Hydronium [H3O]+ Lewis dot structure, molecular geometry or shape, electron geometry, bond angle, hybridization, formal charges, polar vs non-polar
The hydronium ion represented by the chemical formula [H3O]+ is a very important chemical entity for acid-base reactions in chemistry. Acidic substances liberate H+ ions in an aqueous solution. The hydronium [H3O]+ ion, the simplest oxonium ion, is formed by the coordinate covalent bonding of the H+ ion with H2O molecules.
In this way, the presence and the concentration of [H3O]+ ions determine the pH and/or acidity of a solution.
In this article, we will discuss how to draw the Lewis structure of [H3O]+. You will also learn interesting facts about the molecular geometry or shape, electron geometry, bond angle, formal charge, hybridization, polarity, etc., of [H3O]+ ion.
The Lewis structure of hydronium [H3O]+ ion consists of an oxygen (O) atom at the center. It is bonded to three atoms of hydrogen (H) at the sides. There are a total of 4 electron density regions around the central O-atom in [H3O]+. The 4 electron density regions are constituted of 3 bond pairs and 1 lone pair on the central atom.
If you have any confusion regarding the Lewis dot structure of [H3O]+ ion, then your problem is solved because we have split the task into the following simple steps.
So, follow the steps given below and draw the Lewis dot structure of [H3O]+ with us.
Steps for drawing the Lewis dot structure of [H3O]+
1. Count the total valence electrons in [H3O]+
The Lewis dot structure of a molecule is referred to as a simplified representation of all the valence electrons present in it. Therefore, the very first step while drawing the Lewis structure of [H3O]+ is to find the total valence electrons present in the concerned elemental atoms.
As H3O+ consists of atoms from two different elements i.e., oxygen (O) and hydrogen (H) so you just need to look for these elements in the Periodic Table.
Oxygen (O) is present in Group VI A of the Periodic Table so it has a total of 6 valence electrons. On the other hand, hydrogen (H) lies at the top of the Periodic Table of elements containing a single valence electron only.
Total number of valence electrons in Hydrogen = 1
Total number of valence electrons in Oxygen = 6
The [H3O]+ ion consists of 1 O-atom and 3 H-atoms. Therefore, the valence electrons in the Lewis dot structure of [H3O]+ = 1(6) + 3(1) = 9 valence electrons.
However, the twist here is that the [H3O] +ion carries a positive (+1) charge which means 1 valence electron is removed from the total valence electrons initially available for drawing the H3O+ Lewis structure.
∴ Hence, the total valence electrons available for drawing the Lewis structure of [H3O]+ = 9 – 1 = 8 valence electrons.
2. Choose the central atom
While drawing the Lewis structure of a molecule or a molecular ion, usually the least electronegative atom is chosen as the central atom.
This is because electronegativity refers to the ability of an atom to attract a shared pair of electrons from a covalent chemical bond. So, the least electronegative atom is least likely to attract electrons and more likely to share its electrons with other atoms in its vicinity.
However, in the case of H3O+, there are only two types of atoms involved. Out of the two elemental atoms, hydrogen is very much less electronegative than oxygen but it cannot be chosen as the central atom.
A hydrogen (H) atom can accommodate only 2 electrons so it can form a bond with a single adjacent atom only. This denotes that H is always placed as an outer atom in a Lewis structure.
Consequently, oxygen (O) is chosen as the central atom in the Lewis structure of hydronium [H3O]+ while the 3 H-atoms are placed in its surroundings, as shown below.
3. Connect outer atoms with the central atom
Now we need to connect the outer atoms with the central atom of a Lewis structure using single straight lines.
As hydrogen atoms are the outer atoms while the oxygen atom is the central atom in the H3O+ Lewis structure so all three H-atoms are joined to the central O-atom via straight lines, as shown in the diagram below.
Each straight line represents a single covalent bond i.e., a bond pair containing 2 electrons. There are a total of 3 single bonds in the above diagram which means a total of 3(2) = 6 valence electrons are used till this step, out of the 8 initially available.
4. Complete the duplet of outer atoms
Each H-atom needs a total of 2 valence electrons in order to achieve a stable duplet electronic configuration. The Lewis structure obtained till this step already shows 2 electrons around each H-atom.
