How much of a difference, and which subshell is lower in energy, varies by element. The 4s and 3d subshells have nearly the same energy level. I have assumed that the configuration that follows the rules is the ground state, since that is what is usually taught in general chemistry. Note: Nickel has two electron configurations that are commonly encountered over which there is a dispute amongst chemists as to which should be considered the ground state. There is a simple chart that makes learning the filling order easy. He did not know that La is an exception to the rule. Note: in one of his videos, Sal does the configuration of La, but he gets it wrong. However, it is unlikely you would be asked to know these in a general chemistry course. Thus, the actual configuration of the Pd is 4d¹⁰ (the 5s is empty).Īll the other exceptions are in the f block and just have to be learned on a case-by-case basis. Both of these electrons are instead in the outermost d subshell. In Pd (palladium) the outermost s subshell, predicted to have 2 electrons, is actually empty. In the following elements, there is 1 electron, not the predicted 2, in the outermost s subshell, the "missing" electron is instead located in the outermost d subshell: So, rather than going through the far more difficult to understand rules that do correctly predict the electron configurations for all elements, it is easier just to learn the elements that this simplified method doesn't work correctly for. There are similar irregularities in the filling of 5s and 4d as well as some other pairs of subshells. For example, the 4s and 3d have nearly the same energy level and so, in a way that varies according to which element it is, either of the two subshells could be the first to fill and there is no certainty that one will be completely filled before the other actually starts filling. The actual filling order is more complicated. This pattern will give you the correct configuration for all but about 19 elements. There are no known elements that, in their ground state, have electrons in a subshell beyond 7p. What you do is you start assigning electrons to the subshells using the following pattern (you completely fill up one subshell before moving onto the next higher) and you keep going until the total number of electrons assigned is equal to the atomic number:ġs, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p However, in the simplified version of these rules that is understandable to a general chemistry student, there are about 19 elements that are exceptions. Given the atomic number, there is a set of rules that allow you to determine the electron configuration. Without that, you cannot determine the electron configuration. Can you write the generic oxidation reaction for each to give #M^(2+)# and #M^(3+)#? Mass and charge MUST be conserved.You must know the atomic number of the element. Transition metal ions commonly adopt #+II# and #+III# oxidation states (certainly true in the examples above agree?). So if you haven't done the experiment in your coursework, you don't have to consider them. Note that all of these reactions are proposed on the basis of EXPERIMENT. We would typically use permanganate ion, #MnO_4^-# for redox titrations in that permanganate ion #MnO_4^-# is deep-red in colour, and #Mn^(2+)# is almost colourless, and thus an endpoint is built-in to the reaction. In both instances electron transfer is EXPLICIT. When these ions are REDUCED down to their stable ions.electron transfer is explicit. For complex ions, the SUM of the oxidation states is equivalent to the charge on the ion.Ĭonsider #Cr_2O_7^(2-)# we have #Cr(VI+)#+7xxO(-II)=-2# as required.Likewise for #MnO_4^-# we gots #Mn(VII+)#. And thus they tend to be oxidizing, and tend to GAIN electrons, and thus they assume the closed-shell electronic configuration the NEXT Noble Gas, cf:Īnd so when you are quoted a formula such as #Fe_2O_3#, you tend to break it up and assign oxidation states, i.e. On the other hand, non-metals, from the right hand side of the Periodic Table as we face it, have high nuclear charges. Thus metals in Groups 1, 2, and 3, TEND to lose 1, 2, and 3 electrons respectively, and they assume the closed-shell electronic configuration the LAST Noble Gas.
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