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Complete the second column ("rough work") with the calculations of the oxidation state of the atoms for each molecule (skip the molecule triphosphorus monoxide).*the chemical formula of niobium (V) phosphate is Nb2(PO4)5

Complete the second column ("rough work") with the calculations of the oxidation-example-1
User Rnoway
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The question provides a list of chemical formulas and their respective names and requests us to show the "rought work" - the calculations of the oxidation states 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, such as oxygen (O2-) and hydrogen (H+1). Also, it is important to remember that the charge of the entire molecule must be considered - in this case, all molecules are neutral (total charge = 0). At last, we always need to consider the number of atoms of each element in the molecule to calculate the total charge and, if needed, multiply the number of atoms by the charge of the atom.

For copper (II) nitrite, CuN2O4, we can start with the charge of Cu, given in the name of the molecule (+2) and O (-2) and then calculate the required nox of N:

Cu: 1 * +2 = +2

O: 4 * -2 = -8

To achieve total charge = 0, the atoms of N need to balance +2-8=-6

Thus, the charge of each atom of N should be +6/2 = +3

N: 2 * +3 = +6

(total charge = +2 -8 +6 = 0)

For NH4Cl, we start from Cl (-1) and H (+1) and then calculate the charge of N:

Cl: 1 * -1 = -1

H: 4 * +1 = +4

To achieve total charge = 0, the atoms of N must balance +4 - 1 = +3

Thus, we calculate the charge of each atom of N: +3/1 = +3

N: 1 * +3 = +3

For AuI3, the name of the compund 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 Nb2(PO4)5, we can start from O (-2) and Nb (+5, as said in the compound's name). Note that, for this molecule, we also need to consider the number outside the parenthesis!

O: 4*5 * -2 = -40

Nb: 2 * +5 = +10

To achieve total charge = 0, the atoms of P must balance -40+10=-30

Thus, we calculate the charge of each atom of P, knowing that there are 5 of them: +30/5 = +6

P: 5 * +6 = +30

(total charge = -40 + 10 +30 = 0)

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

User Geoff Kendall
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