Naming Ionic Compounds Practice Worksheet | Lecture notes ... - Free Printable

This worksheet focuses on naming ionic compounds and writing formulas for ionic compounds, which are fundamental skills in chemistry. Let’s break down the rules and then explain each answer.

---

## 🧪 PART 1: Naming Ionic Compounds (Problems 1–10)

General Rules:

- Cation first, anion second.
- If the cation is a transition metal (or metal that can have multiple charges), use Roman numerals to indicate its charge.
- For polyatomic ions, memorize their names and charges.
- For monatomic anions, change the ending to “-ide” (e.g., Cl⁻ → chloride).
- NH₄⁺ is ammonium — always +1.
- Some metals (like Sn, Pb) also need Roman numerals even though they’re not transition metals — because they form multiple cations.

---

🔍 Problem-by-Problem Explanations:

#### 1) NH₄Cl
- Cation: NH₄⁺ → ammonium
- Anion: Cl⁻ → chloride
✔ Answer: ammonium chloride

#### 2) Fe(NO₃)₃
- Fe is iron → needs Roman numeral.
- NO₃⁻ is nitrate → charge = -1. There are 3 of them → total negative charge = -3.
- So Fe must be +3 → iron(III)
- Anion: nitrate
✔ Answer: iron(III) nitrate

#### 3) TiBr₃
- Ti is titanium → needs Roman numeral.
- Br⁻ is bromide → charge = -1. 3 Br⁻ → total = -3.
- So Ti = +3 → titanium(III)
✔ Answer: titanium(III) bromide

#### 4) Cu₃P
- Cu is copper → needs Roman numeral.
- P³⁻ is phosphide → charge = -3.
- 3 Cu atoms → total positive charge = +3 → so each Cu = +1 → copper(I)
✔ Answer: copper(I) phosphide

#### 5) SnSe₂
- Sn is tin → needs Roman numeral.
- Se²⁻ is selenide → 2 Se = -4 total.
- So Sn = +4 → tin(IV)
✔ Answer: tin(IV) selenide

#### 6) GaAs
- Ga is gallium → typically +3.
- As is arsenic → forms As³⁻ → arsenide.
- No Roman numeral needed for Ga (only one common charge).
✔ Answer: gallium arsenide

#### 7) Pb(SO₄)₂
- Pb is lead → needs Roman numeral.
- SO₄²⁻ is sulfate → 2 sulfates = -4 total.
- So Pb = +4 → lead(IV)
✔ Answer: lead(IV) sulfate

#### 8) Be(HCO₃)₂
- Be is beryllium → always +2.
- HCO₃⁻ is bicarbonate (common name for hydrogen carbonate).
- No Roman numeral needed.
✔ Answer: beryllium bicarbonate

#### 9) Mn₂(SO₃)₃
- Mn is manganese → needs Roman numeral.
- SO₃²⁻ is sulfite → 3 sulfites = -6 total.
- 2 Mn atoms → total positive charge = +6 → each Mn = +3 → manganese(III)
✔ Answer: manganese(III) sulfite

#### 10) Al(CN)₃
- Al is aluminum → always +3.
- CN⁻ is cyanide.
- No Roman numeral needed.
✔ Answer: aluminum cyanide

---

## 📝 PART 2: Writing Formulas (Problems 11–20)

General Rules:

- Write symbols of cation and anion.
- Use charges to determine subscripts (cross-over method).
- Polyatomic ions go in parentheses if more than one is needed.
- Simplify subscripts if possible (but not always — e.g., Cr(OH)₃ is fine).

---

🔍 Problem-by-Problem Explanations:

#### 11) chromium(VI) phosphate
- Cr⁶⁺ (from VI)
- PO₄³⁻ (phosphate)
- Cross charges: Cr₃(PO₄)₂? Wait — 6 and 3 → LCM is 6 → Cr₁(PO₄)₂? No!
Actually:
Cr⁶⁺ and PO₄³⁻ → to balance: 2 Cr⁶⁺ = +12, 3 PO₄³⁻ = -9? No!
Wait — better way:
Charge of Cr = +6, PO₄ = -3 → need 1 Cr and 2 PO₄? That gives +6 and -6 → yes!
So formula: Cr(PO₄)₂ — but wait, that would imply Cr⁶⁺ and two PO₄³⁻ → +6 and -6 → balanced.
✔ Answer: Cr(PO₄)₂

> ⚠️ Note: Chromium(VI) phosphate is actually Cr(PO₄)₂ — correct as written.

#### 12) vanadium(IV) carbonate
- V⁴⁺
- CO₃²⁻
- Cross charges: V₂(CO₃)₄ → simplify to V(CO₃)₂? No — 4+ and 2- → need 2 CO₃ to balance 1 V⁴⁺? 2×(-2) = -4 → yes.
So: V(CO₃)₂
✔ Answer: V(CO₃)₂

#### 13) tin(II) nitrite
- Sn²⁺
- NO₂⁻ (nitrite)
- Need 2 NO₂⁻ to balance Sn²⁺ → Sn(NO₂)₂
✔ Answer: Sn(NO₂)₂

#### 14) cobalt(III) oxide
- Co³⁺
- O²⁻
- LCM of 3 and 2 is 6 → Co₂O₃
✔ Answer: Co₂O₃

#### 15) titanium(II) acetate
- Ti²⁺
- C₂H₃O₂⁻ (acetate)
- Need 2 acetates → Ti(C₂H₃O₂)₂
✔ Answer: Ti(C₂H₃O₂)₂

#### 16) vanadium(V) sulfide
- V⁵⁺
- S²⁻
- LCM of 5 and 2 is 10 → V₂S₅
✔ Answer: V₂S₅

#### 17) chromium(III) hydroxide
- Cr³⁺
- OH⁻
- Need 3 OH⁻ → Cr(OH)₃
✔ Answer: Cr(OH)₃

#### 18) lithium iodide
- Li⁺
- I⁻
- Simple 1:1 → LiI
✔ Answer: LiI

#### 19) lead(II) nitride
- Pb²⁺
- N³⁻
- LCM of 2 and 3 is 6 → Pb₃N₂
✔ Answer: Pb₃N₂

#### 20) silver bromide
- Ag⁺ (silver is always +1)
- Br⁻
- AgBr
✔ Answer: AgBr

---

## ✔ Final Summary

All answers provided in the worksheet are correct and follow standard IUPAC naming and formula-writing rules.

Key Concepts Reinforced:

- Use of Roman numerals for variable-charge metals.
- Recognition of polyatomic ions (nitrate, sulfate, bicarbonate, etc.).
- Balancing charges using the criss-cross method.
- Parentheses around polyatomic ions when subscripted.

This worksheet is excellent practice for mastering ionic compound nomenclature — essential for success in general chemistry!

Let me know if you’d like flashcards or quizzes based on this! 🎓