Tin and lead form divalent Sn (II) and lead (II) and tetravalent i.e., Tin (IV) and lead. (IV) compounds. Tetravalent and dioxides of lead and Tin are amphoteric in nature. Lead tetra fluorite is ionic solid where as `PbCl_(4)` is covalent. `Pbl_(4)` does not exist. lead (II) halides are white whereas `Pbl_(2)` is yellow in colour.
Q. `SnCl_(2)` is solid whereas `SnCl_(4)` is liquid, this is because?

Options

A. `SnCl_(2)` is ionic whereas `SnCl_(4)` is covalent
B. `SnCl_(2)` is covalent whereas `SnCl_(4)` is ionic
C. `SnCl_(2)` exist in polymeric form
D. `SnCl_(2)` is linear and `SnCl_(4)` is square planar in-structure.

Answer

To understand why \( \text{SnCl}_2 \) is a solid while \( \text{SnCl}_4 \) is a liquid, we can analyze the nature of the bonding in these compounds and the implications of their structures.

### Step 1: Identify the oxidation states
- \( \text{SnCl}_2 \) contains tin in the +2 oxidation state (\( \text{Sn}^{2+} \)).
- \( \text{SnCl}_4 \) contains tin in the +4 oxidation state (\( \text{Sn}^{4+} \)).

**Hint:** The oxidation state of an element can influence its ionic or covalent character.

### Step 2: Compare the sizes of the cations
- The size of \( \text{Sn}^{2+} \) is larger than that of \( \text{Sn}^{4+} \). This is because as electrons are removed to form a cation, the remaining electron cloud experiences a greater effective nuclear charge, leading to a decrease in size.

**Hint:** Remember that cation size decreases with increasing positive charge due to the loss of electrons.

### Step 3: Analyze polarization and covalent character
- The polarization of a cation is inversely proportional to its size and directly proportional to its charge. Thus, \( \text{Sn}^{2+} \) has lower polarization compared to \( \text{Sn}^{4+} \).
- Higher polarization in \( \text{Sn}^{4+} \) leads to greater covalent character in \( \text{SnCl}_4 \).

**Hint:** Polarization refers to the distortion of electron clouds in atoms or ions, which can lead to covalent character in bonds.

### Step 4: Determine the nature of bonding
- \( \text{SnCl}_2 \) is primarily ionic due to the lower covalent character and stronger electrostatic forces of attraction between \( \text{Sn}^{2+} \) and \( \text{Cl}^- \).
- \( \text{SnCl}_4 \), being covalent, has weaker intermolecular forces (Van der Waals forces), which results in it being a liquid at room temperature.

**Hint:** Ionic compounds typically have strong attractions between ions, leading to solid structures, while covalent compounds may have weaker forces, leading to liquid or gaseous states.

### Step 5: Conclusion
- Therefore, \( \text{SnCl}_2 \) exists as a solid due to its ionic nature and strong electrostatic attractions, while \( \text{SnCl}_4 \) exists as a liquid due to its covalent nature and weaker intermolecular forces.

**Final Answer:** \( \text{SnCl}_2 \) is solid because it is ionic, whereas \( \text{SnCl}_4 \) is liquid because it is covalent.