Question

what is the correct formula for the compound tin (iv) nitrate?
○ sn(no₂)₄
○ ti(no₃)₄
○ sn₄(no₃)
○ sn(no₃)₄

Explanation:

Step1: Identify Tin(IV) and Nitrate

Tin(IV) has a charge of \( +4 \) (\( \text{Sn}^{4+} \)). Nitrate is \( \text{NO}_3^- \) (charge \( -1 \)).

Step2: Balance Charges

To balance charges, the number of \( \text{NO}_3^- \) ions needed is \( \frac{4}{1} = 4 \) (since \( \text{Sn}^{4+} \) needs to balance with \( 4 \times \text{NO}_3^- \) to make the compound neutral). So the formula is \( \text{Sn(NO}_3\text{)}_4 \).

Step3: Eliminate Incorrect Options

• \( \text{Sn(NO}_2\text{)}_4 \): Uses nitrite (\( \text{NO}_2^- \)), not nitrate. Eliminate.
• \( \text{Ti(NO}_3\text{)}_4 \): Uses titanium (Ti), not tin (Sn). Eliminate.
• \( \text{Sn}_4(\text{NO}_3) \): Incorrect charge balancing (Sn would have \( +\frac{1}{4} \) charge, impossible). Eliminate.

Answer:

The correct option is \( \text{Sn(NO}_3\text{)}_4 \) (the last option, assuming the first "8n" is a typo for "Sn"). So the answer is the option with formula \( \text{Sn(NO}_3\text{)}_4 \) (e.g., if the last option is D. \( \text{Sn(NO}_3\text{)}_4 \), then D. \( \text{Sn(NO}_3\text{)}_4 \)).