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

Posted: Sat Nov 02, 2019 4:01 pm
by EvaLi_3J
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?

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

Posted: Sat Nov 02, 2019 4:59 pm
by Hannah Lee 2F
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.

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

Posted: Sat Nov 02, 2019 5:03 pm
by Pablo 1K
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.