How to draw XeF4 Lewis Structure? - Science Education and Tutorials

Drawing XeF4 Lewis Structure is very easy to by using the following method. Here in this post, we described step by step method to construct XeF4 Lewis Structure.

Table of Contents

Key Points To Consider When Drawing The XeF4 Structure

A three-step approach for drawing the XeF4 Lewis structure can be used. The first step is to sketch the Lewis structure of the XeF4 molecule, to add valence electrons around the xenon atom; the second step is to add valence electrons to the four fluorine atoms, and the final step is to combine the step1 and step2 to get the XeF4 Lewis Structure.

The XeF4 Lewis structure is a diagram that illustrates the number of valence electrons and bond electron pairs in the XeF4 molecule. The geometry of the XeF4 molecule can then be predicted using the Valence Shell Electron Pair Repulsion Theory (VSEPR Theory), which states that molecules will choose the XeF4 geometrical shape in which the electrons have from one another.

Finally, you must add their bond polarities to compute the strength of the Xe-F bond (dipole moment properties of the XeF4 molecule). The xenon-fluorine bonds in xenon tetrafluoride(XeF4), for example, are polarised toward the more electronegative fluorine, and because both bonds have the same size and located around four fluorine terminals of the square planar with two lone pairs of electrons just out of the plan, their sum is zero due to the XeF4 molecule’s bond dipole moment and two lone pair of electron on the xenon atom. The xenon tetrafluoride(XeF4) molecule is classified as a polar molecule.

The molecule of xenon tetrafluoride (with square planar molecular geometry) is tilted, the bond angles between xenon and fluorine are 90 degrees. It has a difference in electronegativity values between xenon and fluorine atoms, with xenon’s pull being less than fluorine’s terminal in the XeF4 molecule. But they canceled each other due to the symmetrical molecular geometry. As a result, it has the zero dipole moment. The XeF4 molecule has a zero dipole moment due to an equal charge distribution of negative and positive charges. The net dipole moment of the XeF4 molecule is 0 D.

Overview: XeF4 Lewis Structure

The central atom is xenon, which is bordered on four terminals with four fluorine atoms( in square planar geometry), and two lone pairs of electrons on the xenon in the top and bottom of the square planar geometry. Xenon has eight outermost valence electrons, indicating that it possesses eight electrons in its outermost shell, whereas fluorine only has seven valence electrons in its outermost shell. To complete the octet of the fluorine atom, a fluorine terminal atom requires one valence electron. If you’re interested in learning more about the fluorine octet rule, please see in our previous post.

Four fluorine atoms establish covalent connections with the central xenon atom as a result, leaving the xenon atom with two lone pairs. There are two lone pairs of electrons on the xenon central atom that resist the bond pairs of the four Xe-F. According to VSEPR theory, no electronic repulsion of the lone pairs and bond pairs leads the XeF4 molecule to take on a square planar molecular geometry shape.

The XeF4 molecule’s Xe-F bonds are arranged in symmetrical order around the square planar molecular geometry, giving rise to the XeF4 molecular shape. The XeF4 molecule has a square planar molecular geometry because there is no electrical repulsion between two lone pairs and four bond pairs(Xe-F) of the XeF4 molecule.

Electronegative Difference Calculation of XeF4 Molecule:

The xenon atom has an electronegativity of 2.6, while fluorine has an electronegativity of 3.98 in the XeF4 molecule. The difference in electronegativity can be estimated using the method below.

The electronegative value difference between xenon and fluorine
Electronegativity value of xenon = 2.6
Electronegativity value of fluorine= 3.98
Difference of electronegativity value between xenon and fluorine= 3.98 – 2.6 =1.38
Electronegativity difference between Xe-F bond calculation of XeF4 molecule

Due to the difference in electronegativity value of greater than 0.5, the Xe-F bond of the XeF4 molecule becomes polar. Because of this difference in electronegativity, the XeF4 molecule’s Xe-F bond becomes polar. The total net dipole moment of the XeF4 molecule is zero due to the cancellation of the bond dipole moment. The electronegativity of an atom is the strength with which it may attract bound electron pairs to its side. The polarity of XeF4 is discussed in our previous post.

