Naming Ionic & Covalent Compounds Interactive Worksheet - Edform - Free Printable
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Step-by-step solution for: Naming Ionic & Covalent Compounds Interactive Worksheet - Edform
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Step-by-step solution for: Naming Ionic & Covalent Compounds Interactive Worksheet - Edform
Let’s work through each problem step by step.
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Part 1: Name the following ionic compounds
We need to name each compound using standard ionic naming rules:
- For metals with fixed charges (like Group 1, 2, Al, Zn, Ag), just use the metal name.
- For transition metals or metals that can have multiple charges (like Fe, Cu, Sn, Pb, Cr, V, Ti, Mn, etc.), we must include a Roman numeral indicating the charge.
- Nonmetal ions end in “-ide” if monatomic, or keep their polyatomic ion names (like nitrate, sulfate, cyanide, etc.).
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1) NH₄Cl
→ NH₄⁺ is ammonium (a polyatomic cation). Cl⁻ is chloride.
→ Name: Ammonium chloride
2) Fe(NO₃)₃
→ NO₃⁻ is nitrate (charge -1). There are 3 nitrates → total negative charge = -3. So Fe must be +3.
→ Iron(III) nitrate
3) TiBr₄
→ Br⁻ is bromide (-1). Four bromides → total -4. So Ti is +4.
→ Titanium(IV) bromide
4) Cu₃P
→ P³⁻ is phosphide. Three copper atoms balance one P³⁻ → each Cu is +1.
→ Copper(I) phosphide
5) SnSe₂
→ Se²⁻ is selenide. Two selenides → total -4. So Sn must be +4.
→ Tin(IV) selenide
6) GaAs
→ Ga is usually +3, As is usually -3 → they balance. Gallium has only one common oxidation state (+3), so no Roman numeral needed.
→ Gallium arsenide
7) Pb(SO₄)₂
→ SO₄²⁻ is sulfate. Two sulfates → total -4. So Pb must be +4.
→ Lead(IV) sulfate
8) Be(HCO₃)₂
→ Be is always +2. HCO₃⁻ is bicarbonate (or hydrogen carbonate).
→ Beryllium bicarbonate OR Beryllium hydrogen carbonate — both acceptable, but “bicarbonate” is more common in practice.
9) Mn₂(SO₃)₃
→ SO₃²⁻ is sulfite. Three sulfites → total -6. Two Mn atoms → each Mn is +3.
→ Manganese(III) sulfite
10) Al(CN)₃
→ CN⁻ is cyanide. Al is always +3.
→ Aluminum cyanide
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Part 2: Write formulas for the following compounds
Use crisscross method: swap charges to become subscripts (simplify if possible).
11) chromium (VI) phosphate
→ Cr⁶⁺ and PO₄³⁻
→ Cr₂(PO₄)₆ → simplify? No common factor → but wait, actually, we should reduce: 6 and 3 → divide by 3 → Cr₂(PO₄)₆ becomes Cr₂(PO₄)₆? Wait, let's do it properly:
Charges: Cr⁶⁺, PO₄³⁻
Crisscross: Cr₃(PO₄)₆ → now simplify by dividing subscripts by 3 → Cr₁(PO₄)₂ → Cr(PO₄)₂? But that doesn’t make sense because 6+ and 2×3-=6- → yes! Actually, simplest form is Cr(PO₄)₂? Wait — no:
Wait: Cr⁶⁺ and PO₄³⁻ → least common multiple of 6 and 3 is 6. So you need 1 Cr⁶⁺ and 2 PO₄³⁻ → Cr(PO₄)₂
But check: 6+ + 2×(-3) = 0 → correct.
Actually, standard way: Cr⁶⁺ and PO³⁻ → cross: Cr₃(PO₄)₆ → then divide by 3 → Cr(PO₄)₂ → YES.
So formula: Cr(PO₄)₂
BUT — hold on! Chromium(VI) phosphate is often written as CrO₃·P₂O₅ or other forms, but for ionic compound naming purposes, we treat it as Cr⁶⁺ and PO₄³⁻ → so Cr(PO₄)₂ is correct.
However, some sources might write it as Cr₂(P₂O₇)₃ or something else — but based on standard ionic naming, we go with Cr(PO₄)₂.
