Electrons in Lewis Structures

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Mary Gallo 1G
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Electrons in Lewis Structures

Postby Mary Gallo 1G » Wed Nov 04, 2020 10:10 am

When drawing Lewis Structures, what is a good rule of thumb for determining the amount of electrons to use for each element?

Cecilia Cisneros 1F
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Re: Electrons in Lewis Structures

Postby Cecilia Cisneros 1F » Wed Nov 04, 2020 10:17 am

Looking at the row in the periodic table always helps me. When you count the rows [1,2,13,14,15,16,17,18] when you omit the d block. You can see that the number of electrons correspond to the row because there are 8 rows. Hope that makes sense!

Sabina House 2A
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Re: Electrons in Lewis Structures

Postby Sabina House 2A » Wed Nov 04, 2020 10:19 am

To determine the valence electrons for each element, you can look at the group number of that element. For group 1 and 2, they each have 1 and 2 valence electrons respectively, meaning they have the same amount of valence electrons as their group number. For groups 13-18, you have to subtract 10 from their group number to determine the amount of valence electrons. One method I use to determine how many valence electrons an atom has is by counting groups starting at group 1, then counting over groups and skipping over the middle d-block with 10 elements in each row. So, group 14 elements would have 4 valence electrons. Once you determine the number of valence electrons, that would be what you draw on the Lewis structures.

Mary Gallo 1G
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Re: Electrons in Lewis Structures

Postby Mary Gallo 1G » Wed Nov 04, 2020 10:23 am

Thank you both so much for the help! :)

Becca Nelson 3F
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Re: Electrons in Lewis Structures

Postby Becca Nelson 3F » Wed Nov 04, 2020 10:24 am

It helps to look at the groups! For groups 1 and 2 it is pretty simple, but for groups 13 through 18, 10 needs to be subtracted to determine the number of valence electrons. Remember that the periodic table is grouped by quantum numbers and is why 10 needs to be subtracted from the group.

Ava Nickman
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Re: Electrons in Lewis Structures

Postby Ava Nickman » Wed Nov 04, 2020 10:29 am

Like they said above the number valence electrons in each element is the same as the group number up until 13-18. Then you must subtract 10 from the group number. Next, you add up the amount of electrons to get the total number. Don't forget to multiply by the number of each element if more than one is present. For example H20 has two H and one O so the total would be 2(1)+6=8e

LeanneBagood_2F
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Re: Electrons in Lewis Structures

Postby LeanneBagood_2F » Wed Nov 04, 2020 7:35 pm

Cecilia Cisneros 1J wrote:Looking at the row in the periodic table always helps me. When you count the rows [1,2,13,14,15,16,17,18] when you omit the d block. You can see that the number of electrons correspond to the row because there are 8 rows. Hope that makes sense!


Quick question, can you explain why you omit the d-block please?

Cecilia Cisneros 1F
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Re: Electrons in Lewis Structures

Postby Cecilia Cisneros 1F » Wed Nov 04, 2020 9:00 pm

LeanneBagood_2E wrote:
Cecilia Cisneros 1J wrote:Looking at the row in the periodic table always helps me. When you count the rows [1,2,13,14,15,16,17,18] when you omit the d block. You can see that the number of electrons correspond to the row because there are 8 rows. Hope that makes sense!


Quick question, can you explain why you omit the d-block please?


Omitting the d-block just allows you to count 1-8 valence electrons in groups 1,2,13-18. It is the same as skipping over the d block like others previously mentioned. This allows the atoms to form an octet. However, if that is confusing in any way you can always just subtract 10 from the groups 13-18 in order to get the number of valence electrons.

Sydney Jensen 3L
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Re: Electrons in Lewis Structures

Postby Sydney Jensen 3L » Thu Nov 05, 2020 8:44 am

When determining the number of electrons, look at which family they are at on the periodic table, and the second digits corresponds to the amount of valence electrons they have to offer, which will come in handy when drawing these structures and knowing both the amount of bonds necessary, but also the amount of leftover electrons that need to be included.


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