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Question
Chemistry
In K4[Fe(CN)6] the EAN of Fe is
Q. In
K
4
[
F
e
(
CN
)
6
]
the EAN of
F
e
is
1820
221
Coordination Compounds
Report Error
A
33
20%
B
35
19%
C
36
40%
D
26
20%
Solution:
EAN = Atomic number - Oxidation state + 2
×
number of
Ligands
=
26
−
2
+
2
(
6
)
=
36.