Dimethyl ether (CH3OCH3) Lewis structure, molecular geometry or shape, electron geometry, bond angle, hybridization, polar or nonpolar, formal charges,
CH3OCH3 is the chemical formula for dimethyl ether. Less commonly used names for this chemical compound are methoxymethane and dimethyl oxide.
If the above properties have piqued your curiosity about the chemistry behind dimethyl ether (CH3OCH3), then you are at the right place.
In this article, we will discuss how to draw the Lewis dot structure of CH3OCH3, what is its molecular shape or geometry, electron geometry, bond angle, hybridization, formal charges, polarity, etc. So, continue reading!
|Name of Molecule||Dimethyl ether|
|Molecular geometry of CH3OCH3||Bent, angular, or V-shaped|
|Electron geometry of CH3OCH3||Tetrahedral|
∠(O-C-H) = ∠(H-C-H)= 109.5°, ∠(C-O-C)= 111°
|Total Valence electron in CH3OCH3||28|
|Overall Formal charge in CH3OCH3||Zero|
How to draw lewis structure of CH3OCH3 (Dimethyl ether)?
The Lewis structure of dimethyl ether (CH3OCH3) consists of an oxygen (O) atom at the center. It is bonded to two CH3 groups, one on each side. Two lone pairs of electrons are also present on the central O-atom. This makes a total of 4 electron density regions or electron domains around the central O-atom in the CH3OCH3 lewis structure.
There are a few unique aspects that we need to consider while drawing the Lewis dot structure of CH3OCH3.
But don’t worry because we will teach you how to draw the Lewis dot structure of CH3OCH3 step-by-step using the guidelines given below.
Steps for drawing the Lewis dot structure of CH3OCH3
1. Count the total valence electrons in CH3OCH3
The very first step while drawing the Lewis structure of CH3OCH3 is to find the total valence electrons present in its concerned elemental atoms.
As three different elemental atoms are present in CH3OCH3, so you first need to look for the position of these elements in the Periodic Table.
Carbon (C) belongs to Group IV A (or 14), so it has a total of 4 valence electrons. Oxygen (O) is present in Group VI A (or 16), so it has 6 valence electrons, while hydrogen (H) lies at the top of the Periodic Table containing a single 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 CH3OCH3 molecule consists of 2 C-atoms, 1 O-atom, and 6 H-atoms. Therefore, the total valence electrons available for drawing the Lewis dot structure of CH3OCH3 = 2(4) + 1(6) + 6(1) = 20 valence electrons.
2. Choose the central atom
In the second step of drawing the Lewis structure of a molecule, usually, the least electronegativity atom out of all the atoms available is chosen as the central atom.
This is because the least electronegative atom is most likely to share its electrons with the atoms spread around it.
But the CH3OCH3 Lewis structure is an exception.
Although oxygen (O) atom is more electronegative than both C and H-atoms. But the need for an ether (-O-) linkage in CH3OCH3 demands an O-atom at the central position in the CH3OCH3 Lewis structure.
Another interesting fact is that a CH3 group acts as an electron donor. 2 CH3 groups push electrons toward the central highly electronegative O-atom in CH3OCH3, stabilizing the molecule overall.
In short, an O-atom is placed at the center while the two C-atoms and six H-atoms occupy terminal positions, as shown below.
3. Connect outer atoms with the central atom
In this step, the outer C-atoms are joined to the central O-atom using single straight lines.
But you must remember that an H-atom can only form a single covalent bond with an atom present next to it. This is because the H-atom can only accommodate a total of 2 valence electrons. So each H-atom is joined to its adjacent C-atom only and not to the central O-atom, as shown below.
In this way, the above structure has a total of 8 single bonds, which denotes a total of 8(2) = 16 valence electrons consumed out of the 20 initially available.
4. Complete the duplet and/or octet of the outer atoms
As we already identified, the hydrogen and carbon atoms are the outer atoms in the Lewis dot structure of CH3OCH3.
Each hydrogen (H) atom requires a total of 2 valence electrons in order to achieve a stable duplet electronic configuration.
A C-H single bond already represents 2 valence electrons around each H-atom. This means all the H-atoms already have a complete duplet in the Lewis structure drawn till yet. Thus, we do not need to make any changes with regard to the hydrogen atoms in this structure.
In contrast to that, a C-atom needs a total of 8 valence electrons to achieve a stable octet electronic configuration.
