What is the geometry of XeF4?
1)tetrahedral
2)square planar
3)octahedral
4)seesaw
5)bent
The electronic geometry is octahedral; the molecular geometry is square planar.
tetrahedral
To determine the geometry of XeF4, we first need to draw its Lewis structure:
1. Start by writing the chemical symbol for xenon (Xe).
2. Determine the total number of valence electrons for XeF4. Xenon (Xe) is in Group 8A, so it has 8 valence electrons. Fluorine (F) is in Group 7A and has 7 valence electrons. Multiply the number of fluorine atoms (4) by 7 to get 28 valence electrons. Add these two numbers together to get a total of 36 valence electrons.
3. Place the xenon atom in the center and arrange the four fluorine atoms around it.
4. Connect each fluorine atom to the xenon atom with a single bond, using two electrons for each bond.
5. Distribute the remaining valence electrons around the atoms to satisfy the octet rule. Since xenon can expand its octet, place the remaining 28 electrons as three lone pairs and two lone electrons on the xenon atom.
After drawing the Lewis structure for XeF4, we can determine its geometry using the VSEPR (Valence Shell Electron Pair Repulsion) theory, which states that electron pairs around an atom tend to repel each other and adopt a geometry that minimizes this repulsion. The lone pairs and bonded pairs of electrons are considered in the analysis.
In the case of XeF4, the central xenon atom has 4 bonding pairs and 2 lone pairs of electrons. The arrangement of these electron pairs is such that it forms an octahedral geometry, making option 3) octahedral the correct answer.