Ethyl alcohol (CH3CH2OH) Lewis structure, molecular geometry or shape, electron geometry, bond angle, hybridization, formal charges, polarity
CH3CH2OH represents the chemical formula for ethanol, aka ethyl alcohol, a primary alcohol commonly found in alcoholic beverages, cosmetics, perfumes, and spirits.
Ethyl alcohol is a clear, colorless liquid (molar mass = 46.07 g/mol) often used as a solvent for various industrial applications. Chemists are also investigating its potential as a safe fuel for the future.
In this article, we will teach you how to draw the Lewis dot structure of ethyl alcohol (CH3CH2OH), what is its molecular geometry or shape, electron geometry, bond angles, hybridization, formal charges, and polarity.
So continue reading!
Name of molecule | Ethanol or Ethyl alcohol |
Chemical formula | CH3CH2OH |
Molecular geometry of CH3CH2OH | Tetrahedral (w.r.t each C-atom)Bent, angular, or V-shaped (w.r.t O-atom) |
Electron geometry of CH3CH2OH | Tetrahedral |
Hybridization | sp3 |
Bond angles | ∠ H-C-H = 109.28° and ∠ C-O-H = 104.5° |
Nature | Polar molecule |
Total valence electrons in CH3CH2OH | 20 |
The overall formal charge on CH3CH2OH | Zero |
How to draw lewis structure of CH3CH2OH?
The Lewis structure of CH3CH2OH comprises two carbon (C) atoms, single covalently bonded to each other at the center. One C-atom is bonded to 3 H-atoms at the other end, while the second C-atom carries 2 terminal H-atoms, and a hydroxyl (OH) functional group, respectively.
There is no lone pair of electrons on either the C-atoms or the H-atoms. However, 2 lone pairs of electrons are present on the O-atom.
You may follow the simple steps given below to draw the Lewis dot structure of Ethyl alcohol (CH3CH2OH).
Steps for drawing the Lewis dot structure of CH3CH2OH
1. Count the total valence electrons present in CH3CH2OH
The three distinct elements present in CH3CH2OH are carbon, hydrogen, and oxygen.
If we look at the Periodic Table of Elements, carbon (C) is located in Group IV A (or 14) of the Periodic Table, which implies that it has a total of 4 valence electrons.
In contrast, oxygen (O) is present in Group VI A (or 16), containing 6 valence electrons in each atom.
However, hydrogen (H) lies at the top of the Periodic Table, containing 1 valence electron only.
- Total number of valence electrons in hydrogen = 1
- Total number of valence electrons in carbon = 4
- Total number of valence electrons in oxygen = 6
The CH3CH2OH molecule comprises 2 C-atoms, 6 H-atoms and 1 O-atom.
∴ Therefore, the total valence electrons available for drawing the Lewis dot structure of CH3CH2OH = 2(4) + 6(1) + 1(6) = 20 valence electrons.
2. Find the least electronegative atom and place it at the center
By convention, the least electronegative atom out of all those available is chosen as the central atom while drawing the Lewis structure of a molecule.
The least electronegative atom can easily form covalent bonds with other atoms by sharing its electrons.
Among the atoms present in CH3CH2OH, hydrogen (E.N = 2.20) is less electronegative than both carbon (E.N = 2.55) and oxygen (E.N = 3.44).
However, the H-atom is an exception as it cannot be chosen as the central atom in any Lewis structure. It can accommodate a total of 2 valence electrons, forming a single covalent bond with 1 adjacent atom only.
Therefore, we select the second option, i.e., a C-atom as the central atom in the CH3CH2OH Lewis structure.
As both the C-atoms are identical, therefore, any one C-atom can be considered a central atom while the 6 H-atoms and 1 O-atom are spread around the central C-atoms, as shown below.
3. Connect the outer atoms with the central atoms
In this step, all the outer atoms are joined to their adjacent C-atoms using single straight lines. However, the H-atom next to oxygen is only joined to this O-atom as it can form a single covalent bond only.
Also, the two central C-atoms are joined to each other, as shown below.
A straight line represents a single covalent bond, i.e., a bond pair containing 2 electrons.
In the above structure, there are a total of 8 single bonds, i.e., 8(2) = 16 valence electrons are already consumed out of the 20 initially available.
