Concepts of Acids and Bases MCQ Quiz in मल्याळम - Objective Question with Answer for Concepts of Acids and Bases - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

The acidity of compounds is influenced by the stability of their conjugate bases. Key factors include:

• Ring Size: Smaller rings often have higher acidity (lower pKa) due to increased stabilization of the conjugate base from ring strain, which enhances resonance stabilization. Larger rings have reduced acidity and higher pKa due to decreased ring strain and weaker conjugate base stabilization.

• Acyclic vs. Cyclic Compounds: Acyclic compounds are generally less acidic than cyclic compounds because the lack of ring strain leads to reduced stabilization of the conjugate base. This results in a higher pKa for acyclic compounds.

• Resonance Stabilization: Compounds with extended resonance stabilization in their conjugate base are more acidic (lower pKa). For example, β-diketones have extensive resonance stabilization compared to single ketones.

Explanation: 

• Compound III: This is an acyclic β-diketone. While resonance stabilization exists between the two carbonyl groups, the absence of a acyclic structure leads to less stabilization of the conjugate base compared to cyclic compounds. Hence, it has the highest pKa (least acidic).

• Compound I: This is a six-membered cyclic ketone. Its conjugate base is moderately stabilized due to the resonance and cyclic nature of the compound. This makes it more acidic (lower pKa) than compound III, but less acidic than compound II.

• Compound II: This is a five-membered cyclic ketone with higher ring strain than compound I. The increased strain enhances conjugate base stabilization, resulting in the lowest pKa (highest acidity) among the three compounds.

Conclusion:

The correct order of pKa values is: III > I > II. This order reflects that acyclic compounds are the least acidic (highest pKa), and smaller cyclic ketones are the most aci

CONCEPT:

Acidity and Lewis Acids

• Acidity refers to the ability of a substance to donate protons (H⁺) or accept electron pairs.
• Lewis acids are substances that can accept a pair of electrons to form a covalent bond.

EXPLANATION:

• In the given cases:
  1. CuSO₄ is added to a saturated (NH₄)₂SO₄ solution
  2. SbF₅ is added to anhydrous HF
• For the first case:
  • CuSO₄ is a salt of a weak base (Cu(OH)₂) and a strong acid (H₂SO₄).
  • When added to a saturated (NH₄)₂SO₄ solution, it will increase the concentration of H⁺ ions, thereby increasing the acidity.
• For the second case:
  • SbF₅ is a strong Lewis acid, and HF is a weak acid.
  • When SbF₅ is added to anhydrous HF, it will accept F^- ions from HF, increasing the concentration of H⁺ ions and thus increasing the acidity.

Therefore, the addition of CuSO₄ in a saturated (NH₄)₂SO₄ solution and SbF₅ in anhydrous HF both increase the acidity.

Concept:

The formation constant (log K₁) represents the stability of the complex formed between metal ions (Aⁿ⁺, B^m+, C^p+) and ligands such as OH^-, Cl^-, and Br^-. The stability of these complexes depends on the hardness or softness of the metal ions and ligands, as per the Hard and Soft Acids and Bases (HSAB) theory.

• Hard acids and bases: Hard acids prefer to form strong complexes with hard bases. OH^- is a hard base, and hard acids like Aⁿ⁺ will form stronger complexes with OH^-.

• Borderline acids and bases: Borderline acids such as B^m+ prefer borderline bases like Br^-, forming stronger complexes with Br^- than with OH^-.

• Soft acids and bases: Soft acids like C^p+ prefer soft bases like Br^-, and their interaction with hard bases like OH^- is relatively weak.

Explanation: 

• Statement A is correct because Aⁿ⁺ is a hard acid, which forms stronger complexes with OH^-, while B^m+, being a borderline acid, forms stronger complexes with Br^-.

• Statement B is incorrect because Aⁿ⁺ is not a soft acid, and it prefers OH^- over Br^- as it is a hard acid. C^p+ is a soft acid but does not show a preference for Br^- over OH^- in this con.

• Statement C is correct because formation constant of B^m+ for Cl^- is more than Aⁿ⁺ hinting that B^m+ is softer acid than Aⁿ⁺ and soft-soft interaction are more favourable than soft-hard interactions. Also, formation constant of C^p+ is least for OH^- thus C^p+ has the weakest interaction with OH^-

Conclusion:

The correct statements are A and C.

