Sulfuric acid (H2SO4) Lewis structure, molecular geometry or shape, electron geometry, bond angle, hybridization, polar or nonpolar, formal charge
The chemical formula H2SO4 represents dihydrogen sulfate, most commonly known as sulfuric acid, the king of chemicals.
H2SO4 exists as a colorless, odorless, viscous liquid. It is an extremely strong acid (pH 2.75); it is highly water-soluble and is a major component of acid rain.
Our main focus in this article is to learn about the chemistry behind H2SO4, including how to draw its Lewis dot structure, what is its molecular geometry or shape, electron geometry, bond angle, hybridization, formal charges, polarity etc.
So without any further delay, let’s start reading.
The Lewis structure of sulfuric acid (H2SO4) consists of three different elemental atoms. A sulfur (S) atom is present at the center of the molecule. It is directly bonded to two oxygen (O) atoms via double covalent bonds and two hydroxyl (OH) functional groups via single covalent bonds.
In this way, there are a total of 4 electron density regions around the central S-atom in the Lewis structure of H2SO4. All three electron density regions are constituted of bond pairs; thus, there is no lone pair of electrons on the central S-atom in H2SO4.
To draw the Lewis dot structure of H2SO4, grab a piece of paper and a pencil and follow the simple steps given below.
Steps for drawing the Lewis dot structure of H2SO4
1. Count the total valence electrons in H2SO4
The very first step while drawing the Lewis structure of H2SO4 is to calculate the total valence electrons present in its concerned elemental atoms.
In H2SO4, there are atoms from three different elements of the Periodic Table. So you need to look for these elements in the Periodic Table.
Both sulfur (S) and oxygen (O) are present in Group VI A (or 16), so they have 6 valence electrons in each atom, while hydrogen (H) lies at the top of the Periodic Table containing a single valence electron only.
The H2SO4 molecule consists of 2 H-atoms, 1 S-atom, and 4 O-atoms.
∴ Therefore, the total valence electrons available for drawing the Lewis dot structure of H2SO4 = 2(1) + 6 + 4(6) = 32 valence electrons.
2. Choose the central atom
In this second step, usually the least electronegative atom out of all the concerned atoms is chosen as the central atom.
This is because the least electronegative atom is the one that is most likely to share its electrons with the atoms spread around it.
Oxygen is more electronegative than both sulfur and hydrogen, so it cannot be chosen as the central atom. Hydrogen is less electronegative than sulfur, but it can also not be chosen as the central atom because a hydrogen (H) atom can accommodate only 2 valence electrons; hence 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.
As a result, the sulfur (S) atom is chosen as the central atom in the Lewis dot structure of H2SO4, while two H and four O atoms are placed around it as outer atoms, as shown in the figure below.
3. Connect outer atoms with the central atom
In this step, all the outer atoms are joined to the central atom using single straight lines. But as we discussed already, hydrogen can form a single bond with one adjacent atom only.
So the two H-atoms at the sides are first connected to the adjacent O-atoms. Contrarily, all the other outer atoms are directly joined to the central S-atom using straight lines, as shown below.
Each straight line represents a single covalent bond, i.e., a bond pair containing 2 electrons. There are a total of 6 single bonds in the above diagram.
As 6(2) = 12, that means 12 valence electrons are already consumed out of the 32 initially available. But we still have 32 -12 = 20 valence electrons to be accommodated in the Lewis dot structure of H2SO4.
4. Complete the duplet and/or octet of the outer atoms
As we already identified, the hydrogen and oxygen atoms are the outer atoms in the Lewis dot structure of H2SO4.
Each hydrogen (H) atom requires a total of 2 valence electrons in order to achieve a stable duplet electronic configuration.
The O-H single bonds represent 2 valence electrons around each H-atom in H2SO4 Lewis structure. This means the H-atoms already have a complete duplet, and we do not need to make any changes in the Lewis structure obtained so far with regard to the hydrogen atoms.
In contrast to that, an O-atom needs a total of 8 valence electrons in order to achieve a stable octet electronic configuration.
An O-H bond and an S-O bond represent a total of 2 + 2 = 4 electrons around the two oxygen atoms present at the right and left. This denotes that these two O-atoms are still deficient in 4 valence electrons, each that are required to complete their octet. So these 4 electrons are placed as 2 lone pairs around each O-atom, as shown below.
The other two oxygen atoms only form an S-O bond which denotes 2 electrons each. Thus, these are deficient in 6 more electrons to complete their octet. Consequently, these 6 electrons are placed as 3 lone pairs around each of the other two O-atoms, as shown below.
