What is the oxidation number of chromium in K2Cr2O7?

I know that K = 1
O=-2 so 7*-2=-14

I do not know what to do now. Thank you for your help!

That's good, you have a good foundation. Since you're only looking for the oxidation number of chromium at this point, let's look at it like this. The polyatomic ion Cr2O7, or dichromate, has a 2- charge. Hence, the charges of the individual ions have to add up to -2. So, we set up an equation, where x is the oxidation number of chromium. In this case, we'll have 2(x) + 7(-2) = -2. See what I did there? I set up an equation so that the individual charges will all add up to -2, so let's solve for x, and when you do that, you'll get +6, which happens to be the answer!

To find the oxidation number of chromium (Cr) in K2Cr2O7, we can start by assigning oxidation numbers to potassium (K) and oxygen (O).

Since potassium belongs to Group 1 elements in the periodic table, its oxidation number is +1.

Oxygen, on the other hand, usually has an oxidation number of -2 in compounds, unless it is bonded to a more electronegative element or in a peroxide where it has an oxidation number of -1.

There are seven oxygen atoms in K2Cr2O7, so their total oxidation number is -14 (-2 * 7 = -14).

Now, to find the oxidation number of chromium, we can set up the equation:

2(K) + 2(Cr) + 7(O) = 0

Substituting the oxidation numbers we know:

2(+1) + 2(Cr) + 7(-2) = 0

Simplifying, we get:

2 + 2(Cr) - 14 = 0

Combining like terms:

2(Cr) - 12 = 0

Adding 12 to both sides:

2(Cr) = 12

Dividing both sides by 2:

Cr = 6

Therefore, the oxidation number of chromium in K2Cr2O7 is +6.