3.19 6th edition

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alanaarchbold
Posts: 68
Joined: Fri Sep 28, 2018 12:27 am

3.19 6th edition

Postby alanaarchbold » Tue Oct 30, 2018 7:24 pm

I know it says to emit c on this problem but how would you find how many unpaired electrons there are for this problem?

Beatrice Petelo 1F
Posts: 62
Joined: Fri Sep 28, 2018 12:17 am

Re: 3.19 6th edition

Postby Beatrice Petelo 1F » Tue Oct 30, 2018 8:00 pm

The electron configuration for Tungsten is [Xe] 4f^14 5d^4 6s^2. Because it's W2+, we take away two electrons from the s energy level.

The electron configuration is now [Xe] 4f^14 5d^4.

If you were to draw out the orbitals, you'll find that d energy levels have five orbitals. The four electrons will fill out 4 of the 5 orbitals (Hund's rule). These electrons do not have a pair, and therefore there are four unpaired electrons.


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