## Problem 4.50

#### Determine the oxidation number for the indicated element in each of the following compounds:

(a)Co in LiCoO2

To determine the oxidation number of an atom in a compound, we assign to the most electronegative element in the compound a negative oxidation number equal to its charge in simple ionic compounds. For neutral compounds, the oxidation numbers of all elements must add to give zero; for ionic species, the oxidation numbers must add to give the ionic charge of the compound. Here, O is the more electronegative element, and it gets the -2 oxidation number. Li only forms the 1+ ion and therefore has an oxidation number of +1. From the above, all of the oxidation numbers add to zero, so

1+ Co + 2(-2) =0. Co is the oxidation number of Co; solving for Co, we find Co = +3/

Al in NaAlH4

Here, Na + Al +4(H) = 0 (where Na. Al, and H are the oxidation numbers of Na, Al, and H.) We use the rules (p 132) to assign oxidation numbers of +1 to Na and -1 to H (be sure you understand this!!). We have 1 + Al +4(-1) = 0, and Al = +3.

(c)C in CH3OH (methanol)

Again, use the rules!!! O gets a -2 oxidation number, and H gets a +1 (why? See the rules!!!) So we have C + 4 -2 =0, so C = -2.

(d)N in GaN

This a binary compound - since Ga is a group IIIA element we expect it to form only a 3+ ion so Ga has an oxidation number of +3 and N has an oxidation number of -3.

(e)Cl in HClO2

We assign O the -2 charge and H the +1 charge (rules!!!), so, adding up the oxidation numbers, 1 + Cl + 2(-2) = 0 and Cl = 3.

(f)Cr in BaCrO4

Give oxygen the -2 oxidation number and Ba the +2 and add 'em up: 2 + Cr + 4(-2) = 0. We have Cr in the +6 state.