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Posted: Tue Nov 13, 2018 1:13 pm
How should we know which ones are exceptions?
Posted: Tue Nov 13, 2018 2:14 pm
some exceptions to the octet rule can be found in the 3rd period in the p block because they have an unoccupied d block
Posted: Wed Nov 14, 2018 12:29 pm
So just building onto that, the most common exceptions that we deal with are found within the 3rd period, such as Sulfur or Phosphorus, where they violate the octet rule due to having that d block available to bind to more electrons. Additionally, another exception would be for elements like Boron and Aluminum where they do not require an octet within compounds. This is because they have 3 valence electrons which causes them to only have a total of 6 when bonded but they can receive more electrons when bonded with halogens.
Posted: Fri Nov 16, 2018 5:35 pm
In addition to this Xenon also has an expanded octet
Posted: Thu Dec 06, 2018 6:07 pm
jane_ni_3b wrote:How should we know which ones are exceptions?
The most common exceptions we deal with are found in the 3rd period because of an unfilled d-orbital.
Posted: Fri Dec 07, 2018 8:25 pm
elements in the d block are exceptions as they are able to fill their d orbital with more electrons, allowing for an expanded octet
Posted: Sat Dec 08, 2018 10:02 am
Many elements can have expanded orbitals, keep that in mind when creating your lewis structures.
Posted: Sat Dec 08, 2018 10:28 am
Exceptions being in the 3rd period since they can expand into their d orbital
Posted: Sun Oct 27, 2019 1:00 am
If the element has an empty d orbital, then it can expand, which causes it to have more than 8 electrons in the valence shell. These elements can be found in the 3rd period of the p block
Posted: Sun Oct 27, 2019 8:33 pm
Not only are some elements in the d block exceptions to the octet rule, so are the first four elements (hydrogen, Beryllium, Helium, and Lithium.
Posted: Sun Oct 27, 2019 9:02 pm
Going off of that, I know 4 specific chemical structures that have the octet rule (numbers 1/2 are electron deficient exceptions with less than 8 electrons and 3/4 have expanded octets with more than 8 electrons)
N has 5 electrons, O has 6 electrons which totals in 11 electrons. Usually every lewis structure has an even number of electrons but this one doesn't. When writing this structure, there is a double bond between the two of them and the extra electrons around the rest (4 around oxygen and 3 around nitrogen)
B has 3 and H3 has 3 which totals in 6 electrons. Boron would be the central element with the Hydrogens surrounding them but due to the limited number of electrons, there is an electron deficiency and Boron cannot have a completed octet
S has 6 and F has 42 which totals to 48 electrons. S is the central element with 6 F surrounding it. S can have more than 8 electrons so this makes it an expanded octet with each fluorine having an expanded octet
p has 5 and Cl has 35 electrons which totals in 40 electrons. P is the central element with the 5 Cl surrounding it. Each Cl has a complete octet and P has an expanded octet with more than 8 electrons
Hope this helps!