Question

For $K_4[Fe(CN)_6]$, coordination number and oxidation state are respectively

• A. $6$, $6$
• B. $2$, $2$
• C. $6$, $2$
• D. $2$, $6$

Answer 1

Answer: (C)

$6$, $2$

Answer 2

In $K_4[Fe(CN )_6]$, the central metal $Fe$ is surrounded by $6\, CN ^-$ ligands. Hence, coordination no. of $Fe$ is $6$. Oxidation state of $Fe = 4 + x + (-6) = 0$ $\Rightarrow x - 2 = 0$ $\Rightarrow x = + 2$