Question

The shapes of the compounds XeF2, XeF4, XeO3 and XeO4 respectively are:

• (A) Distorted octahedral, Tetrahedral, Square planar, and Trigonal bipyramidal
• (B) See-saw, octahedral, Tetrahedral and Square planar
• (C) Linear, Square planar, Trigonal bipyramidal, and Tetrahedral
• (D) Square planar, octahedral, Pyramidal and See-saw

Answer 1

The Correct Option is C

Explanation:
According to the VSEPR theory, The shape of the molecule is determined by both the total number of electron pairs (bonding and non-bonding) around the molecule's central atom and the orientation of these electron pairs in the space around the central atom.In order to minimize the repulsion forces between them, electron pairs around the molecule's central atom, tend to stay as far away from each other as possible.Xe has electronic configuration 5s² 5p⁶.XeF molecular geometry is linear. It acquires such shape as the lone pairs present around the central atom tend to take up equatorial positions. The bond angle is said to be 180°. The hybridization of XeF is sp³d.XeF4 consists of two lone pair electrons. Now if we follow the VSEPR theory, the net electronic repulsions has to be minimum. With this, they will acquire a stable state. In order to achieve this, the lone pairs lie in a perpendicular plane in an octahedral arrangement opposite (180°) from each other. Therefore, XeF4 molecular geometry is square planar. The hybridization is sp³d².This means there are a total of three bond pairs and one lone pair. Therefore there will be three hybrid orbitals (three for bond pair and one for lone pair). To form these hybrid orbitals one s and three p orbitals will be used. So, the hybridization of this XeO molecule is sp³. Geometry is trigonal bipyramidal. XeO4 has geometry due to sp³ hybridization. The shape is Tetrahedral. Hence, the correct option is (C).