Lingering question from 3/22/19 exam review…

What would the oxidation # be in groups 13 – 15 and 3 – 10?  Would it be the Roman numeral above the group?

Answer: 
If you are writing the formula for a compound (and you’re given the name), the oxidation # for all of these groups (13 – 15 & 3 – 10, except aluminum (ox # = +3), zinc (ox # = +2), cadmium (ox # = +2), or silver (ox # = +1)

Examples:        iron (III) nitrate     à  Fe(NO3)3
                             lead (II) chlorate   
à  Pb(ClO3)2

If you are naming the compound (and you’re given the formula), the oxidation # for these groups will have to be calculated using the equation:

S (# of atoms of element/polyatomic ion . oxid # of element/polyatomic ion) = 0

Examples:       
CrCO3      
à     chromium (II) carbonate
          let x = ox # of Cr               (1 . x) + (1 . -2) = 0
                                                             Cr           CO3

SnS            à     tin (II) sulfide
          let x = ox # of Sn               (1 . x) + (1 . -2) = 0
                                                             Sn             S