Tentukan bilangan oksidasi unsur yang bergaris bawah A. Na2 cr2 o7 ( cr bergaris bawah) B. S2 o3 (s bergaris bawah) C.kmn o4 ( mn bergaris bawah) D. K2 o ( o be

A. Na2Cr2O2
( 2×biloks Na) + ( 2× biloks Cr) + ( 7× biloks O) = 0
( 2 × +1) + ( 2 Cr) + ( 7× -2) = 0
(+2) + (2Cr) + (-14) = 0
(+2) + (2Cr) + = 0+14
2Cr = +14 - 2
Cr = +12/2= +6
B. S2O3
( 2×biloks S)+(3× biloks O)=0
(2S) + (3× -2)=0
2S + -6 =0
2S =0+6
S =+6/2
S =+3
C. KMnO4
(1×biloks K) + (1×biloks Mn) +(4 ×biloks O) =0
(1× (+1) ) + (Mn) + (4× -2)=0
(+1) + Mn + (-8)=0
+1 + Mn =0+8 =+8
Mn =+8 - 1
Mn =+7
D. K2O
(2×biloks K) + (1×biloks O)=0
(2× +1) + ( O) =0
+2 + O =0
O=0-2
=-2