The oxidation state of cr in K2Cr2O7 is
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Answered by
2
Answer:
K2Cr2O7
2(+1) + 2x + 7(-2) = 0
2 + 2x -14 = 0
2x = 12
x = 12/2
x = +6
Therefore oxidtaion no. of Cr in K2Cr2O7 is (+6).
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Answered by
14
Answer:
+ 6
Explanation:
Let O.N. of Cr in K₂Cr₂O₇ be 'y'
We know :
O.N. of K = + 2
O.N. of O = - 2
Here net charge on compound is zero.
= > ( 1 × 2 ) + ( y × 2 ) + ( - 2 × 7 ) = 0
= > 2 + 2 y - 14 = 0
= > 2 y + 2 - 14 = 0
= > 2 y - 12 = 0
= > 2 y = 12
= > y = 12 / 2
= > y = + 6 .
Therefore , O.N. of Cr in K₂Cr₂O₇ is + 6 .
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