Calculate the oxidation number of the underlined element in the following compounds: i) K2Cr2O7 ii) KMnO4 iii) ClO4-.
Understand the Problem
The question is asking to calculate the oxidation number of the underlined element in three specific compounds. This involves understanding the structure and electron configuration of those compounds to determine the oxidation states.
Answer
+6 for Cr in K2Cr2O7, +7 for Mn in KMnO4, +7 for Cl in ClO4-
The oxidation numbers are: i) +6 for Cr in K2Cr2O7, ii) +7 for Mn in KMnO4, iii) +7 for Cl in ClO4-.
Answer for screen readers
The oxidation numbers are: i) +6 for Cr in K2Cr2O7, ii) +7 for Mn in KMnO4, iii) +7 for Cl in ClO4-.
More Information
The oxidation state of chromium in K2Cr2O7 is +6 because each potassium contributes +1, and oxygen contributes -2, leading to the equation 2(+1) + 2(Cr) + 7(-2) = 0. Solving gives Cr = +6. Similarly, in KMnO4, Mn is +7, and in ClO4-, Cl is +7.
Tips
Ensure you account for the charges of all atoms in the molecule correctly, including the charge on the entire polyatomic ion such as ClO4-.
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