Question

what is the electron - domain geometry of sef₄?
tetrahedral
trigonal bipyramidal
octahedral
linear
trigonal planar

Explanation:

To determine the electron - domain geometry of \(SeF_4\), we first find the number of electron domains around the central \(Se\) atom.
The formula for the number of electron domains \(n=\frac{V + M - C + A}{2}\), where \(V\) is the valence electrons of the central atom (\(Se\) has \(6\) valence electrons), \(M\) is the number of monovalent atoms (\(F\) is monovalent, and there are \(4\) \(F\) atoms), \(C\) is the charge for cations (here \(C = 0\) as it is a neutral molecule), and \(A\) is the charge for anions (here \(A=0\)).
So, \(n=\frac{6 + 4-0 + 0}{2}=5\).
When the number of electron domains \(n = 5\), the electron - domain geometry is trigonal bipyramidal according to VSEPR (Valence - Shell Electron - Pair Repulsion) theory.

Answer:

Trigonal bipyramidal