XeO4 Lewis Structure: Drawings, Hybridization, Shape, Charges, Pair, And Detailed Facts


This article contains about the XeO4 lewis structure, Hybridization, bond angle, and other 13 important detailed facts.

The XeO4 lewis structure has tetrahedrally shaped having a bond angle of 109.50. It is a very stable compound of a noble gas which is a very exceptional case.  Due to the tetrahedral geometry, the central Xe is sp3 hybridized. All the valence electrons of Xe are being used for the formation of sigma and π bonds, so there are no lone pairs available for Xe in this molecule.

Bartlett 1st discovered the noble gas as a compound which was XeF. Due to the presence of electronegative substituents O, the positive charge density over Xe increases to such an extent that the energy of 5s and 5p decreases and they become comparable to the O.

Some important facts about XeO4

XeO4 is a yellow solid crystalline molecule that is stable at below temperature, at room temperature it decomposes. The melting point and boiling points are 237.1 K and 273 K respectively. The oxidation state of Xe in the Xeo4 lewis structure is +8 so it has its most stable oxidation state. 

XeO4 is formed by the reaction of concentrated Sulphuric acid on sodium and barium perxenates.

4XeF6 + 18Ba(OH)2 =3Ba2XeO6 + Xe +12BaF2 + 18H2O

Ba2XeO6 + 2H2SO4 = 2BaSO4 + XeO4 + 2H2O

How to draw lewis structure for XeO4 ?

Lewis structure is a very useful tool by the help of this structure we can find out the valence electrons, shape, and lone pairs of any covalent molecule.  So, it’s important to draw the XeO4 lewis structure. There are some techniques we should keep in mind to draw the lewis structure of XeO4.

So, at first, we count the total valence electrons for Xe and O atoms. Then we identify the central atom by its electronegativity, we know Xe is less electronegative than O, so in the XeO4 lewis structure, Xe is the central atom. The molecule bears no extra charge upon it so no need to add extra electrons or subtract electrons here. Now Xe and four O atoms are attached by a single bond as well as a double bond to complete the octet. The lone pairs exist only over O atoms.

XeO4 lewis structure shape

The valence electrons for Xe in the sigma bond formation will be four and there are four O atoms each contributing one electron, so the total electrons count will be, 4+(1*4) =8. According to the VSEPR (Valence Shell Electrons Pair Repulsion) theory of a covalent molecule if the total bonding electrons count will be 8 then the shape of the molecule will be tetrahedral.

XeO4 Lewis Shape

The whole electron density is lie over the Xe in the XeO4 lewis structure as Xe is the central atom here. All the four O atoms are present at the four corners of a tetrahedral geometry. The Bond angle made by central Xe and four O atoms in the tetrahedral moiety is 109.50.

XeO4 lewis structure lone pairs

In the XeO4 lewis structure, Xe forms eight bonds with four O atoms. Out of eight bonds, four bonds are sigma bonds and four bonds are π bonds. So, there are no available of valence electrons for Xe which can exist as lone pair.

XeO4 Lone Pairs

From the XeO4 lewis structure, we can see that the lone pairs are assigned only over O atoms. O is VIA or group 16th  element and from the electronic configuration we can say that there are only six electrons are present at the valence shell of O and those electrons are involved in the bond formation with the central Xe atom. O used only two valence electrons for one sigma and one π bond with Xe. So, it has four electrons remaining in its valence shell which exist as lone pairs. So, the total number of lone pairs in the XeO4 lewis structure is (4*2)= 8 pairs which are only from the O site.

XeO4 lewis structure formal charges

By assuming the same electronegativity of all atoms in a molecule we can calculate the particular charge present over a specific atom or molecule is called a formal charge. So here in the XeO4 lewis structure, we consider the same electronegativity for Xe and O.

The formula we can use to calculate the formal charge of XeO4, F.C. = Nv – Nl.p. -1/2 Nb.p.

Where Nv is the number of electrons in the valence shell or outermost orbital, Nl.p is the number of electrons in the lone pair, and Nb.p  is the total number of electrons that are involved in the bond formation only.

We have to calculate separately the formal charge for Xe as well as O.

The formal charge over Xe, 8-0-(16/2) = 0

The formal charge over O, 6-4-(4/2) = 0

So, we can say that there is no formal charge over Xe as well as O cause the molecule is neutral. So, by calculating the formal charge we can also say whether the molecule is charged or not.

XeO4 lewis structure resonance

Resonance is a hypothetical concept, by which an electron cloud of any molecule can delocalize between different skeleton forms of that molecule In the Xeo4 lewis structure, the electron clouds delocalize between O and Xe atoms in different canonical forms.

XeO4 Resonating Structure

From the XeO4 lewis structure, all five are the different resonating structures of XeO4. Among them, structure V is the most contributing structure as it contains a higher number of covalent bonds so its contribution is the highest. After that structure IV then III, II, and the least contributing is Structure I.

XeO4 lewis structure octet rule

In the XeO4 lewis structure, Xe and O both try to complete their octet by donating or accepting a suitable number of electrons from their valence shell or to the valence shell and gaining the nearest noble gas configuration.

XeO4 Octet Structure

In the XeO4 lewis structure, Xe is a group 18 element so it has a filled valence shell and no need to complete its octet. But O is a group of 16 elements, so it has six electrons in its outermost electrons and its octet is not completed. So, O form a double bond with Xe by sharing two of its electrons and two of Xe electrons and four electrons are present in its lone pairs. This way Oxygen completes its octet.

