The hybridization state of Al in AlH4- is sp3, as it has four electron groups around the central aluminum atom. This leads to the formation of four sigma bonds, resulting in tetrahedral geometry.
The hybridization state of each carbon atom in nemotin is sp3.
Prsumably you mean AlH4- the tetrahydroaluminate anion. The hybridization of the central atom of AlH4 is sp3.
The hybridization state for SiH3 is sp3, which means that the silicon atom is bonded to three hydrogen atoms using four sp3 hybridized orbitals.
The hybridization state of Se in SeCl2 is sp^3 because it has two bonding pairs and two lone pairs around the selenium atom, leading to a tetrahedral electron geometry.
The hybridization state of SiBr4 is sp3 (tetrahedral). Silicon has 4 valence electrons, and in SiBr4, these electrons form 4 sigma bonds with the bromine atoms, resulting in a tetrahedral geometry.
ALH4 doesn't exist; AlH3 is the chemical formula of aluminium hydride.
The hybridization state of each carbon atom in nemotin is sp3.
Prsumably you mean AlH4- the tetrahydroaluminate anion. The hybridization of the central atom of AlH4 is sp3.
The hybridization state for SiH3 is sp3, which means that the silicon atom is bonded to three hydrogen atoms using four sp3 hybridized orbitals.
The hybridization state of Se in SeCl2 is sp^3 because it has two bonding pairs and two lone pairs around the selenium atom, leading to a tetrahedral electron geometry.
An aluminohydride is the univalent anion, AlH4-, present in such compounds as lithium aluminium hydride.
The hybridization state of SiBr4 is sp3 (tetrahedral). Silicon has 4 valence electrons, and in SiBr4, these electrons form 4 sigma bonds with the bromine atoms, resulting in a tetrahedral geometry.
The hybridization of NCl3 is sp3.
The central atom of ammonia is nitrogen and it has 3 bonding pairs and a lone pair around, hence it undergoes sp3 hybridization. The central atom of boron trifluoride is the boron atom, and around it has only three bonding pairs. So it hybridizes as sp2.
The hybridization of Be in BeH2 is sp hybridization. Beryllium has 2 valence electrons and forms 2 bonds with the two hydrogen atoms in BeH2, resulting in sp hybridization.
The hybridization of the carbon atoms in an alkyne is sp.
The silicon atom in SiBr4 has a hybridization state of sp3, forming four sigma bonds with the four bromine atoms. Each bond is formed by overlap between an sp3 hybrid orbital on the silicon atom and a p orbital on each bromine atom.