Respuesta :
Oxidation number of C is a) +2 b) +4 c) +4 d) +3 e) -4 f) 0 g) +4 h) +6 i)+6
j) +4 k) +2 l)2.5 m) -2 n) -1 o) +4
A) CO
x + (-2) = 0
Thus, x = +2
Oxidation number of C = +2
B) CO2
x + 2 (-2) = 0
x = +4
Oxidation number of C = +4
C) Na2CO3
2 (+1) + x + 3(-2) =0
2 + x -6 = 0
x = +4
Oxidation number of C = +4
D) Na2C2O4
2(+1) + 2x + 4(-2)
2x = +
x = +3
Oxidation number of C = +3
E) CH4
x + 4(+1) = -4
Oxidation number of C = -4
F) H2CO
2(+1) + x (-2) =0
x = 0
Oxidation number of C = 0
G) SO2
x + 2 (-2) = 0
x = +4
H) SO3
x + 3(-2) = 0
x = +6
I) Na2SO4
2(+1) + x + 2(--4)
x = +6
K) Na2SO3
2(+1) + x + 3(-2) =0
2 + x -6 = 0
x =+4
K) Na2S2O3
2(+1) + 2x + 3(-2) =0
2x = +4
x = +2
L) Na2S4O6
2(+1) + 4 x + 6(-2) =0
4x = 10
x = 2.5
M)SCl2
x + 2(-1) = 0
x = -2
N) Na2S2
2(+1) + 2x = 0
x = -1
O) SOCl2
x + -2 + 2(-1)
X = +4
An atom's hypothetical charge, or oxidation number, would be zero if all of its links to other atoms were completely ionic. It describes how much an atom in a chemical compound has oxidised (lost electrons). The oxidation state could theoretically be zero, positive, or negative. Owing to the fact that many bonds in nature exhibit high ionicity despite the absence of fully ionic bonds, the oxidation state is a good indicator of charge.
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