The question provides a list of chemical formulas and their respective names and requests us to show the "rought work" - the calculations with the oxidation state for each element in the molecule.
To calculate the charge of the elements in a molecule, we need to keep in mind that some species have a determined number of oxidation that only changes in very specific occasions. Also, it is important to remember that the charge of the entire molecule must be considered - in this case, all molecules are neutral (charge = 0). At last, we always need to consider the number of atoms of each element in the molecule to calculate the total charge.
For, KClO3, we start from K and O, which have +1 and -2 charges, respectively:
K: 1 * +1 = +1
O: 3 * -2 = -6
To achieve total charge = 0: -6 + 1 = +5
Thus, N: 1 * +5 = +5
For NH4Cl, we start from Cl and H, which have -1 and +1 charges, respectively:
Cl: 1 * -1 = -1
H: 4 * +1 = +4
To achieve total charge=0 we need to balance +4 - 1 = +3.
Thus, N: 1 * -3 = -3
For AuI3, the name states that gold has +3 charge (Au III). We could also start from I, which has charge 1:
I: 3 * -1 = -3
Au: 1 * +3 = +3
For NiF2, the name also says that Ni has a charge (+2), but we could also calculate it from the charge of F, which is -1:
F: 2 * -1 = -2
Ni: 1 * +2 = +2
For N2Br4, we can start from Br, which has charge -1:
Br: 4 * -1 = -4
N: 2 * +2 = +4
To achieve total charge = 0, we need that the charge of all atoms of N be equal to +4.
Thus: charge of N: +4/2 = +2
For Li2CrO4, we can start from the charge of O (-2) and Li (+1):
O: 4 * -2 = -8
Li: 2 * +1 = +2
To achieve total charge = 0, we need that Cr has charge = -8 + 2 = +6
Cr: 1 * +6 = +6
For (NH4)2CO3, we can start from O (-2), C (+4) and H (+1). Remember that, in this case, we also need to consider the number outside the parenthesis!
O: 3 * -2 = -6
C: 1 * +4 = +4
H: 4*2 * +1 = +8
Thus, to achieve total charge = 0, we need that the total charge of N atoms to balance -6+8+4
Then, we calculate: (charge of each N atom): N = -6 / 2 = -3
N: 1*2 * -3 = -6