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Chemistry 13 Online
OpenStudy (anonymous):

What is the oxidation number for iodine in Mg(IO3)2 ?

rvc (rvc):

\[\rm Mg(IO_3)_2\] ?

OpenStudy (anonymous):

You should be able to firstly place the oxidation numbers of the main ions. For instance, group 1 ions have +1 Group 2 have +2 Hydrogen has +1 Oxygen has -2 For a given compound, you have to make an equation of the oxidation number of all elements present, equating them to zero. Simplifying that equation, we get \[MgI _{2}O _{6}\] For Mg we have 1 atom, for I we have 2 and for O, 6. The equation then becomes\[+2+2x+(-2*6)=0\]

OpenStudy (anonymous):

\[2x=10\] \[x=5\]

rvc (rvc):

Good Work :)

OpenStudy (anonymous):

Nice hearing that, @rvc.

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