Alkali metals such as sodium always form univalent cations in ionic compounds, and oxygen in oxyanions is assigned an oxidation number of -2. Therefore, to achieve electrical neutrality in Na2CO3, carbon must have an oxidation number of +4.
+1 for Na, +1 for H, +4 for C and -2 for each O
+4
4
Oxidation number of C is +3. Oxidation number of O is -2.
Oxidation number is 4, formula C3O2O=C=C=C=O
0. CO2 is electrically neutral. C is +4 and O is 2(-2)
Oxidation number of Li is +1. It is -2 in O and +4 in C
-2 for each O, +4 for C
Oxidation number of C is +3. Oxidation number of O is -2.
Oxidation number is 4, formula C3O2O=C=C=C=O
The oxidation number is + for C and -2 for O.
0. CO2 is electrically neutral. C is +4 and O is 2(-2)
Oxidation number of Li is +1. It is -2 in O and +4 in C
-2 for each O, +4 for C
-2 for O +4 for C
-2 for each O, +4 for C
+2 for Fe, +4 for C, - 2 for each O
H is +1, O is -2 overall carbon will have an oxidation # of -3
+1 to each H +4 to each C -2 to each O
Na2C03 Oxidation number of Na = + 1 Oxidation number of O=-2, Oxidation number of C=2x1+x+3(-2)=0 so x=4