Question

2. Show the following for the compounds given: BH3(i) Lewis electron dot formula(ii) Lewis structure (draw in 3D- use wedge dash for tetrahedral)(iii) Electronic geometry (name)(iv) Molecular geometry (name)(v) bond angle (ideal)(vi) polar or non-polar

Answer 1

BF₃ has a trigonal planar molecular geometry with 120-degree bond angles. It is non-polar due to the symmetry that cancels out the dipole moments of individual bonds.

(i) Lewis Electron Dot Formula:

BF₃ (boron trifluoride) has three valence electrons for boron and seven for each fluorine. Boron forms three bonds with fluorine, and the Lewis electron dot formula is:

`\[ \text{B} \: \text{F} \: \text{F} \: \text{F} \]`

(ii) Lewis Structure (3D):

The molecule has a trigonal planar shape in 3D. The boron atom is in the center, and three fluorine atoms are arranged around it with 120-degree bond angles, represented using wedge and dash bonds.

(iii) Electronic Geometry (Name):

The electronic geometry around the boron atom is tetrahedral.

(iv) Molecular Geometry (Name):

The molecular geometry around the boron atom is trigonal planar.

(v) Bond Angle (Ideal):

The ideal bond angle between the fluorine atoms in BF₃ is 120 degrees.

(vi) Polar or Non-Polar:

BF₃ is non-polar. Although the individual B-F bonds are polar due to the electronegativity difference, the molecule's symmetry results in the cancellation of dipole moments, making the overall molecule non-polar.

Answer 2

Trihydridoboron, also known as borane is a highly reactive compound and has as chemical formula BH3, from the elements apart we know that Boron has a 3+ charge, that will form bonds with 3 H+, this molecule is also electron deficient and will not follow the octet rule because it has only 6 valence electrons

1. Lewis dot structure will be used to indicate the valence electrons forming bonds

Here we can see that each pair of electrons are being shared from Boron and Hydrogen atoms

2. Lewis structure is basically the same as the dot structures but now with dashes, and since all H atoms are equal, we will have a planar structure

3. BH3 will present a trigonal planar electron geometry, as I previously explained the H atoms will be equally divided into 3 directions, forming the trigonal planar shape, and also because there is no lone pairs, which would affect the electron geometry

4. It Will also present the trigonal planar geometry

5. Bond angle for this compound will be 120°, precisely because we have 3 Hydrogen atoms evenly distributed around the Boron atom

6. Because of the symmetrical shape of the compound, the molecule will not present any polarity, even though B - H have some electronegativity difference, but the even distribution of H around B, will make the polarity disappear