Question

The oxidation state of Cr in K₂Cr₂O₇ is

(A) +6

(B) -7

(C) +2

(D) -2

Answer 1

Correct option is (A) +6

Let the Ox. no. of Cr in K₂Cr₂O₇ be x.

We know that, Ox. no. of K = +1

Ox. no. of O = −2

So, 2(Ox. no. of K) + 2(Ox. no. of Cr) + 7(Ox. no. of O) = 0

2(+1) + 2(x) + 7(−2) = 0

+2 + 2x − 14 = 0

2x = +14 − 2 = +12

x = \(+\frac{12} 2\) = +6

Hence, oxidation number of Cr in K₂Cr₂O₇ is +6.