As a result, the Xe-F bond’s dipole moment is high due to the polarization of the bonds and two lone pairs of electrons, and all Xe-F bonds’ dipoles are arranged in the symmetrical XeF4 molecular geometry. The XeF4 molecule’s total dipole moment is predicted to be 0 D. It has a partial negative charge for the terminal fluorine atoms and a partial positive charge for the central xenon atom.

The electron dot structure of the XeF4 molecule is also known as the XeF4 Lewis structure. It determines the number of outermost valence electrons as well as the electrons engaged in the XeF4 molecule’s bond formation. The outermost valence electrons of the XeF4 molecule must be understood while considering the Lewis structure of the molecule.

The xenon atom is the middle element in XeF4 molecular geometry, with eight electrons in its outermost valence electron shell, whereas the fluorine atom has seven electrons in its outermost valence electron shell.

The XeF4 molecule has a total of 36 valence electrons as a result of the foregoing above said reasoning. With the core central xenon atom, the four terminal with four fluorine atoms forms covalent bonds, leaving the xenon atom with two lone pairs of electrons on the top and bottom of square planar geometry.

Because two lone pairs of electrons on the central xenon atom create interaction with Xe-F bond pairs. The bond angle of the F-Xe-F bond in the square planar molecular geometry is approximately 90 degrees. This angle is less than the CH4 molecule bond angle due to the two lone pairs of electrons on the XeF4 molecule. The Xe-F bond length is 185-198 pm(picometer).

To sketch the XeF4 Lewis structure by following these instructions:

Step-1: XeF4 Lewis dot Structure by counting valence electrons on the xenon atom

To calculate the valence electron of each atom in XeF4, look for its periodic group from the periodic table. The halogen and noble gas families, which are the 17th and 18th groups in the periodic table, are both made up of fluorine and xenon atoms. In their outermost shells, fluorine and xenon have seven and eight valence electrons, respectively.

Because fluorine and xenon are members of the periodic table’s halogen and noble gas family groups, their valence electrons are seven and eight, respectively.

Calculate the total number of valence electrons in the XeF4 molecule’s outermost valence shell. The first step is to determine how many electrons are in the XeF4 Lewis structure’s outermost valence shell. An electron in an atom’s outermost shell is known as a valence electron. It is represented by dots in the XeF4 Lewis diagram. The XeF4 molecule’s core xenon atom can be represented as follows:

Total outermost valence shell electron of xenon atom in XeF4= 8
Total outermost valence shell electron of the fluorine atom in XeF4= 7
The XeF4 molecule has one central xenon and four fluorine atoms. Then the total outermost valence shell electrons can be calculated as follows
∴ Total outermost valence shell electrons available for XeF4 Lewis structure( dot structure) = 8+4*7= 36 valence electrons in XeF4.
calculation of total valence electron of XeF4 molecule

Choose the atom with the least electronegative value atom and insert it in the center of the molecular geometry of XeF4. We’ll choose the least electronegative value atom in the XeF4 molecule to place in the center of the XeF4 Lewis structure diagram in this phase. The electronegativity value in periodic groups grows from left to right in the periodic table and drops from top to bottom.

Step-2: Lewis Structure of XeF4 for counting valence electrons around the terminal fluorine atom

As a result, xenon is the first atom in the periodic table’s noble gas family group. Xenon is the fifth member of the noble gas family. The electronegative value of the xenon atom is lower than that of the fluorine atom in the XeF4 molecule. Furthermore, xenon has an eight electrons limit since fluorine is the most electronegative element in the XeF4 molecule.

In the XeF4 Lewis structure diagram, the xenon atom can be the center atom of the molecule. As a result, central xenon in the XeF4 Lewis structure, with all four fluorine arranged in the square planar geometry.

Add valence electrons around the fluorine atom, as given in the figure.

Step-3: Lewis dot Structure for XeF4 generated from step-1 and step-2

Connect the exterior and core central atom of the XeF4 molecule with three single bonds (Xe-F). In this stage, use four fluorine atoms on the outside of the XeF4 molecule to the central xenon atom in the middle.