5. Complete the octet of the central atom
Total valence electrons used till step 4 = 6 single bonds + 2 (electrons placed around the O-atom forming O-H bond) + electrons placed around other two O-atoms = 6(2) + 2(4) + 2(6) = 32 valence electrons.
Total valence electrons – electrons used till step 4= 32 – 32= 0 valence electrons.
As all the 32 valence electrons initially available are now consumed, so there is no lone pair on the central S-atom in the H2SO4 Lewis structure.
Also, the central S-atom has a total of 4 single bonds around it which denotes a complete octet electronic configuration.
But the final question is, is this structure stable? Let’s find that using the formal charge concept.
6. Check the stability of Lewis’s structure using the formal charge concept
The fewer formal charges present on the atoms of a molecule, the better the stability of its Lewis structure.
The formal charges 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 zero formal charges are present on the O-H bonded oxygen and hydrogen atoms. However, a +2 formal charge is present on the central S-atom, and -1 formal charges are present on each of the two S-O bonded oxygen atoms above and below the Lewis structure.
But as we discussed already, the fewer the formal charges present on the bonded atoms in a molecule, the greater the stability of its Lewis structure. Hence we need to reduce the formal charges present on the above Lewis structure.
We can do that by converting lone pairs on S-O bonded oxygen atoms into bond pairs, as shown in the next step.
7. Minimize the formal charges by converting lone pairs into covalent bonds
A +2 charge represents a deficiency of two electrons, while a -1 formal charge indicates an excess of one electron. So a lone pair present on each negatively charged O-atom is converted into a bond pair between the positively charged S-atom at the center and the respective O-atom.
In the structure obtained below, each of the outer O-atoms has a complete octet. But the central S-atom now has a total of 12 valence electrons around it which means an expanded octet.
Always keep in mind that sulfur can accommodate more than 8 valence electrons during chemical bonding owing to the presence of a 3d subshell in it. After completely filling the 3p orbitals, electrons start occupying 3d atomic orbitals in sulfur.
Now let’s once again check the stability of the above Lewis structure by calculating formal charges.
Non-bonding electrons = no lone pairs = 0 electrons
Formal charge = 1-0-2/2 =1-0-1 = 1-1= 0
The absence of any formal charge on any one bonded atom in the H2SO4 Lewis structure ensures it is a stable structure and that we have drawn it correctly.
What are the electron and molecular geometry of H2SO4?
H2SO4 has an identical electron geometry and molecular geometry or shape, i.e., tetrahedral. There are a total of 4 electron density regions around the central S-atom comprised of 2 oxygen atoms and 2 hydroxyl functional groups.
However, no lone pair of electrons is present on the central S-atom in H2SO4; thus, no distortion is witnessed in its shape and/or geometry.
Molecular geometry of H2SO4
The sulfuric acid (H2SO4) molecule has a tetrahedral molecular geometry or shape.
The two hydroxyls (O-H) functional groups present next to the central S-atom are considered two separate regions of electron density. These two OH groups, in addition to two O-atoms, make a total of 4 electron density regions or electron domains. The four electron domains lie along the four vertices of a tetrahedron, as shown in the figure below.
There is no lone pair of electrons on the central S-atom; thus, no lone pair-bond pair and lone pair-lone pair electronic repulsions are present in the molecule. Consequently, no distortion is witnessed in the shape and geometry of the molecule.
Only a bond pair-bond pair repulsive effect exists that pushes the bonded atoms away. The atoms occupy positions as far apart from one another as possible to minimize the electron-repulsive effect.
Electron geometry of H2SO4
According to the valence shell electron pair repulsion (VSEPR) concept of chemical bonding, the ideal electronic geometry of a molecule containing 4 regions of electron density around the central atom is tetrahedral.
The 2 O-atoms and the 2 OH groups make a total of 4 electron density regions or electron domains around the central S-atom in H2SO4. Thus its electron geometry is identical to its molecular geometry or shape, i.e., tetrahedral.
A more straightforward way of determining the shape and geometry of a 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 shape and geometry of a molecule based on the VSEPR concept.
AXN notation for the H2SO4 molecule
A in the AXN formula represents the central atom. In H2SO4, sulfur (S) acts as the central atom, so A=S.
X denotes the electron domains bonded to the central atom.2 O-atoms and 2 OH groups are directly bonded to the central S-atom in H2SO4; thus, X=4.
N stands for the lone pairs present on the central atom. As per the Lewis dot structure of H2SO4, there is no lone pair present on central sulfur; thus, N=0.
