What is the oxidation state assigned to Cr in zinc dichromate, ZnCr2O7?
To determine the oxidation state assigned to Cr in zinc dichromate (ZnCr2O7), we first need to understand the oxidation state of Zn (zinc) and O (oxygen).
Since Zn is an alkaline earth metal and is commonly found in compounds with a +2 oxidation state, we can assume that Zn in ZnCr2O7 also has an oxidation state of +2.
Oxygen (O) typically has an oxidation state of -2 in compounds, unless it is bonded to fluorine (F). However, in this case, there are two oxygen atoms bonded to each chromium atom, so the total charge from oxygen is -14.
To find the oxidation state of Cr, we can set up the equation:
2(oxidation state of Cr) + 7(-2) = 0
Simplifying this equation, we get:
2(oxidation state of Cr) - 14 = 0
Adding 14 to both sides, we get:
2(oxidation state of Cr) = 14
Dividing both sides by 2, we find that the oxidation state of Cr in ZnCr2O7 is +7.
To determine the oxidation state of Cr in zinc dichromate (ZnCr2O7), we need to know the oxidation states of the other atoms in the compound.
First, let's assign the oxidation state of Zn. Since Zn is a Group 2 element, it usually has an oxidation state of +2 in compounds. So, in ZnCr2O7, Zn has an oxidation state of +2.
Next, let's assign the oxidation state of oxygen (O). Oxygen is typically assigned an oxidation state of -2 in compounds, except when it is combined with fluorine (F) or in certain peroxides and superoxides. Therefore, each oxygen in ZnCr2O7 has an oxidation state of -2.
To calculate the oxidation state of Cr, we need to use the known oxidation states of Zn and O combined with the overall charge of the compound, which is 0 since ZnCr2O7 is neutral.
Step 1: Determine the total oxidation state contributed by Zn and O.
Total oxidation state contributed by Zn: +2
Total oxidation state contributed by O: 7 x (-2) = -14
Step 2: Determine the oxidation state of Cr.
The total oxidation state contributed by Zn and O must cancel out each other since the overall charge is 0. So, the oxidation state of Cr can be calculated as follows:
Cr + 2 x (+2) + 7 x (-2) = 0
Cr + 4 - 14 = 0
Cr - 10 = 0
Cr = +10
Therefore, the oxidation state assigned to Cr in ZnCr2O7 is +10.
Note: This oxidation state is quite high for chromium and indicates that it is in a highly oxidized state.
All compounds are zero.
You know Zn and O, solve for Cr and you want it for EACH Cr atom.