement (4)
The partial pressure of A after 10 minutes is given as 40 mm Hg, which we calculated earlier. Therefore, Statement (4) is correct. Quick Tip: For first-order reactions, use the integrated rate law to calculate the rate constant. The reaction does not go to completion but asymptotically approaches the equilibrium state.

To find the total enthalpy change, we need to consider both the freezing process and the cooling of the substance:

1. Freezing process: The enthalpy change for freezing is \( \Delta_{fus}H = -x \), since freezing is an exothermic process.

2. Cooling of liquid water: The enthalpy change for cooling the water from 10°C to 0°C is calculated using the specific heat capacity of liquid water:
\[ \Delta H_1 = -10y \]
3. Cooling of ice: The enthalpy change for cooling the ice from 0°C to -10°C is calculated using the specific heat capacity of ice:
\[ \Delta H_2 = -10z \]

Thus, the total enthalpy change can be expressed as: \[ \Delta H = -x - 10y - 10z \]

The term \( x \) is given in kJ/mol and should be multiplied by 100 to convert it to J/mol to match the units of \( y \) and \( z \) (which are in J/mol·K). Thus the correct expression for the total enthalpy change is: \[ \Delta H = -10(100x + y + z) \] Quick Tip: Always ensure that the units are consistent when combining enthalpy changes due to different processes. Convert values as necessary.

Step 1: Understanding pH and dilution

The pH of a solution is related to the concentration of hydrogen ions (\([H^+]\)) in the solution by the equation: \[ pH = -\log [H^+] \]
For an aqueous HCl solution with pH 1.0, the concentration of hydrogen ions \([H^+]\) is: \[ pH = 1.0 \quad \Rightarrow \quad [H^+] = 10^{-1} = 0.1 \, M \]

Step 2: Diluting the solution

When an equal volume of water is added to the solution, the concentration of hydrogen ions is halved (since the volume doubles). Therefore, the new concentration of \([H^+]\) will be: \[ [H^+]_{new} = \frac{0.1}{2} = 0.05 \, M \]

Step 3: Calculating the new pH

The pH of the diluted solution is given by: \[ pH_{new} = -\log (0.05) \]
Using the logarithm property \(\log 0.05 = \log (5 \times 10^{-2}) = \log 5 + \log 10^{-2}\), we get: \[ \log 0.05 = \log 5 - 2 = 0.69897 - 2 = -1.30103 \]
Thus: \[ pH_{new} = -(-1.30103) = 1.30103 \approx 1.3 \]

Therefore, the pH increases to 1.3 after dilution. Thus, the correct answer is option (2). Quick Tip: When diluting an acidic solution, the concentration of hydrogen ions is halved, leading to an increase in pH. Always use the logarithmic formula to calculate the change in pH.

Step 1: Analyze Statement I

Dimethyl ether (\(CH_3OCH_3\)) is moderately soluble in water, but not completely. It does exhibit hydrogen bonding but due to the small size of the molecule, its solubility in water is limited.

Diethyl ether (\(C_2H_5OC_2H_5\)) is only slightly soluble in water due to its larger hydrophobic ethyl groups. This statement is partially correct as diethyl ether is indeed soluble in water, but to a very small extent.

Thus, Statement I is true because it correctly describes the solubility of both ethers, although dimethyl ether’s solubility is greater than diethyl ether.


Step 2: Analyze Statement II

Sodium metal is used to dry diethyl ether because it reacts with any water present, forming sodium hydroxide and hydrogen gas.

Ethyl alcohol (ethanol), however, reacts with sodium metal, forming sodium ethoxide and hydrogen, and thus cannot be dried by sodium.

Thus, Statement II is true because sodium can dry diethyl ether but cannot be used for drying ethyl alcohol. Quick Tip: When using sodium metal to dry ethers, ensure that the solution is free of water. Sodium reacts with water to form sodium hydroxide and hydrogen gas.

In the photoelectric effect, electrons are ejected from a material when light of a frequency greater than the threshold frequency strikes it.

Option A: When yellow light (which has a frequency greater than the threshold frequency) is focused on caesium, electrons are ejected, and current flows. Therefore, this statement is correct.

