The Fish Posted March 20, 2011 Report Share Posted March 20, 2011 Hello!I am doing the first year of Chemistry HL. I don't understand how you do this question:A certain element has the electronic configuration 1s2 2s2 sp6 3d10 4s2 4p6 4d6 5s2Which oxidation state(s) would this element most likely show?A. +2 onlyB. +3 onlyC. +2 and +5 onlyD. +2, +3, +4, +5How do you figure it out? Reply Link to post Share on other sites More sharing options...
Drake Glau Posted March 20, 2011 Report Share Posted March 20, 2011 Have to think about the orbitals as if they are literally rings around the atom. For this specific element the outermost layer is 5s2 followed by 4d6 slightly closer to the nucleus. It is very likely that the 2 s electrons will be removed and the d level (for whatever reason) likes to be either half (5) full or completely (10) full so it will likely lose that 1 d electron to make it 4d5 as well resulting in a +3 ion.By power of elimination though I can tell you C and D are wrong because +5 is unreasonable. I am torn between A and B, leaning towards B only because the d level likes to be half full or full. Reply Link to post Share on other sites More sharing options...
Keel Posted March 21, 2011 Report Share Posted March 21, 2011 (edited) There is a problem with this question and that is there is no element with the configurations of [Kr] 5s2 4d6Tc: [Kr]5s2 4d5Ru: [Kr]5s1 4d7Rh: [Kr]5s1 4d8The one element with a similar number of electrons would be Ru but it doesn't have that configuration you mentioned. Also when doing such questions remember that the d-orbitals/sub shell can split into a further sub shell consisting of three orbitals and two orbitals. It all boils down to whether the stability of the atom is possible which determines what oxidation numbers it has. So assuming that the correction reveals that the element indeed was Ru, it could have oxidation states of +1, +2, +3, +4, +5, +6, +7 or +8 as they are all stable. I would choose oxidation states as +3 or +4 as the answer because it seems the most stable with the electrons occupying 4d5 or 5s1 4d3 (lower d-split-orbital filled). So I would agree with Drake on (B) as being the most likely answer. Edited March 21, 2011 by Keel Reply Link to post Share on other sites More sharing options...
Drake Glau Posted March 22, 2011 Report Share Posted March 22, 2011 There is a problem with this question and that is there is no element with the configurations of [Kr] 5s2 4d6Tc: [Kr]5s2 4d5Ru: [Kr]5s1 4d7Rh: [Kr]5s1 4d8The one element with a similar number of electrons would be Ru but it doesn't have that configuration you mentioned. Also when doing such questions remember that the d-orbitals/sub shell can split into a further sub shell consisting of three orbitals and two orbitals. It all boils down to whether the stability of the atom is possible which determines what oxidation numbers it has. So assuming that the correction reveals that the element indeed was Ru, it could have oxidation states of +1, +2, +3, +4, +5, +6, +7 or +8 as they are all stable. I would choose oxidation states as +3 or +4 as the answer because it seems the most stable with the electrons occupying 4d5 or 5s1 4d3 (lower d-split-orbital filled). So I would agree with Drake on (B) as being the most likely answer.I think the configuration they give is without the moving of the electrons into the proper sub shells and stuff (like you would in your first year of chemistry, well before you know about the randomness of the D and F orbitals ). Point of the question was to see if the person knows about them and is able to find the oxidation levels by finding the easy stable state Reply Link to post Share on other sites More sharing options...
Keel Posted March 22, 2011 Report Share Posted March 22, 2011 Yea, that could be possible but its still weird don't you thing? You would expect the question to (a) have a problem which is factually accurate and (b) follow the rules they are trying to teach the students? Reply Link to post Share on other sites More sharing options...
Drake Glau Posted March 25, 2011 Report Share Posted March 25, 2011 I'd agree. I think the point of the question was to see if they understand that concept though. Sort of like translating the weird configuration to the real one because if I remember right the special configuration with D and F are HL level. But they could just be giving a freaky question, certainly would not be the first time IB has done that Reply Link to post Share on other sites More sharing options...
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