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Tardigrade
Question
Chemistry
The effective atomic number of Fe in K 4[ Fe ( CN )6] is
Q. The effective atomic number of
F
e
in
K
4
[
F
e
(
CN
)
6
]
is
1827
164
Report Error
A
35
15%
B
38
5%
C
36
70%
D
34
10%
Solution:
EAN
=
Atomic No.
−
Oxidation state
+
2
×
CN
=
26
−
2
+
2
×
6
=
36