3 C-H and 1 C-O single bond around each C-atom in the above Lewis structure ensures that each C-atom has a total of 4 single bonds, i.e., 4(2) = 8 valence electrons.
Each C-atom thus already has a complete octet. In short, we do not need to make any changes with regard to the C-atoms as well in this structure.
5. Complete the octet of the central atom
- Total valence electrons used till step 4 = 8 single bonds = 8(2) = 16 valence electrons.
- Total valence electrons – electrons used till step 4 = 20 – 16 = 4 valence electrons.
These 4 valence electrons are placed as 2 lone pairs on the central O-atom in the CH3OCH3 Lewis structure, as shown below.
In this way, the central O-atom now has a complete octet with 2 single bonds + 2 lone pairs in the above structure.
As a final step, we need to check the stability of the CH3OCH3 Lewis structure by 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.
- 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 CH3OCH3 atoms.
For hydrogen atoms
- 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
For carbon atoms
- Valence electrons of carbon = 4
- Bonding electrons = 4 single bonds = 4(2) = 8 electrons
- Non-bonding electrons = no lone pairs = 0 electrons
- Formal charge = 4-0-8/2 = 4-0-4 = 4-4 = 0
For 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
The above calculation shows that zero formal charges are present on all the bonded atoms in the CH3OCH3 Lewis structure. Thus it is a stable structure.
Now that we have successfully drawn the Lewis structure of dimethyl ether let’s move ahead and discuss its electron and molecular geometry.
Also check –
What are the electron and molecular geometry of CH3OCH3?
The molecular geometry or shape of a dimethyl ether (CH3OCH3) molecule is bent, angular or V-shaped. Contrarily, its ideal electron geometry is tetrahedral.
The presence of two lone pairs of electrons on the central O-atom in CH3OCH3 leads to strong lone pair-lone pair and lone pair-bond pair repulsions. This makes the molecule adopt a different shape from its ideal electron geometry.
Molecular geometry of CH3OCH3
The molecular geometry or shape of dimethyl ether (CH3OCH3) is bent, angular or V-shaped.
The presence of two lone pairs of electrons on the central O-atom in CH3OCH3 leads to strong lone pair-lone pair and lone pair-bond pair electronic repulsions in addition to a bond pair-bond pair repulsive effect.
This repulsive effect pushes the CH3 groups away from the central O-atom, and the molecule adopts a bent shape, as shown below. The methyl groups lie at the terminals of the inverted V-shape.
An important point to remember is that the molecular geometry or shape of a molecule depends on the different number of lone pairs and bond pairs around the central atom.
Note: When we look at the actual molecular geometry or shape of molecule, the lone pair will be invisible but the influence of lone pair(repelling effect) will be counted on overall shape of molecule.
Contrarily, the ideal electronic geometry just depends upon the total number of electron density regions or electron domains (lone pairs and bond pairs inclusive) around the central atom.
Now let us see how this concept applies to the CH3OCH3 molecule.
Electron geometry of CH3OCH3
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.
While determining the electron geometry of CH3OCH3, each methyl (CH3) group around the central O-atom is considered one region of electron density. 2 CH3 groups and 2 lone pairs of electrons make a total of 4 electron density regions around the central O-atom in CH3OCH3. Hence its ideal electron pair geometry is tetrahedral.
A shortcut to finding the electron and the molecular geometry of a molecule is by using 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 CH3OCH3 molecule
- A in the AXN formula represents the central atom. In CH3OCH3, the oxygen (O) atom act as the central atom, so A= O.
- X denotes the atoms directly bonded to the central atom. 2 C-atoms or 2 CH3 groups are directly bonded to the central O-atom in the CH3OCH3 molecule; thus, X=2.
- N stands for the lone pairs present on the central atom. As 2 lone pairs are present on the central oxygen atom in CH3OCH3, thus N=2.
So, the AXN generic formula for the CH3OCH3 molecule is AX2N2.
Now have a quick look at the VSEPR chart given below to identify where you find AX2N2.
The VSEPR chart confirms that molecules with an AX2N2 generic formula have a bent or V-shape while their ideal electron geometry is tetrahedral, as we already noted for the CH3OCH3 molecule.
Hybridization of CH3OCH3
The central O-atom is sp3 hybridized in CH3OCH3.
The electronic configuration of oxygen is 1s2 2s2 2p4.
