1. Assign known oxidation states:

* Sr (Strontium): Group 2 element, so its oxidation state is +2.

* O (Oxygen): Usually has an oxidation state of -2.

2. Set up an equation:

Since the compound is neutral, the sum of the oxidation states of all the atoms must equal zero. Let "x" represent the oxidation state of Cr:

(+2) + 2(x) + 7(-2) = 0

3. Solve for x:

* 2 + 2x - 14 = 0

* 2x = 12

* x = +6

Therefore, the oxidation state of Cr in SrCr₂O₇ is +6.