How to Find Oxidation Numbers of Elements

Analyze the atomic composition of the molecule and determine to which group in the periodic table each atom belongs.
In Na2CrO4, sodium (Na) belongs to group IA, chromium (Cr) is in group VIB, while oxygen (O) is from group VIA.

Assign the oxidation number +1 for the group IA alkali metal atoms (for example, Na or K) if such atoms are present in the molecule.
In the above example, it is sodium (Na).

Assign the oxidation number +2 for the group IIA alkaline earth atoms (for example, Ca or Mg).
There are no such elements in the molecule in the above example.

Assign the oxidation number -1 for halogens (the group VIIA).
There are no such elements in the above example.

Assign the oxidation number +1 for hydrogen, which doesn't apply to this example.

Assign the oxidation number -2 for oxygen.

In our example, there is oxygen (O) in the molecule Na2CrO4.

Multiply the oxidation number for each element by its quantity in the molecule. If the oxidation number cannot be derived from Steps 2 to 6, denote it as ̶0;X.̶1;

In the example:
Sodium (Na): +1 x 2 = 2
Oxygen (O): -2 x 4 = -8
chromium (Cr): X x 1 = X

Add up all values obtained in Step 7 and make them equal to zero. Zero is the total charge of a molecule that must remain neutral.
In our example, 2 + X - 8 = 0 or X ̵1; 6 =0.

Solve the equation from Step 8 using algebra rules to determine the oxidation number of the element denoted as "X."

In our example, X = 6. Thus, the oxidation number of chromium (Cr) in sodium chromate is +6.