How do you tell which end of a polar bond is positive or negative? My book says that a molecule such as Hf has a center of positive charge a center of negative charge. But in S--O, how do you tell which end is positive, and which is negative.

O attracts electrons from S. Why? which is more electronegative?

You can look up a electronegativity chart on the internet somewhere, should be easy to find. If you have a chemistry book, should be in there as well. O is a more electronegative element than S, so it will have the negative pole.

To determine which end of a polar bond is positive or negative, you need to consider the electronegativity values of the atoms involved. Electronegativity is the ability of an atom to attract electrons towards itself in a covalent bond.

In a polar covalent bond, the electrons are not shared equally between the atoms, leading to a partial positive charge on one end (atom) and a partial negative charge on the other end (atom).

To determine which end of a polar bond is positive or negative:
1. Identify the atoms involved in the bond.
2. Look up the electronegativity values of these atoms on the periodic table.
3. The atom with a higher electronegativity value will be more likely to attract electrons towards itself, developing a partial negative charge.
4. The atom with a lower electronegativity value will have a partial positive charge.

In the examples you mentioned:

1. Hf: Hafnium (Hf) is less electronegative than fluorine (F) on the periodic table. Therefore, in the Hf-F bond, the hafnium (Hf) end will have a partial positive charge, and the fluorine (F) end will have a partial negative charge.

2. S--O: In the oxygen-sulfur bond (S--O), oxygen (O) is more electronegative than sulfur (S). As a result, oxygen will have a partial negative charge, and sulfur will have a partial positive charge.

It's important to note that the direction of the arrow (-->) in a chemical structure often indicates the more electronegative end. So, in S--O, the arrow will point towards the oxygen atom (O), indicating its higher electronegativity and partial negative charge.