Question

Is NaC polar covalent or ionic? I know that the electronegative difference between the two atoms is 1.6. However, Na is a cation and C is an anion and the bonding between cations and anions involve transfer of electrons. The electronegative difference should be greater or equal to 2, for it to be ionic.

Answer 1

NaC or Na2C2 is sodium carbide and it's definitely ionic compound.

Several carbides are assumed to be salts of the acetylide anion $\mathbf{C}_2^{2-}$ (also called percarbide), which has a triple bond between the two carbon atoms. Alkali metals, alkaline earth metals, and lanthanoid metals form acetylides, e.g., sodium carbide $\mathrm{Na}_2\mathrm{C}_2$, calcium carbide $\mathrm{CaC}_2$, and $\mathrm{LaC}_2$. Lanthanoids also form carbides (sesquicarbides, see below) with formula $\mathrm{M}_2\mathrm{C}_3$. Metals from group 11 also tend to form acetylides, such as copper (I) acetylide and silver acetylide. Carbides of the actinide elements, which have stoichiometry $\mathrm{MC}_2$ and $\mathrm{M}_2\mathrm{C}_3$, are also described as salt-like derivatives of $\mathrm{C}_2^{2-}$.

The C-C triple bond length ranges from $109.2\,\mathrm{pm}$ in $\mathrm{CaC}_2$ (similar to ethyne), to $130.3\,\mathrm{pm}$ in $\mathrm{LaC}_2$ and $134\,\mathrm{pm}$ in $\mathrm{UC}_2$. The bonding in $\mathrm{LaC}_2$ has been described in terms of $\mathrm{La}^{\mathrm{III}}$ with the extra electron delocalised into the antibonding orbital on $\mathrm{C}_2^{2-}$, explaining the metallic conduction.

http://en.wikipedia.org/wiki/Carbide