Question

For each of the following compounds, determine the bond angles, electron-domain and molecular geometries, hybridization, and polarity associated with each molecule.

KrF₄

CHBr₃

N0₂^1-

BrF₃

Answer 1

KrF₄:

Bond Angle: 90^o (corresponding to Octahedral Geometry).

Electron Domain: Central atom has 12 electrons.

Molecular Geometry: Octahedral (Axial positions occupied by lone pairs)

Hybridization: d²sp³

Polarity: Non Polar

CHBr₃:

Bond Angle: 109^o 28' (Aprox)(corresponding to Tetrahedral Geometry).

Electron Domain: All atoms have octet fulfilled. Hydrogen has duplet

Molecular Geometry: Tetrahedral (Slightly distorted due to Bromine)

Hybridization: sp³

Polarity: Polar

NO₂^-:

Bond Angle: 120^o (corresponding to Trigonal Planar Geometry).

Electron Domain: All atoms have octet fulfilled.

Molecular Geometry: Trigonal Planar.

Hybridization: sp²

Polarity: Polar

BrF₃:

Bond Angle: 90^o (corresponding to the angle between axial and equatorial substituents in TBP).

Electron Domain: Central atom has 10 electrons.

Molecular Geometry: Trigonal Bipyramidal (Equatorial positions occupied by lone pairs)

Hybridization: sp³d

Polarity: Polar