Hello there! I was wondering if you could check my answers for me? I was having trouble and I finally understand.
1. Indicate the oxidation number of carbon and sulfur in the following compounds.
A) CO= +2
C) Na2CO3= +4
E) CH4= -4
G) SO2= +4
I) Na2SO4= +6
K) Na2S303= +2
M) SCl2= +2
O) SOCl2= +4
2.) Indicate the oxidation number of phosphorous, iodine, nitrogen, tellurium, and silicon.
b) PO3^3-= +3
d) P3O10^5-= +5
f) IO2^-= +3
h) NH4^+= -3
j) NO2^-= +3
l) NO2^+= +5
N) TeO4^2-= +6
3.) Indicate the oxidation number of the metallic elements in the following compounds.
a) Fe2O3= +3
c) CoS= +2
e) K3CoCl6= +3
G) CrO3= +6
I) K2Cr2O7= +6
K) K2MnO4= +6
M) MnO2= +4
O) Pb04= +8
Thank you so much! I have a quiz tomorrow and I want to make sure I am practicing everything correctly.
1. Indicate the oxidation number of carbon and sulfur in the following compounds.
A) CO= +2
C) Na2CO3= +4
E) CH4= -4
G) SO2= +4
I) Na2SO4= +6
K) Na2S303= +2 OK if that is Na2S2O3
M) SCl2= +2
O) SOCl2= +4
2.) Indicate the oxidation number of phosphorous, iodine, nitrogen, tellurium, and silicon.
b) PO3^3-= +3
d) P3O10^5-= +5
f) IO2^-= +3
h) NH4^+= -3
j) NO2^-= +3
l) NO2^+= +5
N) TeO4^2-= +6
3.) Indicate the oxidation number of the metallic elements in the following compounds.
a) Fe2O3= +3
c) CoS= +2
e) K3CoCl6= +3
G) CrO3= +6
I) K2Cr2O7= +6 yes, for each Cr
K) K2MnO4= +6
M) MnO2= +4
O) Pb04= +8 OK if you meant this but I've never heard of PbO4. You may have meant (PbO4)^2-
They look ok to me except for those notes I made.
i meant Pb304…sorry, my mistake
For Pb3O4, Pb is overall a 8/3 for each Pb. You can read about how this happens here. Pb3O4 is a mixture.
https://answers.yahoo.com/question/index?qid=20130525034348AA6MFHx
Certainly! I'd be happy to check your answers and provide explanations.
1. Indicate the oxidation number of carbon and sulfur in the following compounds:
A) CO = +2
The oxidation number of oxygen is typically -2. Since there are no other elements present, the oxidation number of carbon in CO is +2.
C) Na2CO3 = +4
The oxidation number of oxygen is typically -2. The sum of the oxidation numbers in a compound must be equal to the overall charge, which is 0 for Na2CO3. Therefore, we can set up the equation: (2 * Na) + (x * C) + (3 * -2) = 0. Solving for x, we find that the oxidation number of carbon in CO3 is +4.
E) CH4 = -4
Hydrogen typically has an oxidation number of +1. The sum of the oxidation numbers in a compound is 0. Since CH4 is neutral, we can set up the equation: x + (4 * 1) = 0. Solving for x, we find that the oxidation number of carbon in CH4 is -4.
G) SO2 = +4
The oxidation number of oxygen is typically -2. The sum of the oxidation numbers in a compound is 0. Therefore, we can set up the equation: (x * S) + (2 * -2) = 0. Solving for x, we find that the oxidation number of sulfur in SO2 is +4.
I) Na2SO4 = +6
Using similar reasoning as in compound C, we can set up the equation: (2 * Na) + (x * S) + (4 * -2) = 0. Solving for x, we find that the oxidation number of sulfur in Na2SO4 is +6.
K) Na2S2O3 = +2
Using similar reasoning as in compound C, we can set up the equation: (2 * Na) + (2 * S) + (3 * -2) = 0. Solving for x, we find that the oxidation number of sulfur in Na2S2O3 is +2.
