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

1889 263 JIPMERJIPMER 2004 Report Error

A

+1

B

+2

C

+3

D

+4

Solution:

Let, oxidation number of $F e$ be ' $x$ '.
$∴ 3 (+ 1) + x + 6 (- 1) = 0$
$3 + x - 6 = 0$
$x = + 3$