What is the oxidation states for K2CrO4

http://www.chemteam.info/Redox/Redox-Rules.html

K2= +1

Cr=+6
o=-2

To determine the oxidation states for each element in K2CrO4 (potassium chromate), we follow a set of rules:

1. The sum of the oxidation states in a compound must be equal to the overall charge of the compound.
2. The oxidation state of potassium (K) is always +1, as it belongs to Group 1 of the periodic table and tends to lose one electron to achieve a stable configuration.
3. Oxygen's (O) oxidation state is usually -2, except in peroxides like H2O2, where it is -1.
4. The oxidation state of an individual element in a compound can be calculated by assuming that all other elements have their common oxidation states.

With these rules in mind, let's determine the oxidation state for chromium (Cr):

Assume the oxidation state of chromium is "x."
The total charge of K2CrO4 is 0 (since it is neutral).

According to rule 1, we can write the equation:
2(+1) + x + 4(-2) = 0

Simplifying this equation, we have:
2 + x - 8 = 0
x - 6 = 0
x = +6

Thus, the oxidation state of chromium (Cr) in K2CrO4 is +6.
The full set of oxidation states in K2CrO4 can be written as: K(+1), Cr(+6), O(-2) x 4.