# data[].attributes.headline
Calculate the oxidation number of Chromium in K2Cr2O7   (Include the steps)

# data[].attributes.subject
Calculate the oxidation number of Chromium in K2Cr2O7   (Include the steps)

# data[].attributes.body
Oxidation number :1*2+X+7*-2=0
2+2x-14=0
X=6

# data[].attributes.body
K2Cr2O7
2(1) + 2x+ 7(-2)=0
2 + 2x -14=0
2x-12=0
2x=12
x=6

# data[].attributes.body
Taking the Oxidation Number of Chromium as 'x'
+2(+1)+2(x)+7(-2)=0
+2+2x-14=0
2x=12
x=+6

# data[].attributes.body
K2CrO7
1x2 + 2X + 7x-2 = 0
2 + 2X -14 = 0
2x=12
x=+6

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K2Cr207
+2 + x + (-2x7)=0
2x+ -14+2=0
2X=12
x=6

# data[].attributes.body
K2Cr2O7
1(2)2x -2(7)
2+x-14=0
2x=12
x=+6

# data[].attributes.body
K2Cr2O7
1x2 +2x + (-2)x7 =0
2+ 2x-14=0
2x-12=0
X=6

# data[].attributes.body
1×2 + 2x -2×7=0
2+2x-14=0
2x-12=0
2x=12
x=12/2
X=6

# data[].attributes.body
K2Cr2O7
1x2  +  2x  + (-2)x7  = 0
2 + 2x - 14 = 0
2x - 12 = 0
x = 6

# data[].attributes.body
K2Cr2O7:-
2+2x+ (-2*7)= 0
2+2x-14=0
2x=12
x=6
therefore, the oxidation number of chromium is 6.

# data[].attributes.body
2+2x-14=0
-12+2x=0
2x=12
X=6

# data[].attributes.body
Let oxidation no of Cr  be x
(1x2) + 2x+( -2x7) =0
2+ 2x-14=0
2+2x= 14
2x= 14-2
2x= 12
x= 12/2=6

# data[].attributes.body
oxidation number of k2cr2o7
2*1+2x+7*-2=0
2+2x-14=0
2x-12=0
x=

# data[].attributes.body
K2Cr2O7–>
1(2)+2x+7(-2)=0
2+2x-14=0
2x=12
x=6

# data[].attributes.body
K2Cr2O7
+2(+1)+2(x)+7

# data[].attributes.body
K2Cr2O7
1x2+X+(-2)x7=0
2+X-14=0
X-12=0
X=12

# data[].attributes.body
K2Cr2O7
1*2 +2x + (-2)7 = +6

# data[].attributes.body
Oxidation No.
1x2+ 2x +7x-2=0
2x=14-2
x=12/2
x=6

# data[].attributes.body
K2Cr2O7
oxidation no. of  k is  1
oxidation no. of o is -2
oxidation no of Cr is x
2(1)+2x+7(-2)=0
2x=12
x=6

# data[].attributes.body
2(1) + 2x + 7(-2)=0
2 + 2x + (-14) = 0
2x = 14 - 2
2x = 12
x = 6

# data[].attributes.body
K2CR2O7
1*2+2X+ (-2)7 = +6

# data[].attributes.body
K2Cr2O7
1(2) + 2x -2(7)=0
2x + 2 - 14 = 0
2x - 12 = 0
x - 6 = 0
x = 6

# data[].attributes.body
K2Cr2O7
1x2+2x + (-2x7)=0
2+2x-14=0
2x-12=0
2x=12
x=12/2
x=+6

# data[].attributes.body
let x be the oxidation of chromium
+2{+1

# data[].attributes.body
KR2Cr2O

# data[].attributes.body
K2Cr2O7

# data[].attributes.body
K2Cr2O7
1x2+2x+7x-2=0
2+2x-14=0
x=6