Chemistry Test 246

CONCEPT:

Pauli's Exclusion principle: 

• Pauli's Exclusion Principle states that no two electrons in the same atom can have identical values for all four of their quantum numbers. 
• In other words:
  • no more than two electrons can occupy the same orbital
  • two electrons in the same orbital must have opposite spins

EXPLANATION:

• Aufbau Principle: The electrons filled in the orbitals are in acordance with the Aufbau principle , so this rule hasn't been violated here.
• (n + ℓ) Rule: This rule helps determine the order of filling orbitals based on their principal quantum number (n) and azimuthal quantum number (ℓ), but it is not directly violated in this configuration.
• Pauli's Exclusion Principle: States that no two electrons in the same atom can have identical quantum numbers. It is  violated here because each electron in an orbital has same spins.
• Hund's Rule: States that electrons occupy orbitals singly as much as possible before pairing. This rule is not violated as there is no improper pairing in degenerate orbitals.

CONCLUSION:

The correct option is: Option 3 (Pauli's Rule)

CONCEPT:

Solubility of Covalent Compounds

• Covalent compounds are composed of molecules held together by covalent bonds, where atoms share pairs of electrons.
• The solubility of covalent compounds primarily depends on the nature of the solvent and the type of intermolecular forces involved.
• Non-polar solvents dissolve non-polar covalent compounds due to similar types of intermolecular forces (London dispersion forces).

EXPLANATION:

• Statement I: Covalent compounds are usually soluble in non-polar solvents.
• Reason: This is because like dissolves like, meaning that non-polar solvents can dissolve non-polar substances due to similar intermolecular forces.
• Statement II: Covalent compounds dissolve in non-polar solvents due to the similar nature of intermolecular forces, such as London dispersion forces.

The correct answer is: Both Statement I and Statement II are true.

CONCEPT:

Applicability of Bohr's Theory

• Bohr's theory is applicable to species that have only one electron in their system. This includes hydrogen-like atoms and ions where a single electron orbits the nucleus.
• Bohr's theory successfully explains the atomic structure and spectral lines of single-electron systems like H (hydrogen), He⁺ (helium ion with one electron), Li²⁺ (lithium ion with one electron), and similar single-electron species.
• For multi-electron species, Bohr's theory does not work as it cannot account for electron-electron interactions and complex atomic spectra arising from multiple electrons.

EXPLANATION:

• Be³⁺: This ion has a nuclear charge of +4 (from the beryllium nucleus) and a single electron after losing three electrons, making it a hydrogen-like species. Bohr's theory is applicable here.
• Li²⁺: This ion has a nuclear charge of +3 and only one electron after losing two electrons. It is also a hydrogen-like species, so Bohr's theory is applicable.
• He²⁺: This ion has lost both of its electrons, so it has no electrons left. Bohr's theory is not applicable here, as there is no electron orbiting the nucleus.
• H (Hydrogen): Hydrogen has one proton and one electron, making it a single-electron species. Bohr's theory is applicable to hydrogen.

CONCLUSION:

The correct option is: Option 3 (He²⁺)

CONCEPT:

Oxidation Reaction of Iodine with Concentrated Nitric Acid

• When iodine (I₂) reacts with concentrated nitric acid (HNO₃), it undergoes oxidation, where iodine is converted to a higher oxidation state.
• In this reaction, iodine (I) is typically oxidized to form iodic acid (HIO₃), which contains iodine in a higher oxidation state.
• To determine the oxidation state of iodine in the product (Y), we analyze the oxidation state of iodine in HIO₃.

CALCULATION:

• Oxidation State of Iodine in HIO₃:
  • In HIO₃, we have one hydrogen (H) atom, one iodine (I) atom, and three oxygen (O) atoms.
  • Oxidation state of H is +1, and for each O atom, it is -2.
  • Let the oxidation state of iodine in HIO₃ be x .
  • Setting up the equation for oxidation states:
    • (+1) + x + 3(-2) = 0 
    • Simplifying:  1 + x - 6 = 0 
    • x = +5 
  • Thus, the oxidation state of iodine in HIO₃ (Y) is +5.

CONCLUSION:

• The correct option is: Option 1 (5)

Concept:

The freezing point depression (ΔT_f) is a colligative property, which means it depends on the number of solute particles in a solution and not on the nature of the solute. This property is observed when a non-volatile solute is added to a solvent, which causes the freezing point of the solvent to decrease.

The formula for freezing point depression is: ΔT_f = i × K_f × m, where i is the van 't Hoff factor, K_f is the cryoscopic constant, and m is the molality of the solution.

Explanation:

• Assertion (A): Addition of ethylene glycol (non-volatile) to water lowers the freezing point of water.

