Calculate the oxidation number of the above questions pls?
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Na2B4O7
1×2+x×4+(-2×7)=0
2+4x-14=0
-12+4x=0
x=12/4
x=3
oxidation no of B is 3
similarly for all.if there are charges above the compound the the above on 0 will be replaced by that charge for example the h one Cr2O7²-
x×2+(-2×7)=-2
2x-14=-2
2x=12
x=6
oxidation no of Cr is 6
1×2+x×4+(-2×7)=0
2+4x-14=0
-12+4x=0
x=12/4
x=3
oxidation no of B is 3
similarly for all.if there are charges above the compound the the above on 0 will be replaced by that charge for example the h one Cr2O7²-
x×2+(-2×7)=-2
2x-14=-2
2x=12
x=6
oxidation no of Cr is 6
Anonymous:
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i don't have that much of points
hmmm
Answered by
2
Answer:
Explanation:
b answer 3x+ 4*-2 =o. So x=8÷3
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