Apply Coulomb’s Law and electronic structure to compare
electronegativity values for two elements.
Explain why Fluorine (F) is more electronegative than Bromine
(Br)
Fluorine (F) has less shells than bromine (Br), and therefore its valence shell is closer to the nucleus than Br's. As a result it can excert a greater attractive force on additional electrons.
Classify a bond as ionic, non-polar covalent, polar covalent, or
metallic using periodic trends.
Classify each as non-polar covalent, polar covalent, ionic, or metallic: Mg-O, B-F, Na-Na, O-O?
Mg-O: Mg is a metal and O is a non-metal therefore it's best classified an ionic bond.
B-F: atoms are non-metals with different electronegativities therefore it's best classified as a polar covalent bond.
Na-Na: atoms are both metals therefore it's a metallic bond.
O-O: atoms are non-metals with equivalent electronegativities therefore it a non-polar covalent bond.
Assign partial charges to atoms in a bond based on
electronegativity.
Assign the partial charges to the polar covalent molecule carbon
monoxide (CO)
Recognize that ionic and covalent bonds exist on a continuum, with
polar covalent bonds in between.
Think of a covalent bond that has greater ionic character
than a N-O bond?
A greater difference in electronegativity (ΔEN) will result in greater ionic character.
Many possible correct answer but some include: C-O, N-F, C-F.
Describe the nature of bonding in ionic, covalent, metallic bonds.
Match the descriptors to ionic, covalent, or metallic bonding: electron sharing, electron donating, delocalized electrons, electron accepting, non-directional bonding.
Ionic bonding: electron donating, electron accepting.
Covalent bonding: electron sharing.
Metallic bonding:delocalized electrons, non-directional bonding.
