Question

Correct order with respect to the given properties in parentes.

A. H₂O > H₂S > H₂Se > H₂Te (acidic character)

B. HF < HCl < HBr < HI (thermal stability)

C. H₂O > HF > NH₃ (boiling point)

D. H₂O > H₂S > H₂Se > H₂Te (reducing property)

• A. H₂O > H₂S > H₂Se > H₂Te (acidic character)
• B. HF < HCl < HBr < HI (thermal stability)
• C. H₂O > HF > NH₃ (boiling point)
• D. H₂O > H₂S > H₂Se > H₂Te (reducing property)

Answer 1

Answer: (C)

H₂O > HF > NH₃ (boiling point)

Answer 2

The question asks us to identify the correct order with respect to the given properties in parentheses. Let's analyze each option:

A. H₂O > H₂S > H₂Se > H₂Te (acidic character)

• For hydrides of Group 16 elements (H₂X), acidic character increases down the group. This is because as we move down the group, the size of the central atom (O, S, Se, Te) increases, leading to an increase in bond length and a decrease in the bond dissociation enthalpy of the H-X bond. A weaker H-X bond means it is easier to donate a proton (H⁺), thus increasing acidity.
• Therefore, the correct order of acidic character is H₂O < H₂S < H₂Se < H₂Te.
• The given order is incorrect.

B. HF < HCl < HBr < HI (thermal stability)

• For hydrogen halides (HX), thermal stability depends on the bond dissociation enthalpy of the H-X bond. As we move down Group 17, the size of the halogen atom increases, leading to an increase in bond length and a decrease in bond strength. Weaker bonds are less stable thermally.
• Therefore, the correct order of thermal stability is HF > HCl > HBr > HI.
• The given order is incorrect.

C. H₂O > HF > NH₃ (boiling point)

• Boiling points are influenced by intermolecular forces. H₂O, HF, and NH₃ all exhibit hydrogen bonding.
• H₂O: Oxygen is highly electronegative and has two lone pairs and two hydrogen atoms. This allows each H₂O molecule to form, on average, two hydrogen bonds with other H₂O molecules, leading to an extensive 3D network of hydrogen bonds.
• HF: Fluorine is the most electronegative element. Each HF molecule has one hydrogen atom and three lone pairs on fluorine. It forms strong hydrogen bonds, but each molecule can only form one hydrogen bond with another HF molecule (due to only one H atom). This leads to zigzag chains of HF molecules.
• NH₃: Nitrogen is less electronegative than oxygen or fluorine. Each NH₃ molecule has one lone pair and three hydrogen atoms. It forms weaker hydrogen bonds compared to H₂O and HF, and each molecule can form, on average, one hydrogen bond.
• Comparing the boiling points: H₂O (100 °C) > HF (19.5 °C) > NH₃ (-33.35 °C).
• The stronger and more extensive hydrogen bonding in H₂O leads to its highest boiling point. Although HF forms stronger individual hydrogen bonds than H₂O, H₂O forms a more extensive network of hydrogen bonds (two per molecule) compared to HF (one per molecule). NH₃ forms weaker and less extensive hydrogen bonds.
• Therefore, the given order is correct.

D. H₂O > H₂S > H₂Se > H₂Te (reducing property)

• For hydrides of Group 16 elements (H₂X), reducing property increases down the group. This is because as we move down the group, the H-X bond strength decreases, making it easier for the molecule to lose hydrogen and get oxidized (thus acting as a reducing agent).
• Therefore, the correct order of reducing property is H₂O < H₂S < H₂Se < H₂Te.
• The given order is incorrect.

Based on the analysis, only option C is correct.