Chemistry Test 233

CONCEPT:

Properties of Lanthanides

• The lanthanides are a series of 15 metallic elements from lanthanum (La) to lutetium (Lu) in the periodic table.
• They are known for their similar chemical properties, which make them difficult to separate.
• Lanthanides are typically separated by ion-exchange methods due to their similar ionic radii and chemical behavior.
• The ionic radius of trivalent lanthanides (Ln³⁺ generally decreases with an increase in atomic number, a phenomenon known as the lanthanide contraction.
• Most of the lanthanides have high density, making them dense metals.
• The most common oxidation state of lanthanides is +3, although other oxidation states can exist.

Explanation:-

• Consider each statement provided:
  • Statement 1: “Lanthanides are separated from one another by ion-exchange method.”
    • This statement is true because ion-exchange chromatography is commonly used to separate lanthanides.
  • Statement 2: “Ionic radii of trivalent lanthanides steadily increases with increase in atomic number.”
    • This statement is false. The ionic radius of trivalent lanthanides (Ln³⁺) decreases as the atomic number increases due to lanthanide contraction.
  • Statement 3: “All lanthanides are highly dense metals.”
    • This statement is true as lanthanides are generally dense metals.
  • Statement 4: “Most common oxidation state of lanthanides is +3.”
    • This statement is true as +3 is the most common and stable oxidation state for lanthanides.

CONCLUSION:

The correct answer is 2) Ionic radii of trivalent lanthanides steadily increases with increase in atomic number

CONCEPT:

Emission Spectra and Transition of Electrons in He⁺ Ion

• When electrons transition from a higher energy state to a lower energy state, they emit photons.
• The wavelength of the emitted photon is inversely proportional to the energy difference between the two states.
• For a He⁺ ion (which is hydrogen-like with a single electron), the energy levels are given by:
  • E_n = −13.6 × Z²/n²
    eV, where Z = 2 (for helium) and n is the principal quantum number.

Explanation

• Given: Six different photons observed, indicating transitions between multiple levels.
• This suggests the electron transitions between four distinct energy levels ( n = 4, 3, 2, 1 ).
• Transitions possible between these levels:

• The transition that results in the minimum wavelength (maximum energy difference) is from n = 4 to n = 1 .
• Energy difference ( \Delta E ) between these levels is:

CONCEPT:

Identifying a Compound Based on its Reactions

• Reaction with Sodium: Compounds containing active hydrogen atoms (such as alcohols) react with sodium to produce hydrogen gas.
• Reaction with Phosphorus Pentachloride (PCl₅): Compounds with hydroxyl groups (such as alcohols) react with PCl₅ to produce hydrogen chloride (HCl).
• Reduction of Alkaline Copper(II) Salt: Aldehydes reduce alkaline copper(II) salts (e.g., Fehling's solution) to form a red precipitate of copper(I) oxide (Cu₂O).

Explanation:-

• The molecular formula of compound 'X' is C₅H₁₀O.
• 'X' does not react with sodium, indicating it is not an alcohol.
• 'X' does not react with PCl₅, further indicating the absence of a hydroxyl group.
• 'X' reduces alkaline copper(II) salt on gentle warming, suggesting it has a functional group that can be oxidized, typically a carbonyl group (C=O).
• Among carbonyl-containing compounds, aldehydes are known to reduce Fehling's solution.

• The absence of reactions with sodium and PCl₅ indicates that 'X' is not an alcohol (neither primary nor secondary).
• The ability to reduce alkaline copper(II) salt indicates the presence of an aldehyde group, as ketones generally do not reduce Fehling's solution.

CONCLUSION:

The correct answer is 3) an aldehyde

CONCEPT:

Molar Ratio of Two Substances in a Mixture

• Molality calculations determine the quantity of reactants or products in reactions.
• The equivalence between neutralization and oxidation reactions facilitates calculating molar ratios.
• Equivalents represent the molar amount reacting in these processes.

