HW Question 2.43

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Ashley Davis 1I
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Joined: Fri Sep 29, 2017 7:04 am

HW Question 2.43

Postby Ashley Davis 1I » Tue Oct 24, 2017 6:41 pm

For anyone who has done question #2.43 on the homework: Why would tungsten's configuration be [Xe]4f^14 5d^4 6s^2 instead of [Xe]4f^14 5s^2 5p^6 5p^5 5d^4 6s^2?

Ashley Chipoletti 1I
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Joined: Fri Sep 29, 2017 7:04 am

Re: HW Question 2.43

Postby Ashley Chipoletti 1I » Tue Oct 24, 2017 7:23 pm

The [Xe] takes into account all of the electron configurations before it so you would not need to include 5s^2 5p^6 5p^5 for the configuration of tungsten. You only need to include the electrons after Xenon up to Tungsten in the electron configuration.

Rana YT 2L
Posts: 49
Joined: Thu Jul 27, 2017 3:01 am

Re: HW Question 2.43

Postby Rana YT 2L » Wed Oct 25, 2017 10:56 pm

The answer is correct because when you put [Xe] you are saying that the electron configuration already consists of the electron configuration for Xe

Essly Mendoza 1J
Posts: 25
Joined: Fri Sep 29, 2017 7:04 am

Re: HW Question 2.43

Postby Essly Mendoza 1J » Thu Oct 26, 2017 10:57 pm

I'm also confused on the electron configuration for Tungsten and [Xe] on 2.43, can someone clarify how the answer is [Xe] 4f^14 5d^4 6s^2.. thank you!

Jana Sun 1I
Posts: 52
Joined: Sat Jul 22, 2017 3:00 am

Re: HW Question 2.43

Postby Jana Sun 1I » Thu Oct 26, 2017 11:32 pm

Like people have noted before, we can first use the noble gas notation [Xe] to notate all the electron configurations up to Xe. Then, we write 4f^14 to account for all the electrons in the f block (starting with elements 57 and all the way to element 70). We get the number 4 because n-2 gives us our f orbital number (our's is n=6 in this case, so 6-2=4). Then, we write 5d^4 to account for the electrons in the d block. We get the number 5 because n-1 gives us our d orbital number (again, our's is n=6, so 6-1=5). Finally we write 6s^2 to account for the electrons in the s block.

Hope this wasn't too confusing.


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