Oxidation Numbers Worksheet by Olivia Hunter - Issuu - Free Printable

Let's solve each part of this Oxidation Numbers Worksheet step by step. We’ll use the standard rules for assigning oxidation numbers:

🔹 Rules for Oxidation Numbers:

1. The oxidation number of an element in its elemental form is 0.
2. The oxidation number of a monatomic ion equals its charge.
3. Oxygen is usually –2, except in peroxides (–1) or when bonded to fluorine.
4. Hydrogen is +1 when bonded to nonmetals, –1 when bonded to metals.
5. Fluorine is always –1.
6. Group 1 metals: +1, Group 2: +2.
7. The sum of oxidation numbers in a neutral compound is 0; in a polyatomic ion, it equals the ion’s charge.

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## ✔ Problem 1: Give oxidation numbers of all elements

a. SO, SO₂, SO₃, SO₃²⁻, SO₄²⁻

We know:
- O = –2
- Let S = x

#### i. SO
- x + (–2) = 0 → x = +2
- S: +2, O: –2

#### ii. SO₂
- x + 2(–2) = 0 → x – 4 = 0 → x = +4
- S: +4, O: –2

#### iii. SO₃
- x + 3(–2) = 0 → x – 6 = 0 → x = +6
- S: +6, O: –2

#### iv. SO₃²⁻
- x + 3(–2) = –2 → x – 6 = –2 → x = +4
- S: +4, O: –2

#### v. SO₄²⁻
- x + 4(–2) = –2 → x – 8 = –2 → x = +6
- S: +6, O: –2

✔ Summary:
| Compound | S | O |
|--------|---|---|
| SO | +2 | –2 |
| SO₂ | +4 | –2 |
| SO₃ | +6 | –2 |
| SO₃²⁻ | +4 | –2 |
| SO₄²⁻ | +6 | –2 |

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b. ClO₂, ClO⁻, ClO₂⁻, ClO₃⁻, ClO₄⁻

O = –2
Let Cl = x

#### i. ClO₂ (neutral)
- x + 2(–2) = 0 → x – 4 = 0 → x = +4
- Cl: +4, O: –2

#### ii. ClO⁻ (hypochlorite)
- x + (–2) = –1 → x = +1
- Cl: +1, O: –2

#### iii. ClO₂⁻ (chlorite)
- x + 2(–2) = –1 → x – 4 = –1 → x = +3
- Cl: +3, O: –2

#### iv. ClO₃⁻ (chlorate)
- x + 3(–2) = –1 → x – 6 = –1 → x = +5
- Cl: +5, O: –2

#### v. ClO₄⁻ (perchlorate)
- x + 4(–2) = –1 → x – 8 = –1 → x = +7
- Cl: +7, O: –2

✔ Summary:
| Ion/Compound | Cl | O |
|-------------|----|---|
| ClO₂ | +4 | –2 |
| ClO⁻ | +1 | –2 |
| ClO₂⁻ | +3 | –2 |
| ClO₃⁻ | +5 | –2 |
| ClO₄⁻ | +7 | –2 |

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c. N₂O, NO, NO₂, N₂O₄, N₂O₅, NO₂⁻, NO₃⁻

O = –2
Let N = x

#### i. N₂O
- 2x + (–2) = 0 → 2x = 2 → x = +1
- N: +1, O: –2

#### ii. NO
- x + (–2) = 0 → x = +2
- N: +2, O: –2

#### iii. NO₂
- x + 2(–2) = 0 → x – 4 = 0 → x = +4
- N: +4, O: –2

#### iv. N₂O₄
- 2x + 4(–2) = 0 → 2x – 8 = 0 → x = +4
- N: +4, O: –2

#### v. N₂O₅
- 2x + 5(–2) = 0 → 2x – 10 = 0 → x = +5
- N: +5, O: –2

#### vi. NO₂⁻
- x + 2(–2) = –1 → x – 4 = –1 → x = +3
- N: +3, O: –2

#### vii. NO₃⁻
- x + 3(–2) = –1 → x – 6 = –1 → x = +5
- N: +5, O: –2

✔ Summary:
| Compound/Ion | N | O |
|-------------|---|---|
| N₂O | +1 | –2 |
| NO | +2 | –2 |
| NO₂ | +4 | –2 |
| N₂O₄ | +4 | –2 |
| N₂O₅ | +5 | –2 |
| NO₂⁻ | +3 | –2 |
| NO₃⁻ | +5 | –2 |

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## ✔ Problem 2: Oxidation number of sulfur atom

a. H₂S

- H = +1, so 2(+1) + x = 0 → 2 + x = 0 → x = –2
- S: –2

b. S (elemental)