CONCLUSION:

The correct answer is 3:1

CONCEPT:

Lucas Test for Alcohols

• The Lucas test is used to classify alcohols based on their reactivity with Lucas reagent (anhydrous zinc chloride in concentrated hydrochloric acid).

• Primary alcohols react very slowly and do not produce turbidity at room temperature.
• Secondary alcohols produce turbidity within 5 minutes.
• Tertiary alcohols produce immediate turbidity.

EXPLANATION:

• Compound 1: Cyclopropylmethanol (cyclopropane ring with CH₂OH group) – A primary alcohol, does not produce immediate turbidity.
• Compound 2: Ethanol (CH₃CH₂OH) – A primary alcohol, does not produce immediate turbidity.
• Compound 3: 2-Butanol (CH₃CH₂CH(OH)CH₃) – A secondary alcohol, produces turbidity within 5 minutes but not immediately.
• Compound 4: Cyclobutanol (cyclobutane ring with OH group) – A titary alcohol, produces turbidity within a minute.

Conclusion:

So compound 4 will give turbidity immediately.

CONCEPT:

Lassaigne's Test for Nitrogen

• Lassaigne's test is used to detect the presence of nitrogen in an organic compound.
• The test involves the fusion of the compound with sodium metal, which converts any nitrogen present into sodium cyanide (NaCN).
• The resulting solution is treated with ferrous sulfate (FeSO₄) and ferric chloride (FeCl₃), followed by acidification, which forms Prussian blue (ferric ferrocyanide, Fe₄[Fe(CN)₆]₃).

EXPLANATION:

• Compound 1: Phenylhydrazine (C₆H₅NHNH₂) – Contains nitrogen in the hydrazine group, will give a positive test.
• Compound 2: Semicarbazide (H₂N-NH-C(=O)-NH₂) – Contains nitrogen in both hydrazine and amide groups, will give a positive test.
• Compound 3: Hydrazine (NH₂-NH₂) typically does not give a positive test for nitrogen in Lassaigne's test. This is because the structure and chemistry of hydrazine differ significantly from most organic nitrogen compounds for which Lassaigne's test is designed.
• Compound 4: Sulfanilic acid has the chemical formula C6H7NO3S, indicating that it contains nitrogen (N) as one of its elements. During Lassaigne's test, the nitrogen present in sulfanilic acid will form sodium cyanide (NaCN) when fused with sodium metal. 

Conclusion:

Hydrazine (NH₂-NH₂) typically does not give a positive test for nitrogen in Lassaigne's test

CONCEPT:

Paramagnetism and Unpaired Electrons

• Paramagnetism arises from the presence of unpaired electrons in a material.
• The more unpaired electrons an ion or molecule has, the more paramagnetic it is.
• For transition metal complexes, the oxidation state and the nature of the ligands influence the number of unpaired electrons.
• Strong field ligands like CN^- can cause pairing of electrons, reducing paramagnetism, while weak field ligands like H₂O do not.

CONCEPT:

Acidity of Aromatic Carboxylic Acids

• The acidity of aromatic carboxylic acids is influenced by the presence of electron-withdrawing or electron-donating groups attached to the benzene ring.
• Electron-withdrawing groups (such as -NO₂) increase acidity by stabilizing the carboxylate ion through resonance and inductive effects.
• Electron-donating groups (such as -OH or -CH₃) decrease acidity by destabilizing the carboxylate ion.

EXPLANATION:

• Compound 1: Benzoic acid (C₆H₅COOH) – No substituents, baseline acidity.
• Compound 2: 2-Ethylbenzoic acid (C₆H₄CH₂CH₃COOH) – Electron-donating ethyl group decreases acidity.
• Compound 3: 2-Nitrobenzoic acid (C₆H₄NO₂COOH) – Electron-withdrawing nitro group increases acidity.
• Compound 4: 4-Hydroxybenzoic acid (C₆H₄COHCOOH) – Electron-withdrawing aldehyde group increases acidity .

