Chemistry Test 239

CONCEPT:

Balancing Redox Reactions (Disproportionation Reaction)

• A redox reaction involves the transfer of electrons between species, where one substance is oxidized and the other is reduced.
• In a disproportionation reaction, a single species undergoes both oxidation and reduction.
• The balancing of such reactions requires ensuring that the number of atoms and the charge are conserved.

CONCEPT:

Balancing Redox Reactions

• A redox reaction involves both oxidation and reduction. Oxidation refers to the loss of electrons, while reduction refers to the gain of electrons.
• In a redox reaction, the number of electrons lost in oxidation must equal the number of electrons gained in reduction to balance the charges.
• For balancing redox reactions, we often balance atoms and charges separately using the half-reaction method, which involves balancing each half-reaction for mass and charge.

CONCEPT:

Nature and Behavior of Boric Acid (H₃BO₃)

• Boric acid (H₃BO₃) is often classified as a Lewis acid rather than a Bronsted acid.
• Lewis acids are species that accept electron pairs. In the case of H₃BO₃, the boron atom has an empty orbital that can accept an electron pair.
• Instead of donating protons (H⁺), boric acid interacts with water by accepting an OH⁻ ion from a water molecule, forming the complex [B(OH)₄]⁻ and generating H₃O⁺ ions (hydronium).

EXPLANATION:

• Monobasic Nature: Although H₃BO₃ contains three hydroxyl groups (-OH), it only interacts with one molecule of water, releasing one equivalent of hydronium ions (H₃O⁺), which makes it monobasic.
• In an aqueous solution, the reaction can be represented as:
  H₃BO₃ + H₂O → [B(OH)₄]⁻ + H₃O⁺
• Here, boric acid acts by accepting an electron pair from the water's hydroxide ion, forming [B(OH)₄]⁻.
• Weak Lewis Acid: H₃BO₃ is classified as a weak acid because its interaction with water does not lead to a highly dissociative reaction. It does not strongly ionize in solution, making it a weak Lewis acid.

Additional Information

• Lewis Acid Behavior: Boric acid's classification as a Lewis acid is because it accepts an OH⁻ ion rather than donating a proton. Unlike a typical Bronsted acid, which donates a proton (H⁺), boric acid works by accepting an electron pair.
• Monobasic Property: The term "monobasic" indicates that only one hydroxide ion (OH⁻) is involved in the reaction, which corresponds to the formation of one molecule of (H₃O⁺)(hydronium) per molecule of boric acid.

CONCLUSION:

The correct answer is Option 1: Mono basic and weak Lewis acid

CONCEPT:

Boron Compounds and Their Lewis Acid Behavior

• Lewis acids are defined as species that can accept an electron pair. This electron pair acceptance is typically facilitated by the presence of an empty or vacant orbital.
• Boron, being in Group 13 of the periodic table, has three valence electrons, which it uses to form three covalent bonds. This leaves boron with only six electrons in its valence shell, making it electron-deficient.
• This electron deficiency creates a vacant p-orbital on the boron atom, which makes it highly reactive towards species that can donate an electron pair (Lewis bases).

EXPLANATION:

• Vacant Orbital and Electron Deficiency: In compounds like boron trifluoride (BF₃) and boron trichloride (BCl₃), the boron atom forms three covalent bonds with fluorine or chlorine, leaving it with only six electrons in its valence shell. The absence of a complete octet (8 electrons) leaves boron with a vacant p-orbital.
• This vacant orbital makes boron compounds ideal Lewis acids, as they can accept electron pairs from Lewis bases to fill their electron deficiency and complete their octet. For instance, BF₃ readily reacts with molecules like ammonia (NH₃), which have lone pairs of electrons to donate.
• Boron does not act as a Lewis acid because of its covalent or ionic nature but rather due to the availability of this vacant orbital that seeks an electron pair for stabilization.

