Please help me with this and explain thank you
a)
Oxidation number of O = -2
lets the oxidation number of Cl be x
use:
4* oxidation number (O) + 1* oxidation number (Cl) = net charge
4*(-2)+1* x = -1
-8 + 1 * x = -1
x = 7
So oxidation number of Cl = +7
Oxidation number of each element in ClO4-1 is
Cl=+7
O=-2
b)
lets the oxidation number of Cl be x
use:
1* oxidation number (Cl) = net charge
1* x = -1
0 + 1 * x = -1
x = -1
So oxidation number of Cl = -1
Oxidation number of each element in Cl-1 is
Cl=-1
C)
Oxidation number of F = -1
lets the oxidation number of S be x
use:
6* oxidation number (F) + 1* oxidation number (S) = net charge
6*(-1)+1* x = 0
-6 + 1 * x = 0
x = 6
So oxidation number of S = +6
Oxidation number of each element in SF6 is
S=+6
F=-1
d)
This is compound of Au+3 and NO3-
So, oxidation number of Au is +3
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
Oxidation number of each element in NO3-1 is
Au=+3
N=+5
O=-2
Please help me with this and explain thank you Review Problem 5.030 Assign oxidation numbers to the atoms in the follow...
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