Question

The oxidation number of chromium in $K_2Cr_2O_7$ is

• A. -6
• B. 6
• C. 2
• D. -2

Answer 1

Answer: (B)

6

Answer 2

Let the oxidation state of S be x
Oxidation number of Na is +1 and that of oxygen is -2.

Substituting the values

2 x +1 + 4X + 6 x(-2)
4X -10 = 0.
X = 2.5

Therefore the oxidation state of S in the compound is 2.5.