Additional Information

• Examples of Boron as a Lewis Acid:
  • BF₃ (Boron trifluoride): Boron in BF₃ has only six valence electrons, with a vacant p-orbital, making it an excellent Lewis acid. It can accept an electron pair from molecules with lone pairs like ammonia (NH₃).
  • BCl₃ (Boron trichloride): Similar to BF₃, BCl₃ has an electron-deficient boron atom that acts as a Lewis acid by accepting electron pairs to complete its octet.
  • H₃BO₃ (Boric Acid): Although H₃BO₃ is often referred to as an acid, it acts as a Lewis acid by accepting OH⁻ ions from water, forming the tetrahydroxyborate ion.
• Nature of the Lewis Acidity: The Lewis acidity of boron compounds is not due to their acidic nature or bond type (covalent or ionic). Instead, it is fundamentally due to the vacant orbital that enables them to accept electron pairs from donor species (Lewis bases).

CORRECT ANSWER:

The correct answer is Option 4: Vacant orbital

CONCEPT:

Ionic Radii and Isoelectronic Species

• Ionic radius refers to the size of an ion in a crystal lattice, and it is influenced by the ion's charge and the number of electrons compared to protons.
• Anions (negatively charged ions) are larger than their parent atoms because they gain electrons, increasing electron-electron repulsion, which pushes the electrons farther from the nucleus.
• Cations (positively charged ions) are smaller than their parent atoms because they lose electrons, which reduces electron-electron repulsion, and the remaining electrons are pulled closer to the nucleus by the protons.
• For isoelectronic species (ions that have the same number of electrons), the size of the ion decreases as the nuclear charge increases (i.e., the number of protons in the nucleus increases).

EXPLANATION:

• The ions N³⁻, O²⁻, F⁻, and Na⁺ all have 10 electrons, making them isoelectronic species. However, they have different nuclear charges (number of protons), which affects the ionic radius.
• The nuclear charge increases as we move from N^3- (7 protons) to Na⁺ (11 protons), meaning that the nucleus can pull the electrons closer to itself more effectively in species with higher nuclear charge.
• This leads to the following trend in ionic radii:
  • N^3- has the smallest nuclear charge (7 protons) and thus the largest ionic radius because the attraction between the nucleus and the electrons is weakest.
  • O^2- has 8 protons, so it has a smaller ionic radius than N^3- because the nucleus pulls the electrons more tightly.
  • F⁻ has 9 protons, so it has a smaller ionic radius than O^2-.
  • Na⁺ has the highest nuclear charge (11 protons), making it the smallest ion because the nucleus pulls the electrons most tightly. 

Additional Information

• Trend Among Isoelectronic Species: The size of isoelectronic ions decreases as the positive charge on the nucleus increases. Therefore, for isoelectronic species, cations are smaller than anions.
• Factors Influencing Ionic Radius:
  • As the number of protons increases, the effective nuclear charge increases, pulling the electron cloud closer and reducing the size of the ion.
  • For anions, adding electrons increases repulsion between electrons, causing the ionic radius to expand.

Conclusion:

The correct answer is Option 1: N^3- > O^2- > F^- > Na⁺

CONCEPT:

Zero Electron Affinity of Be and Mg

• Electron affinity refers to the amount of energy released when an electron is added to a neutral atom in its gaseous state.
• For elements like Beryllium (Be) and Magnesium (Mg), the electron affinity is nearly zero or very low because they have completely filled s-orbitals in their valence shells.
• The electron configurations of Be and Mg are:
  • Be: [He] 2s²
  • Mg: [Ne] 3s²
• These configurations show that the 2s orbital in Be and the 3s orbital in Mg are completely filled, making it difficult for these elements to accept additional electrons.

EXPLANATION:

• Filled Orbitals: The 2s and 3s orbitals of Be and Mg are filled to their capacity, which provides these atoms with extra stability.
• Electron Configuration Stability: The electron configurations [He] 2s² for Be and [Ne] 3s² for Mg indicate that these elements have a stable arrangement of electrons, making it highly unfavorable to accept additional electrons.
• Inability to Accept Electrons: Because of their stable electron configurations and filled s-orbitals, Be and Mg have zero or very low electron affinity. Adding another electron would disturb this stable arrangement and require placing the electron in a higher energy p-orbital.

CONCLUSION:

• Option 1: Be and Mg have [He]2s² and [Ne]3s² configuration respectively.

