Question

3. Assign an oxidation number to each element in the following compounds or ions, put your answers on the line a) NH4 N= Each of the H: b) NF4 Each of the N: Each of the F: c) CF2Cl C= Each of the F: Each of the C:

Answer 1

a)

Oxidation number of H = +1

lets the oxidation number of N be x

use:

4* oxidation number (H) + 1* oxidation number (N) = net charge

4*(+1)+1* x = +1

4 + 1 * x = +1

x = -3

So oxidation number of N = -3

Oxidation number of each element in NH4+1 is

N=-3

H=+1

b)

Oxidation number of F = -1

lets the oxidation number of N be x

use:

4* oxidation number (F) + 2* oxidation number (N) = net charge

4*(-1)+2* x = 0

-4 + 2 * x = 0

x = 2

So oxidation number of N = +2

Oxidation number of each element in N2F4 is

N=+2

F=-1

C)

Oxidation number of F = -1

Oxidation number of Cl = -1

lets the oxidation number of C be x

use:

2* oxidation number (F) + 2* oxidation number (Cl) + 1* oxidation number (C) = net charge

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

-4 + 1 * x = 0

x = 4

So oxidation number of C = +4

Oxidation number of each element in CF2Cl2 is

C=+4

F=-1

Cl=-1