Option B: Dimming the brightness of the yellow light reduces the number of photons, which in turn decreases the number of electrons ejected. Hence, the current in the ammeter is reduced. This statement is correct.
Option C: Red light has a frequency lower than the threshold frequency of caesium, so it does not have enough energy to eject electrons. Hence, no current will be produced. This statement is incorrect.

Option D: Blue light has a frequency greater than the threshold frequency, so it will eject electrons and form current in the ammeter. This statement is correct.

Option E: White light, which contains frequencies higher than the threshold frequency (including blue light), will also cause the ejection of electrons and formation of current. This statement is correct.

Thus, the correct answer is B, C, and D Only. Quick Tip: In the photoelectric effect, only light with a frequency greater than the threshold frequency can cause the ejection of electrons. Reducing light intensity reduces the current, but light with insufficient frequency (like red light) will not cause any emission of electrons.

Step 1: Identify the longest carbon chain.

The longest continuous chain that includes the double bond consists of 4 carbon atoms, so the base name is "butene" (indicating a 4-carbon alkene).

Step 2: Number the chain.

Number the chain such that the double bond gets the lowest possible number. Since the double bond is between carbon 2 and carbon 3, we assign numbers to the carbon chain as follows: \[ H_3C - CH_3 - CH = CH - H - Br \]
The double bond is between carbons 2 and 3, so the compound is "but-2-ene."

Step 3: Identify the substituents.

There is a methyl group (CH₃) attached to carbon 2.

There is a bromo group (Br) attached to carbon 1.


Step 4: Assign positions to the substituents.
The methyl group is on carbon 2, and the bromo group is on carbon 1. The substituents should be numbered to give the lowest possible numbers, so the correct positions are:

Methyl group on carbon 2

Bromo group on carbon 1


Step 5: Construct the IUPAC name.

The name of the compound is 1-Bromo-2-methylbut-2-ene, where:

"But" indicates the 4-carbon chain (butene).

"2-ene" indicates the double bond between carbons 2 and 3.

"2-methyl" refers to the methyl group on carbon 2.

"1-bromo" refers to the bromo group on carbon 1.


Thus, the correct name is 1-Bromo-2-methylbut-2-ene. Quick Tip: When naming organic compounds, always prioritize giving the lowest possible number to the position of the double bond and substituents. If there are multiple substituents, assign the lowest possible sum of numbers.

Step 1: Understanding Dalton’s Law of Partial Pressures

Dalton’s Law states that the partial pressure of a gas is proportional to its mole fraction in the mixture. The partial pressure is given by: \[ P_{gas} = X_{gas} \times P_{total} \]
Where:

\( P_{gas} \) is the partial pressure of the gas,

\( X_{gas} \) is the mole fraction of the gas,

\( P_{total} \) is the total pressure.


Step 2: Calculate the mole fraction of each gas

We are given the mass percentages of the gases and their molar masses:

Mass percentage of nitrogen (\(N_2\)) = 70.0%

Mass percentage of oxygen (\(O_2\)) = 27.0%

Mass percentage of argon (\(Ar\)) = 3.0%


The moles of each gas are calculated by dividing the mass by the molar mass of each gas: \[ Moles of N_2 = \frac{70}{14} = 5 \, moles \] \[ Moles of O_2 = \frac{27}{16} = 1.6875 \, moles \] \[ Moles of Ar = \frac{3}{40} = 0.075 \, moles \]

Now, the total moles of gases: \[ Total moles = 5 + 1.6875 + 0.075 = 6.7625 \, moles \]

Step 3: Calculate the partial pressures
The mole fraction of each gas is given by: \[ X_{N_2} = \frac{5}{6.7625} = 0.7396, \quad X_{O_2} = \frac{1.6875}{6.7625} = 0.2499, \quad X_{Ar} = \frac{0.075}{6.7625} = 0.0111 \]

Now, calculate the partial pressures: \[ P_{N_2} = X_{N_2} \times P_{total} = 0.7396 \times 1.15 = 0.8516 \, atm \] \[ P_{O_2} = X_{O_2} \times P_{total} = 0.2499 \times 1.15 = 0.2879 \, atm \] \[ P_{Ar} = X_{Ar} \times P_{total} = 0.0111 \times 1.15 = 0.0128 \, atm \]

Step 4: Calculate the ratios
(i) The ratio of the partial pressures of nitrogen to oxygen is: \[ \frac{P_{N_2}}{P_{O_2}} = \frac{0.8516}{0.2879} = 2.96 \]

(ii) The ratio of the partial pressures of oxygen to argon is: \[ \frac{P_{O_2}}{P_{Ar}} = \frac{0.2879}{0.0128} = 11.2 \]

Thus, the correct answer is (4) 2.96, 11.2. Quick Tip: Dalton’s Law of partial pressures allows you to find the partial pressures of gases from their mole fractions. Ensure to use mass percentages and molar masses to first calculate the moles of each gas.