​​​​​CONCEPT:

Basic Strength of Hydroxides in Relation to Periodic Table

• The basic strength of a hydroxide depends on the degree of dissociation in water and its ability to donate hydroxide ions (OH⁻).
• Elements from Group 1 (alkali metals) form hydroxides that dissociate almost completely in water, making them very strong bases.
• Elements from Group 2 (alkaline earth metals) form hydroxides that are less soluble in water, leading to lower dissociation and, consequently, weaker basic strength compared to Group 1 hydroxides.
• Amphoteric hydroxides, like Al(OH)₃, can act both as acids and bases but do not dissociate significantly in water, making them much weaker bases compared to Group 1 and 2 hydroxides.

EXPLANATION:

• Why KOH is the strongest base:
  • KOH (Potassium hydroxide) belongs to the alkali metal group (Group 1), which are known for their strong basicity.
  • KOH is highly soluble in water and dissociates completely, releasing a large number of OH⁻ ions. This complete dissociation makes it a strong base.
  • Therefore, KOH has the highest basic strength among the given hydroxides.
• Why Mg(OH)₂ is less basic than KOH:
  • Mg(OH)₂ (Magnesium hydroxide) belongs to the alkaline earth metal group (Group 2). Alkaline earth metal hydroxides are generally less basic than alkali metal hydroxides.
  • While Mg(OH)₂ does dissociate to produce OH⁻ ions, its solubility in water is much lower compared to KOH. This limits the number of OH⁻ ions released into the solution.
  • Thus, Mg(OH)₂ is weaker than KOH but still a reasonably strong base due to its ability to produce OH⁻ ions.
• Why Al(OH)₃ is the weakest base:
  • Al(OH)₃ (Aluminum hydroxide) is an amphoteric hydroxide, meaning it can react with both acids and bases, but it dissociates very poorly in water.
  • Because of its very low solubility and weak dissociation, Al(OH)₃ releases very few OH⁻ ions, making it much weaker as a base compared to KOH and Mg(OH)₂.
  • In addition, its amphoteric nature allows it to act both as a base and an acid depending on the medium, further reducing its effective basic strength in comparison to other hydroxides.

Additional Information

• Factors influencing basic strength:
  • Solubility in water: Highly soluble hydroxides (like KOH) dissociate completely, making them strong bases.
  • Degree of dissociation: Bases that dissociate more completely in water produce more OH⁻ ions and are stronger bases.
  • Periodic trend: Hydroxides of Group 1 elements are typically stronger bases than hydroxides of Group 2 elements due to the easier dissociation of Group 1 hydroxides.
  • Amphoteric behavior: Amphoteric hydroxides like Al(OH)₃ have lower basic strength because they do not dissociate well and can react both as acids and bases.

CONCLUSION:

• The correct order of basic strength is:
  • KOH > Mg(OH)₂ > Al(OH)₃

CONCEPT:

Thermal Decomposition of Carbonates (MCO₃)

• The decomposition temperature of metal carbonates (MCO₃) depends on the thermal stability of the carbonate ion in the presence of the metal ion.
• The stability of carbonates increases as we move down Group 2 (alkaline earth metals) in the periodic table, meaning carbonates of lighter elements decompose at lower temperatures compared to heavier elements.
• This means BeCO₃ and MgCO₃ decompose at lower temperatures (below 900°C), while SrCO₃ and BaCO₃ decompose at higher temperatures (above 900°C).

EXPLANATION:

• BeCO₃ (Beryllium carbonate):
  • BeCO₃ is highly unstable and decomposes at a much lower temperature because the beryllium ion (Be²⁺) is small and highly polarizing, which weakens the carbonate bond, making it decompose easily.
  • This instability leads to BeCO₃ decomposing well below 900°C.
• MgCO₃ (Magnesium carbonate):
  • MgCO₃ is more thermally stable than BeCO₃, but it still decomposes below 900°C due to the smaller size and higher charge density of Mg²⁺ compared to heavier Group 2 elements like Sr²⁺ and Ba²⁺.
  • This lower stability causes MgCO₃ to decompose at temperatures below 900°C.
• SrCO₃ (Strontium carbonate) and BaCO₃ (Barium carbonate):
  • These carbonates are much more stable because the larger size of Sr²⁺ and Ba²⁺ ions reduces the polarizing power on the carbonate ion, making the bond stronger and more resistant to decomposition.
  • As a result, SrCO₃ and BaCO₃ decompose at temperatures above 900°C.

