To draw the Lewis structure for IF3 (iodine trifluoride), we first need to determine the number of valence electrons in the molecule, then assess its geometry and polarity.
a. How many valence electrons are there?
Iodine has 7 valence electrons, and each fluorine atom has 7 valence electrons. Therefore, for IF3:
Valence electrons = 7 (from I) + 3 * 7 (from 3 F) = 7 + 21 = 28 valence electrons.
b. What is the electron geometry?
The electron geometry for IF3 considers both bonds and lone pairs on the central atom (iodine). Iodine is surrounded by three fluorine atoms, and it has two lone pairs, which gives it a total of five regions of electron density. Thus, the electron geometry is trigonal bipyramidal.
c. What is the molecular geometry?
While the electron geometry is trigonal bipyramidal, the lone pairs of electrons occupy equatorial positions, resulting in a molecular shape that is T-shaped due to the remaining bonds to the fluorine atoms.
d. What are the hybridized orbitals?
In IF3, the central iodine atom undergoes sp3d hybridization. This involves one s orbital, three p orbitals, and one d orbital to form five hybrid orbitals.
e. Is the molecule polar or nonpolar?
IF3 is a polar molecule because of its T-shaped geometry and the difference in electronegativity between iodine and fluorine, which creates a net dipole moment pointing towards the more electronegative fluorine atoms.