1. Show the e- dot structure for CuCl2, I2 and H2O. State the bond type and name. Show charges where necessary.
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To determine the electron dot structure for CuCl2, I2, and H2O, we need to follow a few steps.
1. CuCl2 (Copper (II) chloride):
- Determine the total number of valence electrons for copper (Cu) and chlorine (Cl).
Cu: Group 11, so 1 valence electron.
Cl: Group 17, so 7 valence electrons each.
Total = 1 (Cu) + 2 (2 x Cl) = 1 + 14 = 15.
- Place the least electronegative atom (Cu) in the center, surrounded by chlorine atoms (Cl).
- Connect the atoms with a single bond (one pair of electrons) between each Cl and the Cu atom.
- Assign lone pairs of electrons to each Cl atom to satisfy the octet rule. Cl needs only one additional electron to complete its octet.
- Since there are no remaining electrons, the electron dot structure for CuCl2 would look like this:
Cl: .: Cl :. -- Cu -- .: Cl :.
- Bond type: Covalent bond.
- Name: Copper (II) chloride.
2. I2 (Iodine):
- Determine the total number of valence electrons for iodine (I).
I: Group 17, so 7 valence electrons.
- Place the two iodine (I) atoms side by side.
- Connect the atoms with a sigma bond (single bond) using two electrons shared between the two atoms.
- Assign lone pairs of electrons to each iodine (I) atom to satisfy the octet rule.
- The electron dot structure for I2 would look like this:
I :. -- I :.
- Bond type: Covalent bond.
- Name: Iodine.
3. H2O (Water):
- Determine the total number of valence electrons for hydrogen (H) and oxygen (O).
H: Group 1, so 1 valence electron.
O: Group 16, so 6 valence electrons.
- Place the oxygen (O) atom in the center and connect it with single bonds (one pair of electrons) to two hydrogen (H) atoms.
- Assign lone pairs of electrons to the oxygen (O) atom to satisfy the octet rule. Oxygen needs two additional electrons to complete its octet.
- The electron dot structure for H2O would look like this:
H :. -- O -- .: H
- Bond type: Covalent bonds.
- Name: Water.