On this page, random topics can be posted which are small details we spent little time on in class, or some that are confusing. I'll post two here that definitely need to be done:

There will be Lewis structure questions on your term 3 test that do not show you the connectivity of the molecule (Unlike what I told you earlier). Here is the basic idea for this kind of structure.

Lewis structures show each atom and its position in the structure of the molecule using its chemical symbol. Lines are drawn between atoms that are bonded to one another (pairs of dots can be used instead of lines). Excess electrons that form lone pairs are represented as pairs of dots, and are placed next to the atoms.
Although many of the elements react by gaining, losing or sharing electrons until they have achieved a valence shell electron configuration with a full octet (8) of electrons, there are many noteworthy exceptions to the 'octet rule'. One example is hydrogen (H), which has only a single valence electron and tends to react to attain either 0 or 2 valence electrons. When H has two electrons in its valence shell, it could be said to obey a 'duet rule', and achieves a valence shell electron configuration equivalent to helium (He).
  • The formula of the nitrite ion is NO2−.
  • Step one: Make the least electronegative atom the central atom. Nitrogen is the least electronegative atom, so it is the central atom.
  • Step two: Count valence electrons. Nitrogen has 5 valence electrons; each oxygen has 6, for a total of (6 × 2) + 5 = 17. The ion has a charge of −1, which indicates an extra electron, so the total number of electrons is 18.
  • Step three: Place ion pairs. Each oxygen must be bonded to the nitrogen, which uses four electrons — two in each bond. The 14 remaining electrons should initially be placed as 7 lone pairs. Each oxygen may take a maximum of 3 lone pairs, giving each oxygen 8 electrons including the bonding pair. The seventh lone pair must be placed on the nitrogen atom.
  • Step four: Satisfy the octet rule. Both oxygen atoms currently have 8 electrons assigned to them. The nitrogen atom has only 6 electrons assigned to it. One of the lone pairs on an oxygen atom must form a double bond, but either atom will work equally well. *We therefore must have a resonance structure.
  • *Step five: Tie up loose ends. Two Lewis structures must be drawn: one with each oxygen atom double-bonded to the nitrogen atom. The second oxygen atom in each structure will be single-bonded to the nitrogen atom. Place brackets around each structure, and add the charge (−) to the upper right outside the brackets. Draw a double-headed arrow between the two resonance forms.
*(NOTE: If you come up with just one of these structures that is all you need for structures this chapter. This last step "resonance structures" is not necessary at this point in Chem 11)


-----> indicates an increase in the reactivity of elements as we move in this direction on the periodic table

Metals reactivity trend**
  • In groups, reactivity of metals increases with atomic number because the ionization energy decreases. ^
  • In periods, reactivity of metals decreases when atomic number increases because the ionization energy increases. <-----

Nonmetals reactivity trend
  • In groups, reactivity of non-metals decreases when atomic number increases because the electronegativity decreases. v
  • In periods, reactivity of non-metals increases with atomic number because the electronegativity increases. ----->