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Question: Choose the CORRECT statement(s) about the compounds of xenon. (A) B(OH)3 is planar in solid state a...

Choose the CORRECT statement(s) about the compounds of xenon.

(A) B(OH)3 is planar in solid state and three hydrogen bonds are formed per molecule. X (B) POF3 exists but NOF3 does not since N cannot form 5 bonds (C) N3- and NO2+ are isostructural and isosteric (D) Hybridized orbitals always form π\pi-bonds

A

B(OH)3 is planar in solid state and three hydrogen bonds are formed per molecule.

B

POF3 exists but NOF3 does not since N cannot form 5 bonds

C

N3- and NO2+ are isostructural and isosteric

D

Hybridized orbitals always form π\pi-bonds

Answer

B, C

Explanation

Solution

The question asks to identify the correct statement(s) among the given options (A), (B), (C), and (D). The line "8. Choose the CORRECT statement(s) about the compounds of xenon." appears to be a misplaced or unrelated part of the question, as none of the statements (A)-(D) are about xenon compounds. We will evaluate each statement individually.

Statement (A): B(OH)₃ is planar in solid state and three hydrogen bonds are formed per molecule.

  • Planarity: The B(OH)₃ molecule itself is trigonal planar because the boron atom is sp² hybridized. In the solid state, these planar B(OH)₃ units are linked together by hydrogen bonds to form a two-dimensional layered structure. So, the individual molecule is planar.
  • Hydrogen bonds: In the solid state, each B(OH)₃ molecule participates in hydrogen bonding through its three -OH groups. Each oxygen atom acts as a hydrogen bond acceptor, and each hydrogen atom (bonded to oxygen) acts as a hydrogen bond donor. Therefore, each B(OH)₃ molecule forms 3 donor hydrogen bonds and accepts 3 hydrogen bonds, making it involved in a total of 6 hydrogen bonds. The statement "three hydrogen bonds are formed per molecule" is incorrect if it refers to the total number of hydrogen bonds.
  • Thus, statement (A) is incorrect.

Statement (B): POF₃ exists but NOF₃ does not since N cannot form 5 bonds.

  • POF₃ (Phosphoryl fluoride): Phosphorus is a third-period element and possesses vacant 3d-orbitals. It can expand its octet by utilizing these d-orbitals. In POF₃, phosphorus forms one double bond with oxygen and three single bonds with fluorine, resulting in 5 bonds (10 electrons around P). This is a stable molecule, and POF₃ exists.
  • NOF₃ (Nitrogen trifluoride oxide): Nitrogen is a second-period element and lacks vacant d-orbitals. Consequently, it cannot expand its octet beyond 8 electrons. If NOF₃ were to exist with a structure analogous to POF₃ (N=O, 3 N-F bonds), nitrogen would need to form 5 bonds, which would require 10 valence electrons around it. This is not possible for nitrogen. Nitrogen's maximum covalency is 4.
  • Thus, statement (B) is correct.

Statement (C): N₃⁻ and NO₂⁺ are isostructural and isosteric.

  • Isostructural: Having the same shape/geometry.
  • Isosteric: Having the same number of atoms and the same number of valence electrons.
  • N₃⁻ (Azide ion):
    • Total valence electrons = (3 × 5) + 1 (for charge) = 16 valence electrons.
    • The central nitrogen atom is sp hybridized, and the ion has a linear geometry.
    • Resonance structures: \ceN−=N+=N−↔N≡N+−N2−↔2−N−N+≡N\ce{N^-=N^+=N^- \leftrightarrow N\equiv N^+-N^{2-} \leftrightarrow ^{2-}N-N^+\equiv N}
  • NO₂⁺ (Nitronium ion):
    • Total valence electrons = 5 (for N) + (2 × 6) (for O) - 1 (for charge) = 16 valence electrons.
    • The central nitrogen atom is sp hybridized, and the ion has a linear geometry.
    • Structure: \ceO=N+=O\ce{O=N^+=O}
  • Comparison: Both N₃⁻ and NO₂⁺ have 3 atoms and 16 valence electrons, making them isosteric. Both have a linear geometry, making them isostructural.
  • Thus, statement (C) is correct.

Statement (D): Hybridized orbitals always form π-bonds.

  • Hybridized orbitals: These orbitals are formed by the mixing of atomic orbitals (s, p, d) on the same atom. They are primarily involved in forming sigma (σ) bonds and holding lone pairs of electrons.
  • π-bonds: These are formed by the lateral (sideways) overlap of unhybridized p-orbitals (or d-orbitals) that are perpendicular to the internuclear axis.
  • Therefore, hybridized orbitals form sigma bonds, not pi bonds.
  • Thus, statement (D) is incorrect.

Conclusion: Statements (B) and (C) are the correct statements.

Explanation of the solution:

  • (A) B(OH)₃ is planar but forms 6 hydrogen bonds per molecule in the solid state (3 donor, 3 acceptor), not 3. Hence, incorrect.
  • (B) POF₃ exists because P can expand its octet using vacant d-orbitals. NOF₃ does not exist because N is a second-period element and cannot expand its octet beyond 8 electrons (cannot form 5 bonds). Hence, correct.
  • (C) N₃⁻ and NO₂⁺ both have 3 atoms and 16 valence electrons. Both have a linear geometry. Therefore, they are both isostructural and isosteric. Hence, correct.
  • (D) Hybridized orbitals form sigma (σ) bonds and accommodate lone pairs. Pi (π) bonds are formed by the lateral overlap of unhybridized p-orbitals. Hence, incorrect.