Step 1: Analyzing Statement I

Mohr's salt is composed of three ions: ferrous \( Fe^{2+} \), ammonium \( NH_4^+ \), and sulphate \( SO_4^{2-} \). This statement is true because Mohr's salt (also known as ammonium ferrous sulphate) contains only these three types of ions. Thus, Statement I is correct.

Step 2: Analyzing Statement II

The molar conductance of a salt at infinite dilution is the sum of the molar conductances of the ions it dissociates into. For Mohr's salt, the ions involved are:

Ferrous ion \( Fe^{2+} \), with a molar conductance of \( x_1 \),

Ammonium ion \( NH_4^+ \), with a molar conductance of \( x_2 \),

Sulphate ion \( SO_4^{2-} \), with a molar conductance of \( x_3 \).

At infinite dilution, the total molar conductance \( \lambda_{\infty} \) of Mohr's salt should be the sum of the conductances of these ions. However, Statement II suggests the wrong coefficient for the sulphate ion. The sulphate ion \( SO_4^{2-} \) is a monoatomic ion and should contribute \( x_3 \) to the total conductance, not \( 2x_3 \). Thus, the correct expression for the molar conductance should be: \[ \lambda_{\infty} = x_1 + x_2 + x_3 \]
Therefore, Statement II is incorrect because it incorrectly doubles the contribution of the sulphate ion. Quick Tip: When calculating molar conductance at infinite dilution for ionic compounds, remember that each ion contributes to the total conductance according to its molar conductance, without any extra multiplication unless the ion is polyatomic and contributes differently.

Step 1: Atomic Configuration and Valence Electrons


Chromium (Cr): Atomic number = 24. Its electron configuration is \([ Ar ] 3d^5 4s^1\), giving it 6 valence electrons.
Cobalt (Co): Atomic number = 27. Its electron configuration is \([ Ar ] 3d^7 4s^2\), giving it 9 valence electrons.
Iron (Fe): Atomic number = 26. Its electron configuration is \([ Ar ] 3d^6 4s^2\), giving it 8 valence electrons.
Nickel (Ni): Atomic number = 28. Its electron configuration is \([ Ar ] 3d^8 4s^2\), giving it 10 valence electrons.


Step 2: Relation to Enthalpy of Atomisation

Enthalpy of atomisation generally decreases as the number of valence electrons increases, because more electrons lead to stronger metallic bonding and thus require more energy to break the bonds. Therefore, the element with fewer valence electrons will have the lowest enthalpy of atomisation.

Since Cr has 6 valence electrons, it will have the lowest enthalpy of atomisation compared to Co, Fe, and Ni.

Thus, the correct answer is 6 valence electrons. Quick Tip: In general, for transition metals, a lower number of valence electrons leads to weaker metallic bonding, hence a lower enthalpy of atomisation.

Step 1: Understand the reaction with sodium hydroxide
When a salt, such as ammonium chloride (\( NH_4Cl \)), reacts with sodium hydroxide (\( NaOH \)), ammonia gas (\( NH_3 \)) is evolved: \[ NH_4Cl + NaOH \rightarrow NH_3 + NaCl + H_2O \]
Thus, gas X is ammonia (\( NH_3 \)).

Step 2: Analyze the reagent Y
When ammonia gas is passed through potassium tetraiodomercurate(II) (\( K_2HgI_4 \)) in the presence of potassium hydroxide (KOH), a brown precipitate of mercury(I) iodide (\( Hg_2I_2 \)) is formed: \[ NH_3 + K_2HgI_4 + KOH \rightarrow Hg_2I_2 + KCl + H_2O \]
The brown precipitate is mercury(I) iodide (\( Hg_2I_2 \)).

