The ratio of oxygen atoms present in one molecule of cupric nitrite and ferric sulphite is

Options

A. `4:9`
B. `2:3`
C. `1:2`
D. `1:3 `

Explanation

To find the ratio of oxygen atoms present in one molecule of cupric nitrite and ferric sulphite, we will follow these steps:

### Step 1: Determine the formula for cupric nitrite.
- Cupric ion has the formula \( \text{Cu}^{2+} \).
- Nitrite ion has the formula \( \text{NO}_2^{-} \).
- To form cupric nitrite, we need to combine these ions. We use the crisscross method to balance the charges:
- The cupric ion has a charge of +2, and the nitrite ion has a charge of -1. Thus, we need 2 nitrite ions to balance the +2 charge of one cupric ion.
- Therefore, the formula for cupric nitrite is \( \text{Cu(NO}_2\text{)}_2 \).

### Step 2: Count the number of oxygen atoms in cupric nitrite.
- In the formula \( \text{Cu(NO}_2\text{)}_2 \), there are 2 nitrite ions.
- Each nitrite ion contains 2 oxygen atoms, so:
- Total oxygen atoms in cupric nitrite = \( 2 \text{ (from NO}_2\text{)} \times 2 \text{ (number of NO}_2\text{ ions)} = 4 \) oxygen atoms.

### Step 3: Determine the formula for ferric sulphite.
- Ferric ion has the formula \( \text{Fe}^{3+} \).
- Sulphite ion has the formula \( \text{SO}_3^{2-} \).
- To form ferric sulphite, we again use the crisscross method:
- The ferric ion has a charge of +3, and the sulphite ion has a charge of -2. Thus, we need 3 sulphite ions to balance the +3 charge of one ferric ion.
- Therefore, the formula for ferric sulphite is \( \text{Fe(SO}_3\text{)}_2 \).

### Step 4: Count the number of oxygen atoms in ferric sulphite.
- In the formula \( \text{Fe(SO}_3\text{)}_2 \), there are 2 sulphite ions.
- Each sulphite ion contains 3 oxygen atoms, so:
- Total oxygen atoms in ferric sulphite = \( 3 \text{ (from SO}_3\text{)} \times 2 \text{ (number of SO}_3\text{ ions)} = 6 \) oxygen atoms.

### Step 5: Calculate the ratio of oxygen atoms.
- We have determined:
- Oxygen atoms in cupric nitrite = 4
- Oxygen atoms in ferric sulphite = 6
- The ratio of oxygen atoms in cupric nitrite to ferric sulphite is:
\[ \text{Ratio} = \frac{4}{6} = \frac{2}{3} \]

### Final Answer:
The ratio of oxygen atoms present in one molecule of cupric nitrite to ferric sulphite is \( 2:3 \).

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