sp3d2 Br hybridizes 4s, 4p and 4d If I'm not mistaken. Seems kind of strange for a hybridization since it involves two 4d orbital when you'd expect it to hybridize with only one 4d orbital, but that's what I found on a few websites.
The bromine atom in BrF5 has sp3d2 hybridization. This means that it has one 3s, three 3p, and two 3d orbitals that combine to form six equivalent sp3d2 hybrid orbitals.
fl / fl -- Br-- fl fl fl
I might not know the answer to my question but I do not think it is SP^3.
sp3d2, this accounts for all 6 electron domains found around the central Br atom. Verify by checking the Lewis dot configuration. Br has 4 bonded electron domains and 2 lone pairs. Therefore the hybridization of the central atom in BrF4 is sp3d2
The bond angles in BrF5 are approximately 90 degrees.
Out of SiCl4, BrF5, AsF5, BrF3, only SiCl4 has sp3 hybridization on the central atom, which is silicon. SiCl4 has four regions of electron density around the central silicon atom, leading to sp3 hybridization. The other compounds have different geometries and hybridizations: BrF5 and AsF5 have sp3d2 hybridization, while BrF3 has sp3d hybridization.
The bromine atom in BrF5 has sp3d2 hybridization. This means that it has one 3s, three 3p, and two 3d orbitals that combine to form six equivalent sp3d2 hybrid orbitals.
The ABE notation for BrF5 is AX5E. This indicates that BrF5 has a central atom (Br) bonded to five surrounding atoms (F) and one lone pair (E).
Molecular Geometry is T-shaped -F Br -F -F Two lone e- pairs to left of Br There's three bonds then on the left side of Br there are two lone electron pairs that repel against each other Hybridization at Central atom is sp^3d
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fl / fl -- Br-- fl fl fl
The smallest bond angle in BrF5 is approximately 90 degrees, which occurs between the axial Br-F bonds and the equatorial Br-F bonds.
In BRF5, there are 5 sigma bonds formed through sp3d hybridization by overlapping of orbitals. The fluorine atoms each donate one electron for bonding, while the bromine atom donates five electrons. This results in a trigonal bipyramidal molecular geometry around the central bromine atom.
I might not know the answer to my question but I do not think it is SP^3.
BrF5 is a polar molecule. The presence of the lone pair on the central atom (Br) and the arrangement of the surrounding fluorine atoms create an uneven distribution of charge, leading to a net dipole moment.
The approximate bond angles for BrF5 is approximately 90 degrees because there would be one lone pair of electrons left over, making the molecular shape square pyramidal... This gives an approximate bond angle of 90 degrees. AX5E, sp3d2 hybridized.
sp3d2, this accounts for all 6 electron domains found around the central Br atom. Verify by checking the Lewis dot configuration. Br has 4 bonded electron domains and 2 lone pairs. Therefore the hybridization of the central atom in BrF4 is sp3d2