Question

The oxidation number of Cr in $K_{2}Cr_{2}O_{7}$ is

• A.
  • 6
• B. – 7
• C.
  • 2
• D. – 2

Answer 1

Answer: (A)

• 6

Answer 2

$K_{2}{\underline{Cr}}_{2}O_{7}$

$2 + 2x - 2 \times 7 = 0$or $2x - 14 + 2 = 0$or $2x = 12$or $x = \frac{12}{2} = + 6$

So, oxidation state of Cr is +6.