Oxidation WS - Chapter 1 - Redox Reactions - Practice Problems ... - Free Printable

Let's solve each compound step by step to determine the oxidation numbers of each element. We'll use standard oxidation number rules:

Rules for Oxidation Numbers:

1. The oxidation number of an atom in its elemental form is 0.
2. For ions, the oxidation number equals the charge.
3. Oxygen usually has an oxidation number of –2, except in peroxides (–1) or when bonded to fluorine.
4. Hydrogen is +1 when bonded to nonmetals and –1 when bonded to metals.
5. Alkali metals (Group 1) = +1, alkaline earth metals (Group 2) = +2.
6. Halogens (F, Cl, Br, I) are usually –1, unless bonded to oxygen or another halogen.
7. The sum of oxidation numbers in a neutral compound is 0; in a polyatomic ion, it equals the charge.

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Now let’s go through each one:

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a. SnCl₄

- Cl is more electronegative → oxidation state = –1
- There are 4 Cl atoms → total = –4
- Compound is neutral → Sn must be +4
- Sn = +4, Cl = –1

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b. Ca₃P₂

- Ca is Group 2 → +2
- Total from Ca: 3 × (+2) = +6
- Neutral compound → P must balance to –6
- Two P atoms → each P = –3
- Ca = +2, P = –3

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c. SnO

- O is usually –2
- One O → –2
- Neutral compound → Sn = +2
- Sn = +2, O = –2

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d. Ag₂S

- S is more electronegative → typically –2
- Two Ag atoms → total must be +2 → each Ag = +1
- Ag = +1, S = –2

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e. HI

- H is bonded to nonmetal (I) → H = +1
- Iodine must be –1
- H = +1, I = –1

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f. N₂H₄ (Hydrazine)

- H = +1 (bonded to nonmetal)
- Four H atoms → total = +4
- Neutral molecule → two N atoms must total –4 → each N = –2
- N = –2, H = +1

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g. Al₂O₃

- O = –2
- Three O atoms → –6
- Two Al atoms → total = +6 → each Al = +3
- Al = +3, O = –2

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h. S₈

- Elemental sulfur → 0
- S = 0

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i. HNO₂ (Nitrous acid)

- H = +1
- O = –2 (each)
- Two O atoms → –4
- Let oxidation state of N = x
- Sum: (+1) + x + (–4) = 0 → x – 3 = 0 → x = +3
- H = +1, N = +3, O = –2

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j. O₂

- Elemental oxygen → 0
- O = 0

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k. H₃O⁺ (Hydronium ion)

- H = +1 (three H atoms → +3)
- O = x
- Total charge = +1 → 3(+1) + x = +1 → 3 + x = 1 → x = –2
- H = +1, O = –2

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l. ClO₃⁻ (Chlorate ion)

- O = –2 (three O atoms → –6)
- Let Cl = x
- Total charge = –1 → x + (–6) = –1 → x = +5
- Cl = +5, O = –2

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m. S₂O₃²⁻ (Thiosulfate ion)

- O = –2 (three O atoms → –6)
- Total charge = –2
- Two S atoms → let oxidation states be x₁ and x₂
- But note: In thiosulfate, one S is central (like sulfate), other is terminal (like sulfide)

However, we can assign average oxidation state:
Let both S have same oxidation state x
Then: 2x + 3(–2) = –2 → 2x – 6 = –2 → 2x = 4 → x = +2

But actually, the two S atoms have different oxidation states:
- Central S: +5
- Terminal S: –1
- Average: +2

But since question asks for "oxidation number of each element", and doesn't specify which S, we report average or commonly accepted values.

In most textbooks, S = +2 (average)

So:
- S = +2, O = –2

> Note: More accurately, one S is +5, the other is –1, but since they're not distinguished, average is used.

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n. KMnO₄ (Potassium permanganate)

- K = +1 (Group 1)
- O = –2 (four O → –8)
- Let Mn = x
- Total charge = 0 → (+1) + x + (–8) = 0 → x – 7 = 0 → x = +7
- K = +1, Mn = +7, O = –2

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o. (NH₄)₂SO₄ (Ammonium sulfate)

This is made of NH₄⁺ and SO₄²⁻ ions.

#### First, NH₄⁺:
- H = +1 (four H → +4)
- Charge = +1 → N + 4 = +1 → N = –3
- So: N = –3, H = +1

#### Then, SO₄²⁻:
- O = –2 (four O → –8)
- Let S = x
- Total charge = –2 → x + (–8) = –2 → x = +6
- So: S = +6, O = –2

Thus:
- N = –3, H = +1, S = +6, O = –2

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✔ Final Answers:

| Compound | Element | Oxidation Number |
|--------|--------|------------------|
| a. SnCl₄ | Sn | +4 | Cl | –1 |
| b. Ca₃P₂ | Ca | +2 | P | –3 |
| c. SnO | Sn | +2 | O | –2 |
| d. Ag₂S | Ag | +1 | S | –2 |
| e. HI | H | +1 | I | –1 |
| f. N₂H₄ | N | –2 | H | +1 |
| g. Al₂O₃ | Al | +3 | O | –2 |
| h. S₈ | S | 0 |
| i. HNO₂ | H | +1 | N | +3 | O | –2 |
| j. O₂ | O | 0 |
| k. H₃O⁺ | H | +1 | O | –2 |
| l. ClO₃⁻ | Cl | +5 | O | –2 |
| m. S₂O₃²⁻ | S | +2 (avg) | O | –2 |
| n. KMnO₄ | K | +1 | Mn | +7 | O | –2 |
| o. (NH₄)₂SO₄ | N | –3 | H | +1 | S | +6 | O | –2 |

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Summary Table (as requested):

```
a. SnCl₄ Sn: +4 Cl: –1
b. Ca₃P₂ Ca: +2 P: –3
c. SnO Sn: +2 O: –2
d. Ag₂S Ag: +1 S: –2
e. HI H: +1 I: –1
f. N₂H₄ N: –2 H: +1
g. Al₂O₃ Al: +3 O: –2
h. S₈ S: 0
i. HNO₂ H: +1 N: +3 O: –2
j. O₂ O: 0
k. H₃O⁺ H: +1 O: –2
l. ClO₃⁻ Cl: +5 O: –2
m. S₂O₃²⁻ S: +2 O: –2
n. KMnO₄ K: +1 Mn: +7 O: –2
o. (NH₄)₂SO₄ N: –3 H: +1 S: +6 O: –2
```

✔ All oxidation numbers determined using standard rules and charge balancing.