Question

determine the chemical formula of the ionic compounds listed below with both a transition metal and a polyatomic ion: 1. copper (ii) hydroxide 2. platinum (iv) chlorate 3. iron (iii) acetate 4. cobalt (ii) chlorite 5. tin (iv) sulfate

Explanation:

Step1: Recall ion - charge rules

For transition metals, the Roman numeral indicates the charge of the metal ion. Polyatomic ions have fixed charges. Hydroxide ($OH^-$) has a - 1 charge, acetate ($C_2H_3O_2^-$) has a - 1 charge, chlorite ($ClO_2^-$) has a - 1 charge, chlorate ($ClO_3^-$) has a - 1 charge, and sulfate ($SO_4^{2 - }$) has a - 2 charge.

Step2: Determine formula for Copper(II) hydroxide

Copper(II) has a charge of $Cu^{2+}$. Using the criss - cross method to balance charges, the formula is $Cu(OH)_2$.

Step3: Determine formula for Iron(III) acetate

Iron(III) has a charge of $Fe^{3+}$. Acetate is $C_2H_3O_2^-$. Using the criss - cross method, the formula is $Fe(C_2H_3O_2)_3$.

Step4: Determine formula for Cobalt(II) chlorite

Cobalt(II) has a charge of $Co^{2+}$. Chlorite is $ClO_2^-$. Using the criss - cross method, the formula is $Co(ClO_2)_2$.

Step5: Determine formula for Platinum(IV) chlorate

Platinum(IV) has a charge of $Pt^{4+}$. Chlorate is $ClO_3^-$. Using the criss - cross method, the formula is $Pt(ClO_3)_4$.

Step6: Determine formula for Tin(IV) sulfate

Tin(IV) has a charge of $Sn^{4+}$. Sulfate is $SO_4^{2 - }$. Using the criss - cross method, the formula is $Sn(SO_4)_2$.

Answer:

1. $Cu(OH)_2$
2. $Fe(C_2H_3O_2)_3$
3. $Co(ClO_2)_2$
4. $Pt(ClO_3)_4$
5. $Sn(SO_4)_2$