Can you account for the Periodic Table given electronic structure?

1 Answer
Jan 31, 2016

The pattern follows the Periodic Table: complete valence shells of electrons have special stabilities.


Look at the left of the Periodic Table, the alkali metals, Group I, and the alkaline earths,Group II. Now metals are electron rich, they tend to lose electrons and form CATIONS. Group I metals such as sodium and potassium, form #M^+# ions, #Na^+#, and #K^+#; Group II metals have 2 electrons in their valence shell, and routinely form #M^(2+)# ions: #Mg^(2+)#, #Ca^(2+)#.

As we go the right along a PERIOD, the atomic charge gets incrementally larger, and unfilled electron shells shield this charge very poorly. Nuclear charge becomes dominant, and the electronic radii of the ATOMS (NOT the ions) get smaller, they contract. Atoms to the right of the Periodic Table thus tend to gain electrons to form ANIONS: i.e. #F^-#, #Cl^-#, and #O^(2-)#; in other words, non metals OXIDIZE.

So for this periodic perspective METALS tend to be electron-rich (because their valence electrons are removed from the nuclear charge), and NON-METALS tend to be electron-poor, because of their enhanced nuclear charge. In other words, METALS REDUCE (donate electrons) and NON-METALS OXIDIZE (ACCEPT ELECTRONS). This can all be rationalized on the basis of nuclear charge. Remember that the argument is based on the NEUTRAL ATOMS, metal and non-metal, and NOT their redox products.

Sorry for all the SHOUTING!!