Question

complete the table below by writing

ionic compound	cation	anion
mn₂s₃
mncl₄
znbr₂
mni₂

Explanation:

Step1: Determine cations and anions for $Mn_2S_3$

In an ionic compound, the total positive charge from cations equals the total negative charge from anions. Sulfur forms a - 2 anion ($S^{2 - }$). For $Mn_2S_3$, if the charge of $Mn$ is $x$, then $2x+3\times(- 2)=0$. Solving for $x$ gives $x = + 3$. So the cation is $Mn^{3+}$ and the anion is $S^{2 - }$.

Step2: Determine cations and anions for $MnCl_4$

Chlorine forms a - 1 anion ($Cl^{-}$). Let the charge of $Mn$ be $y$. Then $y + 4\times(-1)=0$, so $y=+4$. The cation is $Mn^{4+}$ and the anion is $Cl^{-}$.

Step3: Determine cations and anions for $ZnBr_2$

Bromine forms a - 1 anion ($Br^{-}$). Zinc typically has a + 2 charge. So the cation is $Zn^{2+}$ and the anion is $Br^{-}$.

Step4: Determine cations and anions for $MnI_2$

Iodine forms a - 1 anion ($I^{-}$). Let the charge of $Mn$ be $z$. Then $z+2\times(-1)=0$, so $z = + 2$. The cation is $Mn^{2+}$ and the anion is $I^{-}$.

Answer:

ionic compound	cation	anion
$MnCl_4$	$Mn^{4+}$	$Cl^{-}$
$ZnBr_2$	$Zn^{2+}$	$Br^{-}$
$MnI_2$	$Mn^{2+}$	$I^{-}$