M) SCl2 = +2
Using similar reasoning as in compound G, we can set up the equation: (x * S) + (2 * -1) = 0. Solving for x, we find that the oxidation number of sulfur in SCl2 is +2.
O) SOCl2 = +4
Using similar reasoning as in compounds G and M, we can set up the equation: (x * S) + (2 * -1) + (1 * -2) = 0. Solving for x, we find that the oxidation number of sulfur in SOCl2 is +4.
2. Indicate the oxidation number of phosphorous, iodine, nitrogen, tellurium, and silicon:
b) PO3^3- = +3
The overall charge of the polyatomic ion is -3. Since oxygen typically has an oxidation number of -2, we can set up the equation: (x * P) + (3 * -2) = -3. Solving for x, we find that the oxidation number of phosphorous in PO3^3- is +3.
d) P3O10^5- = +5
Using similar reasoning as in compound b, we can set up the equation: (3 * P) + (10 * -2) = -5. Solving for x, we find that the oxidation number of phosphorous in P3O10^5- is +5.
f) IO2^- = +3
Using similar reasoning as in compound b, we can set up the equation: (x * I) + (2 * -2) = -1. Solving for x, we find that the oxidation number of iodine in IO2^- is +3.
h) NH4^+ = -3
The overall charge of the ammonium ion is +1. Since hydrogen typically has an oxidation number of +1, we can set up the equation: (x * N) + (4 * 1) = +1. Solving for x, we find that the oxidation number of nitrogen in NH4^+ is -3.
j) NO2^- = +3
Using similar reasoning as in compound b, we can set up the equation: (x * N) + (2 * -2) = -1. Solving for x, we find that the oxidation number of nitrogen in NO2^- is +3.
l) NO2^+ = +5
Using similar reasoning as in compound j, we can set up the equation: (x * N) + (2 * -2) = +1. Solving for x, we find that the oxidation number of nitrogen in NO2^+ is +5.
N) TeO4^2- = +6
Using similar reasoning as in compound b, we can set up the equation: (x * Te) + (4 * -2) = -2. Solving for x, we find that the oxidation number of tellurium in TeO4^2- is +6.
3. Indicate the oxidation number of the metallic elements in the following compounds:
a) Fe2O3 = +3
The sum of the oxidation numbers in a compound is zero. Since oxygen typically has an oxidation number of -2, we can set up the equation: (2 * Fe) + (3 * -2) = 0. Solving for x, we find that the oxidation number of iron in Fe2O3 is +3.
c) CoS = +2
Using similar reasoning as in compound a, we can set up the equation: (x * Co) + (-2) = 0. Solving for x, we find that the oxidation number of cobalt in CoS is +2.
e) K3CoCl6 = +3
Using similar reasoning as in compound a, we can set up the equation: (3 * K) + (x * Co) + (-6 * 1) = 0. Solving for x, we find that the oxidation number of cobalt in K3CoCl6 is +3.
G) CrO3 = +6
Using similar reasoning as in compound a, we can set up the equation: (x * Cr) + (3 * -2) = 0. Solving for x, we find that the oxidation number of chromium in CrO3 is +6.
I) K2Cr2O7 = +6
Using similar reasoning as in compound a, we can set up the equation: (2 * K) + (2 * x * Cr) + (7 * -2) = 0. Solving for x, we find that the oxidation number of chromium in K2Cr2O7 is +6.
K) K2MnO4 = +6
Using similar reasoning as in compound a, we can set up the equation: (2 * K) + (x * Mn) + (4 * -2) = 0. Solving for x, we find that the oxidation number of manganese in K2MnO4 is +6.
M) MnO2 = +4
Using similar reasoning as in compound a, we can set up the equation: (x * Mn) + (2 * -2) = 0. Solving for x, we find that the oxidation number of manganese in MnO2 is +4.
O) PbO4 = +8
Using similar reasoning as in compound a, we can set up the equation: (x * Pb) + (4 * -2) = 0. Solving for x, we find that the oxidation number of lead in PbO4 is +8.
I hope this clears up any confusion and helps you study for your quiz! Let me know if you have any more questions or need further explanations. Good luck on your quiz!