Wait — let me double-check: Phosphate is PO₄³⁻. To balance Cr⁶⁺, you need two PO₄³⁻ → total charge -6. So yes, Cr(PO₄)₂
12) vanadium (IV) carbonate
→ V⁴⁺ and CO₃²⁻
→ Cross: V₂(CO₃)₄ → simplify by 2 → V(CO₃)₂
Check: 4+ + 2×(-2) = 0 → correct.
Formula: V(CO₃)₂
13) tin (II) nitrite
→ Sn²⁺ and NO₂⁻
→ Sn(NO₂)₂
Check: 2+ + 2×(-1) = 0 → correct.
Formula: Sn(NO₂)₂
14) cobalt (III) oxide
→ Co³⁺ and O²⁻
→ Cross: Co₂O₃
Check: 2×3+ = 6+, 3×2- = 6- → correct.
Formula: Co₂O₃
15) titanium (III) acetate
→ Ti³⁺ and CH₃COO⁻ (acetate is C₂H₃O₂⁻ or CH₃COO⁻, same thing)
→ Ti(C₂H₃O₂)₃
Check: 3+ + 3×(-1) = 0 → correct.
Formula: Ti(C₂H₃O₂)₃
16) vanadium (IV) sulfide
→ V⁴⁺ and S²⁻
→ Cross: V₂S₄ → simplify by 2 → VS₂
Check: 4+ + 2×(-2) = 0 → correct.
Formula: VS₂
17) chromium (III) hydroxide
→ Cr³⁺ and OH⁻
→ Cr(OH)₃
Check: 3+ + 3×(-1) = 0 → correct.
Formula: Cr(OH)₃
18) lithium iodide
→ Li⁺ and I⁻
→ LiI
Formula: LiI
19) lead (II) nitride
→ Pb²⁺ and N³⁻
→ Cross: Pb₃N₂
Check: 3×2+ = 6+, 2×3- = 6- → correct.
Formula: Pb₃N₂
20) silver bromide
→ Ag⁺ and Br⁻
→ AgBr
Silver always +1, bromide -1 → simple 1:1.
Formula: AgBr
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Now, compiling all answers:
Final Answer:
1) Ammonium chloride
2) Iron(III) nitrate
3) Titanium(IV) 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
11) Cr(PO₄)₂
12) V(CO₃)₂
13) Sn(NO₂)₂
14) Co₂O₃
15) Ti(C₂H₃O₂)₃
16) VS₂
17) Cr(OH)₃
18) LiI
19) Pb₃N₂
20) AgBr
---
Part 1: Name the following ionic compounds
We need to name each compound using standard ionic naming rules:
- For metals with fixed charges (like Group 1, 2, Al, Zn, Ag), just use the metal name.
- For transition metals or metals that can have multiple charges (like Fe, Cu, Sn, Pb, Cr, V, Ti, Mn, etc.), we must include a Roman numeral indicating the charge.
- Nonmetal ions end in “-ide” if monatomic, or keep their polyatomic ion names (like nitrate, sulfate, cyanide, etc.).
---
1) NH₄Cl
→ NH₄⁺ is ammonium (a polyatomic cation). Cl⁻ is chloride.
→ Name: Ammonium chloride
2) Fe(NO₃)₃
→ NO₃⁻ is nitrate (charge -1). There are 3 nitrates → total negative charge = -3. So Fe must be +3.
→ Iron(III) nitrate
3) TiBr₄
→ Br⁻ is bromide (-1). Four bromides → total -4. So Ti is +4.
→ Titanium(IV) bromide
4) Cu₃P
→ P³⁻ is phosphide. Three copper atoms balance one P³⁻ → each Cu is +1.
→ Copper(I) phosphide
5) SnSe₂
→ Se²⁻ is selenide. Two selenides → total -4. So Sn must be +4.
→ Tin(IV) selenide
6) GaAs
→ Ga is usually +3, As is usually -3 → they balance. Gallium has only one common oxidation state (+3), so no Roman numeral needed.
→ Gallium arsenide
7) Pb(SO₄)₂
→ SO₄²⁻ is sulfate. Two sulfates → total -4. So Pb must be +4.
→ Lead(IV) sulfate
8) Be(HCO₃)₂
→ Be is always +2. HCO₃⁻ is bicarbonate (or hydrogen carbonate).
→ Beryllium bicarbonate OR Beryllium hydrogen carbonate — both acceptable, but “bicarbonate” is more common in practice.
9) Mn₂(SO₃)₃
→ SO₃²⁻ is sulfite. Three sulfites → total -6. Two Mn atoms → each Mn is +3.
