How do you find the charge or an element on the periodic table?

1 Answer
Jul 3, 2017

Answer:

Well, metals are electron-rich materials..........

Explanation:

And tend to lose electrons. 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. On the other hand, non-metals, from the right hand side of the Periodic Table as we face it, have high nuclear charges. 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:

#1/2F_2+e^(-) rarr F^-#

#1/2O_2+2e^(-) rarr O^(2-)#

And so when you are quoted a formula such as #Fe_2O_3#, you tend to break it up and assign oxidation states, i.e. #Fe_2O_3# is #stackrel(+III)"Fe"_2stackrel(-II)O_3#. For complex ions, the SUM of the oxidation states is equivalent to the charge on the ion.....

Consider #Cr_2O_7^(2-)#; we have #Cr(VI+)# #+7xxO(-II)=-2# as required......Likewise for #MnO_4^-# we gots #Mn(VII+)#. When these ions are REDUCED down to their stable ions....electron transfer is explicit.....

#Cr_2O_7^(2-) + 14H^+ + 6e^(-) rarr 2Cr^(3+) + 7H_2O#

#MnO_4^(-) +8H^+ + 5e^(-) rarr Mn^(2+) + 4H_2O#

In both instances electron transfer is EXPLICIT. 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. Note that all of these reactions are proposed on the basis of EXPERIMENT. So if you haven't done the experiment in your coursework, you don't have to consider them.......

Transition metal ions commonly adopt #+II# and #+III# oxidation states (certainly true in the examples above; agree?). Can you write the generic oxidation reaction for each to give #M^(2+)# and #M^(3+)#? Mass and charge MUST be conserved.