Now let’s see in the next steps where to place the remaining 4 valence electrons in the CH3CH2OH Lewis dot structure.
4. Complete the octet and/or duplet of the outer atoms
An O-atom needs a total of 8 valence electrons in order to achieve a stable octet electronic configuration.
A C-O and an O-H bond represent 4 valence electrons already present around the terminal O-atom in the CH3CH2OH Lewis dot structure.
Therefore, the remaining 4 valence electrons are placed as 2 lone pairs around the oxygen atom to complete its octet.
In contrast, each H-atom already has a complete duplet, possessing 2 valence electrons, so we do not need to make any changes w.r.t the H-atoms in the above structure.
5. Complete the octets of the central atoms
- Total valence electrons used till step 4 = 8 single bonds + electrons placed around O-atom, shown as dots = 8(2) + 4 = 20 valence electrons.
- Total valence electrons – electrons used till step 4 = 20 – 20 = 0 valence electrons.
As all the valence electrons initially available for drawing the CH3CH2OH Lewis structure are already consumed so there is no lone pair on either of the carbon atoms at the center.
However, we need not worry because these central C-atoms already have complete octets with a total of 4 single bonds, i.e., 8 valence electrons surrounding each carbon.
Hence, the final step is to check the stability of the CH3CH2OH Lewis structure by applying the formal charge concept.
6. Check the stability of Lewis’s 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 charges can be calculated using the formula given below.
- Formal charge = [valence electrons- nonbonding electrons- ½ (bonding electrons)].
Now let us use this formula and the Lewis structure obtained in step 5 to determine the formal charges present on the CH3CH2OH bonded atoms.
For each carbon atom
- Valence electrons of carbon = 4
- Bonding electrons = 4 single bonds = 4(2) = 8 electrons
- Non-bonding electrons = no lone pair = 0 electrons
- Formal charge = 4-0-8/2 = 4-0-4 = 4-4 = 0
For the oxygen atom
- Valence electrons of oxygen = 6
- Bonding electrons = 2 single bonds = 2(2) = 4 electrons
- Non-bonding electrons = 2 lone pairs = 2(2) = 4 electrons
- Formal charge = 6-4-4/2 = 6-4-2= 6-6 = 0
For each hydrogen atom
- Valence electrons of hydrogen = 1
- Bonding electrons = 1 single bond = 2 electrons
- Non-bonding electrons = no lone pairs = 0 electrons
- Formal charge = 1-0-2/2 = 1-0-1= 1-1 = 0
Zero or no formal charges on either of the atoms present in CH3CH2OH mark the incredible stability of the Lewis dot structure obtained below.
Now let us move ahead and discuss its molecular and electron geometry.
Also check –
What are the electron and molecular geometry of CH3CH2OH?
The molecular geometry or shape of CH3CH2OH w.r.t each C-atom is tetrahedral, while w.r.t the O-atom is bent, angular or V-shaped. In contrast, the electronic geometry of CH3CH2OH w.r.t both C-atoms and the O-atom is tetrahedral.
The presence of 2 lone pairs of electrons on the O-atom leads to strong lone pair-lone pair and lone pair-bond pair electronic repulsions, thus distorting the molecular shape of Ethyl alcohol w.r.t this oxygen atom.
Molecular geometry of CH3CH2OH
The molecular geometry or shape of Ethyl alcohol (CH3CH2OH) w.r.t each C-atom is tetrahedral while that w.r.t the O-atom is bent, angular or V-shaped.
To a carbon atom at the center, four different bond pairs are attached like four corners of a tetrahedron, and there is no lone pair of electrons on the respective C-atom.
Contrarily, to an O-atom, 1 C-atom and 1 H-atom are attached as bond pairs, and it also carries 2 lone pairs of electrons. The lone pairs lead to lone pair-lone pair and lone pair-bond pair electronic repulsions.
The bonded atoms tilt away from the center to minimize the electron-repulsive effect. Thus, the molecule occupies a bent or Inverted-V shape w.r.t this oxygen atom, as shown below.
Electron geometry of CH3CH2OH
According to the valence shell electron pair repulsion (VSEPR) theory of chemical bonding, the ideal electron geometry of a molecule containing a total of 4 electron density regions around the central atom is tetrahedral.
