how do you relate an element's periodic table position to its quantum electron configuration and reactivity

Answer 1

All elements withing the same group (Alkali metals for example) will have an electron configuration that ends the same way (s1 for alkali metals, s2 for Alkaline earth, Groups 3-8 end in p1, p2, p3, p4... so on. The trasition metals end with d orbitals of an energy level one less than the period the element is found in. Lanthanides and actinides end in f orbitals with an energy level equal to its period # minus 2.)

The periodic table can be divided up into a "block" for each of the orbital shapes. Groups 1and 2 make the s block. 3-8 make the p block. Trans metals make the d block. Lant/Actinides make the f block. Simply count over the # of spaces that the element is within that block to determine the number of electrons in the last orbital of the electron config.

I'm not sure what you mean by "reactivity." Lots of things affect that. But elements are usually more stable with full or half full orbitals in the OUTER (highest number) energy levels. So, elements ending in s1 will generally lose that electron to gain stability forming a +1 cation. Elements in group 7, whose configurations end in p5 will usually gain 1 electron either by sharing in a covalent bond, or by "taking" them outright and becoming a -1 anion.

Answer 2

This is too complex to write here. Please see the great tutorial below:

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch6/quantum.html

Once you do one or two, it is really easy.