Thus, reagent Y is \( K_2HgI_4 + KOH \). Quick Tip: Ammonia (\( NH_3 \)) reacts with potassium tetraiodomercurate(II) and potassium hydroxide to form a brown precipitate of mercury(I) iodide (\( Hg_2I_2 \)).

Step 1: Understanding Ionization Enthalpy

Ionization enthalpy refers to the energy required to remove an electron from a neutral atom in its gaseous state. Lower ionization enthalpy indicates that the element can lose electrons more easily.

The elements A and B have the lowest ionization enthalpy values in their group. This suggests that they are relatively easier to ionize compared to other members of their group.



Step 2: Analyzing the Nature of Their Ions
Element A: Since A has a low ionization enthalpy, it will lose electrons easily and thus, the A\(^{2+}\) ion will readily accept electrons (reduced). Hence, element A behaves as a reducing agent.


Element B: On the other hand, element B has a slightly higher ionization enthalpy compared to A. The B\(^{4+}\) ion tends to attract electrons, which makes it a strong oxidizing agent.


Thus, the nature of their ions is reducing (for A\(^{2+}\)) and oxidizing (for B\(^{4+}\)). Quick Tip: In general, elements with low ionization enthalpies are reducing agents because they can easily lose electrons, while elements with high ionization enthalpies tend to attract electrons and thus act as oxidizing agents.

Step 1: Identifying the Metal 'M'

The first transition series metal with the highest enthalpy of atomisation is iron (Fe). Iron has a relatively high enthalpy of atomisation compared to other metals in the first transition series.

Step 2: Analyzing the Aquated Ion

The green colour of the aquated ion \( M^{n+} \) indicates that it is likely to be a transition metal ion, as many transition metal ions exhibit characteristic colours due to d-d transitions. For iron, the \( Fe^{2+} \) ion is known to be green in colour when aquated in water. Therefore, the metal \( M \) is iron.

Step 3: Nature of the Oxide Formed

Iron forms oxides in both the \( +2 \) and \( +3 \) oxidation states. The oxide formed in the \( +2 \) oxidation state is iron(II) oxide (FeO), which is basic in nature. Basic oxides tend to react with acids to form salts and water.

Therefore, the oxide formed by the \( Fe^{2+} \) ion (a metal in the \( +2 \) oxidation state) is basic. Quick Tip: For transition metals in the lower oxidation states (such as \( +2 \)), the oxide formed is usually basic. These oxides react with acids to form salts.

In this question, we are determining which compound is least likely to produce effervescence of CO\(_2\) when reacted with aqueous NaHCO\(_3\). Effervescence occurs when an acid reacts with NaHCO\(_3\), producing CO\(_2\).


Compound (1) contains a hydroxyl group (-OH) and nitro groups (-NO\(_2\)) which will likely result in an acidic environment and cause effervescence with NaHCO\(_3\).

Compound (2) contains a carboxyl group (-COOH), a strong acid, which will react with NaHCO\(_3\) and release CO\(_2\).

Compound (3) contains an amine group (-NH\(_3\)), which is basic and does not typically react with NaHCO\(_3\) to produce CO\(_2\).

Compound (4) contains a nitro group (-NO\(_2\)) but lacks a strongly acidic functional group that would promote CO\(_2\) production. Thus, it is the least likely to produce CO\(_2\) effervescence in the presence of NaHCO\(_3\).


Therefore, the compound least likely to give effervescence of CO\(_2\) is compound (4). Quick Tip: In reactions with NaHCO\(_3\), look for acidic functional groups (like -COOH or -OH) that readily react to release CO\(_2\). Nitrogenous compounds without strong acidic groups generally do not react.

Let's analyze the molecular structures:

ICl\(^-\): This ion has a central iodine atom with 2 bonding pairs and 3 lone pairs, resulting in a 2 : 3 ratio (option III).

H\(_2\)O: The oxygen atom has 2 lone pairs and 2 bonding pairs, which gives the ratio 2 : 2 (option IV).

SO\(_2\): The sulfur atom has 2 bonding pairs and 1 lone pair, leading to a 4 : 1 ratio (option II).