There is no distinction between lone pairs and bond pairs when considering the total electron density regions and, thus, the electronic geometry of a molecule.
Therefore, the electron geometry of CH3CH2OH w.r.t both the C-atoms and the O-atom is tetrahedral as all these atoms are surrounded by 4 electron density regions at the sides.
An easy trick to finding a molecule’s electron and molecular geometry is using the AXN method.
AXN is a simple formula representing the number of bonded atoms and lone pairs present on the central atom.
It is used to predict the shape and geometry of a molecule using the VSEPR concept.
As we are considering the C-atoms as central atoms in the CH3CH2OH Lewis structure so we will study the AXN formula w.r.t the C-atoms only.
AXN notation for CH3CH2OH molecule
- A in the AXN formula represents the central atom. In the CH3CH2OH molecule, a carbon (C) atom is present at the center, so A = C.
- X denotes the atoms bonded to the central atom. In CH3CH2OH, 1 C-atom, 2 H-atoms, and 1 O-atom are directly bonded to the central C-atom. So X =4 for CH3CH2OH.
- N stands for the lone pairs present on the central atom. As per the Lewis structure of CH3CH2OH, the central C-atom has no lone pair of electrons. Thus, N = 0 for CH3CH2OH.
As a result, the AXN generic formula for CH3CH2OH is AX4N0 or simply AX4.
Now, you may have a look at the VSEPR chart below.
The VSEPR chart confirms that a molecule with AX4 generic formula possesses an identical electron and molecular geometry or shape, i.e., tetrahedral, as we already noted down for the Ethyl alcohol (CH3CH2OH) molecule.
Hybridization of CH3CH2OH
Both the C-atoms and the O-atom are sp3 hybridized in CH3CH2OH.
The electronic configuration of carbon is 1s2 2s2 2p2.
The electronic configuration of oxygen is 1s2 2s2 2p4.
During chemical bonding in CH3CH2OH, one of the two 2s electrons of carbon shifts to its empty 2p atomic orbital. As a result, the 2s and three 2p atomic orbitals of carbon hybridize to produce four sp3 hybrid orbitals.
Each sp3 hybrid orbital possesses a 25 % s-character and a 75 % p-character. All four sp3 hybrid orbitals are equivalent and contain a single unpaired electron only.
The C-atoms use these sp3 hybrid orbitals to form C-H, C-C, and C-O sigma bonds with the adjacent atoms.
In contrast, two of the four sp3 hybrid orbitals of oxygen contain paired electrons which are situated as 2 lone pairs on the O-atom in CH3CH2OH.
The other two sp3 hybrid orbitals of oxygen-containing a single electron only form C-O and O-H sigma bonds by sp3-sp3 and sp3-s orbital overlap, respectively.
Another shortcut to finding the hybridization present in a molecule is using its steric number against the table given below.
The steric number of both the C-atoms and the O-atom in CH3CH2OH is 4, so it has sp3 hybridization.
Steric number | Hybridization |
2 | sp |
3 | sp2 |
4 | sp3 |
5 | sp3d |
6 | sp3d2 |
The Bond angles of CH3CH2OH
It is due to the tetrahedral shape of CH3CH2OH that each H-C-H bond angle is equal to 109.28° in the Ethyl alcohol molecule.
However, the lone pairs present on the O-atom lead to electronic repulsion and thus molecular distortion which reduces the C-O-H bond angle from an ideal value of 109° to approx. 104.5°.
Also check:- How to find bond angle?
Is CH3CH2OH polar or nonpolar?
As per Pauling’s electronegativity scale, a polar covalent bond is formed between two dissimilar atoms having an electronegativity difference between 0.4 to 1.6 units.
The four types of covalent chemical bonds present in CH3CH2OH are C-C, C-H, C-O and O-H.
The C-C bond is purely non-polar as it is formed between two identical carbon atoms having zero or no electronegativity differences.
In contrast, a small electronegativity difference of 0.35 units exists between the carbon (E.N = 2.55) and hydrogen (E.N = 2.20) atoms in the C-H bond. Thus, the C-H bond is also only weakly polar, almost non-polar, as per Pauling’s electronegativity scale.
However, the C-O and O-H covalent bonds are strongly polar as per the electronegativity differences of 0.89 units and 1.24 units between the bonded carbon and oxygen (E.N = 3.44) and oxygen and hydrogen atoms, respectively.
