Question

Structure of $Xe{F_6}$ is octahedron with $s{p^3}{d^3}$ hybridisation of $Xe$.
A) True
B) False

Answer 1

To answer this question, you must understand the structure of xenon hexafluoride. Xenon has eight electrons in its valence shell and it will use six electrons for bonding with six fluoride ions. Thus, one pair of electrons remains non-bonded and this will be a lone pair. Lone pairs will cause distortion of the regular geometry.

Complete step by step solution:
There are a set of postulates given by VSEPR theory on which structure of a molecule can be drawn. One postulate says that the shape of the molecule depends on the number of valence electron pairs, including both bonding and non bonding electron pairs, around the central atom. Pairs of electrons in the valence shell repel each other since the electron clouds are negatively charged. Thus these pairs of electrons tend to occupy such positions in space that minimise repulsion and thus maximise distance between them. In the given molecule, $Xe{F_6}$ , Xe has $s{p^3}{d^3}$ hybridisation. Xenon (Xe) is the central atom and thus around this fluorine atom will be bonded with it. Xenon, an element of group 18, has eight valence electrons in the valence shell. It forms six bonds with the six fluorine atoms because each fluorine atom requires one electron to complete its octet. Thus, there are six bonding pairs in the structure. Now, fluorine is left with one electron pair (that is two electrons) which will act as a lone pair (non-bonding pair). Consequently, $Xe{F_6}$ molecule will contain six bonding pairs and one non-bonding pair or lone pair. This one lone pair tends to occupy that position in space that minimises repulsion between electron clouds. Due to six bonding pairs, $Xe{F_6}$ should have octahedral geometry but it does not due to the lone pair which is located at $s{p^3}{d^3}$ hybrid orbital. Hence, $Xe{F_6}$has distorted geometry from regular octahedron or $Xe{F_6}$has distorted octahedral geometry.

Thus, the correct option is B.

Note: Xenon hexafluoride, $Xe{F_6}$ is a noble gas compound. It is one of the three binary compounds of xenon. The other two fluorides of xenon are xenon tetrafluoride, $Xe{F_4}$ and xenon difluoride, $Xe{F_2}$ . $Xe{F_6}$ is the strongest fluorinating agent of the series of its fluorides.