Compound 3 (2-Nitrobenzoic acid) is the most acidic due to the presence of the electron-withdrawing nitro group which stabilizes the carboxylate ion more effectively than CHO group any other substituent present in the given compounds.

Conclusion: 

So, Most acidic compound is compound 3

CONCEPT:

Determination of Structure Through Chemical Reactions

• An optically active compound has at least one chiral center, meaning a carbon atom bonded to four different groups.
• Ethanolic KOH induces elimination reactions, forming alkenes from alkyl halides through the E2 mechanism.
• Positional isomers differ in the position of a functional group (e.g., a double bond) but have the same molecular formula.
• A meso compound is achiral and contains multiple chiral centers with an internal plane of symmetry, resulting in a meso-diol when treated with cold, dilute alkaline KMnO₄.

Explanation:-

• Optically Active Alkyl Bromide (C₆H₁₃Br, X):
  • X must have a chiral center. This rules out symmetrically substituted compounds.
• Formation of Two Alkenes (Y and Z):

• • Y and Z are positional isomers with the molecular formula C₆H₁₂. They are formed via an elimination reaction (E2 mechanism).
• Formation of Meso-Diol:
  • Z must be capable of forming a meso-diol, meaning Z is a symmetrical alkene leading to symmetrical hydroxylation.

OPTIONS EVALUATION:

• Option 1: 2-Bromohexane
  • Would form 2-hexene and 1-hexene upon elimination, which are not symmetrical, hence not forming a meso-diol.
• Option 2: 1-Bromo-2,3-dimethylbutane
  • Would not produce alkenes that fit the criteria for forming a meso-diol.
• Option 3: 3-Bromohexane
  • Would form 3-hexene and 2-hexene, Y and Z are positional isomers, where 3-hexene can undergo oxidation to form a meso-diol, satisfying the criteria.
• Option 4: 2-Bromo-2,3-dimethylbutane
  • Would form 2,3-dimethylbut-2-ene and 2,3-dimethylbut-1-ene, which do not fit the criteria for forming a meso-diol.

The correct answer is 3-bromohexane

CONCEPT:

Acids of Phosphorus: H₃PO₂, H₃PO₃, and H₃PO₄

• Phosphorous acids vary in their structure and properties depending on the number of hydrogen atoms bonded directly to the phosphorus atom and their oxidation states.
• H₃PO₂ (Hypophosphorous acid): Also written as HOP(O)H₂. It has one P-H bond and two P-OH bonds, making it a monobasic acid.
• H₃PO₃ (Phosphorous acid): Also written as HPO(OH)₂. It has one P-H bond and two P-OH bonds, and is dibasic.
• H₃PO₄ (Phosphoric acid): Also written as PO(OH)₃. It has three P-OH bonds and no P-H bonds, making it tribasic.

Explanation:-

• Statement 1: "Their acidic nature is H₃PO₄ < H₃PO₃ < H₃PO₂"
  • This statement is incorrect. The increasing acidic strength (considering basicity and usual Ka values) is H₃PO₂ < H₃PO₃ < H₃PO₄, given increasing effective proton donations regarding equivalent structures essential accordingly.
• Statement 2: "All of them are reducing in nature"
  • This is incorrect. H₃PO₂ and H₃PO₃ are reducing agents, but H₃PO₄ does not have reducing properties.
• Statement 3: "All of them are tribasic acids"
  • This is incorrect. H₃PO₂ is monobasic, H₃PO₃ is dibasic, and H₃PO₄ is tribasic.
• Statement 4: "The geometry of phosphorus is tetrahedral in all three"
  • This is the true statement. In all three acids, phosphorus is surrounded by four substituents (either P-H or P-O), leading to a tetrahedral geometry around the phosphorus atom.

The correct answer is the geometry of phosphorus is tetrahedral in all the three