During chemical bonding, the 2s atomic orbital of oxygen hybridizes with its three 2p atomic orbitals to produce four sp3 hybrid orbitals.
Each sp3 hybrid orbital possesses a 25% s-character and a 75% p-character. Out of the four sp3 hybrid orbitals, two sp3 hybrid orbitals of oxygen contain paired electrons.
These electrons are situated as two lone pairs on the central O-atom in the CH3OCH3 molecule.
The other two sp3 hybrid orbitals contain a single electron. These overlap with the sp3 hybrid orbitals of adjacent C-atoms to form the C-O sigma (σ) bonds by sp3-sp3 overlap.
Refer to the figure drawn below.
A short trick for finding the hybridization present in a molecule is to memorize the table given below. You can find the steric number of a molecule and use that against this table to find its hybridization.
The steric number of central oxygen (O) in CH3OCH3 is 4, so it has sp3 hybridization.
The CH3OCH3 bond angle
The ideal bond angle in a symmetrical tetrahedral shape is 109.5°. Therefore, each H-C-H and O-C-H bond angle is 109.5° in CH3OCH3.
However, it is due to the distortion present in the shape and geometry of CH3OCH3 that the C-O-C bond angle changes. Thus the C-O-C bond angle is approx. 111° in CH3OCH3.
Each C-O bond length is 141.4 pm in CH3OCH3, while each C-H bond length is close to 110 pm.
Also check:- How to find bond angle?
Is CH3OCH3 polar or nonpolar?
Pauling’s electronegativity scale states that a covalent chemical bond is polar if the bonded atoms have an electronegativity difference between 0.5 to 1.6 units.
In CH3OCH3, two different types of bonds are present, i.e., a C-O bond and a C-H bond.
A small electronegativity difference of 0.35 units exists between the carbon (E.N = 2.55) and hydrogen (E.N = 2.20) atoms. So each C-H bond is only weakly polar.
Contrarily, a higher electronegativity difference of 0.89 units exists between a carbon and an oxygen (E.N = 3.44) atom. So, both C-O bonds are strongly polar in CH3OCH3.
It is due to the asymmetric bent shape of CH3OCH3 that the dipole moments of individually polar C-H and C-O bonds do not get canceled in the molecule overall.
A partial negative (δ–) charge develops on the central O-atom while both the C –atoms and all six H-atoms acquire partial positive (δ+) charges.
The electron cloud stays non-uniformly distributed over the molecule; thus, CH3OCH3 is polar (net µ = 1.30 D).
Read in detail–
What is the Lewis structure for CH3OCH3?
Both lone pairs are present on the central O-atom in the CH3OCH3 Lewis structure. There is no lone pair on any of the C or H-atoms.
What is the molecular geometry or shape of CH3OCH3?
The CH3OCH3 molecule has a bent shape. It is also known as angular or V-shape.
Why is the shape of CH3OCH3 different from its ideal electronic geometry?
The ideal electronic geometry of CH3OCH3 is tetrahedral. It is due to two lone pairs of electrons present on the central O-atom that lone pair-lone pair and lone pair-bond pair repulsions distort the shape and geometry of the molecule. It thus adopts a bent shape.
How is the shape of CH3OCH3 different from that of CH3CH2OH?
Dimethyl ether (CH3OCH3) and ethanol (CH3CH2OH) are functional group isomers that have the same molecular formula (C2H6O). However, each possesses a different structural arrangement.
CH3OCH3 has a bent shape, while CH3CH2OH has a tetrahedral shape.
There are two lone pairs on the central O-atom in CH3OCH3, so it has a different shape from its ideal electron geometry.
Contrarily, there are no lone pairs on the central C-atom in CH3CH2OH; hence its shape is identical to its ideal electron geometry.
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- The total number of valence electrons available for drawing dimethyl ether (CH3OCH3) Lewis structure is 20.
- The ideal electron pair geometry of CH3OCH3 is tetrahedral.
- The molecular geometry or shape of CH3OCH3 is bent or V-shaped.
- The CH3OCH3 molecule has sp3 hybridization.
- Each H-C-O and H-C-H bond angle is 109.5° while the C-O-C bond angle is 111°.
- The C-O bond length is 141.4 pm. The C-H bond length is 110 pm in CH3OCH3.
- CH3OCH3 is a polar molecule (net µ = 1.30 D).
- Zero formal charges present on all the bonded atoms account for the extraordinary stability of the CH3OCH3 Lewis structure drawn in this article.
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