6th Edition, 3.5

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Tyra Nguyen 4H
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Joined: Fri Sep 28, 2018 12:25 am

6th Edition, 3.5

Postby Tyra Nguyen 4H » Thu Nov 01, 2018 3:05 pm

I understand how Cu+ is [Ar]3d10, but I don't understand parts b, c, and d? I think the f-block throws me off, but can anybody explain them? Thanks!

Hai-Lin Yeh 1J
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Re: 6th Edition, 3.5

Postby Hai-Lin Yeh 1J » Thu Nov 01, 2018 5:31 pm

How I looked at these types of problems is I first wrote the electron configuration out. Then, I rearranged it according to the charge and number.

For (b), Bi3+: if you look at the periodic table, the noble gas notation is [Xe]. From there, I just list it from left to right: so, first is 6s2. When you get to the f-block, you need to know that there are 15 electrons in the f orbital. So, after 6s2 is 4f14. Then, go to the d-block which is 5d10, then p-block until you reach bismuth, which is 6p3. The charge of Bi is 3+, which means remove 3 electrons and it's easier to remove it from the p block since there are already 3.
So far: [Xe] 6s2 4f15 5d10 6p3 -> remove 3 electrons -> [Xe] 6s2 4f15 5d10. Then rearrange according to lowest energy:
[Xe] 4f15 5d10 6s2

For (c), Ga3+, do the same. There isn't an f-block so it should be relatively easier. Noble gas notation is [Ar].
[Ar] 4s2 3d10 4p1. The charge of Ga is 3+ so that means, remove 3 electrons. It would be easier to remove from the s and p orbital, but you can rearrange it first before you remove, whichever is easier
The answer should be: [Ar] 3d10

For (d), Tl3+, the noble gas notation is [Xe] again. Same as b, I would go from left to right before rearranging
[Xe] 6s2 4f15 5d10 6p1. The charge indicates to remove 3 electrons. It's again, easier to remove from s and p orbital.
What you get: [Xe] 4f15 5d10

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