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Tardigrade
Question
Chemistry
Oxidation number of P in P4,PH3 &NaH2PO2 respectively are :-
Q. Oxidation number of
P
in
P
4
,
P
H
3
&
N
a
H
2
P
O
2
respectively are :-
107
154
NTA Abhyas
NTA Abhyas 2020
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A
0
,
−
3
,
+
1
B
0
,
−
3
,
−
1
C
0
,
+
3
,
−
1
D
0
,
+
3
,
+
1
Solution:
P
4
→
O
⋅
N
=
0
(
P
H
)
3
→
x
+
3
(
+
1
)
=
0
x
=
−
3
(
N
a
H
)
2
(
PO
)
2
→
(
+
1
)
+
2
(
+
1
)
+
x
+
2
(
−
2
)
=
0
x
=
+
1