XeF\(_4\): The xenon atom has 4 bonding pairs and 2 lone pairs, giving the ratio 4 : 2 (option I).


Therefore, the correct matching is:
A-III, B-IV, C-II, D-I. Quick Tip: In molecular structures, use the VSEPR theory to determine the number of bonding and lone pairs on the central atom, which helps in predicting the molecular geometry and bond pair to lone pair ratios.

Step 1: Understanding Bacterial Growth Before Treatment

The initial growth of the bacteria follows 1st order kinetics. This means that the rate of bacterial growth is proportional to the current number of bacteria present.

In first-order kinetics, the graph of the number of bacteria \( N \) over time \( t \) (shown as \( \frac{N}{N_0} \)) would exhibit an exponential increase. This is characterized by a rapidly rising curve as bacteria reproduce over time.

Since the rate of decay \( r \) is proportional to the square of the number of bacteria at any instant, \( r \propto N^2 \). Hence, the rate \( r \) increases as \( N \) increases.

Step 2: Introducing the Medicine

Once the medicine is introduced, its effect is to slow down the bacterial growth by increasing the decay rate. The rate of bacterial decay is now proportional to the square of the number of bacteria, i.e., \( r \propto N^2 \).

As the medicine works, the number of bacteria no longer follows the exponential increase. Instead, it starts to flatten, and the growth rate slows down because of the increased bacterial decay rate.

Thus, the growth of bacteria is hindered, and the graph showing bacterial growth should start to flatten out after a certain time.

Step 3: Interpreting the Graphs

Before treatment: The graph of \( \frac{N}{N_0} \) versus \( t \) should show an upward exponential curve (exponential growth), as described earlier for first-order kinetics.

After treatment: Once the medicine is applied, the bacterial growth curve flattens. The graph of rate of decay \( r \) versus the number of bacteria \( N \) should show a sharp decline, as the bacteria are no longer able to grow rapidly.

Step 4: Evaluating the Graphs

Option (1): The graphs do not correctly show the flattening of bacterial growth after the application of the medicine.

Option (2): This option correctly represents the situation, where:

The "before" graph shows an exponential increase of \( \frac{N}{N_0} \) over time.

The "after" graph shows a declining rate of decay \( r \), reflecting the effect of the antibacterial treatment.

Option (3): The "after" graph incorrectly shows an increasing rate of decay, which is not correct after the medicine is applied.

Option (4): This graph shows a linear relationship for the "after" scenario, which is not consistent with bacterial decay behavior under first-order kinetics.


Therefore, Option (2) is the correct set of graphs that represents the situation before and after the application of the medicine. Quick Tip: In first-order kinetics, the growth curve exhibits exponential increase. When an antibacterial treatment is applied, the growth rate slows down, resulting in a flattening of the growth curve.

Step 1: Understanding Statement I

Statement I is incorrect. The reaction described in Statement I is wrong because it says that D-(+)-glucose and D-(+)-fructose combine to form sucrose, which is incorrect. The correct reaction should be the formation of sucrose from glucose and fructose, not the other way around.

Therefore, Statement I is false because it incorrectly describes the formation of sucrose.


Step 2: Understanding Statement II

Statement II is correct. Invert sugar is indeed formed during the hydrolysis of sucrose. When sucrose undergoes hydrolysis (by the action of water), it breaks down into glucose and fructose. This mixture of glucose and fructose is called invert sugar.

Therefore, Statement II is true because invert sugar is indeed formed during sucrose hydrolysis.


Step 3: Conclusion

Statement I is false because it incorrectly describes the synthesis of sucrose.

Statement II is true because it correctly describes the formation of invert sugar during sucrose hydrolysis. Quick Tip: In biochemical processes, the hydrolysis of sucrose results in the formation of invert sugar (a mixture of glucose and fructose), but the synthesis of sucrose is the reverse reaction of this process.

Step 1: Understanding the molecular composition:

The complex is given as Co.5NH\(_3\).Cl.SO\(_4\). It consists of a central Co\(^3+\) ion surrounded by 5 ammonia molecules (NH\(_3\)) and one chloride (Cl) ion as well as a sulfate (SO\(_4^{2-}\)) ion.