Xe is itself a noble gas and it participates in bond formation with O involving all the eight electrons in the valence shell.

XeO4 hybridization

In the XeO4 lewis structure, the molecular orbital energy of Xe and O are not the same, so Xe and O must undergo hybridization to form a hybrid orbital of equivalent energy and form a covalent bond.

We calculate the XeO4 hybridization by using the following formula,

H = 0.5(V+M-C+A), where H= hybridization value, V is the number of valence electrons in the central atom, M = monovalent atoms surrounded, C=no. of cation, A=no. of the anion.

For the XeO4 lewis structure, Xe has four electrons that are forming a sigma bond and four O atoms are surrounded.

So, the hybridization of central Xe in XeO4 is, ½(4+4+0+0)= 4 (sp3)

Structure      Hybridization value  State of hybridization of central atom   Bond angle
Linear   2  sp /sd / pd  1800
Planner trigonal 3  sp2                    1200
Tetrahedral     4sd3/ sp3     109.50
Trigonal bipyramidal5sp3d/dsp3       900 (axial), 1200(equatorial)
Octahedral   6 sp3d2/ d2sp3    900
Pentagonal bipyramidal7sp3d3/d3sp3      900,720

              

From the above table of hybridization, we can say that if the hybridization value of central atom is 4 then the central atoms is obviously sp3 hybridized involving one s and three p orbitals via mixing.

XeO4 Hybridization

From the box diagram of the XeO4 lewis structure, it is evident that we only consider the sigma bond in hybridization, not the π bond or multiple bonds. In the covalent bond, we also consider the sigma bond.

Xe has eight electrons in its outermost orbital in the ground state. When it gets excited four electrons are transferred to the vacant 5d orbitals. Then Xe has four unpaired electrons in 5s and 5p orbital. These four unpaired electrons undergo hybridization with four O atoms and make a covalent bond.

Here One 5s orbital and 3 5p orbitals are involved in the hybridization. So, the hybridization is sp3. This sp3 hybridized orbital is of equivalent energy of Xe as well as O so that they can make bonds. The remaining four electrons of Xe in the 5d orbital are involved in π bond formation with O and they are not involved in hybridization.

From the hybridization value, we can predict the bond angle of this molecule. In the XeO4 lewis structure, the hybridization is sp3, so here the % of s character is 1/4th or 25%. Now from Bent’s rule bond angle of a molecule is, COSθ =s/s-1, where s is the % of s character in hybridization and θ is the bond angle.

Now using the formula, we get, COSθ = (¼)/(1/4)-1

Θ= 1090, so we can say that from the hybridization value we can evaluate the bond angle of a molecule or vice versa. Again, from VSEPR  theory the bond angle for a tetrahedral-shaped molecule will be 109.50.

So VSEPR and hybridization theory can be correlated. The bond angle is perfectly 109.50 which is ideal for tetrahedral molecules. Actually, in the XeO4 lewis structure, there is no deviation factor is present for bond angle. Xe has no lone pairs and the size of the Xe is very large compared to O. Though four O make the double bond with central Xe there is enough space in tetrahedral moiety to minimize the lone pair-bond pair repulsion. So, there is no deviation in the bond angle.

Is XeO4 polar?

we know that the polarity of any molecule depends on the value of the resultant dipole moment. To understand the polarity of the XeO4 lewis structure we have to find out its dipole moment value of it.

In the XeO4 lewis structure, the shape of the molecule is tetrahedral. The shape is symmetrical which means all the O atoms are present they are exactly neutralizing the dipole-moment value of each other. So, in the XeO4 lewis structure, the final resultant dipole moment value is zero, which makes the entire molecule nonpolar.

Being Xe is noble gas though it can form compounds like XeO4. Why?

From the values of Ionization energy, it can be said that the threshold of chemical reactivity is reached at Kr and it has been found that Xe can form quite a large number of compounds with highly electronegative substituents like O and F atoms.

This is due to the presence of electronegative substituents F and O the positive charge density over Xe increases to such an extent that the energy of 5s and 5p decreases and they become comparable to the O and F. Thus, the 5s and 5p orbital of Xe may overlap 2p of the F and O.

The bond length Xe-O is expected to be shorter. why?

The single Xe-O bond length is near about 160 pm. But in the XeO4 lewis structure, the molecule possesses a double bond character, so the bond length is decreased.

Xe is group 18th element which is a noble gas and generally, the valence shell is fully occupied. From the electronic configuration, Xe has eight electrons in its valence shell which completed its outermost orbital, and all of the electrons are involved in bond formation with four O atoms (Sigma as well as π bonds) with O. O can form a stable double bond with suitable substituents like Xe and for this reason, there are no lone pairs or single bond character present between Xe-O bond.

Xe forms four sigma bonds as well as four π bonds with O atoms. Making the double bond makes the molecule more stable. The double bond is always shorter than the single bond and stronger than the single bond also.

Conclusion

From the above discussion of the XeO4 lewis structure, we can say that noble gas can also participate in reaction in the presence of strong electronegative substituents. The molecule is most stable due to the formation of a double bond so it cannot further participate in any other reaction. There is no lone pair-bond repulsion factor is present so the bond angle does not deviate and ideal for tetrahedral that is 109.50.

Biswarup Chandra Dey

Hi......I am Biswarup Chandra Dey, I have completed my Master's in Chemistry. My area of specialization is Inorganic Chemistry. Chemistry is not all about reading line by line and memorizing, it is a concept to understand in an easy way and here I am sharing with you the concept about chemistry which I learn because knowledge is worth to share it.

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