Select the correct value for the indicated bond angle in each of the compounds. O−O−OO−O−O angle of O3 90° 109.5° <109.5° 120° <120° 180° F−B−FF−B−F angle of BF3 180° <109.5° <120° 120° 109.5° 90° F−O−FF−O−F angle of OF2 <120° 120° 90° 109.5° 180° <109.5° Cl−Be−ClCl−Be−Cl angle of BeCl2 90° 109.5° 180° 120° <109.5° <120° F−P−FF−P−F angle of PF3 90° 109.5° <109.5° 180° 120° <120° H−C−HH−C−H angle of CH4 90° <109.5° 180° 120° <120° 109.5°

O3-<120°

BF3-120°

OF2- <109.5°

BeCl2-180°

PF3-<109.5°

CH4-109.5°

Explanation:

Ozone is a bent molecule. The bent structure of the molecule owes to the fact that the central atom which is oxygen possesses lone pairs of electrons which causes the structure to bend due to repulsive forces. The actual bond angle of O3 is about 116°

For BF3, the three bond pairs of electrons are directed towards the corners of an equilateral triangle. According to the valence shell electron pair repulsion theory, a trigonal planar molecular geometry is expected with an F-B-F bond angle of 120°.

According to the valence shell electron pair repulsion theory, OF2 has four electron domains hence it is expected to exhibit a tetrahedral geometry with a bond angle of about 109°. The observed geometry of the molecule is bent due to repulsive forces of lone pairs of electrons. The observed bond angle is about 103°.

BeCl2 is a linear molecule with a bond angle of 180°. Beryllium bonds to chlorine atoms using two sp hybridized orbitals to give a linear molecule.

PF3 has four electron domains and an expected tetrahedral geometry based on the number of electron pairs present on the outermost shell of the central atom. However, the presence of a lone pair on the central atom causes the observed molecular geometry to be trigonal pyramidal. The actual bond angle is about 97.7° due to repulsive forces.

CH4 comprises of four electron domains all of which are bond pairs. This minimizes the repulsion of electron pairs on the valence shell of the central atom in the molecule. Hence the molecule has a bond angle of 109.5° which signifies a perfect tetrahedron.

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