Question

$sp^3d^2$ hybridization is not displayed by :

• A. ${BrF5}$
• B. ${SF6}$
• C. ${[CrF6]^{3-}}$
• D. ${PF_5}$

Answer 1

The Correct Option is D

To determine which molecule does not exhibit sp^3d^2 hybridization, we need to analyze each given molecule's hybridization based on its electronic configuration and molecular geometry.

1. {BrF_5}:
   • Bromine (Br) has 7 valence electrons.
   • Each fluorine (F) atom forms a single bond with bromine, using 5 electrons, leaving 2 electrons as a lone pair on Br.
   • BrF₅ has 5 bond pairs and 1 lone pair, indicating sp^3d^2 hybridization, forming a square pyramidal shape.
2. {SF_6}:
   • Sulfur (S) has 6 valence electrons.
   • Each fluorine (F) atom forms a bond with sulfur using all 6 electrons, resulting in no lone pairs.
   • SF₆ shows 6 bond pairs, indicating sp^3d^2 hybridization with an octahedral shape.
3. {[CrF_6]^{3-}}:
   • Chromium acts as the central atom here. The charge on the complex is -3.
   • Considering its electronic configuration and bonding with 6 fluorine atoms, Cr exhibits a high oxidation state compatible with sp^3d^2 hybridization.
   • The complex has no lone pairs but involves strong dative bonds from fluorine, supporting sp^3d^2 hybridization.
4. {PF_5}:
   • Phosphorus (P) has 5 valence electrons, and each fluorine atom forms a bond using an electron from P.
   • PF₅ has 5 bond pairs and no lone pairs. This fits the description of sp^3d hybridization, not sp^3d^2.
   • The geometry is trigonal bipyramidal, which comes from sp^3d hybridization.

Therefore, the molecule that does not exhibit sp^3d^2 hybridization is PF_5, which instead shows sp^3d hybridization.