The following data are available.
i) `%" of "Mg" in MgO and in "MgCl_(2)`
ii) `%" of "C" in "CO&CO_(2)`
iii) `%" of "Cr" in "K_(2)Cr_(2)O and K_(2)CrO_(4)`
iv `%" of Cu isotopes in Cu metal."`
The law of multiple proportions may be illustrated by data.

Options

A. `i&ii`
B. Only ii
C. `i,ii&iii`
D. Only iii

Explanation

To illustrate the law of multiple proportions using the provided data, we will analyze each part step by step.

### Step 1: Understanding the Law of Multiple Proportions
The law of multiple proportions states that when two elements combine to form more than one compound, the ratios of the masses of the second element that combine with a fixed mass of the first element will be in small whole numbers.

### Step 2: Analyzing the Data
We have four parts to analyze:

1. **% of Mg in MgO and in MgCl₂**
2. **% of C in CO and CO₂**
3. **% of Cr in K₂Cr₂O₇ and K₂CrO₄**
4. **% of Cu isotopes in Cu metal**

### Step 3: Calculating the Percentages

#### Part 1: % of Mg in MgO and MgCl₂
- **Molar mass of MgO**:
- Mg = 24 g/mol
- O = 16 g/mol
- Molar mass of MgO = 24 + 16 = 40 g/mol
- % of Mg in MgO = (24/40) * 100 = 60%

- **Molar mass of MgCl₂**:
- Mg = 24 g/mol
- Cl = 35.5 g/mol (2 Cl atoms = 71 g/mol)
- Molar mass of MgCl₂ = 24 + 71 = 95 g/mol
- % of Mg in MgCl₂ = (24/95) * 100 = 25.26%

#### Part 2: % of C in CO and CO₂
- **Molar mass of CO**:
- C = 12 g/mol
- O = 16 g/mol
- Molar mass of CO = 12 + 16 = 28 g/mol
- % of C in CO = (12/28) * 100 = 42.86%

- **Molar mass of CO₂**:
- C = 12 g/mol
- O = 16 g/mol (2 O atoms = 32 g/mol)
- Molar mass of CO₂ = 12 + 32 = 44 g/mol
- % of C in CO₂ = (12/44) * 100 = 27.27%

#### Part 3: % of Cr in K₂Cr₂O₇ and K₂CrO₄
- **Molar mass of K₂Cr₂O₇**:
- K = 39 g/mol (2 K = 78 g/mol)
- Cr = 52 g/mol (2 Cr = 104 g/mol)
- O = 16 g/mol (7 O = 112 g/mol)
- Molar mass of K₂Cr₂O₇ = 78 + 104 + 112 = 294 g/mol
- % of Cr in K₂Cr₂O₇ = (104/294) * 100 = 35.41%

- **Molar mass of K₂CrO₄**:
- K = 39 g/mol (2 K = 78 g/mol)
- Cr = 52 g/mol
- O = 16 g/mol (4 O = 64 g/mol)
- Molar mass of K₂CrO₄ = 78 + 52 + 64 = 194 g/mol
- % of Cr in K₂CrO₄ = (52/194) * 100 = 26.84%

#### Part 4: % of Cu Isotopes in Cu Metal
- Copper has two stable isotopes: Cu-63 and Cu-65.
- The average atomic mass of Cu is approximately 63.55 g/mol.
- The % of Cu-63 is around 69.17% and Cu-65 is around 30.83%.

### Step 4: Applying the Law of Multiple Proportions
Now we will check the ratios of the second elements (O, C, Cr) that combine with a fixed mass of the first element (Mg, C, K).

1. **For Mg in MgO and MgCl₂**:
- Ratio of % of O to % of Cl = 60% Mg in MgO : 25.26% Mg in MgCl₂ does not yield a simple whole number ratio.

2. **For C in CO and CO₂**:
- Ratio of % of O in CO to % of O in CO₂ = 57.14% O in CO : 72.73% O in CO₂.
- This gives a ratio of approximately 2:1, which is a simple whole number.

3. **For Cr in K₂Cr₂O₇ and K₂CrO₄**:
- Ratio of % of O in K₂Cr₂O₇ to % of O in K₂CrO₄ does not yield a simple whole number ratio.

4. **For Cu isotopes**:
- The isotopes do not form compounds in the traditional sense, so this part does not apply.

### Conclusion
From our analysis, the only part that illustrates the law of multiple proportions correctly is the second part concerning carbon in CO and CO₂.

### Final Answer
The law of multiple proportions is illustrated by the data in part 2: the percentage of carbon in CO and CO₂.