Question

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
agbr	$square$	$square$
$ce{cr2s3}$	$square$	$square$
$ce{mnf2}$	$square$	$square$
$ce{nii2}$	$square$	$square$

Explanation:

Step1: Analyze AgBr

Ionic compound AgBr is formed by silver ion and bromide ion. Silver has a +1 charge, bromide has -1. So cation is $\text{Ag}^+$, anion is $\text{Br}^-$.

Step2: Analyze $\text{Cr}_2\text{S}_3$

In $\text{Cr}_2\text{S}_3$, let Cr charge be $x$. $2x + 3(-2)=0\Rightarrow x = +3$. So cation is $\text{Cr}^{3+}$, anion is $\text{S}^{2-}$ (sulfide ion, charge -2).

Step3: Analyze $\text{MnF}_2$

For $\text{MnF}_2$, let Mn charge be $y$. $y + 2(-1)=0\Rightarrow y = +2$. Cation is $\text{Mn}^{2+}$, anion is $\text{F}^-$ (fluoride ion, charge -1).

Step4: Analyze $\text{NiI}_2$

In $\text{NiI}_2$, let Ni charge be $z$. $z + 2(-1)=0\Rightarrow z = +2$. Cation is $\text{Ni}^{2+}$, anion is $\text{I}^-$ (iodide ion, charge -1).

Answer:

Ionic Compound	Cation	Anion
$\text{Cr}_2\text{S}_3$	$\text{Cr}^{3+}$	$\text{S}^{2-}$
$\text{MnF}_2$	$\text{Mn}^{2+}$	$\text{F}^-$
$\text{NiI}_2$	$\text{Ni}^{2+}$	$\text{I}^-$