XeO3 is sp3 since Xe has three bonds to oxygens and one long pair of electrons. These 4 together conclude a tetrahedral shape and hybridize as sp3 where 1 of the four electrons reside in the s and each p orbital. so, s gets one leaving 3 for the p orbital, thus sp3.
XeO4 is kind of the same except it is bonded to 4 Oxygens and those 4 bonds spread out into the s and p orbitals in such that one is in the s and 3 in the p orbitals. together they make sp3.
Now why are they both the same is because long pair electrons in this case count as a bond so the XeO3 that had one long pair electron in fact is like if it had 4 bonds.
I hope this enlightened you and helped.
The hybridization of XeO4 is sp3. This means that xenon is surrounded by four electron pairs, giving it a tetrahedral geometry with bond angles of approximately 109.5 degrees.
The hybridization of NCl3 is sp3.
The XeO4 molecule uses sp3 hybrid orbitals from xenon for sigma-bonding. This allows xenon to form four sigma bonds, each with one oxygen atom in XeO4.
The hybridization of PH3 is sp3, as the phosphorus atom is bonded to three hydrogen atoms and has one lone pair of electrons in the valence shell. This results in four regions of electron density, leading to sp3 hybridization.
The hybridization of CH3 is sp3. Each carbon atom forms four sigma bonds with hydrogen atoms, resulting in a tetrahedral geometry and sp3 hybridization.
The hybridization of XeO4 is sp3. This means that xenon is surrounded by four electron pairs, giving it a tetrahedral geometry with bond angles of approximately 109.5 degrees.
The hybridization of NCl3 is sp3.
sp3
sp3
The XeO4 molecule uses sp3 hybrid orbitals from xenon for sigma-bonding. This allows xenon to form four sigma bonds, each with one oxygen atom in XeO4.
The hybridization of PH3 is sp3, as the phosphorus atom is bonded to three hydrogen atoms and has one lone pair of electrons in the valence shell. This results in four regions of electron density, leading to sp3 hybridization.
The hybridization of CH3 is sp3. Each carbon atom forms four sigma bonds with hydrogen atoms, resulting in a tetrahedral geometry and sp3 hybridization.
The hybridization of NF3 is sp3. This means that the nitrogen atom in NF3 forms four equivalent sp3 hybrid orbitals when it bonds with the three fluorine atoms.
The carbon atom in CF4 has a hybridization of sp3.
The hybridization of the central atom in NCl3 is sp3.
The carbon in CH3CHCH2 has sp3 hybridization. Each carbon atom forms four sigma bonds, leading to the tetrahedral geometry characteristic of sp3 hybridization.
in XeO3 ,Xe shows sp3 but shape is pyramidal because of the presence of a lone pair of electrons on the central xenon atom. This lone pair distorts the shape of the molecule making it pyramidal.