XeF6 Lewis Structure :Drawings, Hybridization, Shape,Charges,Pair And Detailed Facts


In this article,”xef6 Lewis structure”, different facts like Lewis structure drawing, hybridization, shape, charges, pair and formal charges with some detailed explanations are described below.

XeF6 is one of the noble gas compounds which is also known as xenon hexafluoride.  The binary fluoride of xenon is XeF2, XeF4, and XeF6 all are stable at room temperature out of the above three binary fluorides XeF6 is the strongest fluorinating agent It is a colorless solid that readily sublimes into intensely yellow vapors.

How to draw Lewis structure for XeF6 ?

In the XeF6 Lewis structure, Xenon (Xe) is a noble gas and Fluorine (F) is a halogen. So, Xenon has 8 valence electrons and Fluorine has 7 valence electrons. Considering there are 6 fluorine atoms in XeF6 molecule hence total, 6 x 7 = 42 valence electrons of Fluorine atom. So, in total we have 42 + 8 = 50 valence electrons. 

out of the two atoms, Xenon is the least electronegative atom, so while drawing Lewis’s structure of xenon fluoride we can put Xe in the centre and six fluorine atoms around it. Then connect the Xe and six fluorine with single bonds. We’ve used 12 electrons so far (one bond = two electrons).

Then we can put 3 more lone pairs on each F atom fulfilling the octet rule. So, 3 x 2 x 6 = 36 and with the 12 we used earlier, that’s 48 in total. We can put the two remaining electrons on Xe Also, Xe is relatively bigger. Thus, it has less steric hindrance.

XeF6 Lewis structure shape

According to VSEPR theory Xenon hexafluoride (XeF6) shows sp3d3 hybridization due to which XeF6 shows pentagonal bipyramidal structure but due to one lone pair it shows distorted octahedral structure.

XeF6 Lewis structure wikipedia

XeF6 Lewis structure formal charges

The Lewis structure formal charges of XeF6 can be calculated by the following formula

 FC = V – N – B/2

Where V = no. of valence electrons

               N = no. of non–bonding electrons

               B = no. of bonding electrons 

The formal charge of the XeF6 Lewis structure

FC of Xe in XeF6 Lewis structure = 8 – 2 -12/2 = 0

FC of F in XeF6 Lewis structure = 7 – 6 – 2/2 = 0

XeF6 Lewis structure lone pairs

The lone pair of electrons do not participate in chemical bonding which is also called a nonbonding electron or unshared electron. In the XeF6 molecule total of 25 electron pairs are present, including both bonding and non-bonding electron pairs.

Out of these 25 electron pairs, six electron pairs form Xe-F bonds remaining 19 electron pairs are lone pairs of the electron, as we know that xenon is placed in the 5th period in the periodic table and fluorine placed in the 2nd period of the periodic table so

xenon has more than eight electrons in its last shell while fluorine atoms last shell cannot occupy more than eight electrons.  therefore, start filling the lone pair of electrons from the outer atom in the XeF6 molecule outer atom is fluorine so each fluorine has three lone pairs and the xenon atom has one line pair.

XeF6 hybridization

In xenon hexafluoride (XeF6) molecule xenon atom undergoes sp3d3 hybridization as,

Xe: [Kr] 4d105s25px15py15pz15dxy15dyz15dxz1

 Out of seven sp3d3 hybrid orbitals, one sp3d3 hybrid orbital is occupied by one lone pair electron (lp) and six sp3d3 hybrid orbitals are half-filled which overlap with six 2pz half-filled orbitals of six F-atoms. Due to sp3d3 hybridization, it is expected to be pentagonal bipyramidal but due to one lone pair of electrons, its geometry is distorted octahedral.

XeF6 Lewis structure resonance

Resonance is a chemical phenomenon, in which all properties of a molecule are not able to explain with a single structure. There are many canonical structures involved. But not every molecule can exhibit resonance.

 Due to the asymmetrical structure of XeF6, it does not show Lewis structure In the XeF6 molecule there are six Xe-F single bonds. so, there is no movement of electrons. Therefore, XeF6 does not show a resonance structure.

XeF6 Lewis structure octet rule

In the XeF6 Lewis structure, Xenon (Xe) is a noble gas and Fluorine (F) is a halogen, Xenon has 8 valence electrons and Fluorine has 7 valence electrons. Considering there are 6 fluorine atoms in XeF6 molecule hence total, 6 x 7 = 42 valence electrons of Fluorine atom. So, in total we have 42 + 8 = 50 valence electrons. 

out of the two atoms, Xenon is the least electronegative atom, so while drawing Lewis’s structure of xenon fluoride we can put Xe in the centre and six fluorine atoms around it. Then connect the Xe and six fluorine with single bonds. We’ve used 12 electrons so far (one bond = two electrons). Then we can put 3 more lone pairs on each F atom fulfilling the octet rule.

What is the molecular geometry of XeF6?

Answer: The molecular geometry of XeF6 is distorted octahedral geometry, it has six bond pair and one lone pair. It shows Sp3d3 hybridization.

How many lone pairs are in XeF6 ?

Answer: In XeF6 molecule xenon has eight valence electrons, it forms six bonds with six fluorine atom and one lone pair of electron.

What is the hybridization of xenon in XeF6?

Answer: XeF6 molecule shows sp3d3 hybridization.

Conclusion

In the above we can concluded that Xenon hexafluoride (XeF6) molecule shows Lewis dot structure with sp3d3 hybridization , due to which it shows distorted octahedral geometry. xenon forms six bonds with six fluorine atoms also it has one lone pair.

Darshana Fendarkar

Hi..I am Darshana Fendarkar, I have completed my PhD from University of Nagpur and am pursuing M.Sc in Chemistry from Nagpur University. My area of specialization is on Inorganic Chemistry.

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