Author Question: How to find the valence electron of transitional elements? (Read 1319 times)

stevenposner · **on:** Sep 20, 2013

how to find the valence electron of MO?
is there a formula to find their valence electron?

clippers! · **Reply #1 on:** Sep 20, 2013

Translation:

How to find the number of valence electrons for a transitional element?

How many valence electrons does MO (Molybdenum) have?

http://answers.yahoo.com/question/index?qid=20071203134233AASWPwh

There are six of them

Melanie · **Reply #2 on:** Sep 20, 2013

The periodic table tells it all. The transition elements start with their outer electrons being in a 4s orbital in which there are 2 electrons which are donated. Then the whole row has a valence of 2+ to start with. Then B Groups start with 3d1 electrons, which also can be donated. That gives Sc two valeneces of 2+ and 3+. Next is Ti with 2+ and 4+ V tells you its valence! etc
There is also another factor to consider and that is when the 3d orbitals are completely half-filled it has more stablitlity as all the electrons have the same spin
Cu can be one or two + and Zn is 2+ only because the 3d orbital is filled, and it can donmate the two 4s2 electrons. Good luck. Isn't this interesting?.

bobbysung · **Reply #3 on:** Sep 20, 2013

It is easy to determine the number of valence e? a TM metal has (provided you have a PT!). Just find it in the PT and look which group it is in and that gives you the number of valence e?. Ti (Z= 22) is in Group 4 and so has four valence e?. Re (Z= 75) is in Group 7 and has seven valence e?, Ag (Z = 47) is in Group 11 and hence has eleven valence e? , and so on. Up to Group 7 the maximum oxidation state exhibited by a TM is its Group number. Ti(IV) as in TiO2; and Mn(VII) is found in [MnO4]^-. Fe does not form any Fe(VIII) cmpds, but Ru(VIII) and Os(VIII) are found in the tetroxides RuO4 and OsO4. This is the maximum oxdn number known.
The problem comes in how these valence e? occupy the nd and (n+1)s valence atomic orbitals. Let's deal with with the 1st Row TM (Ti ?Cu) first. In the gas phase the uncharged atoms have the e? configuration [Ar] 3d^n 4s^2 We will not discuss why the higher energy 4s AOs are occupied first but it is fully understood. The exceptions are Cr: [Ar] 3d^5 4s^1 and Cu [Ar] 3d^10 4s^1. A hand-waving argument of extra stability associated with half-filled and completely-filled d shells is used. Just be content that it is a quantum mechanical effect called spin exchange. When the 1st Row TM forms the M^2+ or M^3+ ion (the common oxidation states of 1st Row TM) the higher energy 4s e? are lost first. Thus Cr(III) is [Ar] 3d^3. Take Fe(0)(g) : Group 8 so eight v e? [Ar] 3d^6 4s^2;
Fe(II): [Ar] 3d^6; Fe(II): [Ar] 3d^5
When you get to the 2nd Row you throw all these "rules" out the window. You can look up the Wikipedia entries (RH panel) for Ru, Rh and Pt and see if the half-filled d shell works. So only an expert will know what the configuration is in the gas phase and questions like this should not be given to undergrads until 4th yr. However Mo is in Group 6 and has an e? config [Kr] 4d^5 5s^1 so the half -illed d shell works here (but W the third row member of Group 6 is [Xe] 5d^4 6s^2 !). So the best answer quoted by vekkus4 is wrong on two counts. Once again when forming the cations the 5s e? are lost first: Mo(III) is [Kr] 4d^3.