Count how many electrons from the outermost valence shell have been used in the XeF4 structure so far. Each Xe-F bond carries two electrons because each xenon atom is connected to four fluorine atoms by four Xe-F bonds. Bond pairings of Xe-F are what they’re called.

So, out of the total of 36 valence electrons available for the XeF4 Lewis structure, we used 8 for the XeF4 molecule’s four Xe-F bonds. The XeF4 molecule has two lone pairs of electrons in the central xenon atom. We need to put extra electrons in the molecular geometry of XeF4. Where to place the extra electron in the XeF4 molecule?

Place the valence electrons in the Xe-F bond pairs starting with the core xenon, four fluorine, and two lone pairs of electrons in the XeF4 molecule. In the XeF4 Lewis structure diagram, we always begin by introducing valence electrons from the central xenon atom(in step1). As a result, wrap around the central xenon atom’s bond pair valence electrons first (see figure for step1).

Xenon requires 8 electrons in its outermost valence shell to complete the molecular stability, 8 electrons bond pairs in Xe-F bonds. Then four electrons as a lone pair of electrons on the xenon atom of the XeF4 molecule are placed in a square planar geometry. Xenon already shares 8 electrons to the four Xe-F bonds. Then place the valence electron in the fluorine atom, it placed around seven electrons(step-2). Totally, 28 valence electrons placed on the four fluorine atoms of the XeF4 molecule.

We’ve positioned 8 electrons around the central xenon atom(step-3), which is represented by a dot, in the XeF4 molecular structure above. The xenon atom completes its molecular stability in the XeF4 molecule because it possesses 8 electrons in its (Xe-F) bond pairs with four fluorine in the outermost valence shell.

Count how many outermost valence shell electrons have been used so far using the XeF4 Lewis structure. Four electron bond pairs are shown as dots in the XeF4 chemical structure, whereas three single bonds each contain two electrons. The outermost valence shell electrons of the XeF4 molecule are eight as a result of the calculation.

So far, we’ve used eight of the XeF4 Lewis structure’s total 8 outermost valence shell electrons. Two lone pairs of electrons on the xenon atom in the square planar of the XeF4 molecule.

Complete the middle xenon atom stability and, if necessary, apply a covalent bond. The central xenon atom undergoes extra octet stability. Because it has a total of 12 electrons in the outermost valence shell. Eight electrons come from bond pairs of Xe-F and two lone pairs of electrons on the xenon central atom of XeF4.

The core atom in the XeF4 Lewis structure is xenon, which is bonded to the four fluorine atoms by single bonds (Xe-F). With the help of four single bonds, it already shares eight electrons. As a result, xenon follows the extra octet rule and has 12 electrons (8 bond pairs+2 lone pairs) surrounding it on the four terminals of the XeF4 molecule’s square planar geometry.

How to calculate the formal charge on a xenon atom in XeF4 Lewis Structure?

The formal charge on the XeF4 molecule’s xenon central atom often corresponds to the actual charge on that xenon central atom. In the following computation, the formal charge will be calculated on the central xenon atom of the XeF4 Lewis dot structure.

To calculate the formal charge on the central xenon atom of the XeF4 molecule by using the following formula:
The formal charge on the xenon atom of XeF4 molecule= (V. E(Xe)– L.E(Xe) – 1/2(B.E))
V.E (Xe) = Valence electron in a xenon atom of XeF4 molecule
L.E(Xe) = Lone pairs of an electron in the xenon atom of the XeF4 molecule.
B.E = Bond pair electron in Xe atom of XeF4 molecule
calculation of formal charge on xenon atom in XeF4 molecule

The xenon core atom (four single bonds connected to four fluorine atoms ) of the XeF4 molecule has eight valence electrons, two lone pairs of electrons(four electrons), and eight bonding electrons. Put these values for the xenon atom in the formula above.

Formal charge on xenon atom of XeF4 molecule = (8- 4-(8/2)) =0

In the Lewis structure of XeF4, the formal charge on the central xenon atom is zero.

Summary:

In this post, we discussed the method to construct the XeF4 Lewis structure. Need to remember that, if you follow above said method, you can construct molecular dot structure very easily.