For K4[Fe(CN)6], coordination number and oxidation state are respectively

Question

For K4[Fe(CN)6], coordination number and oxidation state are respectively (A) 6, 6 (B) 2, 2 (C) 6, 2 (D) 2, 6

Tardigrade Admin · Accepted Answer

In K4[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 ⇒ x - 2 = 0 ⇒ x = + 2

Q. For $K_{4} [F e (CN)_{6}]$ , coordination number and oxidation state are respectively

$6$ , $6$

$2$ , $2$

$6$ , $2$

$2$ , $6$

In $K_{4} [F e (CN)_{6}]$ , the central metal $F e$ is surrounded by $6 C N^{-}$ ligands.
Hence, coordination no. of $F e$ is $6$ .
Oxidation state of $F e = 4 + x + (- 6) = 0$
$\Rightarrow x - 2 = 0$
$\Rightarrow x = + 2$

Q. For K4​[Fe(CN)6​], coordination number and oxidation state are respectively

6, 6

2, 2

6, 2

2, 6