The first thing we need to do when drawing a Lewis structure is determine the total number of valence electrons in the molecule. Remember, valence electrons are those in the outermost principal energy level. For example: Na – 1s22s22p63s1, Cl – 1s22s22p63s23p5
The number of valence electrons, for main group elements, corresponds to their group number in the periodic table:
For d block elements, the outermost d electrons are also counted as valence electrons (ns + (n-1)d). For example, iron has eight valence electrons: Fe – 1s22s22p63s23p64s23d6.
So, oxygen is in group 6A, and therefore, it has 6 valence electrons, thus O3 has 3 x 6 = 18 valence electrons.
Next, we need to connect the atoms in the correct order and add the electrons as bonds and lone pairs.
In short, these are the steps you need to follow for drawing a Lewis structure:
1. Write the correct skeletal structure for the molecule.
* Hydrogen atoms are always terminal (only one bond)
* Put more electronegative elements in terminal positions
2. Sum the valence electrons from all the atoms.
3. Use a pair of electrons to form a bond between each pair of bound atoms.
4. Add the remaining electrons to satisfy the octet for a more electronegative atom first.
5. If any atoms lack an octet, make a double or triple bond to give them an octet.
Sulfur is the central atom, so we can draw the skeletal structure:
There are 6 + 6 + 2×7 = 26 electrons, and 6 are used to make 3 bonds. Two chlorines and the oxygen take 3 lone pairs, and the remaining one goes to the sulfur:
All the atoms have an octet, however, there are three lone pairs on the oxygen which gives it a -1 formal charge. Therefore, one of the lone pairs on the oxygen is used to make another bond with the sulfur, and even if it exceeds the octet, that is acceptable for elements in the third row and below because they have d orbitals.
The central atom has 3 atoms and a lone pair (SN = 4). Therefore, the electron geometry is tetrahedral while the molecular geometry is trigonal pyramidal:
Steric number 4 indicates sp3-hybridization with idealized angles of 109.5o. However, because the groups on the central atom are not identical, the angles are not going to be exactly 109.5o.
Now, the polarity: The first thing here is to determine if the S-O and S-Cl bonds are polar. Depending on the difference in the electronegativity values, covalent bonds can be polar and nonpolar.
- If the difference in electronegativity is less than 0.5, the electrons are about equally shared between the two atoms, forming a nonpolar a covalent bond.
- If the difference in electronegativity is between 0.5 and 1.7, we have a polar covalent bond.
- A difference of 1.7 or higher is so large that the electrons are no longer shared, and an ionic bond is formed. Ionic bonds are formed between metals and nonmetals.
Both S-O and S-Cl bonds are polar, and therefore, whether the molecule is polar or not depends on the orientation of the three bonds.
Remember, the molecule is polar if it has a dipole moment, and the molecular dipole is the vector sum of all the dipoles. All the dipoles are pointing down to the sides as drawn, and therefore, the molecular dipole will be pointing down perhaps a bit more towards the left because the O-S bond is more polar thus the dipole is greater:
Check this 99-question multiple-choice quiz on Geometry and Hybridization:
- The VSEPR Model
- VSEPR Theory Practice Problems
- Hybridization of Atomic Orbitals
- sp, sp2, sp3, sp3d, and sp3d2 Hybridization Practice Problems