## How do we draw the electron configurations for transition metal atoms?

EvaLi_3J
Posts: 53
Joined: Wed Oct 02, 2019 12:16 am

### How do we draw the electron configurations for transition metal atoms?

On the textbook 2A exercise 1, we are asked to write the valence electron of Sb, and the answer is 5. I thought that it would be 15. Can anyone explain why the valence electron is 5 and explain the electronic configuration of transition atoms in general?

Hannah Lee 2F
Posts: 99
Joined: Thu Jul 11, 2019 12:15 am

### Re: How do we draw the electron configurations for transition metal atoms?

The electron configuration of Sb is [Kr]4d105s25p3.

Because valence electrons are used in bonding, they are always in the highest-energy, outermost orbitals. In the case of Sb, this would be n = 5, and we would exclude the completely filled 4d-subshell (which would not participate in bonding). 5s2 and 5p3 hold 5 electrons total, so Sb has 5 valence electrons.

Overall, when counting the valence electrons for p-block elements, you don't necessarily go by group number. Group 15 elements like Sb have 5 valence electrons in their highest-energy orbitals, ns2np3.

D-block elements like Mn in part (c) usually have valence e- equal to their group number. Let's consider the electron configuration of Mn: [Ar]3d54s2. Their outermost orbitals are 3d5 and 4s2, so it would have 7 valence electrons total.

Pablo 1K
Posts: 103
Joined: Sat Feb 02, 2019 12:15 am

### Re: How do we draw the electron configurations for transition metal atoms?

I believe you would simply exclude the d block, and count over them so for example, as mentioned above the number of valence e- for group 15 would all be 5.