Concept:

Liquid ammonia (NH₃) is a useful solvent, particularly in the field of inorganic chemistry. It has unique properties that make it different from water and other common solvents. Here are some important properties of liquid ammonia:

• Low Oxidizing Power: Liquid ammonia significantly reduces the oxidizing power of various potential oxidizing agents, making it an excellent medium for reactions where strong oxidizing environments are undesirable.

• Solvating Ability: It dissolves alkali metals, forming blue-colored solutions due to the solvated electrons. These solutions exhibit unique electrical and magnetic properties depending on the concentration of the dissolved metal.

• Conductivity: The conductivity of metal-ammonia solutions can vary with the concentration of the metal. Initially, the conductivity decreases as the concentration increases, but beyond a certain point, it increases again due to the metallic properties of the solution.

• Paramagnetism: Liquid ammonia solutions of alkali metals display paramagnetic behavior, and this property also varies with the concentration of the dissolved metal.

Explanation: 

• Statement A: The oxidizing power of different potential oxidizing agents is reduced in liquid ammonia because liquid ammonia stabilizes solvated electrons. This makes the solution less capable of undergoing oxidation-reduction reactions, hence reducing the oxidizing power.

• Statement B: The color of the metal-ammonia solution does not depend on the nature of the metal but on the presence of solvated electrons. These solvated electrons are responsible for the deep blue color of the solution, which remains consistent regardless of the metal dissolved.

• Statement C: As the concentration of the alkali metal increases, the interaction between solvated electrons increases, causing a change in color from blue to bronze. This transition reflects the onset of metallic bonding, giving the solution some metallic properties.

• Statement D: Initially, the conductivity and paramagnetism decrease as the concentration of the metal increases because the number of free solvated electrons decreases. However, once a certain concentration threshold is crossed, metallic bonding begins to dominate, and the conductivity and paramagnetism increase again, passing through a minimum.

Conclusion:

All the statements are correct, making option 4 the correct answer.

Concept:

Factors Affecting Acidity of Carboxylic Acids

• The acidity of carboxylic acids is influenced by the electron-withdrawing or electron-donating effects of substituents.
• Inductive Effect: Electron-withdrawing groups (EWGs) such as halogens increase acidity by stabilizing the conjugate base (carboxylate anion).
• Alkyl Groups: Electron-donating alkyl groups decrease acidity by destabilizing the carboxylate ion through their inductive effect.
• Number and Position of Halogens: The more electronegative and closer the halogen atoms are to the -COOH group, the stronger the acid.

Explanation:

• Compound A (CH3COOH - Formic Acid): No alkyl or halogen groups, so it has moderate acidity.
• Compound B (CH3CH₂CH2COOH - Butanoic Acid): Contains an alkyl group that has an electron-donating effect, reducing acidity.
• Compound C (ClCH2COOH - Dichloroacetic Acid): The chlorine atoms provide a strong electron-withdrawing inductive effect, increasing acidity.
• Compound D (Cl₂CH2COOH - Trichloroacetic Acid): Two chlorine atoms increase the electron-withdrawing effect further, making it the most acidic.

Acidity Order: D > C > A > B.

Therefore, the correct answer is: D > C > A > B.

The correct answer is  \({H}^+ > {Mg}^{2+} > {Fe}^{2+} > {Cu}^+\)

CONCEPT:

Hard and Soft Acids and Bases (HSAB) Theory

• Hard Acids: Tend to have small ionic radii, high positive charges, and are less polarizable.
• Soft Acids: Tend to have larger ionic radii, lower positive charges, and are more polarizable.
• Borderline Acids: Fall between hard and soft acids in terms of their properties.

Explanation:-

• H⁺: Hard acid (small and highly charged).
• Mg²⁺: Hard acid (small and highly charged).
• Fe²⁺: Borderline acid.
• Cu⁺: Soft acid (larger and more polarizable).:
• 1. H⁺: Hardest acid.
• 2. Mg²⁺: Hard acid.
• 3. Fe²⁺: Borderline acid.
• 4. Cu⁺: Softest acid.

CONCLUSION:

• The correct order from hard acid to soft acid is:
• H⁺ > Mg²⁺ > Fe²⁺ > Cu⁺.

Thus, the correct arrangement is \({H}^+ > {Mg}^{2+} > {Fe}^{2+} > {Cu}^+\)

The correct answer is  AHard acids bond in the order \( {I}^- < {Br}^- < {Cl}^- < {F}^-\); Soft acids bond in the order \({F}^- < {Cl}^- < {Br}^- < {I}^- .\) .