Step 2: Identifying the isomers:

Isomer A: The solution of A gives a white precipitate with AgNO\(_3\) solution. This indicates the presence of chloride ions (Cl\(^-\)) in the solution because AgNO\(_3\) reacts with chloride ions to form a white precipitate of AgCl.

Isomer B: The solution of B gives a white precipitate with BaCl\(_2\) solution. This indicates the presence of sulfate ions (SO\(_4^{2-}\)) in the solution because BaCl\(_2\) reacts with sulfate ions to form a white precipitate of BaSO\(_4\).


Step 3: Analyzing the isomerism:

The two isomers A and B are ionisation isomers.

In ionisation isomerism, two or more compounds have the same molecular formula but release different ions in solution.

In this case, the only difference between the two isomers is that in A, chloride (Cl\(^-\)) is the counter-ion, while in B, sulfate (SO\(_4^{2-}\)) is the counter-ion. These two isomers give different ions in solution, which is the defining characteristic of ionisation isomerism.


Step 4: Conclusion:
Since the two isomers differ in the ions that are released into the solution (chloride in A and sulfate in B), this is a classic example of ionisation isomerism. Quick Tip: In ionisation isomerism, even though the molecular formula is the same, the arrangement of ligands around the metal ion and the counter-ions can differ, leading to different ions being released in solution.

Step 1: Reviewing Reaction A:

Reaction A involves the NaOEt base, which induces an E2 elimination reaction. This reaction forms an alkene by eliminating the bromine from the cyclohexane ring. Thus, Reaction A is valid for elimination.

Step 2: Reviewing Reaction B:

Reaction B involves the aqueous KOH, which can induce E2 elimination on the alkyl halide to form an alkene. However, Reaction B is invalid because KOH (aqueous) is more likely to induce nucleophilic substitution (SN2) rather than elimination under these conditions.

Reaction B is not valid for elimination because aqueous KOH typically favors nucleophilic substitution over elimination.


Step 3: Reviewing Reaction C:

Reaction C involves sodium methoxide (\( NaOMe \)), which can induce an E2 elimination reaction to form an alkene by abstracting a proton from the \(\beta\)-carbon. This reaction is valid for elimination.

Reaction C is valid for elimination.


Step 4: Reviewing Reaction D:

Reaction D involves the oxidation of phenol (\( C_6H_5OH \)) with \( Na_2Cr_2O_7 \) and \( H_2SO_4 \), which leads to the formation of a quinone, not an alkene. This is an oxidation reaction, not an elimination reaction.

Reaction D is not valid for elimination.


Step 5: Reviewing Reaction E:
Reaction E involves the use of Cu at 573K, which can dehydrogenate the alcohol to form an alkene via E1 elimination. This reaction works and forms an alkene.
Reaction E is valid for elimination.

Conclusion:

Reactions B and D are the reactions that cannot be applied to prepare an alkene by elimination. Reaction B favors nucleophilic substitution, and Reaction D involves oxidation rather than elimination. Quick Tip: In elimination reactions, strong bases such as NaOEt or NaOMe can induce E2 eliminations, while aqueous KOH typically favors SN2 reactions. Additionally, oxidation reactions like in Reaction D do not lead to alkenes via elimination.

Step 1: Determine the amount of carbon in CO\(_2\):

When the compound undergoes complete combustion, carbon from the compound reacts with oxygen to produce CO\(_2\). The molar mass of CO\(_2\) is: \[ Molar mass of CO_2 = 12 + 2 \times 16 = 44 g/mol \]
The molar mass of carbon (C) is 12 g/mol. In 44 g of CO\(_2\), 12 g is carbon. Therefore, the mass of carbon in 220 mg of CO\(_2\) can be calculated as: \[ Mass of C = \left(\frac{12}{44}\right) \times 220 = 60 mg \]


Step 2: Calculate the percentage of carbon in the compound:

The mass of carbon in the original organic compound is 60 mg. The total mass of the compound is 500 mg. The percentage of carbon in the compound is given by: \[ Percentage of C = \left(\frac{60}{500}\right) \times 100 = 12% \] Quick Tip: When calculating the percentage composition, use the ratio of the mass of the element to the total mass of the compound, then multiply by 100.

