Question

case 2. ionic compounds: transition metal + non - metal
in general, it is not possible to use the periodic table to predict what ions are formed by transition metals. you have to learn the formulas for the ions of those transition elements that were assigned in lecture. some of these metal elements also form more than one kind of stable ion. these kind of metals (and their cations) are called type ii metals. you will see that you can figure out the charge on a type ii metal cation in a given compound either from its formula or from its name.

if the transition metal forms only one ion, name the compound as in case 1 (page 1). examples: zncl₂ zinc chloride; ag₂s silver sulfide; nif₂ nickel fluoride

if the transition metal forms two ions, the naming system is as follows:

modern (stock) system: a roman numeral after the metal name indicates the charge on the metal ion. this is the system to learn for class.
examples: fe³⁺ is iron (iii) sn⁴⁺ is tin (iv) cu⁺ is copper (i)

old system: -ous ending refers to the ion with lower charge. -ic ending refers to the ion with the higher charge.
fe²⁺ ferrous fe³⁺ ferric cr²⁺ chromous cr³⁺ chromic
cu⁺ cuprous cu²⁺ cupric hg₂²⁺ mercurous hg²⁺ mercuric
sn²⁺ stannous sn⁴⁺ stannic pb²⁺ plumbous pb⁴⁺ plumbic

note: the charge on a transition metal ion can be determined from its compound.
examples: in fef₂ there are 2 f⁻ so it’s fe⁺² name: ferrous fluoride or iron (ii) fluoride in fe₂o₃ there are 3 o²⁻ so here the ion is fe⁺³ name: ferric oxide or iron (iii) oxide

note: you must be able to give the modern (stock) name only for this class on quizzes and exams but should be able use the old system using this worksheet and the book.

exercise: name the following compounds (two names if applicable):
agcl ____________ febr₃ ____________
cu₃n ____________ cr₂s₃ ____________

exercise: give formulas for the following compounds:
chromium (iii) oxide ____________ stannous fluoride __________ ferrous iodide ____________
zinc nitride ____________ cupric bromide __________ cobalt (ii) oxide ____________

Explanation:

Step1: Analyze AgCl

Silver (Ag) forms only one stable ion ($Ag^+$). Chloride is $Cl^-$. So the name is silver chloride.

Step2: Analyze $FeBr_3$

Bromide is $Br^-$. There are 3 bromide ions, so iron has a charge of +3. In the modern (Stock) system, it is iron (III) bromide. In the old - system, it is ferric bromide.

Step3: Analyze $Cu_3N$

Nitride is $N^{3 - }$. For the compound to be neutral, each copper ion has a charge of +1. In the modern (Stock) system, it is copper (I) nitride. In the old - system, it is cuprous nitride.

Step4: Analyze $Cr_2S_3$

Sulfide is $S^{2 - }$. There are 3 sulfide ions, so each chromium ion has a charge of +3. In the modern (Stock) system, it is chromium (III) sulfide. In the old - system, it is chromic sulfide.

Step5: Find formula for chromium (III) oxide

Chromium has a charge of +3 ($Cr^{3+}$) and oxide is $O^{2 - }$. To balance the charges, the formula is $Cr_2O_3$.

Step6: Find formula for stannous fluoride

Stannous means tin with a +2 charge ($Sn^{2+}$), fluoride is $F^-$. The formula is $SnF_2$.

Step7: Find formula for ferrous iodide

Ferrous means iron with a +2 charge ($Fe^{2+}$), iodide is $I^-$. The formula is $FeI_2$.

Step8: Find formula for zinc nitride

Zinc forms a +2 ion ($Zn^{2+}$), nitride is $N^{3 - }$. To balance the charges, the formula is $Zn_3N_2$.

Step9: Find formula for cupric bromide

Cupric means copper with a +2 charge ($Cu^{2+}$), bromide is $Br^-$. The formula is $CuBr_2$.

Step10: Find formula for cobalt (II) oxide

Cobalt has a +2 charge ($Co^{2+}$), oxide is $O^{2 - }$. The formula is $CoO$.

Answer:

AgCl: silver chloride
$FeBr_3$: iron (III) bromide, ferric bromide
$Cu_3N$: copper (I) nitride, cuprous nitride
$Cr_2S_3$: chromium (III) sulfide, chromic sulfide
chromium (III) oxide: $Cr_2O_3$
stannous fluoride: $SnF_2$
ferrous iodide: $FeI_2$
zinc nitride: $Zn_3N_2$
cupric bromide: $CuBr_2$
cobalt (II) oxide: $CoO$