6.2 Covalent Bonding form molecules

Nonmetals make a compound through covalent bond (sharing e-)

Covalent Bonds

Sharing e-

  • H has 1 e- wants to have 2 e- like He  2 H atoms share e-

Hydrogen gas: H-Hcompleted 2 e- w/ single bond (share 1 pair of e-)

  • All halogen atoms have valence 7 e-. So, they can share 1 pair of e- to achieve “stable 8”.

Clorine gasCl-Clcompleted 8 e- w/ single bond (share 1 pair of e-)

  • Oxygen atom has 6 valence e-.  2 O atoms need to share 2 pairs of e-.

O=O

oxygen gasO=Ocompleted 8 e- w/ double bond

(share 2 pair of e-)

  • N has only 5 covalent e- so, 2 N atoms share 3 pairs of e-triple bond

N N

Molecules of elements

  • Molecule = group of atoms joined by one or more covalent bonds
  • Diatomic molecules – 2 atoms joined together – H2, O2, N2, Cl2, Br2, I2, F2…(Halogen gases have single bonds, (remember Mr. Rendell’s neighbor “BrINClHOF”?)
  • Attractions between shared e- and p+ hold the molecules together.

Polar covalent bond -unequal sharing of e (cf. not ionic bond – no transfer of e-)

  • e-‘s not shared equally -b/c of electronegativity - atoms w/ partial (+) charge and atoms w/ partial (-) charges
  • Types of atoms + molecule shape determine polar or nonpolar bond.

Polar covalent bond (ex.H2O) / Non-polar covalent bond (ex. CO2)

  • forces of attraction for shared e- are not equal between O and H
  • O attract e- stronger than H  O becomes a little negative; H becomes a little positive.
  • So O side  (-) pole; H side (+) pole (So one end of the molecule has (+) charge; the other end of the molecule has (-) charge)
  • Shape - bended
  • (+)side of one molecule is attracted to (-) side of another molecule  special arrangement

  • Attractions among molecules stronger
/ (CO2)
  • O pulls harder than C but both sides pull w/ same force so net force.


  • So no special way of arranging molecules
  • Weaker attractions among molecules than polar bond molecules
  • Shape -straight

Exercices for molecule structures (Lewis dot diagrams)-

  1. Know the valence e- for each element (from the periodic table) – Know # of bond each element needs (SEE THE TABLE AT THE BOTTOM OF THIS PAGE)
  2. Know the total of all valence e-.
  3. Find the symmetrical structure.
  4. Make sure all e- (total #) are there and each element has 8 (2 for H and He) after sharing.

SF2 6+7*2=
F S F / F27+7
F F / O2 6+6
O O
CO 4+6
C O / HBr 1+7
H Br / N2 5+5
N N
SiBr4 4+7*4=
Br
Br Si Br
Br / CH4 4+1*4=
H
H C H
H / CS2 4+2*6
S C S
NCl3 5+3*7
Cl
Cl N Cl / SO2 6+2*6
O S O / CCl4 4+4*7
Cl
Cl C Cl
Cl

(yellow elements – nonmetals)

I (only H) / IV (C) / V (N – P) / VI / VII
# val e- / 1 / 4 / 5 / 6 / 7
#bond needed / 1 / 4 / 3 / 2 / 1