To draw the Lewis structure for selenium difluoride (SeF₂), we start by counting the total number of valence electrons. Selenium (Se) has 6 valence electrons, and each fluorine (F) atom has 7 valence electrons. Thus, the total number of valence electrons for SeF₂ is:
6 (Se) + 2 × 7 (F) = 20 valence electrons
Next, we place selenium in the center with the two fluorine atoms bonded to it. Each single bond (Se-F) uses 2 electrons, so we have:
20 – 2 (for 2 bonds) = 16 electrons remaining
We then distribute the remaining electrons to fulfill the octet rule for the fluorine atoms. Each fluorine will get 3 lone pairs (6 electrons) since they only need one bond to complete their octet:
- Fluorine 1: 2 (from Se-F bond) + 6 (3 lone pairs) = 8 electrons (octet achieved)
- Fluorine 2: 2 (from Se-F bond) + 6 (3 lone pairs) = 8 electrons (octet achieved)
Now, we place the remaining electrons around the selenium. Since selenium can expand its octet, we place the remaining 4 electrons as 2 lone pairs on the selenium atom:
This results in the Lewis structure looking like:
F:
\
Se:
/
F:
Now let’s predict the shape of the molecule. The selenium atom has 2 bonded pairs (the Se-F bonds) and 2 lone pairs. Using VSEPR theory, we identify the electron geometry as tetrahedral because of the 4 regions of electron density. However, the molecular shape is determined by only the bonded pairs, which leads to a bent (angular) shape.
To determine whether SeF₂ is polar or nonpolar, we need to look at the individual bond dipoles and the shape. Fluorine is much more electronegative than selenium, resulting in a dipole moment directed towards the fluorine atoms. Because of the bent shape, the dipoles do not cancel out, and therefore:
SeF₂ is a polar molecule.