Use VSEPR theory to predict the bond angles of sulfur hexafluoride.
You can read more about it here but the bond angle is 90o (and 180o.
http://en.wikipedia.org/wiki/Molecular_geometry
To predict the bond angles of sulfur hexafluoride (SF6) using VSEPR (Valence Shell Electron Pair Repulsion) theory, we need to follow these steps:
Step 1: Determine the central atom
In SF6, sulfur (S) is the central atom as it is less electronegative than fluorine (F) and can form more bonds.
Step 2: Count the total number of valence electrons
Sulfur (S) is in Group 16 of the periodic table, so it has 6 valence electrons. Fluorine (F) is in Group 17, so each fluorine atom brings 7 valence electrons. Considering there are 6 fluorine atoms in SF6, the total number of valence electrons in SF6 is:
6 (from sulfur) + 6 × 7 (from fluorine) = 48
Step 3: Determine the electron geometry
To determine the electron geometry, we will use the VSEPR theory, which states that electron pairs—bonding pairs and lone pairs—around the central atom repel each other and try to maximize their distance. The number of electron pairs determines the electron geometry.
In SF6, sulfur has 6 bonding pairs (6 bonds) and no lone pairs, so the electron geometry is octahedral (a central atom surrounded by six bonding pairs).
Step 4: Determine the molecular geometry
The molecular geometry is based on the arrangement of only the bonding pairs (not the lone pairs). In the case of SF6, as all bonding pairs are bonding with fluorine, there is no repulsion between bonding pairs, and the molecular geometry is also octahedral.
Step 5: Determine the bond angles
In an octahedral molecular geometry, there are six identical bond angles formed between the central atom and the surrounding atoms. These angles are 90 degrees.
Therefore, the bond angles in sulfur hexafluoride (SF6) are all 90 degrees.