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Skyler Wild - Final Chemistry Notebook

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2. What is the oxidation number of Na in Na2O?<br />

Since O = -2, the two Na must each be +1.<br />

3. What is the oxidation number of Cl in ClO¯?<br />

The O is -2, but since a -1 must be left over, then the Cl is +1.<br />

4. What is the oxidation number for each element in KMnO4?<br />

K = +1 because KCl exists. We know the Cl = -1 because HCl exists.<br />

O = -2 by definition<br />

Mn = +7. There are 4 oxygens for a total of -8, K is +1, so Mn must be the rest.<br />

5. What is the oxidation number of S in SO4 2¯<br />

O = -2. There are four oxygens for -8 total. Since -2 must be left over, the S must = +6.<br />

Please note that, if there is no charge indicated on a formula, the total charge is taken to be zero.<br />

Practice Problems<br />

Find oxidation numbers<br />

1. N in NO3¯<br />

N = +5<br />

2. C in CO3 2¯<br />

N = +4<br />

3. Cr in CrO4 2¯<br />

Cr = +6<br />

4. Cr in Cr2O7 2¯<br />

Cr = +6<br />

5. Fe in Fe2O3<br />

Fe = +3<br />

6. Pb in PbOH +<br />

Pb = +2<br />

7. V in VO2 +<br />

V = +3<br />

8. V in VO 2+<br />

V = 4<br />

9. Mn in MnO4¯<br />

Mn = 7<br />

10. Mn in MnO4 2¯<br />

Mn = 6<br />

Notes:<br />

The objective is to balance the equation and get the charge that it is asking for, if there is no charge on<br />

an element, than its charge is automatically equal to 0.<br />

- Follow the 7 steps (Rules for Oxidation Numbers) and the examples given and practiced.<br />

Example:<br />

Cr in Cr2O72- : O has a charge equal to -2, 7 times -2 is -14, and the aim is to get negative two. In order to<br />

get -2, the 2 in Cr2 must be multiplied by 6. Therefore, Cr in Cr2O52- is 6.

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