NCl3 + 3H2O => NH3 + 3HOCl
Find the oxidation state for each element
There are two rules to follow to determine the oxidation states:
1. The bonding electrons of a bond A-B are allocated to the atom with the highest electronegativity. The oxidation state of this atom is then the negative sum of all allocated electrons drawn from the bonded atoms.
2. The sum of the oxidation states of all atoms in a molecule equals the charge of the molecule.
Let’s look at NCl3. The central atom, N, is bonded to three Cl atoms via a single bond. Cl has a higher electronegativity than N.
It follows: All Cl have an oxidation state of -1. N has an oxidation state of +3. The sum of oxidations states is zero.
H2O: H atoms both +1, O atom -2.
NH3: N -3, H’s +1.
HOCl: It is H-O-Cl: EN(H) < EN(O), therefore H +1 and O in the H-O part: -1. Then consider the bond O-Cl: EN(O) < EN(Cl) therefore Cl -1 and oxidation state of O is zero. O “gains” one electron from the H-O bond and “loses” on electron in the O-Cl bond. Another way of determining the oxidation state of O is applying rule 2: Since HOCl is a neutral molecule and the oxidations states of H and Cl are zero when added, the oxidation state of O must be zero.