What Is The Lewis Structure Of Xenon Hexafluoride?

The Lewis structure of xenon hexafluoride (XeF6) can be determined by following the octet rule, which states that atoms tend to gain, lose, or share electrons in order to achieve a full outer shell of eight valence electrons.

Xenon has eight valence electrons and each fluorine atom has seven valence electrons. Therefore, six fluorine atoms will contribute a total of 42 valence electrons to the XeF6 molecule.

To begin constructing the Lewis structure, we start by placing the Xenon atom in the center and arranging the six fluorine atoms around it. Each fluorine atom should be bonded to the Xenon atom with a single bond, which accounts for six of the 42 valence electrons.

Next, we add lone pairs of electrons to the remaining fluorine atoms until they each have an octet. After distributing all of the remaining 36 electrons as lone pairs on the fluorine atoms, we find that the Xenon atom still has two unpaired electrons.

To satisfy the octet rule for Xenon, we must form multiple bonds between the Xenon atom and some of the surrounding fluorine atoms. The most stable arrangement involves two double bonds between the Xenon atom and two of the fluorine atoms. In this final Lewis structure, the Xenon atom has a total of eight valence electrons and each fluorine atom has a complete octet.

The resulting Lewis structure of XeF6 is represented as follows:

F F

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F--Xe--F

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F F