Deducing the ions in a polyatomic ionic compound from its empir... Complete the table below by writing the symbols for the cation and anion that make up each ionic compound. TI ionic compound NaCl V(NO3)2 VCl5 NH4I Mn2O3 cation anion Na+Cl−V5+Cl−NH4+I−
The Correct Answer and Explanation is:
I−
Let’s complete the missing entries in the table from the image and explain the reasoning behind each.
Completed Table:
| Ionic Compound | Cation | Anion |
|---|---|---|
| NaCl | Na⁺ | Cl⁻ |
| V(NO₃)₂ | V³⁺ | NO₃⁻ |
| VCl₅ | V⁵⁺ | Cl⁻ |
| NH₄I | NH₄⁺ | I⁻ |
| Mn₂O₃ | Mn³⁺ | O²⁻ |
Explanation
To deduce the ions in a polyatomic ionic compound, we need to understand the rules of ionic bonding and oxidation states. Ionic compounds consist of positively charged ions (cations) and negatively charged ions (anions) combined in ratios that result in an overall neutral compound.
- NaCl: Sodium (Na) is an alkali metal and forms a +1 cation (Na⁺). Chloride (Cl) is a halogen and forms a -1 anion (Cl⁻). Their 1:1 ratio balances the charges.
- V(NO₃)₂: The nitrate ion (NO₃⁻) is a polyatomic ion with a -1 charge. There are two nitrate ions, contributing a total of -2 charge. Therefore, the vanadium (V) must have a +3 charge to balance it (V³⁺).
- VCl₅: There are five chloride ions (Cl⁻), each with a -1 charge, totaling -5. To balance this, vanadium must have a +5 charge (V⁵⁺).
- NH₄I: Ammonium (NH₄⁺) is a common polyatomic cation with a +1 charge. Iodide (I⁻) is a halogen anion with a -1 charge. Their charges cancel out in a 1:1 ratio.
- Mn₂O₃: Oxygen (O) forms a -2 anion (O²⁻). Three oxide ions give a total of -6 charge. To balance this, two manganese (Mn) ions must contribute +6 total, so each Mn is +3 (Mn³⁺).
Understanding these principles allows us to systematically determine the correct cations and anions in any given compound using the total charge balance and known ion charges.