CONCLUSION:

• The correct answer is Option 1: BeCO₃, MgCO₃, as these decompose below 900°C.

CONCEPT:

Hydration Energy of Ions

• Hydration energy refers to the amount of energy released when an ion interacts with water molecules during the formation of a hydrated ion.
• The hydration energy is inversely related to the size of the ion. Smaller ions have higher charge density, allowing them to attract more water molecules and release more energy upon hydration.
• In the case of alkali metal ions, the hydration energy decreases as we move down the group because the ionic radius increases, and the charge density decreases.
• Thus, Li⁺ (the smallest ion) has the highest hydration energy, followed by Na⁺, K⁺, Rb⁺, and Cs⁺ in decreasing order.

EXPLANATION:

• Li⁺ (Lithium ion):
  • Li⁺ is the smallest alkali metal ion, so it has the highest charge density. This allows it to attract water molecules more effectively, resulting in the highest hydration energy among the alkali metals.
• Na⁺ (Sodium ion) and K⁺ (Potassium ion):
  • Na⁺ is larger than Li⁺ but smaller than K⁺, so its hydration energy is lower than Li⁺ but higher than K⁺.
  • K⁺ has a larger ionic radius than both Li⁺ and Na⁺, leading to a lower charge density and a lower hydration energy compared to both.
• Cs⁺ (Cesium ion) and Rb⁺ (Rubidium ion):
  • Cs⁺ and Rb⁺ are the largest ions among alkali metals, which makes their charge densities very low. As a result, they have the lowest hydration energies in the alkali metal group.

CONCLUSION:

• The correct answer is Option 1: Li⁺ > Na⁺ > K⁺.

CONCEPT:

Bond Formation Involving p and s Orbitals

• π bond is formed when two orbitals overlap laterally (side-by-side). In this case, the electron density lies above and below the bond axis.
• σ bond is formed when orbitals overlap head-on along the bond axis, concentrating electron density directly along the axis.
• The overlap between an s-orbital and a p_y orbital will result in a π bond if the bond axis is perpendicular to the orbital overlap (such as along the y-axis).

EXPLANATION:

• Statement 1: s-orbital of another atom produces a π bond when y is the bond formation axis.
  • When the p_y orbital is oriented along the y-axis, the overlap between the s-orbital of another atom and the p_y orbital can occur laterally. This lateral overlap leads to the formation of a π bond.
  • Since the y-axis is the bond formation axis, the electron density will be above and below the bond axis, which characterizes a π bond.
  • Thus, this statement is correct because the s-orbital can form a π bond with a p_y orbital when the overlap is lateral.
• Statement 2: P_y-orbital of another produces a σ bond when x is the bond formation axis.
  • If the bond axis is along the x-axis, the p_y orbital would overlap laterally with another p_y orbital, resulting in a π bond (not a σ bond).
  • For a σ bond to form, there must be head-on overlap along the bond axis. Since the overlap between p_y orbitals is lateral, this statement is incorrect.
• Statement 3: P_z-orbital of another atom produces a π bond when x is the bond formation axis.
  • A p_z orbital is oriented along the z-axis. If the bond axis is along the x-axis, lateral overlap between p_z orbitals would result in a π bond, but this has no involvement with a p_y orbital.
  • This statement is incorrect because it describes the interaction of p_x orbitals, not p_y orbitals.
• Statement 4: dxy-orbital of another atom produces a π bond when x is the bond formation axis.
  • The d_xy orbital lies in the xy-plane. If the bond axis is along the x-axis, lateral overlap could occur, but this overlap does not directly involve the p_y orbital in forming a π bond in this case.
  • Therefore, this statement is also incorrect as it does not involve the correct orbital interactions.

CONCLUSION:

• The correct answer is Option 1: s-orbital of another atom produces a π bond when y is the bond formation axis.