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Tardigrade
Question
Chemistry
The oxidation state of Fe in the brown ring complex: [Fe(H2O)5 NO]SO4 is
Q. The oxidation state of
F
e
in the brown ring complex :
[
F
e
(
H
2
O
)
5
NO
]
S
O
4
is
3805
167
KCET
KCET 2009
Coordination Compounds
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A
+2
23%
B
+1
37%
C
+3
32%
D
0
8%
Solution:
Let the oxidation state of
F
e
in
[
F
e
(
H
2
O
)
5
NO
]
S
O
4
is
x
.
[
F
e
(
H
2
O
)
5
NO
]
2
+
⇒
x
+
0
+
1
=
2
∴
x
=
+
1
Here
NO
exists as nitrosyl ion
(
N
O
+
)