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9. complete the following table to write the chemical formula for each …

Question

  1. complete the following table to write the chemical formula for each compound
name of compoundcationanionchemical formula
(b) lead(iv) bromide
(c) iron(iii) sulfide
(d) nickel(iii) fluoride
(e) manganese(iv) sulfide

analyze

  1. explain why you must use roman numerals to name ionic compounds containing transition metals. in which cases do you not need to use roman numerals to name an ionic compound? explain your answer.
  1. there is something wrong with the name written beside each of the following compounds. in each case, explain why the name is wrong, and then write the correct name.
formulaincorrect nameexplanationcorrect name
(b) naclsodium(i) chloride
(c) cucl₂copper chloride
(d) no₂nitrogen(iv) oxide

Explanation:

Response
First Table: Complete the Chemical Formula Table

Step1: Identify cation/anion charges

For each compound, extract the cation (with charge from Roman numeral) and anion (fixed charge):
(a) Copper(I) oxide: $\text{Cu}^+$, $\text{O}^{2-}$
(b) Lead(IV) bromide: $\text{Pb}^{4+}$, $\text{Br}^-$
(c) Iron(III) sulfide: $\text{Fe}^{3+}$, $\text{S}^{2-}$
(d) Nickel(III) fluoride: $\text{Ni}^{3+}$, $\text{F}^-$
(e) Manganese(IV) sulfide: $\text{Mn}^{4+}$, $\text{S}^{2-}$

Step2: Balance charges for neutrality

Use crisscross method to get subscripts:
(a) $2\times(+1) + 1\times(-2) = 0$ → $\text{Cu}_2\text{O}$
(b) $1\times(+4) + 4\times(-1) = 0$ → $\text{PbBr}_4$
(c) $2\times(+3) + 3\times(-2) = 0$ → $\text{Fe}_2\text{S}_3$
(d) $1\times(+3) + 3\times(-1) = 0$ → $\text{NiF}_3$
(e) $1\times(+4) + 2\times(-2) = 0$ → $\text{MnS}_2$

Brief Explanations

Transition metals can form multiple cations with different charges; Roman numerals specify the cation's charge to avoid ambiguity (e.g., iron can be $\text{Fe}^{2+}$ or $\text{Fe}^{3+}$). Roman numerals are not needed for ionic compounds containing group 1/2 metals (alkali/alkaline earth metals) or aluminum, as these elements only form one fixed-charge cation (e.g., $\text{Na}^+$, $\text{Mg}^{2+}$).

Step1: Assess each formula/name pair

For each entry, identify naming rules violated:
(a) $\text{Ti}_2\text{O}_3$: Ti is a transition metal; use Roman numerals, not prefixes.
(b) $\text{NaCl}$: Na is group 1, fixed charge; no Roman numeral needed.
(c) $\text{CuCl}_2$: Cu is a transition metal; charge must be specified.
(d) $\text{NO}_2$: It is a covalent compound; use prefixes, not Roman numerals.

Step2: Calculate charges (ionic) or apply covalent rules

(a) Ti charge: $\frac{2\times(-3)}{2}=+3$ → titanium(III) oxide
(b) Na only forms $\text{Na}^+$ → sodium chloride
(c) Cu charge: $+2$ (from $2\times(-1)$) → copper(II) chloride
(d) Covalent: 1 N, 2 O → nitrogen dioxide

Answer:

Name of compoundCationAnionChemical Formula
(b) lead(IV) bromide$\text{Pb}^{4+}$$\text{Br}^-$$\text{PbBr}_4$
(c) iron(III) sulfide$\text{Fe}^{3+}$$\text{S}^{2-}$$\text{Fe}_2\text{S}_3$
(d) nickel(III) fluoride$\text{Ni}^{3+}$$\text{F}^-$$\text{NiF}_3$
(e) manganese(IV) sulfide$\text{Mn}^{4+}$$\text{S}^{2-}$$\text{MnS}_2$

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Analyze 1: Roman Numerals in Ionic Naming