Question

How can I find valence electrons of transition metals?

Answer 1

Most transition metals have 2 valence electrons.

Valence electrons are the sum total of all the electrons in the highest energy level (principal quantum number n). Most transition metals have an electron configuration that is `ns^2 (n-1)d`, so those `ns^2` electrons are the valence electrons.

For example. How many valence electrons are there in Fe?

Solution: 2 valence electrons.

Reason: The electron configuration of Fe is `1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5`. The two 4s electrons are in the highest principal quantum number, n = 4, so they are the valence electrons.

Copper and chromium have one valence electron (they are exceptions), because they have one 4s electron. Chromium has an electron configuration of `[Ar] 4s^1 3d^5` because having a half filled 3d subshell is more stable, so it has one valence electron. Copper has one valence electron (the 4s electron) because it has electron configuration of `[Ar] 4s^1 3d^10`. Having a filled 3d and a half fille 4s subshell is more stable than `[Ar] 4s^2 3d^9`.

Answer 2

It is not obvious. Valence electrons are those that are important in chemical bonding. For transition metals, the word "important" will vary depending on the context.

It is easier and more practical to describe which orbitals are valence orbitals when it comes to transition metals (although it gets difficult with lanthanides and actinides).

In general, the first-row transition metals have a set of valence orbitals that include their `4s` and `3d`'s, but the number of valence electrons will vary.

For example...

• Scandium makes sense to have up to three valence electrons, since a `"Sc"^(+3)` oxidation state exists (e.g. `"ScCl"_3`), but not `"Sc"^(+4)` or higher. A `+3` oxidation state would have required transferring three valence electrons if it were to form a pure cation.
• Chromium could have up to six valence electrons, which would include its `3d` electrons, since it can accomplish a `+6` oxidation state (i.e. in `"Cr"_2"O"_7^(2-)`, or in `"CrO"_4^(2-)`).
• Copper tends to have a `+1` oxidation state (e.g. `"CuCl"`), so it makes sense that it uses its one `4s` electron most often as its valence electron(s). But a `+2` oxidation state is also known (e.g. `"CuCl"_2`), and so it is capable of taking from its `bb(3d)` electrons as well for its valence electron(s).

On the other hand, we could easily say that the valence ORBITALS of the first-row transition metals are the `4s` and `3d` orbitals.

So in general, I would say the number of valence electrons for transition metals (and lanthanide and actinides) vary in an unpredictable way, but the valence orbitals could sometimes be predicted with enough chemical intuition.

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DISCLAIMER: It is, however, in general difficult for the lanthanides and actinides to predict which orbitals are valence.

For example, the actinides have `5f` and `6d` orbitals very close in energy to their `7s` orbital, so we may GUESS and include the `7s`, `6d`, AND the `5f` in the valence space (even if the `6d` orbitals are empty) when performing atomic energy computations.