In the redox conversion of ni2+ to nio4−, the oxidation number of ni goes from (−2, 0, +2) to (−1, +1, +7, +9). recall that the oxidation number of oxygen is typically −2. 1. 0; +9 2. +2; +7 3. +2; −1 4. −2; +9 5. −2; +7 6. 0; +1 7. 0; +7 8. +2; +9 9. −2; +1

Oxidation number of an atom is the charge that atom would have if the compound is composed of ions. In neutral substances that contains atoms of one element the oxidation number of an atom is zero. Thus atoms in O2, Ni2, and aluminium all have oxidation number of zero.
In this case, Ni2, the oxidation number of Ni atom is zero, 
for NiO4-, assuming oxidation number of Ni is x 
 (x ×1) + (-2 × 4) = -1
  x = + 7
Therefore, the oxidation number goes from 0 to +7

Because electrons have negligible mass, to account for the mass of the isotope, there must be 16 neutrons, 16 neutrally charged, massive, fundamental particles present in the sulfur nucleus. and thus finally, if we have the 32s2− ion, there are 18 electrons, 16 protons, and 16 neutrons.