Naming Compounds with Transition Metals worksheet with practice exercises and a highlighted periodic table section.

Let’s go step by step to solve this worksheet.

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Part 1: Vanadium with Chloride Ion (Cl⁻)

Chloride ion has a charge of -1. So, to make a neutral compound, the total positive charge from vanadium must balance the total negative charge from chloride ions.

We’re given oxidation states for vanadium: V²⁺, V³⁺, V⁴⁺, V⁺

For each, we write the formula by balancing charges:

- V²⁺ → needs two Cl⁻ to balance → VCl₂
- V³⁺ → needs three Cl⁻ → VCl₃
- V⁴⁺ → needs four Cl⁻ → VCl₄
- V⁵⁺ → needs five Cl⁻ → VCl₅

✔ So fill in:
- V²⁺ → VCl₂
- V³⁺ → VCl₃
- V⁴⁺ → VCl₄
- V⁵⁺ → VCl₅

*(Note: The first one is already filled as example.)*

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Part 2: Iron Compounds — Name them using Roman numerals

Iron can have different charges, so we use Roman numerals to show which one it is.

Given formulas:

- FeCl₂ → Iron(II) chloride *(already done as example)*
Why? Because Cl is -1, two of them = -2, so Fe must be +2 → Iron(II)

Now do the rest:

- FeCl₃ → Three Cl⁻ = -3 total → Fe must be +3 → Iron(III) chloride
- FeCl₂ → Already done → Iron(II) chloride
- FeCl → One Cl⁻ = -1 → Fe must be +1 → Iron(I) chloride

Wait — let’s double-check that last one. Is FeCl common? Actually, iron doesn’t usually form +1 compounds, but since the problem gives us FeCl, we follow the rule: charge of Fe = +1 → Iron(I) chloride.

So:

- FeCl₃ → Iron(III) chloride
- FeCl₂ → Iron(II) chloride *(example)*
- FeCl → Iron(I) chloride

*(Note: In real chemistry, FeCl is rare or unstable, but for this exercise, we go by the math.)*

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Part 3: Practice — Write formulas from names

Rule: Use the Roman numeral to find the metal’s charge. Then balance with the anion’s charge.

Anions:
- bromide = Br⁻
- chloride = Cl⁻
- oxide = O²⁻
- phosphide = P³⁻

Now solve each:

#### 1. Chromium(VI) bromide
- Cr⁶⁺ and Br⁻
- Need 6 Br⁻ to balance one Cr⁶⁺ → CrBr₆

#### 2. Nickel(II) chloride
- Ni²⁺ and Cl
- Need 2 Cl → NiCl₂

#### 3. Zinc(II) oxide
- Zn²⁺ and O²⁻
- Charges are equal and opposite → ZnO

#### 4. Zirconium(IV) phosphide
- Zr⁴⁺ and P³⁻
- Find least common multiple of 4 and 3 → 12
- So: 3 Zr⁴⁺ = +12, 4 P³⁻ = -12 → Zr₃P₄

#### 5. Palladium(IV) oxide
- Pd⁴⁺ and O²⁻
- Two O²⁻ = -4 → balances one Pd⁴⁺ → PdO₂

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Final Answers:

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Final Answer:
Vanadium chlorides:
V²⁺ → VCl₂
V³⁺ → VCl₃
V⁴⁺ → VCl₄
V⁵⁺ → VCl₅

Iron compound names:
FeCl₃ → Iron(III) chloride
FeCl₂ → Iron(II) chloride
FeCl → Iron(I) chloride

Practice formulas:
1. Chromium(VI) bromide → CrBr₆
2. Nickel(II) chloride → NiCl₂
3. Zinc(II) oxide → ZnO
4. Zirconium(IV) phosphide → Zr₃P₄
5. Palladium(IV) oxide → PdO₂