→ Manganese(III) sulfite
10) Al(CN)₃
→ CN⁻ is cyanide. Al is always +3.
→ Aluminum cyanide
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Part 2: Write formulas for the following compounds
Use crisscross method: swap charges to become subscripts (simplify if possible).
11) chromium (VI) phosphate
→ Cr⁶⁺ and PO₄³⁻
→ Cr₂(PO₄)₆ → simplify? No common factor → but wait, actually, we should reduce: 6 and 3 → divide by 3 → Cr₂(PO₄)₆ becomes Cr₂(PO₄)₆? Wait, let's do it properly:
Charges: Cr⁶⁺, PO₄³⁻
Crisscross: Cr₃(PO₄)₆ → now simplify by dividing subscripts by 3 → Cr₁(PO₄)₂ → Cr(PO₄)₂? But that doesn’t make sense because 6+ and 2×3-=6- → yes! Actually, simplest form is Cr(PO₄)₂? Wait — no:
Wait: Cr⁶⁺ and PO₄³⁻ → least common multiple of 6 and 3 is 6. So you need 1 Cr⁶⁺ and 2 PO₄³⁻ → Cr(PO₄)₂
But check: 6+ + 2×(-3) = 0 → correct.
Actually, standard way: Cr⁶⁺ and PO³⁻ → cross: Cr₃(PO₄)₆ → then divide by 3 → Cr(PO₄)₂ → YES.
So formula: Cr(PO₄)₂
BUT — hold on! Chromium(VI) phosphate is often written as CrO₃·P₂O₅ or other forms, but for ionic compound naming purposes, we treat it as Cr⁶⁺ and PO₄³⁻ → so Cr(PO₄)₂ is correct.
However, some sources might write it as Cr₂(P₂O₇)₃ or something else — but based on standard ionic naming, we go with Cr(PO₄)₂.
Wait — let me double-check: Phosphate is PO₄³⁻. To balance Cr⁶⁺, you need two PO₄³⁻ → total charge -6. So yes, Cr(PO₄)₂
12) vanadium (IV) carbonate
→ V⁴⁺ and CO₃²⁻
→ Cross: V₂(CO₃)₄ → simplify by 2 → V(CO₃)₂
Check: 4+ + 2×(-2) = 0 → correct.
Formula: V(CO₃)₂
13) tin (II) nitrite
→ Sn²⁺ and NO₂⁻
→ Sn(NO₂)₂
Check: 2+ + 2×(-1) = 0 → correct.
Formula: Sn(NO₂)₂
14) cobalt (III) oxide
→ Co³⁺ and O²⁻
→ Cross: Co₂O₃
Check: 2×3+ = 6+, 3×2- = 6- → correct.
Formula: Co₂O₃
15) titanium (III) acetate
→ Ti³⁺ and CH₃COO⁻ (acetate is C₂H₃O₂⁻ or CH₃COO⁻, same thing)
→ Ti(C₂H₃O₂)₃
Check: 3+ + 3×(-1) = 0 → correct.
Formula: Ti(C₂H₃O₂)₃
16) vanadium (IV) sulfide
→ V⁴⁺ and S²⁻
→ Cross: V₂S₄ → simplify by 2 → VS₂
Check: 4+ + 2×(-2) = 0 → correct.
Formula: VS₂
17) chromium (III) hydroxide
→ Cr³⁺ and OH⁻
→ Cr(OH)₃
Check: 3+ + 3×(-1) = 0 → correct.
Formula: Cr(OH)₃
18) lithium iodide
→ Li⁺ and I⁻
→ LiI
Formula: LiI
19) lead (II) nitride
→ Pb²⁺ and N³⁻
→ Cross: Pb₃N₂
Check: 3×2+ = 6+, 2×3- = 6- → correct.
Formula: Pb₃N₂
20) silver bromide
→ Ag⁺ and Br⁻
→ AgBr
Silver always +1, bromide -1 → simple 1:1.
Formula: AgBr
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Now, compiling all answers:
Final Answer:
1) Ammonium chloride
2) Iron(III) nitrate
3) Titanium(IV) 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
11) Cr(PO₄)₂
12) V(CO₃)₂
13) Sn(NO₂)₂
14) Co₂O₃
15) Ti(C₂H₃O₂)₃
16) VS₂
17) Cr(OH)₃
18) LiI
19) Pb₃N₂
20) AgBr
Parent Tip: Review the logic above to help your child master the concept of naming ionic compound worksheet.