#57. Assign Oxidation States to the highlighted atoms in the following compounds:

a) CH₄:  the H's are bound only to the non-metal, C...

Each H is therefore +1.  In order to maintain "neutrality" the C must be in the -4 oxidation state if only in a formal sense.

b) SF₄: here the fluorines must be -1, so the S must be in the +4 oxidation state in order to maintain neutrality.

c) Na₂O₂: an interesting problem...  the Na's must be +1.  While the oxygen atoms should normally be -2 each this would violate our netrality principle.  Each O must be in the -1 oxidation state.  Without knowing the structure, but knowing the formula, one can assume that the oxygen atoms are bound to each other and they are!

d) C₂H₃O₂^-: this is an ion, so the sum of the oxidation states must be -1.  I assume each H is +1 and each O is -2, meaning each carbon must be in the 0 oxidation state (on average).  This is the acetate ion, however, and the carbon atoms are not equivalent.  In fact, the C's are bound to each other, one is bound to all three H's and the other is bound to the two O's.  One C is formally +3, while the other is -3...  hence the average is 0.

OS(Fe) + 4 x OS(O) = -2
OS(Fe) + 4 x (-2) = -2
OS(Fe) - 8 = -2
OS(Fe) = +6