write the oxidation state of CR in K2 cr2 O7 white Oxidation state of CR in K2 cr2 O7 also describe the steps you have taken to get the answer
Answers
Answer:
Explanation:
First you have to ‘break up’ the potassium dichromate into its elements.
K = potassium
O = oxygen
Cr = chromium.
Dismissing the uncharged neutral elements as possibilities (oxidation state = 0), figure out the most common (or in the case of potassium the only) ionic form(s) or oxidation numbers of each element in the formula.
K + = Potassium ion (positive one charge)
O 2− = Oxide ion (negative two charge)—always true except for peroxides (-1) or OF 2 (oxygen difluoride, +2)
K 2 Cr 2 O 7 is a neutral salt overall; net charge = 0
Only chromium is a ‘wildcard’ here as it can exhibit multiple oxidation states: II, III, or VI (there are others but I, IV, or V are seldom seen outside inorganic research). Same as Cr 2+ , Cr 3+ , or Cr 6+ .
Multiply the charges according to the formula, the sum being set as zero. “Cr” is being used as the algebraic unknown variable in the math below.
0 = 2 x (K + ) + 7 x (O 2− ) + 2 x Cr(?)
0 = 2(+1) - 7(-2) + (2 x Cr) = 2 - 14 + 2Cr = -12 + 2Cr
Rearrange the order of the equation.
2Cr - 12 = 0
Add 12 to both sides.
2Cr = 12
Finally, divide both sides by 2.
Cr = 6
That means the chromium ions have formal charges of +6 or an oxidation state of VI in potassium dichromate.