Question

what is the formula when silver (i) bonds with chlorate?
○ ag₂(clo₃)
○ agcl
○ ag(clo₃)₂
○ ag(clo₃)

Explanation:

To determine the formula when Silver (I) bonds with Chlorate, we follow these steps:

Step 1: Identify the ions and their charges

• Silver (I) ion: \( \text{Ag}^+ \) (charge of +1)
• Chlorate ion: \( \text{ClO}_3^- \) (charge of -1)

Step 2: Balance the charges

For a neutral compound, the total positive charge must equal the total negative charge.

• The charge of \( \text{Ag}^+ \) is +1.
• The charge of \( \text{ClO}_3^- \) is -1.

To balance the charges, we need 1 \( \text{Ag}^+ \) ion and 1 \( \text{ClO}_3^- \) ion because \( (+1) + (-1) = 0 \).

Step 3: Write the formula

Combining the ions in the ratio determined (1:1), the formula is \( \text{Ag(ClO}_3\text{)} \).

Now let's analyze the given options:

• \( \text{Ag}_2(\text{ClO}_3) \): Incorrect. This would imply a charge imbalance (2 \( \text{Ag}^+ \) and 1 \( \text{ClO}_3^- \) gives a net charge of +1).
• \( \text{AgCl} \): Incorrect. This is Silver Chloride, not Silver Chlorate (Chlorate is \( \text{ClO}_3^- \), Chloride is \( \text{Cl}^- \)).
• \( \text{Ag(ClO}_3\text{)}_2 \): Incorrect. This would imply a charge imbalance (1 \( \text{Ag}^+ \) and 2 \( \text{ClO}_3^- \) gives a net charge of -1).
• \( \text{Ag(ClO}_3\text{)} \): Correct. This has a 1:1 ratio of \( \text{Ag}^+ \) and \( \text{ClO}_3^- \), resulting in a neutral compound.

Answer:

D. \( \text{Ag(ClO}_3\text{)} \) (assuming the last option is D, as per the order: 1. \( \text{Ag}_2(\text{ClO}_3) \), 2. \( \text{AgCl} \), 3. \( \text{Ag(ClO}_3\text{)}_2 \), 4. \( \text{Ag(ClO}_3\text{)} \))