ratio of lone pair on surrounding atom to central atom in Xeo2F2?
To determine the ratio of lone pair on the surrounding atom to the central atom in XeO2F2, we need to analyze the Lewis structure of the molecule.
1. Start by determining the total number of valence electrons.
- Xenon (Xe) is in Group 8A and has 8 valence electrons.
- Oxygen (O) is in Group 6A and has 6 valence electrons.
- Fluorine (F) is in Group 7A and has 7 valence electrons.
XeO2F2 consists of one xenon atom, two oxygen atoms, and two fluorine atoms. So the total number of valence electrons is:
1 (Xe) + 2 (O) + 2 (F) = 1 + 2 + 2 = 5 + 2 = 7 + 2 = 9.
2. Assign lone pairs and bonds to the atoms.
- The central atom, Xenon (Xe), forms single bonds with both oxygen atoms and double bonds with both fluorine atoms.
- Each oxygen atom has 2 lone pairs of electrons.
- Each fluorine atom has 3 lone pairs of electrons.
3. Distribute the remaining electrons.
- In XeO2F2, all the remaining electrons are located on the central atom, Xenon (Xe).
- Xenon has 3 lone pairs of electrons.
Therefore, the ratio of lone pairs on the surrounding atoms (oxygen and fluorine) to the central atom (xenon) in XeO2F2 is:
Lone pair on surrounding atom: Lone pair on central atom
(2 + 2 * 3) : 3
(2 + 6) : 3
8 : 3
So the ratio is 8:3.