How do you do the hybridization of CBr4?
To determine the hybridization of CBr4 (carbon tetrabromide), we need to follow a few steps:
Step 1: Determine the central atom.
In CBr4, the central atom is carbon (C).
Step 2: Count the total number of valence electrons.
Carbon (C) has four valence electrons, and each bromine (Br) atom has seven valence electrons. Since we have four bromine atoms, the total number of valence electrons is:
4 (C) + 4 (Br) = 32 valence electrons
Step 3: Determine the molecular geometry.
In CBr4, the carbon atom is the central atom bonded to four bromine atoms. Therefore, the molecular geometry is tetrahedral.
Step 4: Determine the hybridization.
For tetrahedral geometry, the hybridization of the central atom is sp3. The "s" orbital and all three "p" orbitals of carbon (C) hybridize to form four sp3 hybrid orbitals.
So, in conclusion, the hybridization of the carbon atom in CBr4 is sp3.
To determine the hybridization of CBr4, we need to first identify the central atom, which is carbon (C). The molecular formula CBr4 tells us that there are four bromine (Br) atoms attached to the central carbon atom.
Now, let's determine the hybridization of carbon using the following steps:
Step 1: Count the total number of valence electrons.
C (carbon) has 4 valence electrons, and each Br (bromine) atom has 7 valence electrons. So, the total number of valence electrons in CBr4 is 4 + (4 × 7) = 32.
Step 2: Calculate the number of electron pairs around the central atom.
Each covalent bond contributes one electron pair. Carbon forms four covalent bonds with the four bromine atoms, resulting in 4 electron pairs around the central carbon atom.
Step 3: Determine the steric number of the central atom.
The steric number is the sum of the number of sigma bonds and the number of lone pairs around the central atom. In this case, since carbon forms four bonds and has no lone pairs, the steric number is 4.
Step 4: Use the steric number to determine the hybridization.
In CBr4, the steric number of 4 indicates that the carbon atom undergoes sp3 hybridization. In sp3 hybridization, the 2s orbital and three 2p orbitals of the carbon atom mix to form four new hybrid orbitals.
So, the hybridization of the carbon atom in CBr4 is sp3.
C is 1s2 2s2 2p2
Take 1 s electron, promote it to the p level, then hybridize the 1s and 3p orbitals to make sp3 hybrid.