Q. Oxidation number of Fe in $K_{3} [F e (CN)_{6}]$ is:

1295 212 BVP MedicalBVP Medical 2002 Report Error

A

$+ 1$

B

$+ 2$

C

$+ 3$

D

$+$

Solution:

Let oxidation number of Fe in $K_{3} [F e (N)_{6}]$ is x. $(+ 1 \times 3) + x + (- 1 \times 6) = 0$ $3 + x - 6 = 0$ $x = 3$