Practice worksheet for naming ionic compounds and writing their chemical formulas.
Naming Ionic Compounds Practice Worksheet with questions to name ionic compounds and write formulas for given compounds.
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Step-by-step solution for: Naming Ionic Compounds Practice Worksheet | Science 9th Grade
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Step-by-step solution for: Naming Ionic Compounds Practice Worksheet | Science 9th Grade
Here's the complete solution to the Naming Ionic Compounds Practice Worksheet, with clear explanations for each answer.
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## ✔ PART 1: Name the following ionic compounds
Answer: Ammonium chloride
Explanation: NH₄⁺ is the ammonium ion (polyatomic cation), Cl⁻ is chloride. No Roman numeral needed since ammonium has a fixed +1 charge.
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Answer: Iron(III) nitrate
Explanation: Fe is iron, which can have multiple charges. NO₃⁻ is nitrate (-1). Three nitrates = -3 total charge → Fe must be +3 → Iron(III).
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Answer: Titanium(III) bromide
Explanation: Br⁻ is bromide (-1). Three bromides = -3 → Ti must be +3 → Titanium(III).
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Answer: Copper(I) phosphide
Explanation: P³⁻ is phosphide (-3). Three Cu ions balance it → each Cu must be +1 → Copper(I). (Note: Cu can also be +2, so we specify.)
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Answer: Tin(IV) selenide
Explanation: Se²⁻ is selenide. Two selenides = -4 → Sn must be +4 → Tin(IV).
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Answer: Gallium arsenide
Explanation: Ga is gallium (+3), As is arsenide (-3). Charges balance directly. No Roman numeral needed — gallium only has one common oxidation state (+3).
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Answer: Lead(IV) sulfate
Explanation: SO₄²⁻ is sulfate (-2). Two sulfates = -4 → Pb must be +4 → Lead(IV).
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Answer: Beryllium bicarbonate
Explanation: Be²⁺ and HCO₃⁻ (bicarbonate, -1). Two bicarbonates balance beryllium’s +2. “Bicarbonate” is the common name for HCO₃⁻. No Roman numeral — Be is always +2.
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Answer: Manganese(III) sulfite
Explanation: SO₃²⁻ is sulfite (-2). Three sulfites = -6. Two Mn atoms → each Mn is +3 → Manganese(III).
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Answer: Aluminum cyanide
Explanation: Al³⁺ and CN⁻ (cyanide, -1). Three cyanides balance aluminum’s +3. Aluminum has only one common charge (+3), so no Roman numeral.
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## ✔ PART 2: Write the formulas for the following compounds
Answer: Cr(PO₄)₂
Explanation: Cr⁶⁺ and PO₄³⁻. To balance: LCM of 6 and 3 is 6 → 1 Cr⁶⁺ and 2 PO₄³⁻ → Cr(PO₄)₂.
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Answer: V(CO₃)₂
Explanation: V⁴⁺ and CO₃²⁻. To balance: 1 V⁴⁺ needs 2 CO₃²⁻ → V(CO₃)₂.
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Answer: Sn(NO₂)₂
Explanation: Sn²⁺ and NO₂⁻. Two nitrites needed → Sn(NO₂)₂.
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Answer: Co₂O₃
Explanation: Co³⁺ and O²⁻. LCM of 3 and 2 is 6 → 2 Co³⁺ and 3 O²⁻ → Co₂O₃.
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Answer: Ti(CH₃COO)₂ or Ti(C₂H₃O₂)₂
Explanation: Ti²⁺ and CH₃COO⁻ (acetate, -1). Two acetates needed → Ti(CH₃COO)₂.
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Answer: V₂S₅
Explanation: V⁵⁺ and S²⁻. LCM of 5 and 2 is 10 → 2 V⁵⁺ and 5 S²⁻ → V₂S₅.
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Answer: Cr(OH)₃
Explanation: Cr³⁺ and OH⁻. Three hydroxides needed → Cr(OH)₃.
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Answer: LiI
Explanation: Li⁺ and I⁻. Charges balance 1:1 → LiI.
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Answer: Pb₃N₂
Explanation: Pb²⁺ and N³⁻. LCM of 2 and 3 is 6 → 3 Pb²⁺ and 2 N³⁻ → Pb₃N₂.
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Answer: AgBr
Explanation: Ag⁺ and Br⁻. Silver is almost always +1 → AgBr.
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## 📝 Final Answer Key:
Part 1: Names
1) Ammonium chloride
2) Iron(III) nitrate
3) Titanium(III) bromide
4) Copper(I) phosphide
5) Tin(IV) selenide
6) Gallium arsenide
7) Lead(IV) sulfate
8) Beryllium bicarbonate
9) Manganese(III) sulfite
10) Aluminum cyanide
Part 2: Formulas
11) Cr(PO₄)₂
12) V(CO₃)₂
13) Sn(NO₂)₂
14) Co₂O₃
15) Ti(CH₃COO)₂ or Ti(C₂H₃O₂)₂
16) V₂S₅
17) Cr(OH)₃
18) LiI
19) Pb₃N₂
20) AgBr
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✔ This worksheet covers naming binary ionic compounds, polyatomic ions, transition metals with variable charges, and writing formulas using criss-cross method. Always check charges and use Roman numerals for metals that can have more than one charge!