  • This statement is correct. Adding ethylene glycol to water introduces more solute particles in the solution, resulting in a lower freezing point for the solution.

• Reason (R): Addition of any substance to water lowers the freezing point of water.

  • This statement is incorrect. Only the addition of a non-volatile solute will cause freezing point depression. Some substances might not dissolve or might even raise the freezing point under certain conditions.

Conclusion:

Therefore, the correct statement is: A is correct but R is wrong

CONCEPT:

Intramolecular Aldol Condensation

• Aldol condensation is a reaction in which two carbonyl compounds (aldehydes or ketones) react in the presence of a base to form a β-hydroxy carbonyl compound, which can dehydrate to give an α,β-unsaturated carbonyl compound.
• In an intramolecular aldol condensation, the reaction occurs within a single molecule that has two carbonyl groups. This typically leads to the formation of a cyclic compound.
• The size of the ring formed depends on the number of carbons between the two reacting carbonyl groups. Five- or six-membered rings are generally favored due to ring stability.

EXPLANATION:

• The given compound is a 1,6-diketone (hexane-2,6-dione), which has two carbonyl groups separated by four carbon atoms.
• Under basic conditions (HO^-) and heating (Δ), an intramolecular aldol condensation occurs between these carbonyl groups.
• The reaction proceeds by the formation of an enolate ion from one of the carbonyl groups, which attacks the other carbonyl carbon, leading to cyclization and the formation of a six-membered ring.
• After cyclization, dehydration occurs to yield the final product, a six-membered α,β-unsaturated ketone, which is a cyclohexenone derivative.

CONCLUSION:

The correct product (A) is Option 1, which shows a six-membered cyclic ketone (cyclohexenone).

CONCEPT:

Reaction Order Determination

• The order of a reaction refers to the power to which the concentration of a reactant is raised in the rate law equation.
• To determine the order of reaction, we analyze the concentration of reactants at different times.
• Common orders of reactions are zero-order, first-order, and second-order, each with characteristic rate laws and half-life expressions.

Explanation:-

• Given:
  • The initial amount of L: 10 g
  • Amount of L after 30 minutes: 5 g
  • Amount of L after 60 minutes: 1.25 g
• Calculate the rate constant for different orders of reactions:

• Initial Rate Method: The order of reaction can also be determined experimentally by measuring the initial rates at different initial concentrations.

CONCLUSION

The given reaction, L → M, is determined to be a first-order reaction based on the concentration data at different times. 

CONCEPT:

Haloform Reaction (Formation of Carboxylic Acid)

• The haloform reaction occurs when a methyl ketone (a compound containing a -COCH₃ group) is treated with a base such as KOH in the presence of halogens (such as chlorine).
• In this reaction, the methyl group attached to the carbonyl carbon is fully halogenated, forming a trihalomethyl group (-CCl₃), which is then cleaved to yield a carboxylate ion and a haloform (CHCl₃ in this case).
• When this reaction is followed by acidification (H₃O⁺), the carboxylate ion is protonated to form a carboxylic acid.

CALCULATION:

• Starting compound: The given molecule is a methyl ketone with a trichloromethyl group attached to the carbonyl carbon.
• Step 1: Treatment with aqueous KOH leads to the haloform reaction. The trichloromethyl (CCl₃) group is cleaved, forming chloroform (CHCl₃) and a carboxylate ion.
• Step 2: Acidification with H₃O⁺ converts the carboxylate ion into the carboxylic acid.
• Reaction:

• The product formed is benzoic acid, where the methyl group of the original ketone has been converted to a -COOH group.

CONCLUSION:

The correct structure of product P is Option 1, which is benzoic acid (C₆H₅COOH).

CONCEPT:

Nitration of Aromatic Compounds

• Nitration is an electrophilic aromatic substitution reaction where a nitro group (-NO₂) is introduced onto an aromatic ring.
• The reaction typically requires a nitrating mixture, consisting of concentrated nitric acid (HNO₃) and concentrated sulfuric acid (H₂SO₄), which generates the nitronium ion (NO₂⁺), the active nitrating species.
• In cases with multiple aromatic rings, the nitro group is directed to the more electron-rich or activated ring, as it is more susceptible to electrophilic attack.

EXPLANATION:

• The given compound is a benzophenone ether, with two benzene rings connected by a -CO-O- linkage.
• In the nitration reaction, the nitro group is most likely to substitute on the more activated benzene ring.
• Since there is no strong electron-donating or withdrawing group on the benzene rings, the nitration can occur on either ring. However, for simplicity, the product often depicted is where nitration happens on one of the benzene rings without specific directing effects.
• Reaction:

CONCLUSION:

The major product of this nitration reaction is Option 1, where the nitro group is added to one of the benzene rings, forming a monosubstituted nitro derivative of benzophenone ether.