So, the AXN generic formula for H2SO4 is AX4.
Now have a quick look at the VSEPR chart given below to identify where you find AX4.
The VSEPR chart confirms that the molecular geometry or shape of a molecule with an AX4 generic formula is identical to its electron pair geometry, i.e., tetrahedral, as we already noted for H2SO4.
Hybridization of H2SO4
The central S atom is sp3 hybridized in the H2SO4 molecule.
The electronic configuration of sulfur (S) is 1s2 2s2 2p6 3s2 3p4.
During chemical bonding, one 3s and one 3p electron of sulfur shifts to two empty 3d atomic orbitals. Consequently, the half-filled 3s atomic orbital of sulfur hybridizes with three half-filled 3p orbitals to yield three equivalent sp3 hybrid orbitals.
Each sp3 hybrid orbital possesses a 25% s-character and a 75% p-character, and each contains a single electron only.
In H2SO4, the sp3 hybrid orbitals of the sulfur atom overlap with the sp2 and sp3 hybrid orbitals of oxygen atoms to form the S=O and S-O sigma (σ) bonds, respectively.
The unhybridized d-orbitals of sulfur then overlap with the unhybridized p-orbital of the oxygen atoms to form the required pi (π) bonds in each S=O double bond, as shown below.
A short trick for finding the hybridization present in a molecule is to memorize the table given below. You can determine the steric number of a molecule and use that against this table to find its hybridization.
The steric number of central sulfur (S) in H2SO4 is 4, so it has sp3 hybridization.
Steric number
Hybridization
2
sp
3
sp2
4
sp3
5
sp3d
6
sp3d2
The H2SO4 bond angle
The bonded atoms form a mutual bond angle of 109.5° in the perfectly symmetrical tetrahedral H2SO4 molecule. However, there are three different bond lengths in the molecule.
The S=O bond length is 142.2 pm, the S-O bond length is 157.4 pm, and the O-H bond length is 97 pm.
Pauling’s electronegativity scale states that a covalent bond is polar if the bonded atoms possess an electronegativity difference between 0.5 to 1.6 units.
An electronegativity difference of 0.86 units exists between the bonded sulfur (E.N = 2.58) and oxygen (E.N = 3.44) atoms in each of the S=O and S-O bonds in the H2SO4 molecule. Thus both S-O and the S=O bonds are polar in nature.
Similarly, an electronegativity difference of 1.24 units exists between the bonded hydrogen (E.N = 2.20) and oxygen atoms (E.N = 3.44). So the O-H bond is also extremely polar and possesses a high dipole moment value (symbol µ).
The dipole moments of these polar bonds do not get canceled equally in the molecule overall; thus, the H2SO4 molecule is polar with a non-uniformly distributed electron cloud (net µ = 2.7 Debye).
H2SO4 is extremely soluble in polar solvents such as water (H2O), which is also evidence of the molecule’s polar nature. The presence of two hydroxyls (OH) functional groups allows H2SO4 to develop strong hydrogen bonding with H2O molecules and get solubilized quickly at all concentrations.
The Lewis dot structure of sulfuric acid (H2SO4) displays a total of 32 valence electrons i.e., 32/2 = 16 electron pairs.
Out of the 16 electron pairs, there are 8 bond pairs and 8 lone pairs.
A sulfur (S) atom is present at the center. It is bonded to 2 O-atoms and 2 OH functional groups.
There is no lone pair of electrons on the central S-atom. All four oxygen atoms carry 2 lone pairs of electrons each.
What is the molecular geometry of H2SO4 as per VSEPR theory?
As per the VSEPR concept, the AXN generic formula for H2SO4 is AX4. There is a sulfur (S) atom at the center, it is bonded to four other atoms, and there is no lone pair of electrons on this central S-atom. Thus, the molecular geometry or shape of Sulfuric acid (H2SO4) is tetrahedral.
What is the electron geometry of H2SO4?
There are a total of 4 electron density regions or electron domains around the central atom in H2SO4.
So its ideal electron pair geometry is identical to its molecular geometry or shape i.e., tetrahedral.
What are the shapes of these sulfur-containing molecules: H2SO4, SO2, and SO3?
Sulfuric acid (H2SO4) has a tetrahedral shape. There is no lone pair on the central S-atom in H2SO4, so there is no distortion present in its shape and geometry.
The shape of the sulfur dioxide (SO2) molecule is bent or angular. One lone pair on the central S-atom leads to a lone pair-bond pair repulsive effect. Thus the molecule adopts a different shape from its ideal electronic geometry, i.e., tetrahedral.
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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