Step 1: Count atoms from structure

Carbon (C) = 15 atoms
Hydrogen (H) = 11 atoms
Oxygen (O) = 4 atoms
Nitrogen (N) = 1 atom
Iodine (I) = 4 atoms




Step 2: Compute mass contribution
\[ \begin{aligned} Mass of C &= 15 \times 12 = 180 \, g/mol
Mass of H &= 11 \times 1 = 11 \, g/mol
Mass of O &= 4 \times 16 = 64 \, g/mol
Mass of N &= 1 \times 14 = 14 \, g/mol
Mass of I &= 4 \times 127 = 508 \, g/mol \end{aligned} \]



Step 3: Total molar mass of thyroxine
\[ Molar mass = 180 + 11 + 64 + 14 + 508 = 777 \, g/mol \]



Step 4: Percentage of Iodine
\[ Percentage of Iodine = \frac{508}{777} \times 100 \approx 65.37% \]



Correct Answer: \fbox{65% Quick Tip: To calculate the percentage composition, divide the mass of the element by the molar mass of the compound and multiply by 100.

Step 1: Standard Electrode Potentials

The given standard electrode potentials are:
\[ E^\circ_{Cu^{2+}/Cu} = 0.34 \, V \]
\[ E^\circ_{Ag^+/Ag} = 0.80 \, V \]

Step 2: Applying the Nernst Equation

The Nernst equation gives the relationship between the emf of the electrochemical cell, the standard electrode potentials, and the concentrations of the ions involved: \[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \]
Where:

\(E^\circ_{cell}\) is the standard cell potential,

\(n\) is the number of electrons involved,

\(Q\) is the reaction quotient.


Step 3: Determining the Standard Cell Potential

The standard cell potential is the difference between the two half-cell potentials: \[ E^\circ_{cell} = E^\circ_{Ag^+/Ag} - E^\circ_{Cu^{2+}/Cu} = 0.80 \, V - 0.34 \, V = 0.46 \, V \]

Step 4: Reaction Quotient (Q)

The reaction quotient \(Q\) is given by the ratio of the concentrations of the products over the reactants. The concentrations of Cu\(^{2+}\) and Ag\(^+\) ions change due to the passage of electricity:

After 1 Faraday is passed through the Cu\(^{2+}\) cell, the concentration of Cu\(^{2+}\) is reduced by a factor based on the number of moles reduced. Since 1 Faraday corresponds to 1 mole of electrons, the concentration of Cu\(^{2+}\) decreases.

Similarly, after 0.1 Faraday is passed through the Ag\(^+\) cell, the concentration of Ag\(^+\) decreases.


Now, using the changes in concentration:

Initially, the concentration of Cu\(^{2+}\) is 1.5 M, and after 1 Faraday, the concentration of Cu\(^{2+}\) decreases as per the stoichiometry.

Initially, the concentration of Ag\(^+\) is 0.2 M, and after 0.1 Faraday, the concentration of Ag\(^+\) decreases accordingly.
\[ Q = \frac{[Cu^{2+}]}{[Ag^+]} \]

Step 5: Substituting Values into the Nernst Equation

Substitute the concentrations and other known values into the Nernst equation: \[ E_{cell} = 0.46 \, V - \frac{0.0591}{1} \log \left( \frac{1.5 \, M}{0.2 \, M} \right) \]
Now, calculate the logarithmic term: \[ \log \left( \frac{1.5}{0.2} \right) = \log (7.5) \approx 0.875 \]
\[ E_{cell} = 0.46 \, V - 0.0591 \times 0.875 \] \[ E_{cell} = 0.46 \, V - 0.0518 \, V \] \[ E_{cell} = 0.4082 \, V \approx 0.40 \, V \] Quick Tip: The Nernst equation helps to adjust the standard electrode potential by accounting for the concentration of the ions involved in the reaction. For each 10-fold change in concentration, the potential changes by approximately 0.0591 V at 298 K.

Step 1: Understanding the Dissociation of Salt MX\(_3\)

The salt MX\(_3\) dissociates in water as follows:

\[ MX_3 \rightarrow M^{3+} + 3X^- \]
For each mole of MX\(_3\), it dissociates to give one mole of M\(^{3+}\) and three moles of X\(^-\).

Step 2: van't Hoff Factor (i)

The van't Hoff factor \(i\) is given as \(i = 2\). The van't Hoff factor represents the total number of particles in solution per formula unit of solute.

