To determine which of the given compounds is most ionic in character, we can apply Fajans' rules, which help us understand the conditions that favor ionic character in compounds. Here’s a step-by-step solution:
### Step 1: Identify the Cations and Anions
First, we need to identify the cations and anions in the compounds provided. The cations are typically metals, and the anions are usually nonmetals.
### Step 2: Determine the Oxidation States
Next, we need to determine the oxidation states of the cations in the compounds. For example:
- In PbCl₄, lead (Pb) has an oxidation state of +4.
- In SnCl₄, tin (Sn) also has an oxidation state of +4.
- In PbCl₂, lead (Pb) has an oxidation state of +2.
- In SnCl₂, tin (Sn) has an oxidation state of +2.
### Step 3: Analyze Cation Charge
According to Fajans' rules, a higher charge on the cation increases the ionic character. Here, we have:
- Pb⁴⁺ and Sn⁴⁺ (both +4)
- Pb²⁺ and Sn²⁺ (both +2)
Since the charges are the same for Pb and Sn in their respective oxidation states, we need to consider the size of the cations.
### Step 4: Compare Cation Sizes
The ionic character is also influenced by the size of the cation. Smaller cations tend to have a higher ionic character. As we move down a group in the periodic table, the size of the elements increases due to the addition of electron shells. Therefore:
- Tin (Sn) is smaller than lead (Pb) because Sn is higher up in the group.
### Step 5: Compare the Compounds
Now, we compare the compounds:
- For PbCl₄ and SnCl₄, both have cations with a +4 charge, but Sn is smaller than Pb, making SnCl₄ more ionic.
- For PbCl₂ and SnCl₂, the same reasoning applies; Sn²⁺ is smaller than Pb²⁺, making SnCl₂ more ionic.
### Conclusion
Among the compounds analyzed, SnCl₄ (tin(IV) chloride) is the most ionic due to the smaller size of the Sn⁴⁺ cation compared to the Pb⁴⁺ cation.
### Final Answer
The compound that is most ionic in character is **SnCl₄**.
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