Properties and Trends of Group 15 Compounds

To determine the correct statement(s), let us analyze each option individually:

1. Option (A): $\text{Bi}_2\text{O}_5$ is more basic than $\text{N}_2\text{O}_5$

   • As we move down Group 15 (from $\text{N}$ to $\text{Bi}$), the metallic character of the elements increases and their electronegativity decreases.
   • Higher metallic character leads to an increase in the basic nature (or decrease in the acidic nature) of the oxides. $\text{N}_2\text{O}_5$ is strongly acidic, whereas $\text{Bi}_2\text{O}_5$ is basic in comparison (or significantly less acidic).
   • Hence, $\text{Bi}_2\text{O}_5$ is more basic than $\text{N}_2\text{O}_5$.
   • Statement (A) is correct.

2. Option (B): $\text{NF}_3$ is more covalent than $\text{BiF}_3$

   • Nitrogen is a small, highly electronegative non-metal, forming purely covalent bonds with fluorine in $\text{NF}_3$.
   • Bismuth is a large metallic element with low electronegativity. According to Fajan's rules, the larger $\text{Bi}^{3+}$ cation results in greater ionic character in its fluoride, making $\text{BiF}_3$ predominantly ionic in nature.
   • Therefore, $\text{NF}_3$ is significantly more covalent than $\text{BiF}_3$.
   • Statement (B) is correct.

3. Option (C): $\text{PH}_3$ boils at a lower temperature than $\text{NH}_3$

   • Due to the high electronegativity and small size of nitrogen, $\text{NH}_3$ molecules are associated through extensive intermolecular hydrogen bonding.
   • In contrast, phosphorus has low electronegativity and larger atomic size, so $\text{PH}_3$ does not exhibit hydrogen bonding and molecules are held together only by weak van der Waals forces.
   • Consequently, the boiling point of $\text{NH}_3$ ($238.5\text{ K}$) is significantly higher than that of $\text{PH}_3$ ($185.5\text{ K}$).
   • Statement (C) is correct.

4. Option (D): The $\text{N}-\text{N}$ single bond is stronger than the $\text{P}-\text{P}$ single bond

   • The $\text{N}-\text{N}$ single bond is relatively weak because the small size of nitrogen atoms brings the non-bonding lone pairs into close proximity, leading to strong interelectronic repulsions.
   • In phosphorus, the larger atomic size reduces the lone-pair-lone-pair repulsion, making the single $\text{P}-\text{P}$ bond stronger ($\text{Bond enthalpy} \approx 201\text{ kJ mol}^{-1}$) than the single $\text{N}-\text{N}$ bond ($\text{Bond enthalpy} \approx 160\text{ kJ mol}^{-1}$).
   • Statement (D) is incorrect.

Thus, the correct statements are A, B, and C.