f4chemoxidation no.的問題

is there something wrong with your question
Firstly, NO2, N is +4 and O is -2
Since O has a charge of -2, so it will have a oxidation no. of -2
whereas N, it has to balance the change of the 2 O, so it has to be +4
as for CCl4, the Cl is -1 and C is +4
MnO2: there is 2 O, so Mn needed to be +4
Mn2O3: there is 3 O and 2 Mn, so for the Mn to balance out the O, each Mn needed to be +3

hope this would help you

NO2 is covalent bond, since there is no +/- charge in covalent bond, so we need to assume covalent bond has +/- to calculate the oxidation no.
As we know oxdiation no. of O is -2, and the total O.N. = 0, so
N + 2(-2)=0 , then N = +4
CCl4 is also covalent bond , as Cl is group 7 , so its O.N. = -1
C+4(-1)=0, therefore, C= +4
the method of calculation is the same in these 2 cases.

if you understand the above method of calculation, u can find that no matter the bond is ionic or covalent , we can all clasify them as having +/- charge.
So, Mn + 2 (-2)=0, therefore, Mn = +4
and 2Mn + 3(-2)= 0, 2Mn = +6 , therefore, Mn = +3 in this case.

Maybe you forgot some of the rules of calculate the O.N. ,
for group 1 element, the O.N. = +1 , group 2 =+2 , group 3 = +3
group 6= -2, group 7 = -1
if the compound don't have any charge , like Cr2O7-2 , the overall O.N. = 0 , otherwise like the e.g. is -2


hope can help u ~!!!