Answer:
[tex]MgI_2\\\\MgO\\\\CrI_4\\\\ CrO_2[/tex]
Explanation:
Hello!
In this case, we can notice that among the given ions, Mg2+ and Cr4+ are cations and I- and O2- are anions, thus, for each resulting compound we switch the oxidation states as subscripts to get:
[tex]Mg^{2+}I^- \rightarrow MgI_2\\\\Mg^{2+}O^{2-} \rightarrow MgO\\\\Cr^{4+}I^-\rightarrow CrI_4\\\\Cr^{4+}O^{2-}\rightarrow CrO_2[/tex]
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