## Homework problem 2.43

$c=\lambda v$

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Joined: Fri Sep 25, 2015 3:00 am

### Homework problem 2.43

For the electron configuration of tungsten I got [Xe] 5d4 6s2. Why does the answer include 4f14?

Vanessa A 3F
Posts: 29
Joined: Fri Sep 25, 2015 3:00 am

### Re: Homework problem 2.43

Hi Madeline. I can see how you got the answer you did by looking at the periodic table. However, notice that Tungsten's (W) atomic number is 74. Now, do you notice how in period 6, the atomic numbers skip from #57 (Lanthanum) to #72 (Hafnium)? That's because a series of f-block elements actually go in between.

F-block elements' atomic numbers range from 58-71 and 90-103. So when you're looking at the main part of the periodic table and writing out the orbital configuration, be careful when the element's atomic # is 58 or higher. That means it will involve the F orbital as well.

So Tungsten's electron configuration is
[Xe] 6s2 4f14 5d4
(From 6s to the 5d block, you should see the jump in atomic numbers and look down at the f-block)

And since we write the orbitals in increasing numerical order, your final answer is
[Xe]4f14 5d4 6s2