- 0

c. H₂SO₄

- H = +1, O = –2
- 2(+1) + x + 4(–2) = 0 → 2 + x – 8 = 0 → x = +6
- S: +6

d. S²⁻

- Ion charge = –2 → S: –2

e. HS⁻

- H = +1, total charge = –1
- (+1) + x = –1 → x = –2
- S: –2

f. SO₂

- From earlier: S = +4

g. SO₃

- From earlier: S = +6

✔ Answers:
a. –2
b. 0
c. +6
d. –2
e. –2
f. +4
g. +6

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## ✔ Problem 3: Oxidation number of phosphorus

Use: H = +1, O = –2

a. HPO₃

- H = +1, O₃ = 3(–2) = –6
- +1 + x + (–6) = 0 → x – 5 = 0 → x = +5

b. H₃PO₂

- 3(+1) + x + 2(–2) = 0 → 3 + x – 4 = 0 → x = +1

c. H₃PO₃

- 3(+1) + x + 3(–2) = 0 → 3 + x – 6 = 0 → x = +3

d. H₃PO₄

- 3(+1) + x + 4(–2) = 0 → 3 + x – 8 = 0 → x = +5

e. H₄P₂O₇ (pyrophosphoric acid)

- 4(+1) + 2x + 7(–2) = 0 → 4 + 2x – 14 = 0 → 2x = 10 → x = +5

f. H₅P₃O₁₀ (tripolyphosphate)

- 5(+1) + 3x + 10(–2) = 0 → 5 + 3x – 20 = 0 → 3x = 15 → x = +5

✔ Answers:
a. +5
b. +1
c. +3
d. +5
e. +5
f. +5

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## ✔ Problem 4: Oxidation numbers for underlined atoms

We'll go one by one.

a. Cs₂O

- Cs = +1 (Group 1), O = –2
- 2(+1) + (–2) = 0 → ✔️
- Cs: +1

b. PtCl₆²⁻

- Cl = –1, 6 × (–1) = –6
- Total charge = –2 → Pt + (–6) = –2 → Pt = +4
- Pt: +4

c. CaI₂

- Ca = +2 (Group 2), I = –1
- Ca: +2

d. SnF₂

- F = –1, 2×(–1) = –2 → Sn = +2
- Sn: +2

e. Al₂O₃

- Al = +3, O = –2
- 2(+3) + 3(–2) = 0 → ✔️
- Al: +3

f. ClF₃

- F = –1, 3×(–1) = –3
- Cl + (–3) = 0 → Cl = +3
- Cl: +3

g. H₃AsO₃

- H = +1, O = –2
- 3(+1) + x + 3(–2) = 0 → 3 + x – 6 = 0 → x = +3
- As: +3

h. SbF₆⁻

- F = –1, 6×(–1) = –6
- Total charge = –1 → Sb + (–6) = –1 → Sb = +5
- Sb: +5

i. TiO₂

- O = –2, 2×(–2) = –4 → Ti = +4
- Ti: +4

j. P₄

- Elemental form → 0
- P: 0

k. MoO₄²⁻

- O = –2, 4×(–2) = –8
- Total charge = –2 → Mo + (–8) = –2 → Mo = +6
- Mo: +6

l. MnO₄⁻

- O = –2, 4×(–2) = –8
- Mn + (–8) = –1 → Mn = +7
- Mn: +7

m. PtCl₄²⁻

- Cl = –1, 4×(–1) = –4
- Total charge = –2 → Pt + (–4) = –2 → Pt = +2
- Pt: +2

n. O₂

- Elemental form → 0
- O: 0

o. O₃

- Elemental form → 0
- O: 0

✔ Final answers for Problem 4:
| Letter | Atom | Oxidation Number |
|-------|------|------------------|
| a | Cs | +1 |
| b | Pt | +4 |
| c | Ca | +2 |
| d | Sn | +2 |
| e | Al | +3 |
| f | Cl | +3 |
| g | As | +3 |
| h | Sb | +5 |
| i | Ti | +4 |
| j | P | 0 |
| k | Mo | +6 |
| l | Mn | +7 |
| m | Pt | +2 |
| n | O | 0 |
| o | O | 0 |

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## 📌 Final Answer Summary

Problem 1

All oxidation numbers calculated above.

Problem 2

a. –2
b. 0
c. +6
d. –2
e. –2
f. +4
g. +6

Problem 3

a. +5
b. +1
c. +3
d. +5
e. +5
f. +5

Problem 4

a. +1
b. +4
c. +2
d. +2
e. +3
f. +3
g. +3
h. +5
i. +4
j. 0
k. +6
l. +7
m. +2
n. 0
o. 0

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