The strongly electronegative O-atom thus gains a partial negative charge (δ–) while the adjacent C and H-atoms obtain partial positive charges (δ+) in CH3CH2OH.
This O-atom not only attracts the C-O and O-H bonded electrons but also attracts the shared electron cloud from each C-H bond. This leads to an overall non-uniformly distributed electron cloud over Ethyl alcohol.
It is due to the asymmetric bent shape of CH3CH2OH w.r.t the O-atom that the unequal C-O and O-H dipole moments stay uncancelled to yield an overall polar molecule (net µ > 0).
Ethyl alcohol (CH3CH2OH) is readily water-soluble as it can form strong hydrogen bonds, which is further evidence of its strongly polar nature.
Read in detail–
FAQ
What is the Lewis dot structure for CH3CH2OH? |
The Lewis dot structure of Ethyl alcohol (CH3CH2OH) displays a total of 20 valence electrons i.e., 20/2 = 10 electron pairs.
|
What is the molecular geometry of CH3CH2OH? |
The molecular geometry or shape of CH3CH2OH w.r.t each C-atom is tetrahedral while that w.r.t the O-atom is bent, angular or V-shaped. |
What is the electron geometry of CH3CH2OH? |
The ideal electronic geometry of CH3CH2OH is tetrahedral. |
Why is the shape of CH3CH2OH w.r.t O-atom different from that of its electronic geometry? |
The presence of 2 lone pairs of electrons on the oxygen atom leads to strong lone pair-lone pair and lone pair-bond pair electronic repulsions. To minimize this strong repulsive effect, the H-atom and the CH3CH2 group tilt away from the O-atom. This results in a bent, angular or V-shape of ethanol w.r.t the oxygen atom. |
How is the shape of CH3OH the same or different from that of CH3CH2OH? |
Both methanol (CH3OH) and Ethyl alcohol (CH3CH2OH) possess a similar shape, i.e., tetrahedral w.r.t the C-atoms and bent, angular or V-shape w.r.t the O-atoms. |
How is the shape of CH3OCH3 the same or different from that of CH3CH2OH? |
Dimethyl ether (CH3OCH3) and Ethyl alcohol (CH3CH2OH) are isomers possessing the same molecular formula but a different structural arrangement. In CH3OCH3, an O-atom at the center is surrounded by two CH3 groups at the sides. Contrarily, in CH3CH2OH, a C-atom at the center is bonded to 1 CH3 group, 2 H-atoms and an OH group at the sides. Therefore, the shape of CH3OCH3 w.r.t the central O-atom is bent, angular or V-shaped, while that of CH3CH2OH w.r.t the central C-atom is tetrahedral. However, both the molecules possess a similar shape w.r.t the oxygen atoms. |
Also Read:-
- CH3NH2 lewis structure and its molecular geometry
- IF5 lewis structure and its molecular geometry
- CH2Cl2 lewis structure and its molecular geometry
- CH3COOH lewis structure and its molecular geometry
- C2H2Cl2 lewis structure and its molecular geometry
- CHCl3 lewis structure and its molecular geometry
- CH3F lewis structure and its molecular geometry
- CF2Cl2 lewis structure and its molecular geometry
- CH3CN lewis structure and its molecular geometry
- CH2O lewis structure and its molecular geometry
Summary
- The total number of valence electrons available for drawing the CH3CH2OH Lewis structure is 20.
- The molecular geometry or shape of CH3CH2OH w.r.t each C-atom is tetrahedral while that w.r.t the O-atom is bent, angular or V-shaped.
- The ideal electronic geometry of Ethyl alcohol is tetrahedral.
- Both the C-atoms and the O-atom are sp3 hybridized in CH3CH2
- The H-C-H bond angle is equal to 109.28° while the C-O-H bond angle is equal to 104.5° in CH3CH2
- Ethyl alcohol (CH3CH2OH) is a polar molecule (net µ > 0) as the unequal dipole moments of C-O and O-H bonds do not get canceled equally in the asymmetric bent shape of CH3CH2OH w.r.t the O-atom.
- Zero or no formal charges on the covalently bonded atoms in CH3CH2OH ensure the extraordinary stability of the Lewis structure drawn in this article.
About the author
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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