The hybridization of the lead atom in PbCl4 is...?
sp3
I would think sp3
To determine the hybridization of the lead atom in PbCl4, we need to follow a few steps:
Step 1: Determine the total number of valence electrons for lead (Pb) and chlorine (Cl) atoms.
Pb: Atomic number of Pb = 82
Number of valence electrons for Pb = Group number of Pb = 14
Cl: Atomic number of Cl = 17
Number of valence electrons for Cl = Group number of Cl = 7
Step 2: Calculate the total number of valence electrons for the molecule PbCl4.
Total valence electrons = Valence electrons of Pb + Valence electrons of Cl x number of Cl atoms
Total valence electrons = 14 + (7 x 4) = 42
Step 3: Determine the electron pair geometry and molecular geometry of the molecule.
PbCl4 has a tetrahedral electron pair geometry and a tetrahedral molecular geometry.
Step 4: Apply the hybridization formula.
Hybridization = (Number of sigma bonds) + (Number of lone pairs)
For a tetrahedral geometry, there are 4 sigma bonds and 0 lone pairs.
Hybridization = 4 + 0 = 4
The hybridization of the lead atom in PbCl4 is sp3.
To determine the hybridization of the lead (Pb) atom in PbCl4, we need to follow a few steps.
Step 1: Determine the number of valence electrons for the lead atom (Pb).
Lead (Pb) is an element from Group 14 of the periodic table, so it has four valence electrons.
Step 2: Calculate the total number of sigma bonds formed by the lead atom.
In PbCl4, there are four chlorine (Cl) atoms bonded to the lead atom. Each bond between Pb and Cl involves sharing two electrons, making it a sigma bond. Therefore, Pb forms four sigma bonds.
Step 3: Determine the number of lone pairs on the lead atom.
In PbCl4, the lead atom does not have any lone pairs.
Step 4: Use the formula: hybridization = (number of sigma bonds) + (number of lone pairs)
In the case of PbCl4, the hybridization would be equal to the number of sigma bonds because there are no lone pairs on the lead atom.
Therefore, the hybridization of the lead (Pb) atom in PbCl4 is sp³, as it forms four sigma bonds.