Draw a Lewis Structure and determine the electron domain geometry and molecular geometry of XeO4
The Correct Answer and Explanation is:
Lewis Structure of XeO₄ (Xenon Tetroxide):
To draw the Lewis structure for XeO₄, follow these steps:
- Count the valence electrons:
- Xenon (Xe) is in Group 18: 8 valence electrons
- Oxygen (O) is in Group 16: 6 valence electrons × 4 = 24
- Total = 8 + 24 = 32 valence electrons
- Draw the skeletal structure:
Place xenon in the center (least electronegative), and connect four oxygen atoms to it with single bonds. - Distribute remaining electrons:
- 4 single bonds use 8 electrons (4 × 2).
- 32 − 8 = 24 electrons left to place as lone pairs.
- Place 6 electrons (3 lone pairs) on each of the 4 oxygen atoms → 24 electrons.
- Check octet and formal charges:
- Each oxygen has 8 electrons (satisfied).
- Xenon currently has only 8 electrons; it can expand its octet (period 5 element).
- To reduce formal charges and be more stable, make double bonds with each oxygen:
- Now, each Xe=O bond is a double bond (4 × 2 = 8 bonds = 16 electrons around Xe)
Final Lewis Structure:
mathematicaCopyEdit O
||
O = Xe = O
||
O
Each O is double-bonded to Xe, and each O has 2 lone pairs.
Electron Domain Geometry:
- There are 4 bonding domains around xenon (4 double bonds = 4 regions of electron density).
- No lone pairs on Xe.
- Electron geometry = Tetrahedral
Molecular Geometry:
- Since there are no lone pairs on Xe, the molecular geometry is determined only by the positions of bonded atoms.
- Molecular geometry = Tetrahedral
Explanation
Xenon tetroxide (XeO₄) is a molecule in which xenon, a noble gas, forms stable bonds with four oxygen atoms. While noble gases are often considered inert, xenon can form compounds due to its ability to expand its octet and use d-orbitals in bonding, especially with highly electronegative atoms like oxygen.
To understand XeO₄, we begin by counting valence electrons. Xenon contributes 8 valence electrons, and each of the four oxygen atoms contributes 6, resulting in a total of 32 valence electrons. Drawing single bonds between xenon and each oxygen atom accounts for 8 electrons. The remaining 24 electrons are placed as lone pairs on oxygen atoms, ensuring each oxygen completes its octet.
However, this structure leads to a formal charge of +2 on xenon and −1 on each oxygen. To reduce formal charges, we form double bonds between xenon and each oxygen. This results in each oxygen having two lone pairs, and xenon having 16 electrons in its valence shell — a permissible situation due to its position in period 5 of the periodic table.
With four bonding pairs and no lone pairs on xenon, the molecule adopts a tetrahedral electron domain and molecular geometry. The bond angles are approximately 109.5°, typical of a tetrahedral arrangement. The structure is symmetrical, and the equal sharing of electrons (due to equivalent bonds) makes the molecule nonpolar despite the highly electronegative oxygen atoms.
Thus, the Lewis structure of XeO₄ shows xenon at the center with four double-bonded oxygens arranged in a tetrahedral shape. Its electron and molecular geometry are both tetrahedral, with xenon exhibiting an expanded octet — a clear example of how noble gases can form stable, hypervalent molecules under the right conditions.