This means each H-atom already has a complete duplet and we do not need to make any changes regarding H-atoms in this structure. Also, there is no lone pair of electrons around any H-atoms in the [H3O]+ Lewis dot structure.
5. Complete the octet of the central atom
Total valence electrons available – electrons used till step 4 = 8 – 6 = 2 valence electrons.
These 2 electrons are thus placed as a lone pair on the central O-atom, as shown below.
In this way, in addition to a complete duplet of the outer H-atoms, the central O-atom also has a complete octet electronic configuration with 3 single bonds + 1 lone pair on it.
The final step is to check the stability of the final Lewis structure of H3O+ obtained in step 5. Let us do that using the formal charge concept.
6. Check the stability of the [H3O]+ Lewis structure using the formal charge concept
The less the formal charge on the atoms of a molecule, the better the stability of its Lewis structure.
The formal charge can be calculated using the formula given below.
Non-bonding electrons = no lone pairs = 0 electrons
Formal charge = 1-0-2/2 = 1-0-1 = 1-1 = 0
The above calculation shows that a zero formal charge is present on each H-atom in the H3O+ Lewis structure. However, a +1 formal charge is present on the central O-atom which is equivalent to the charge present on the hydronium ion overall.
Therefore, the Lewis structure obtained is the correct and the most stable Lewis representation of the hydronium (H3O+) ion. This Lewis structure is enclosed in square brackets and a +1 charge is placed at the top right corner, as shown below.
What are the electron and molecular geometry of H3O+?
The hydronium [H3O]+ ion has a trigonal pyramidal shape or molecular geometry. In contrast to that, the ideal electron pair geometry of the ion is tetrahedral.
The presence of a lone pair of electrons on the central O-atom distorts the shape of the molecule and makes it occupy a different molecular geometry from its ideal electron pair geometry.
Molecular geometry of [H3O]+
The hydronium ion [H3O]+ ion has a trigonal pyramidal molecular geometry and shape. The molecular ion has a triangular base and a pyramid at the top.
A lone pair of electrons present on the central O-atom in the H3O+ ion leads to lone pair-bond pair repulsions in the molecular ion in addition to bond pair-bond pair electronic repulsions. This repulsive effect influences the ion to adopt a different shape from its ideal electron pair geometry.
An important point to remember is that the molecular geometry or shape of a molecule or molecular ion is strongly influenced by the different number of lone pairs and bond pairs around the central atom.
Contrarily, the ideal electronic geometry only depends on the total number of electron density regions around the central atom no matter whether it’s a bond pair or a lone pair.
Let’s see how this concept applies to the shape and geometry of the hydronium ion.
Electron geometry of [H3O]+
According to the valence shell electron pair repulsion (VSEPR) theory of chemical bonding, the ideal electronic geometry of a molecule or a molecular ion containing a total of 4 electron density regions around the central atom is tetrahedral.
In the hydronium [H3O]+ ion, there are 3 single bonds and 1 lone pair around the central oxygen atom which makes a total of 3+1 = 4 electron density regions. Thus, its electron geometry is tetrahedral.
An easy way to find the shape and geometry of the molecule is to use the AXN method.
AXN is a simple formula to represent the number of atoms bonded to the central atom in a molecule and the number of lone pairs present on it.
It is used to predict the geometry or shape of a molecule using the VSEPR concept.
AXN notation for [H3O]+ molecular ion
A in the AXN formula represents the central atom. In the [H3O]+ ion, oxygen (O) is present at the center so A = Oxygen.
X denotes the atoms bonded to the central atom. In [H3O]+, 3 hydrogens (H) atoms are bonded to the central O so X = 3.
N stands for the lone pairs present on the central atom. As per the Lewis structure of [H3O]+, there is one lone pair on central oxygen so N=1.
Thus, the AXN generic formula for the [H3O]+ ion is AX3N1.
Now, you may have a look at the VSEPR chart below.
The VSEPR chart confirms that the ideal electron geometry of a molecule with AX3N1 generic formula is tetrahedral while its molecular geometry or shape is trigonal pyramidal, as we already noted down for the [H3O]+ ion.
Hybridization of [H3O]+
The central oxygen (O) atom has sp3 hybridization in the hydronium [H3O]+ ion.
During chemical bonding, one 2s orbital of oxygen mix with three 2p atomic orbitals to yield four sp3 hybrid orbitals. Each sp3 hybrid orbital possesses a 75% p-character and a 25 % s-character.
