Question

1. Determine the oxidation number for each element indicated.
   1. Cl in Cl₂
2. 2. Cr in K₂Cr₂O₇
   3. N in NO₃^-

Answer 1

a)

lets the oxidation number of Cl be x

use:

2* oxidation number (Cl) = net charge

2* x = 0

0 + 2 * x = 0

x = 0

So oxidation number of Cl = 0

Answer: 0

b)

Oxidation number of K = +1

Oxidation number of O = -2

lets the oxidation number of Cr be x

use:

2* oxidation number (K) + 7* oxidation number (O) + 2* oxidation number (Cr) = net charge

2*(+1)+7*(-2)+2* x = 0

-12 + 2 * x = 0

x = 6

So oxidation number of Cr = +6

Answer: +6

c)

Oxidation number of O = -2

lets the oxidation number of N be x

use:

3* oxidation number (O) + 1* oxidation number (N) = net charge

3*(-2)+1* x = -1

-6 + 1 * x = -1

x = 5

So oxidation number of N = +5

Answer: +5