Q. The oxidation number of Cr in $K_{2} C r_{2} O_{7}$ is

1258 237 Redox Reactions Report Error

A

$- 2$

2%

B

$- 7$

3%

C

$+ 2$

3%

D

$+ 6$

92%

Solution:

$K_{2} C * r_{2} O_{7}$
$2 + 2 x - 2 \times 7 = 0; 2 x - 14 + 2 = 0$
$2 x = 12; x = \frac{12}{2} = + 6$