12 right angles

two planes

The sulfur atom in the SF6 molecule has to form 6 bonds using all of its 6 3rd orbit electrons, so it promotes electrons to two d orbitals (so as to make them unpaired) and hybridizes one s, three p and two d orbitals to form 6 sp3d2 orbitals.

In chemistry, octahedral molecular geometry describes the shape of compounds with six atoms or groups of atoms or ligands symmetrically arranged around a central atom, defining the vertices of an octahedron. The octahedron has eight faces, hence the prefix octa.

Square pyramidal is a molecular shape that results when there are five bonds and one lone pair on the central atom in the molecule. This molecule is made up of six equally spaced sp3d2 (or d2sp3) hybrid orbitals arranged at 90° angles. The shape of the orbitals is octahedral.

Square pyramidal is a molecular shape that occurs when there are five bonds and one lone pair on the central atom in the molecule. The shape of iodine pentafluoride is AX5E because there are six regions of electron density, but one corresponds to a lone pair.

The geometry is common for certain main group compounds that have a stereochemically active lone pair, as described by VSEPR theory. Certain compounds crystallize in both the trigonal bipyramidal and the square pyramidal structures, notably [Ni(CN)5]3−….

Bromine has 35 electrons in atomic structure he can share his 5 electron with fluorine n makes brf5 fluorine has 7 electrone in his outer orbital so he can share 1 eleceone with bromine n complete his orbital by taking 1 electron from bromine but bromine is more powerful than fluorine so he can not gain his electron …

Transcript: This is the BrF5 Lewis structure. For BrF5, we have a total of 42 valence electrons. Bromine is the least electronegative, we’ll put that in the center and then we’ll put 5 Fluorines around the outside. We’ll draw single bonds between the atoms for a total of 5 single bonds, so 10 valence electrons.

So, there are no 90-degree F−Br−F angles in BrF5​. Was this answer helpful?

Hybridization of BrF5 (Bromine Pentafluoride)

View Live. Chlorine trifluoride has 5 regions of electron density around the central chlorine atom (3 bonds and 2 lone pairs). These are arranged in a trigonal bipyramidal shape with a 175° F(axial)-Cl-F(axial) bond angle. The two lone pairs take equatorial positions because they demand more space than the bonds.

The geometry of the BF 3 molecule is called trigonal planar (see Figure 5). The fluorine atoms are positioned at the vertices of an equilateral triangle.

Xenon trioxide

xeo3 is an unstable compound of xenon in its +6 oxidation state . It is very powerful oxidising agent and liberates oxygen from water slowly (and xenon),accelerated by exposure to sunlight. It ia dangerously explosive upon contact with organic material.

The central Xe atom in XeO3 has three bonding domains and one lone pair of electrons. Hence, the electron geometry is tetrahedral and molecular geometry is pyramidal.

. This anion has octahedral molecular geometry, as determined by Raman spectroscopy, having O–Xe–O bond angles varying between 87° and 93°. The Xe–O bond length was determined by X-ray crystallography to be 1.875 Å.

So XeO3 is sp3 hybridized.

To form these hybrid orbitals one s and three p orbitals will be used. So, hybridization of this XeO3 molecule is sp3. So the correct answer is option B that is sp3.

XeO3 – Xenon Trioxide: Electron geometry: tetrahedral. Hybridization: sp3. Then draw the 3D molecular structure using VSEPR rules: Decision: The molecular geometry of XeO3 is trigonal pyramidal with asymmetric charge distribution on the central atom.

XeF6 i.e xenon hexafluoride is one of the three fluorides formed by xenon. According to VSEPR theory this molecule has seven electron pairs ( 6 bonding pairs and 1 lone pair) and thus,it has distorted octahedral structure. The hybridisation in this molecule is sp3d3.

Here’s a shortcut for how to determine the hybridization of an atom in a molecule. Count the number of atoms connected to it (atoms – not bonds!) Count the number of lone pairs attached to it. Add these two numbers together.

The central atom has a hybridization of sp3d3. Thus, its structure should be pentagonal bipyramidal.

There are six fluorine atoms in xenon hexafluoride, six covalent bonds are formed. – Due to the presence of one lone pair of electrons, the will be lone-pair –bond-pair repulsion in the molecule and the geometry will be distorted octahedral.

Xenon hexafluoride is a noble gas compound with the formula XeF6. It is one of the three binary fluorides of xenon, the other two being XeF2 and XeF4. All known are exergonic and stable at normal temperatures. XeF6 is the strongest fluorinating agent of the series.

Octahedral

The Jahn-Teller effect is a geometric distortion of a non-linear molecular system that reduces its symmetry and energy. This distortion is typically observed among octahedral complexes where the two axial bonds can be shorter or longer than those of the equatorial bonds.