Michael L. answered • 05/27/22
25 years experience teaching Chemistry (including AP)
Let's do Sb (Antimony). We find it in row 5 of the periodic table, so it must have all the good stuff from the first four row (I'm going to put in spaces for readability): 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6. The transition metals of row 5 add on 4d10, Rb and Sr add on 5s2. Sb is 3 atoms into the p orbitals of row 5 so that adds on 5p3. Putting that all together: 1s22s22p63s23p63d104s24p64d105s25p3. This matches answer 1 (except that there is an error and everything repeats after it first reaches 5p3). We might also have looked for an answer that ended in 5p3 and found it even quicker that way.
Taking a look at V (Vanadium), we first have to find the wretched thing in the period table (often the most annoying part! :) It's in the transition metals of row 4 and it's the 3rd transition metal. So let's look at the list of answers for one that contains 3d3 at the end and voila! It's number 2 (again with the weird repetition after it first reaches 3d3).
Hopefully this gives you a sense of how to do these. For example, P is in row 3 in the section of the periodic table that contains the p orbitals and it's the third one in. What might we expect the electron configuration to end with?
Sm is in one of those oddball rows at the bottom of the table made up of the 4f and 5f orbitals. It's the 5th entry but we have to count it as number 6 because that row officially begins with La (element 57). So its electron configuration should end with 4f6.
Let me know if you still need help working these out.
