Chemistry Test 244

Concept:

Non-metallic character refers to an element's ability to gain electrons and form negative ions (anions). It is typically associated with properties such as higher electronegativity, higher ionization energy, and higher electron affinity. In the periodic table, non-metallic character generally increases across a period from left to right and decreases down a group.

Explanation:

• Fluorine (F): The most non-metallic element in the periodic table.

• Nitrogen (N): Highly non-metallic, though less so than fluorine.

• Carbon (C): Non-metallic but to a lesser extent than nitrogen.

• Boron (B): A metalloid with some non-metallic properties.

• Silicon (Si): Also a metalloid, but with more metallic character than boron.

Order of Non-metallic Character

Based on the trends in the periodic table, the correct order of their non-metallic character from highest to lowest is: F > N > C > B > Si

Conclusion:

The correct order of non-metallic character for the elements B, C, N, F, and Si is: F > N > C > B > Si

CONCEPT:

Bond Angle and Hybridization in Ammonia and Phosphine

• In molecules like ammonia (NH₃) and phosphine (PH₃), the central atom has one lone pair and forms three sigma bonds with hydrogen atoms.
• The bond angle in such molecules depends on the hybridization of the central atom, which is influenced by the s- and p -character of the orbitals involved.
• The greater the p -character in the hybrid orbitals, the closer the bond angle will be to 90º, as p -orbitals are orthogonal to each other.
• Ammonia (NH₃) has a bond angle of 107.6º, while phosphine (PH₃) has a bond angle of 93.5º, suggesting that NH₃ has more s-character in its hybrid orbitals compared to PH₃.

EXPLANATION:

• Analyze the bond angles:
  • The H–N–H bond angle in ammonia is 107.6º, which is closer to the tetrahedral angle of 109.5º, indicating more s-character in the hybrid orbitals of nitrogen.
  • The H–P–H bond angle in phosphine is 93.5º, closer to 90º, indicating more p-character in the hybrid orbitals of phosphorus.
• Infer the p-character of the lone pairs:
  • Since ammonia has a larger bond angle than phosphine, the nitrogen in NH₃ has less p-character and hence more s-character compared to phosphorus in PH₃.
  • Therefore, the p -character of the lone pair on ammonia is **less** compared to that in phosphine.

CONCLUSION:

The correct option is: Option 1) Less

CONCEPT:

Ionic Radius vs. Atomic Radius

• The ionic radius of a cation is typically smaller than its atomic radius.
• When an atom loses one or more electrons to form a cation, the loss of negatively charged electrons reduces electron-electron repulsion.
• The remaining electrons are pulled closer to the nucleus due to the unchanged positive nuclear charge, resulting in a smaller ionic radius.

EXPLANATION:

• Assertion A: The ionic radius of a cation is smaller than its atomic radius.
• Explanation: This is due to the reduction in electron-electron repulsion and the increased effective nuclear charge per electron after electron loss.
• Reason R: A cation is formed by the loss of one or more electrons, leading to a decrease in electron-electron repulsion and a contraction of the electron cloud.
• Explanation: This reason aligns with the explanation given in A, as the loss of electrons and reduction in repulsion cause the electron cloud to contract, resulting in a smaller radius.

The correct answer is: Both A and R are true, and R is the correct explanation of A.

Concept:

The elevation in boiling point (ΔT_b) of a solution is a colligative property that depends on the number of solute particles in the solution. For ionic compounds, the number of particles depends on the degree of dissociation.

The formula for boiling point elevation is: ΔT_b = i × K_b × m, where i is the van 't Hoff factor (number of particles), K_b is the ebullioscopic constant, and m is the molality of the solution.

Explanation:

Assertion (A): One molar BaCl₂ solution gives double the elevation in boiling point than one molar NaCl solution.

This statement is incorrect. Although BaCl₂ dissociates into three ions and NaCl into two, the elevation in boiling point is not simply doubled but the elevation in boiling point for one molar BaCl₂ will be 1.5 times than one molar NaCl.

Reason (R): Ba²⁺ ions carry double the charge than Na⁺ ions.

This statement is correct but does not relate directly to the assertion about boiling point elevation.

Conclusion:

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

CONCEPT:

Molality Calculation from Molarity

• Molarity (M) is the number of moles of solute per liter of solution.
• Molality (m) is the number of moles of solute per kilogram of solvent.
• To convert molarity to molality, we need to know the density (or specific gravity) of the solution, which helps us determine the mass of the solution and subsequently the mass of the solvent.

CONCEPT:

C-O Bond Lengths in Different Molecules

• The length of a C-O bond in a molecule is influenced by the bond order, which refers to the number of shared electron pairs (bonds) between carbon and oxygen atoms. Higher bond orders usually result in shorter bond lengths due to the increased electron density between the atoms, pulling them closer together.
• Each molecule has a specific bond order based on its structure:

CONCEPT:

Aldol Condensation and Product Formation

• Aldol condensation is a reaction where an enolate ion, formed from a compound with alpha-hydrogens, reacts with another carbonyl compound to form a new carbon-carbon bond.
• In this case, 5-membered ring undergoes deprotonation by sodium ethoxide (EtONa) to form a stable anion ion.
• This anion then acts as a nucleophile and reacts with the acetone, resulting in the formation of the desired product.

Reaction and mechanism:

CONCLUSION:

• The correct option is: Option 3

CONCEPT:

Reaction Sequence: Conversion of Carboxylic Acid to Nitrile

• The reaction sequence involves converting a carboxylic acid group (-COOH) into a nitrile group (-CN) through intermediate steps.
• This process typically involves:
  • Converting the carboxylic acid to an acyl chloride using thionyl chloride (SOCl₂).
  • Reacting the acyl chloride with ammonia (NH₃) to form an amide.
  • Dehydrating the amide to a nitrile using phosphorus pentoxide (P₂O₅) and heat.

CONCEPT:

Rate of Decarboxylation of Compounds with Sodamide

• The decarboxylation reaction involves the removal of a carboxyl group (-COOH) as carbon dioxide (CO₂), usually facilitated by a base such as sodamide (NaNH₂).
• The presence of electron-withdrawing groups (such as NO₂) on the benzene ring increases the rate of decarboxylation by stabilizing the carbanion intermediate formed after CO₂ is released.
• Electron-withdrawing groups exert their effects more strongly based on their position relative to the carboxyl group. The ortho position (adjacent to -COOH) has the greatest impact on stabilization, followed by the para position, and finally, the meta position has the least effect.

EXPLANATION:

• Examine each compound based on the position of the nitro (NO₂) group:
  • Compound (IV): The NO₂ group is at the ortho position relative to -COOH. This proximity allows it to strongly stabilize the carbanion through electron withdrawal via resonance and inductive effects, making this compound decarboxylate at the fastest rate.
  • Compound (I): The NO₂ group is at the para position relative to -COOH. While not as close as the ortho position, the para position still allows effective electron withdrawal through resonance, enhancing decarboxylation but to a lesser extent than the ortho position.
  • Compound (III): The NO₂ group is at the meta position, which does not provide significant resonance stabilization to the carbanion due to its position. The electron-withdrawing effect is mainly inductive and weaker compared to ortho and para positions, resulting in a slower decarboxylation rate.
  • Compound (II): This compound also has the NO₂ group at the meta position but with additional steric hindrance that reduces the efficiency of the electron-withdrawing effect even further. As a result, this compound has the slowest rate of decarboxylation among the four.

CONCLUSION:

The correct order for the rate of decarboxylation of the given compounds with sodamide is Option 1: IV > I > III > II.