Two electrons are lost and the sum of the oxidation states of the atoms increases by 2 units. In the conversion of water to 1/2 equivalent of molecular oxygen, we can see that the oxidation state of the oxygen atom changes from -II to 0. Overall, the addition of the 2 electrons causes the total of the oxidation states to decrease by 2 units. One of the carbons bonded to nitrogen goes from I to 0 oxidation state. The carbon that bonds to the additional proton goes from -I to -II in oxidation state. The oxidation state of the nitrogen atom doesn't change but 2 of the carbon atoms decrease in oxidation state.

Let's look at the transformation of NADP to NADPH and focus on the 3 carbon atoms and the nitrogen atom that seem to change their bonds (in red).

Use Roman Numerals to indicate the oxidation state.

total < number of valence electrons, the atom has a positive oxidation state (+1 for every electron below the valence number).

total > number of valence electrons, the atom has a negative oxidation state (-1 for every additional electron above the valence number).

total = number of valence electrons, oxidation state is zero.

Compare that number to the number of valence electrons of that atom.

Break all 2-electron bonds and give both electrons to the more electronegative of the bonding pair.

In the oxidation state formalism, we consider that each atom in a molecule or ion is bonded to the others through ionic bonds. The oxidation state of an atom gives us an indication of the electron density around the atom and it helps to keep track of the electron change in oxidation-reduction reactions.

Assigning oxidation numbers