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Question:Is there a way? Also, what are the charges that occur in the ions of the common transition metals? (like iron +2 and +3) Is there a way to tell how many there are or do you just have to memorize it?

Answers:The answer depends a lot on what kind of compounds you're talking about. Most sources will tell you that Cr, for example, exists as +2 +3 and +6 ions (or, more correctly, that it exists in oxidation states of 2, 3, and 6, since no Cr(VI) compound truly contains a discrete +6 ion). Which is true if you're talking about aqueous environments at middling pHs and some oxygen. But I've made Cr compounds will oxidations states of 0 1 and 4. If you really know what you're doing, you can make something as crazy as a Zn(I) complex, but you get to publish in "Science" when you pull it off. But let's assume you're just interested in "common" oxidation states of "common" transition metals. No, there's no good way to tell, you have to just know them, or keep a periodic table handy that lists them. Many of them do stuff that you might predict based on their position in the table and electronic configuration. Ti is Gr 4, it's usually +4. Zn is Gr 12, it's usually +2. But even then, not always, because Hg is the same column as Zn, and it shows up as a +1 ion as well as +2, and most of them can be stable in multiple states. There's a pretty good table at: Which ones do you consider common? Some ones worth remembering: V: 2,3,4,5 Cr: 2,3,6 Mn: 2,4,7 Fe: 2,3 Co: 2,3 Ni: 2 Cu: 1,2 Zn: 2 Mo and W: 2,3,4,6 Ru: 2,4; Os: 4,6,8 Rh and Ir: 1,3 Pd and Pt: 2,4 Ag: 1; Au 1,3 Hg: 1,2

Question:I'm confused as to how to determine how many valence electrons a transition metal has. I know that the electrons in the Highest shell are only considered. That's why Oxygen has 6 valence electrons for example. Since transition metals are known for filling up an s orbital before the d orbital (4s2 3d6 for Fe) then transition metals should only have 2 valence electrons almost always though there are some variations. But when using the lewis structure apparently the transition metals follow an "18 electron rule" as opposed to the octet rule. Im just confused as why this is and why there's 18 electrons. Zn for example is 4s2 3d10 though so shouldnt it be 12 electrons to be stable? If you could expand on what ive said, itd be really helpful for me. i feel like im missing something very important :| maybe i confused highest shell and highest shell in terms of energy. I also read that once an electron is added to the d orbital that their energies switch so 4s2 3d10 would have 2 valence electrons? thank you

Answers:Transitions metals, in addition to s and p orbitals, have d orbitals as well. The "highest shell" means the highest set of orbitals. This means that transition metals' "highest shell" would include an s orbital, a p orbital, and a d orbital. S holds 2 electrons, p holds 6 electrons, and d holds 10 electrons; so if you add those up, you get 18 valence electrons total. PS Notice how the rows above the transition metals span 8 elements (except for the first row) and the rows with the transition metals span 18 elements.



Question:For example, F^- or fluoride gains one electron, and "looks like" neon. do I write [Ne] and state that there are 0 valence electrons? or 8 valence written as [He] 2s^2 2p^6??? Also, for transition metals like cobalt, the notation is [Ar] 4s^2 3d^7. from the superscripts, do I write 9 valence electrons??? or do I write something else?

Answers:F- would look like Ne as you stated as far as the valence electrons go, but remember the protons and neutrons do not look like Ne. So a lewis-dot structure for f- would be F with 8 valance electrons. As far as Cobalt, it would not be [Ar]4s^2 3d^7 because there is more stability if an energy level is half full or completely full so the 1 electron would jump for the 4s level to fill the 3d level. Cobalt would actually be written [Ar]4s^1 3d^8 giving it 8 valance electrons.