Of course, there are, some exceptions: very small atoms H, Be and B have less than an octet, and some main group atoms in the third period and below P, S, Cl, Br, and I may have more than an octet but most elements still strive for the completion of their outer valence shell with 8 electrons. Drawing correct Lewis structures takes practice but the process can be simplified by following a series of steps:. Step 1. Count all the valence electrons for each atom. Add or subtract electrons if the structure is an anion or cation, respectively.
If we add all the electrons together we get 32 valence electrons with which to make bonds and lone pairs around the atoms in the ion. Step 2. Determine which atoms are bonded to one another. Draw a skeletal structure.
A rule of thumb is that the central atom atom which all the other atoms will be bound is the one furthest to the left or bottom in the periodic table. Sulfur and oxygen are in the same group but sulfur is below oxygen so it will most likely be the central atom in the structure:. How do we know that all of the oxygen atoms are bound to the sulfur and none to each other? Well, oxygen only needs to share two electrons to complete its octet so it really does not want or need to make more than one bond.
Also, the structure shown is the most symmetrical way to build the molecule and with a few rare exceptions, nature tends to make molecules in a symmetrical manner. Step 3. Connect the atoms with a pair of electrons in each bond. Subtract the number of bonding electrons from the total number of valence electrons.
Each line in the drawing above represents the use of 2 electrons. Knowing we started with 32 and have now used 8 leaves us with 24 electrons to distribute. Step 4. Add electron pairs to complete octets for all peripheral atoms attached to the central atom. While making bonds electrons were shared only. The ion is not one S atom and 4 O atoms together.
It is not as though there was one sulphur and four neutral oxygen atoms, and they became negatively charged during bond formation. The bond must have formed in some reaction, and in that reaction the charge would have been conserved.
Tho sulfur and oxygen are not neutral atoms. It slightly helps to think about it as being bonded ionically, even though the bond nature is rather covalent. Together it all ends up being Oxygen is electro-negative. It has a partial negative charge. With oxygen pulling the electrons toward itself, the sulfur gets a partial positive charge. Sulfur can have either a partial positive charge or a partial negative charge.
We would need 32 valence electrons for 4 full outer shells 8x4 but only have Thus the charge is 2- for the ion. Sign up to join this community.
The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group. That means we will have to add 2 electrons into our Lewis dot structure. So we have a sulfur atom with 4 oxygen atoms surrounding it.
Oxygen is in period 2 and sulfur is in period 3. Sulfur is directly below oxygen on the periodic table and they are both in group 6A. Both sulfur and oxygen have 6 valence electrons. We begin drawing the Lewis dot structure by drawing the 6 valence electrons around each sulfur and oxygen atom. In this step we added an extra electron to each of the 2 oxygen atoms as illustrated.
Now all of the electrons are accounted for and 2 of the oxygen atoms are electro-negative. In this step we drew a single bond between each atom. The oxygen atoms on the left and right each have 8 valence electrons and are happy but rather than forming a double bond, each of them have acquired an extra electron which will satisfy the octet rule but will make the atom electro-negative. This explains the negative 2 charge of the sulfate ion. The top and bottom oxygen molecule each have an unpaired electron which is available to join in a bond with sulfur.
Here we draw a bond between the unpaired electrons. If we count the electrons, each of the oxygen atoms have 8 electrons in their valence shell so the octet rule is satisfied.
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