For CuI. We deserve to see that the steel in the ionic link is Cu (copper), which is a change metal. This method it has many possible charges.

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The lack of a subscript signifies the copper has acharge of +1. This way the cation isCu+, i beg your pardon is namedcopper (I). Top top the other hand, the anion in the formula isI–which corresponds toiodine. However, since I– is an anion, we add the suffix–ide, make itiodide.

This means the name of CuI is copper (I) iodide.

For Fe2O3. We have the right to see the the steel in the ionic link is Fe (iron), which is a transition metal. This method it has many possible charges.


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Problem Details

Give the systematic names of these compounds. Assignment counts!

(a) CuI(s)

(b) Fe2O3(s)

(c) CoF2(s)

(d) FeO(s)

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