1) N in NO3¯
The O is -2 and three of them makes -6. Since -1 is left over, the N must be +5
2) C in CO32¯
The O is -2 and three of them makes -6. Since -2 is left over, the C must be +4
3) Cr in CrO42¯
The O is -2 and four of them makes -8. Since -2 is left over, the Cr must be +6
4) Cr in Cr2O72¯
The O is -2 and seven of them makes -14. Since -2 is left over, the two Cr must be +12What follows is an important point. Each Cr is +6. Each one. We do not speak of Cr2 being +12. Each Cr atom is considered individually.
5) Fe in Fe2O3
The O is -2 and three of them makes -6. Each Fe must then be +3
The answers for 6-10 are given below. If you are not sure how a value was determined, check with your teacher.
6) +2
7) +5
8) +4
9) +7
10) +6