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## ✔ PART 1: Name the following ionic compounds
1) NH₄Cl
Answer: Ammonium chloride
Explanation: NH₄⁺ is the ammonium ion (polyatomic cation), Cl⁻ is chloride. No Roman numeral needed since ammonium has a fixed +1 charge.
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2) Fe(NO₃)₃
Answer: Iron(III) nitrate
Explanation: Fe is iron, which can have multiple charges. NO₃⁻ is nitrate (-1). Three nitrates = -3 total charge → Fe must be +3 → Iron(III).
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3) TiBr₃
Answer: Titanium(III) bromide
Explanation: Br⁻ is bromide (-1). Three bromides = -3 → Ti must be +3 → Titanium(III).
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4) Cu₃P
Answer: Copper(I) phosphide
Explanation: P³⁻ is phosphide (-3). Three Cu ions balance it → each Cu must be +1 → Copper(I). (Note: Cu can also be +2, so we specify.)
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5) SnSe₂
Answer: Tin(IV) selenide
Explanation: Se²⁻ is selenide. Two selenides = -4 → Sn must be +4 → Tin(IV).
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6) GaAs
Answer: Gallium arsenide
Explanation: Ga is gallium (+3), As is arsenide (-3). Charges balance directly. No Roman numeral needed — gallium only has one common oxidation state (+3).
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7) Pb(SO₄)₂
Answer: Lead(IV) sulfate
Explanation: SO₄²⁻ is sulfate (-2). Two sulfates = -4 → Pb must be +4 → Lead(IV).
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8) Be(HCO₃)₂
Answer: Beryllium bicarbonate
Explanation: Be²⁺ and HCO₃⁻ (bicarbonate, -1). Two bicarbonates balance beryllium’s +2. “Bicarbonate” is the common name for HCO₃⁻. No Roman numeral — Be is always +2.
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9) Mn₂(SO₃)₃
Answer: Manganese(III) sulfite
Explanation: SO₃²⁻ is sulfite (-2). Three sulfites = -6. Two Mn atoms → each Mn is +3 → Manganese(III).
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10) Al(CN)₃
Answer: Aluminum cyanide
Explanation: Al³⁺ and CN⁻ (cyanide, -1). Three cyanides balance aluminum’s +3. Aluminum has only one common charge (+3), so no Roman numeral.
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## ✔ PART 2: Write the formulas for the following compounds
11) chromium(VI) phosphate
Answer: Cr(PO₄)₂
Explanation: Cr⁶⁺ and PO₄³⁻. To balance: LCM of 6 and 3 is 6 → 1 Cr⁶⁺ and 2 PO₄³⁻ → Cr(PO₄)₂.
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12) vanadium(IV) carbonate
Answer: V(CO₃)₂
Explanation: V⁴⁺ and CO₃²⁻. To balance: 1 V⁴⁺ needs 2 CO₃²⁻ → V(CO₃)₂.
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13) tin(II) nitrite
Answer: Sn(NO₂)₂
Explanation: Sn²⁺ and NO₂⁻. Two nitrites needed → Sn(NO₂)₂.
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14) cobalt(III) oxide
Answer: Co₂O₃
Explanation: Co³⁺ and O²⁻. LCM of 3 and 2 is 6 → 2 Co³⁺ and 3 O²⁻ → Co₂O₃.
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15) titanium(II) acetate
Answer: Ti(CH₃COO)₂ or Ti(C₂H₃O₂)₂
Explanation: Ti²⁺ and CH₃COO⁻ (acetate, -1). Two acetates needed → Ti(CH₃COO)₂.
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16) vanadium(V) sulfide
Answer: V₂S₅
Explanation: V⁵⁺ and S²⁻. LCM of 5 and 2 is 10 → 2 V⁵⁺ and 5 S²⁻ → V₂S₅.
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17) chromium(III) hydroxide
Answer: Cr(OH)₃
Explanation: Cr³⁺ and OH⁻. Three hydroxides needed → Cr(OH)₃.
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18) lithium iodide
Answer: LiI
Explanation: Li⁺ and I⁻. Charges balance 1:1 → LiI.
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19) lead(II) nitride
Answer: Pb₃N₂
Explanation: Pb²⁺ and N³⁻. LCM of 2 and 3 is 6 → 3 Pb²⁺ and 2 N³⁻ → Pb₃N₂.
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20) silver bromide
Answer: AgBr
Explanation: Ag⁺ and Br⁻. Silver is almost always +1 → AgBr.
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## 📝 Final Answer Key:
Part 1: Names
1) Ammonium chloride
2) Iron(III) nitrate
3) Titanium(III) bromide
4) Copper(I) phosphide
5) Tin(IV) selenide
6) Gallium arsenide
7) Lead(IV) sulfate
8) Beryllium bicarbonate
9) Manganese(III) sulfite
10) Aluminum cyanide
Part 2: Formulas
11) Cr(PO₄)₂
12) V(CO₃)₂
13) Sn(NO₂)₂
14) Co₂O₃
15) Ti(CH₃COO)₂ or Ti(C₂H₃O₂)₂
16) V₂S₅
17) Cr(OH)₃
18) LiI
19) Pb₃N₂
20) AgBr
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✔ This worksheet covers naming binary ionic compounds, polyatomic ions, transition metals with variable charges, and writing formulas using criss-cross method. Always check charges and use Roman numerals for metals that can have more than one charge!
Parent Tip: Review the logic above to help your child master the concept of ionic nomenclature worksheet.
