Question

Which of the following compounds are ionic

• A. SnF_4
• B. SnCl_4
• C. PbF_2
• D. CF_4

Answer 1

Answer: (C)

PbF_2

Answer 2

Ionic compounds are typically formed between metals and non-metals with a large electronegativity difference, leading to electron transfer and the formation of an ionic lattice. Covalent compounds involve electron sharing, usually between non-metals or metalloids.

• $SnF_4$: Tin (Sn) is a metalloid/metal, and Fluorine (F) is a non-metal. The high polarizing power of $Sn^{4+}$ causes significant covalent character. Experimentally, $SnF_4$ is a molecular solid with a low sublimation point, characteristic of covalent compounds.
• $SnCl_4$: Tin (Sn) is a metalloid/metal, and Chlorine (Cl) is a non-metal. The $Sn^{4+}$ ion has high polarizing power, and the chloride ion ($Cl^-$) is larger and more polarizable than the fluoride ion, strongly favoring covalent bonding. $SnCl_4$ is a volatile liquid at room temperature, indicating it is a covalent compound.
• $PbF_2$: Lead (Pb) is a metal, and Fluorine (F) is a non-metal. The electronegativity difference is significant, and the $Pb^{2+}$ ion has lower polarizing power compared to $Sn^{4+}$. Combined with low anion polarizability, this leads to strong ionic bonding. $PbF_2$ is a solid with a high melting point, characteristic of an ionic compound.
• $CF_4$: Carbon (C) is a non-metal, and Fluorine (F) is a non-metal. Both elements are non-metals, forming polar covalent bonds. The molecule is symmetrical, making it non-polar overall. $CF_4$ is a gas at room temperature, characteristic of covalent compounds.

Based on these analyses, only $PbF_2$ is classified as an ionic compound.