Question

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

ionic compound	cation	anion
mn(co₃)₂
vbr₅
(nh₄)₂s
crf₃

Explanation:

Step1: Analyze $Mn(CO_3)_2$

In $Mn(CO_3)_2$, the carbonate ion $CO_3^{2 - }$ has a charge of - 2. Since there are 2 carbonate ions, the total negative charge is - 4. So the manganese ion $Mn$ has a charge of + 4, and the cation is $Mn^{4+}$, the anion is $CO_3^{2 - }$.

Step2: Analyze $VBr_5$

Bromide ion $Br^-$ has a charge of - 1. Since there are 5 bromide ions, the total negative charge is - 5. So the vanadium ion $V$ has a charge of + 5, and the cation is $V^{5+}$, the anion is $Br^-$.

Step3: Analyze $(NH_4)_2S$

The ammonium ion $NH_4^+$ has a charge of + 1 and the sulfide ion $S^{2 - }$ has a charge of - 2. So the cation is $NH_4^+$, the anion is $S^{2 - }$.

Step4: Analyze $CrF_3$

Fluoride ion $F^-$ has a charge of - 1. Since there are 3 fluoride ions, the total negative charge is - 3. So the chromium ion $Cr$ has a charge of + 3, and the cation is $Cr^{3+}$, the anion is $F^-$.

Answer:

Ionic compound	cation	anion
$VBr_5$	$V^{5+}$	$Br^-$
$(NH_4)_2S$	$NH_4^+$	$S^{2 - }$
$CrF_3$	$Cr^{3+}$	$F^-$