Exercise 2 solution

The steps are the ones in the chapter: count valence electrons, join the atoms, put
the rest on as lone pairs, then check every shell. Applying them to these molecules
is my own work, not copied from a page (check).

HCl
1. Valence electrons (group rule): hydrogen group 1 -> 1; chlorine group 17 -> 7.
   Total 1 + 7 = 8.
2. One bond joins the atoms: 2 electrons used.
3. 8 - 2 = 6 electrons left = 3 lone pairs, all on chlorine (hydrogen is full with 2).
4. Check: hydrogen shares 1 pair = 2 electrons. Chlorine has 1 bond pair (2) plus 3
   lone pairs (6) = 8. Electrons drawn = 8, which matches the count.

      ··
  H — Cl ··
      ··

(three lone pairs on Cl; the drawing puts one above, one to the right, one below)

CH4
1. Valence electrons: carbon group 14 -> units digit 4; four hydrogens at 1 each = 4.
   Total 4 + 4 = 8.
2. Carbon in the middle with a single bond to each of four hydrogens:
   4 bonds x 2 = 8 electrons used.
3. 8 - 8 = 0 left, so no lone pairs.
4. Check: carbon has 4 bond pairs = 8 electrons; each hydrogen has 2.

        H
        |
    H — C — H
        |
        H

The flat drawing is not the real shape. The Methane article describes the molecule as
tetrahedral with four equivalent C-H bonds, so the four hydrogens are not in one plane.