Explanation:

 Hard and soft acids and bases are identified empirically by the trends in stabilities of  the complexes that they form: hard acids tend to bind to hard bases and soft acids tend to bind to soft bases

The bonding order of hard and soft acids with halide ion bases is determined by the stability of the complexes they form. Hard acids prefer to bond with smaller, more electronegative halides, thus the bonding order for hard acids is \( {I}^- < {Br}^- < {Cl}^- < {F}^-\) . Conversely, soft acids prefer to bond with larger, less electronegative halides, leading to the bonding order \({F}^- < {Cl}^- < {Br}^- < {I}^- .\) This empirical observation helps classify acids and bases based on their interaction strengths and resulting complex stability.

Explanation:

 The molecular orbital  representation of the orbital interactions  responsible for formation of a complex 
between Lewis acid A and Lewis base :B.

 Lewis acids and bases undergo a variety of characteristic reactions. The simplest Lewis 
acid–base reaction in the gas phase or noncoordinating solvents is complex formation:
 A :B→ A B
 Two examples are

 Both reactions involve Lewis acids and bases that are independently stable in the gas phase  or in solvents that do not form complexes with them. Consequently, the individual species  (as well as the complexes) may be studied experimentally.
The fig shows the interaction of orbitals responsible for bonding in Lewis complexes. The exothermic character of the formation of the complex stems from the fact that the newly formed bonding orbital is populated by the two electrons supplied by the base whereas the newly formed antibonding orbital is left unoccupied. As a result, there is a net lowering of energy when the bond forms.

illustrates the molecular orbital representation for the interaction between a Lewis acid (A) and a Lewis base (B). In this interaction:

The Lewis acid (A) has a Lowest Unoccupied Molecular Orbital (LUMO).
The Lewis base (B) has a Highest Occupied Molecular Orbital (HOMO).
For the formation of the complex (A-B), the HOMO of the Lewis base donates electron density to the LUMO of the Lewis acid. This interaction leads to the formation of a bond between the Lewis acid and the Lewis base, resulting in the complex. Therefore, the correct statement is that the LUMO of the Lewis acid interacts with the HOMO of the Lewis base to form the complex.

The correct answer is  \(({H}_3{Si})_2{O}\) can expand its valence shell by delocalizing lone pairs.

CONCEPT:

Lewis Base Strength and Valence Shell Expansion

• Lewis bases donate electron pairs.
• The availability of lone pairs affects the strength of a Lewis base.
• Elements in Period 3 (like silicon) and beyond can expand their valence shells, reducing the effective lone pairs available for bonding.

EXPLANATION:

• (H₃Si)₂O: Silicon (Si) is in Period 3 and can expand its valence shell by delocalizing lone pairs.
• (H₃C)₂O: Carbon (C) is in Period 2 and cannot expand its valence shell.
• This delocalization in silicon compounds reduces the availability of lone pairs on oxygen for bonding, making (H₃Si)₂O a weaker Lewis base compared to (H₃C)₂O.

• Statement 1: Incorrect, because oxygen's electronegativity does not change significantly between the compounds.
• Statement 2: Correct, because silicon can expand its valence shell, reducing the availability of lone pairs on the oxygen atom.
• Statement 3: Incorrect, because silicon has a larger atomic radius than carbon, not smaller.
• Statement 4: Incorrect, because hydrogen bonding is not the primary factor affecting Lewis base strength in this context.

CORRECT ANSWER:

• Option 2: (H₃Si)₂O can expand its valence shell by delocalizing lone pairs.

The correct answer is BrF3

CONCEPT:

Acidity in Non-Aqueous Solvents and Autoionization Behavior

• In non-aqueous solvents, the concept of acidity and basicity can be extended from the autoionization behavior of the solvent.
• The autoionization of the solvent can produce distinct cations and anions.
• A solute that increases the concentration of the cations in the solvent is considered an acid in that solvent.

Explanation:-

• Solvent System: Identify the solvent that matches the autoionization products with the cation and anion produced by the solute.
• BrF₃: Autoionizes to form BrF₂⁺ and BrF₄⁻.
• Solute: BrF₂AsF₆ produces BrF₂⁺ and AsF₆⁻ upon dissolving.
• The dissolution of BrF₂AsF₆ increases the concentration of BrF₂⁺ in the solvent.
• The solvent whose autoionization products match those of the solute species is BrF₃.
• BrF₃ autoionizes to BrF₂⁺ and BrF₄⁻.
• Since BrF₂AsF₆ increases the concentration of BrF₂⁺ in the solvent BrF₃, it behaves as an acid in BrF₃.

The correct answer is BrF3