Question 13.13: Prediction of Molecular Structure III Because the noble gase...
Prediction of Molecular Structure III
Because the noble gases have filled s and p valence orbitals, they are not expected to be chemically reactive. In fact, for many years these elements were called inert gases because of this supposed inability to form any compounds. However, in the early 1960s, several compounds of krypton, xenon, and radon were synthesized. For example, a team at the Argonne National Laboratory produced the stable colorless compound xenon tetrafluoride (Xe F_{4} ). Predict its structure and determine whether it has a dipole moment.
The traditional Lewis structure for Xe F_{4} is
The xenon atom in this molecule is surrounded by six pairs of electrons, requiring an octahedral arrangement:
The structure predicted for this molecule depends on how the lone pairs and bonding pairs are arranged. Consider the two possibilities shown in Fig. 13.20. The bonding pairs are indicated by the presence of fluorine atoms. Since the structure predicted differs in the two cases, we must decide which of these arrangements is preferable. The key is to look at the lone pairs. In the structure in part (a), the lone pair–lone pair angle is 90 degrees; in the structure in part (b), the lone pairs are separated by 180 degrees. Since lone pairs require more room than bonding pairs, a structure with two lone pairs at 90 degrees is unfavorable. Thus the arrangement in Fig. 13.20(b) is preferred, and the molecular structure is predicted to be square planar. Note that this molecule is not described as being octahedral. There is an octahedral arrangement of electron pairs, but the atoms form a square planar structure.
Although each Xe—F bond is polar (fluorine has a greater electronegativity than xenon), the square planar arrangement of these bonds causes the polarities to cancel.
Thus Xe F_{4} has no dipole moment.