In this case, for MX\(_3\), the dissociation would produce 4 particles (1 M\(^{3+}\) and 3 X\(^-\)) per formula unit of MX\(_3\). However, since \(i = 2\), this suggests that the dissociation is not complete, and the actual number of particles formed is only double the initial number of formula units.


Step 3: Using the Formula for Percentage Dissociation

The formula for the percentage dissociation (\(\alpha\)) is given by:

\[ i = 1 + \alpha (n - 1) \]
Where:

\(i\) is the van't Hoff factor (2 in this case),

\(\alpha\) is the degree of dissociation,

\(n\) is the number of ions produced per formula unit of solute (which is 4 for MX\(_3\)).

\[ 2 = 1 + \alpha (4 - 1) \] \[ 2 = 1 + 3\alpha \] \[ 3\alpha = 1 \] \[ \alpha = \frac{1}{3} \]

Step 4: Calculating the Percentage Dissociation

The percentage dissociation is given by:

\[ Percentage dissociation = \alpha \times 100 = \frac{1}{3} \times 100 = 33.33% \]
To the nearest integer, the percentage dissociation is 33%. Quick Tip: The van't Hoff factor \(i\) is a key parameter in determining the degree of dissociation. When \(i\) is less than the theoretical number of particles formed from dissociation, the salt has not fully dissociated.

To determine the number of paramagnetic complexes, we need to check the number of unpaired electrons in each complex. Paramagnetic complexes have at least one unpaired electron, while diamagnetic complexes have all paired electrons.


1. \([FeF_6]^{3-}\):

Iron in \(Fe^{3+}\) has the electron configuration \([Ar] 3d^5\).

Fluoride (\(F^{-}\)) is a weak field ligand and does not cause electron pairing.

\(Fe^{3+}\) undergoes \(d^2sp^3\) hybridization, leaving 5 unpaired electrons in the \(d\)-orbitals.

Paramagnetic.


2. \([Fe(CN)_6]^{3-}\):

Iron in \(Fe^{3+}\) has the electron configuration \([Ar] 3d^5\).

Cyanide (\(CN^{-}\)) is a strong field ligand that causes electron pairing.

\(Fe^{3+}\) undergoes \(d^2sp^3\) hybridization, resulting in no unpaired electrons.

Diamagnetic.


3. \([Mn(CN)_6]^{3-}\):

Manganese in \(Mn^{3+}\) has the electron configuration \([Ar] 3d^4\).

Cyanide (\(CN^{-}\)) is a strong field ligand and causes pairing of electrons.

\(Mn^{3+}\) undergoes \(d^2sp^3\) hybridization, resulting in 2 unpaired electrons.

Paramagnetic.


4. \([Co(C_2O_4)_3]^{3-}\):

Cobalt in \(Co^{3+}\) has the electron configuration \([Ar] 3d^6\).

Oxalate (\(C_2O_4^{2-}\)) is a weak field ligand.

\(Co^{3+}\) undergoes \(d^2sp^3\) hybridization, resulting in 3 unpaired electrons.

Paramagnetic.


5. \([MnCl_6]^{3-}\):

Manganese in \(Mn^{3+}\) has the electron configuration \([Ar] 3d^4\).

Chloride (\(Cl^{-}\)) is a weak field ligand and does not cause electron pairing.

\(Mn^{3+}\) undergoes \(d^2sp^3\) hybridization, leaving 4 unpaired electrons.

Paramagnetic.


6. \([CoF_6]^{3-}\):

Cobalt in \(Co^{3+}\) has the electron configuration \([Ar] 3d^6\).

Fluoride (\(F^{-}\)) is a weak field ligand and does not cause electron pairing.

\(Co^{3+}\) undergoes \(d^2sp^3\) hybridization, leaving 2 unpaired electrons.

Paramagnetic.



Step 2: Conclusion.

The number of paramagnetic complexes is 2. These are:

\([FeF_6]^{3-}\)

\([Mn(CN)_6]^{3-}\)


Thus, the number of paramagnetic complexes is 2. Quick Tip: To determine paramagnetism, look for unpaired electrons in the complex. Strong field ligands usually pair up electrons, while weak field ligands leave electrons unpaired.