One of the four sp3 hybrid orbitals contains paired electrons while the remaining three sp3 hybrid orbitals contain a single electron only.
The sp3 hybrid orbital containing paired electrons is situated as a lone pair on the central O-atom. Conversely, the other three sp3 hybrid orbitals overlap with the s-orbitals of hydrogen to form the sp3-s O-H sigma (σ) bond on each side of the hydronium ion, as shown below.
A shortcut to finding the hybridization present in a molecule or a molecular ion is by using its steric number against the table given below. The steric number of central O in [H3O]+ is 4 so it has sp3 hybridization.
Steric number
Hybridization
2
sp
3
sp2
4
sp3
5
sp3d
6
sp3d2
The [H3O]+ bond angle
The ideal bond angle in a tetrahedral molecule is approx. equal to 109.5°. But it is due to a lone pair of electrons on the central O-atom in [H3O]+ that lone pair-bond pair repulsions push the left and right O-H bonds away from the lone pair.
The internal H-O-H bond angle decreases to 107.5 ° while the external bond angle increases up to an experimentally determined value of 113°, as shown in the figure below.
The O-H bond lengths are approximately equal to 103 pm.
Oxygen is a highly electronegative element. Pauling’s electronegativity scale states that a covalent chemical bond is polar if it has an electronegativity difference greater than 0.5 units between the bonded atoms.
A high electronegativity difference of 1.24 units exists between oxygen (E.N = 3.44) and hydrogen (E.N = 2.20) atoms. Thus, each O-H bond is polar in the hydronium [H3O]+ ion. Oxygen gains a partial negative (δ–) charge while each H-atom obtains a partial positive (δ+) charge.
The asymmetric shape of the hydronium ion further endorses the polarity effect. The dipole moments of polar O-H bonds do not get canceled and the electron cloud stays non-uniformly distributed in the molecule overall. Consequently, the hydronium [H3O]+ ion is polar (net µ = 1.84 D).
The Lewis structure of hydronium [H3O]+ ion consists of an oxygen (O) atom at the center and three atoms of hydrogen (H).
There are a total of 8 valence electrons i.e., 8/2 = 4 electron pairs in this Lewis structure.
Out of the 4 electron pairs, there are 3 bond pairs shared between the central O-atom and each H-atom.
The remaining one electron pair is present as a lone pair on oxygen. However, there is no lone pair of electrons on any outer atom in the H3O+ Lewis structure.
According to VSEPR theory, what is the shape of [H3O]+?
According to the VSEPR theory, [H3O]+ ion has an AX3N1 generic formula. The 1 lone pair of electrons on the central O-atom in H3O+ makes it occupy a different shape from its ideal electron geometry i.e., trigonal pyramidal versus tetrahedral respectively.
Why does NH3 have a pyramidal shape like H3O+ but NH4+ has a tetrahedral shape similar to CH4?
In the ammonia (NH3) molecule, there is one lone pair of electrons present on the central N atom similar to a lone pair on the central O-atom in H3O+. Thus, NH3 has a trigonal pyramidal shape, similar to H3O+.
In the ammonium (NH4+) ion, the electron pair on nitrogen is used in coordinate covalent bonding with the H+ In short, there is no lone pair on the central N-atom in NH4+, all the four electron density regions are constituted of N-H bond pairs.
Therefore, NH4+ maintains a shape identical to its ideal electron pair geometry i.e., tetrahedral, similar to the methane (CH4) molecule.
There are two different bond angles reported in the [H3O]+ ion i.e., 107.5° and 113°. These bond angles are different from the ideal bond angle of a tetrahedral molecule i.e., 109.5° due to lone pair-bond pair electronic repulsions present in the hydronium ion.
Vishal Goyal is the founder of Topblogtenz, a comprehensive resource for students seeking guidance and support in their chemistry studies. He holds a degree in B.Tech (Chemical Engineering) and has four years of experience as a chemistry tutor. The team at Topblogtenz includes experts like experienced researchers, professors, and educators, with the goal of making complex subjects like chemistry accessible and understandable for all. A passion for sharing knowledge and a love for chemistry and science drives the team behind the website. Let's connect through LinkedIn: https://www.linkedin.com/in/vishal-goyal-2926a122b/
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