Explaining Why SnI4 Bonding is More Covalent Than SnF4

[Miike012]

What is the explanation for the fact that the bonding in SnI4 is more covalent than the bonding in SnF4 ?


My thought was... because A polar bond is a covalent bond in which there is a separation of charge between one end and the other.
Thus... I am trying to prove that SnI4 is more polar than SnF4... can some one please help me explain...? thank you.

 

[kamas]

Hi Miike,
A covalent bond is one in which electrons are shared between two atoms. Fluorine is more electronegative than iodine, meaning that the Sn-F bond is more polar than the Sn-I bond. Because iodine does not pull on the electrons as strongly as fluorine does, the Sn-I bond has more covalent character.

Be careful in the way you think about the polarity of a molecule... A molecule can have polar bonds but still be nonpolar. This is true for both of the molecules here. Both molecular structures are symmetric (SnI4 is tetrahedral and SnF4 is square planar), and so although the Sn-I and Sn-F bonds are polar, the molecules are not.

Cheers,
Kamas