Questions: Atoms, Ions and Molecules Deducing the ions in a binary ionic compound from its empirical formula 0/5 Complete the table below by writing the symbols for the cation and anion that make up each ionic compound. The first row has been completed for you. ionic compound cation anion ------------------------------- NaCl Na+ Cl- CrCl3 CrS2 CuF ZnBr2

Atoms, Ions and Molecules
Deducing the ions in a binary ionic compound from its empirical formula
0/5

Complete the table below by writing the symbols for the cation and anion that make up each ionic compound. The first row has been completed for you.

ionic compound cation anion
-------------------------------
NaCl Na+ Cl-
CrCl3
CrS2
CuF
ZnBr2

Transcript text: Atoms, Ions and Molecules Deducing the ions in a binary ionic compound from its empirical formula 0/5 Complete the table below by writing the symbols for the cation and anion that make up each ionic compound. The first row has been completed for you. \begin{tabular}{|c|c|c|} \hline \begin{tabular}{c} ionic \\ compound \end{tabular} & cation & anion \\ \hline NaCl & $\mathrm{Na}^{+}$ & $\mathrm{Cl}^{-}$ \\ \hline $\mathrm{CrCl}_{3}$ & $\square$ & $\square$ \\ \hline $\mathrm{CrS}_{2}$ & $\square$ & $\square$ \\ \hline CuF & $\square$ & $\square$ \\ \hline $\mathrm{ZnBr}_{2}$ & $\square$ & $\square$ \\ \hline \end{tabular}

Solution

Solution Steps

Step 1: Determine the Cation and Anion for $\mathrm{CrCl}_{3}$

Identify the metal in the compound: Chromium (Cr).
Chromium can have multiple oxidation states; in $\mathrm{CrCl}_{3}$, it is typically $\mathrm{Cr}^{3+}$.
Chlorine (Cl) is a halogen and forms $\mathrm{Cl}^{-}$ ions.
Therefore, the cation is $\mathrm{Cr}^{3+}$ and the anion is $\mathrm{Cl}^{-}$.

Step 2: Determine the Cation and Anion for $\mathrm{CrS}_{2}$

Identify the metal in the compound: Chromium (Cr).
In $\mathrm{CrS}_{2}$, chromium is typically in the $\mathrm{Cr}^{4+}$ state to balance two $\mathrm{S}^{2-}$ ions.
Sulfur (S) forms $\mathrm{S}^{2-}$ ions.
Therefore, the cation is $\mathrm{Cr}^{4+}$ and the anion is $\mathrm{S}^{2-}$.

Step 3: Determine the Cation and Anion for $\text{CuF}$

Identify the metal in the compound: Copper (Cu).
Copper in $\text{CuF}$ is typically in the $\mathrm{Cu}^{+}$ state.
Fluorine (F) is a halogen and forms $\mathrm{F}^{-}$ ions.
Therefore, the cation is $\mathrm{Cu}^{+}$ and the anion is $\mathrm{F}^{-}$.

Final Answer

\[ \begin{array}{|c|c|c|} \hline \text{ionic compound} & \text{cation} & \text{anion} \\ \hline \mathrm{CrCl}_{3} & \mathrm{Cr}^{3+} & \mathrm{Cl}^{-} \\ \hline \mathrm{CrS}_{2} & \mathrm{Cr}^{4+} & \mathrm{S}^{2-} \\ \hline \text{CuF} & \mathrm{Cu}^{+} & \mathrm{F}^{-} \\ \hline \mathrm{ZnBr}_{2} & \mathrm{Zn}^{2+} & \mathrm{Br}^{-} \\ \hline \end{array} \]

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