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Question
Chemistry
Oxidation number of nitrogen in NaNO2 is
Q. Oxidation number of nitrogen in
N
a
N
O
2
is
1903
239
Redox Reactions
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A
+
2
13%
B
+
4
10%
C
+
3
69%
D
−
3
8%
Solution:
Let the oxidation number of N in
N
a
N
O
2
be x
+
1
+
x
+
(
−
2
)
×
2
=
0
1
+
x
−
4
=
0
;
x
=
+
3