XeF4
No it is not a tetrahedral!
Remember, the geometry of a molecule has to do with how many electron pairs AND lone pairs there are around the CENTRAL atom. Basically, count up the lines and the pairs of dots. Notice that 'Xe' has 8 electrons in its valence shell, and 'F' has only 7. When we put 4 'F' atoms around 1 'Xe' atom, each Flourine atom is going to want to 'share' one of Xenon's electrons so that it can have 8 electrons. Also remember that when an atom shares electrons, it shares in pairs. Each shared bond has 2 electrons. Now count up the number of electron PAIRS around the CENTRAL atom. We have 4 bonds (one with each Fluorine atom), and 2 lone pairs of of electrons bringing our count up to 12 electrons around Xenon, or 6 pairs total. (This violates the 'Octet Rule', but it's ok in this situation) Here is a list of geometries according to the number of Electron Domains: 2 'ED's (electron domains) = Linear shape 3 'ED's = Trigonal Planar 4 'ED's = Tetrahedral 5 'ED's = Trigonal Bi-pyramid 6 'ED's = Octahedral The electron-domain geometry of XeF4 would be Octahedral. Hope this helped!
XeF4 Xe has 8 valance electrons. F has 7 valance electrons * 4 = 28 valance electrons 8 + 28 = 36 valance electron total. Now, there are 4 bonds between Xe and the 4 F's, so that is a total of 8 electrons shared. 36 - 8 = 28 valance electrons left over. That means that 6 each go around the fluorine atoms as three lone pair per atom and one electron for the exon atom, unless this is a charged molecule.
After taking a moment to appreciate seeing a compound with a noble gas in it, for once, we then start to consider some facts.Xe has a valence of 8. Fluorine has a valence of 7.Fluorine really wants 1 electron each to fill that octet. So we can imagine 4 covalent bonds to Xe and Fluorine is nice and happy.So Xe has given away 4 of its electrons, it now has 4 left. That's two lone pairs.Fluorines are going to repel each other equally and so must be equally spaced apart if at all possible. The same goes for the lone pairs. Lone pairs will repel fluorine and vice versa, and all by the same degree.Sounds confusing but actually this one is really kind. We see a square planar molecule with two lone pairs, one above and one below the plane.Each Fluorine is equidistant from another fluorine. And those lone pairs are equidistant from each other and all of the fluorines.Practice this kind of thing, it'll soon become really easy.For further reading, I suggest trying to look up VESPR theory. Also getting a model kit can help a lot when thinking about geometries. Think about the repulsions involved and use your imagination. You'll start seeing patterns when you practice it, and soon it will become second nature.
The molecular shape of XeF4 is square planar. Xenon (Xe) has 4 bonding pairs and 2 lone pairs around it, resulting in a square planar geometry.
Square Planar, Like XeF4
XeF4 is isostructural with ICl4. Both compounds have a square planar molecular geometry with bond angles of 90 degrees and are considered isostructural due to their similar arrangement of atoms in the molecule.
XeF4 is square planar
90 degress
XeF4
XeF4 is a molecular compound. It is composed of xenon and fluorine atoms held together by covalent bonds.
XeF4 is a molecule with 6 electron pairs around the central Xenon atom. Based on VSEPR theory, this would result in an octahedral molecular geometry with 4 bonding pairs and 2 lone pairs.
The correct name for the compound XeF4 is xenon tetrafluoride.
square planar
No, XeF4 does not have geometric isomers because it has a linear molecular shape due to the presence of four bonding pairs of electrons and two lone pairs on the xenon atom. Geometric isomers arise from different arrangements of atoms around a double bond or an atom with a lone pair.
Yes, XeF4 is sp3d2 hybridized. This is because xenon in XeF4 has a total of six electron domains (four bonding pairs and two lone pairs), requiring it to adopt sp3d2 hybridization to accommodate these domains in its molecular structure.