Hydroxide (OH-) ion Lewis dot structure, molecular geometry
or shape, electron geometry, bond angles, hybridization,
formal charges, polar vs nonpolar
OH– represents the hydroxide ion, a versatile chemical superhero that neutralizes acids and regulates pH. Alkaline solutions (such as NaOH, KOH, Mg(OH)2) release OH– ions in water. The greater the concentration of OH– ions in an aqueous solution, the higher its basicity.
In this article, you will learn to draw the Lewis structure of the hydroxide (OH–) ion, its molecular geometry or shape, electron geometry, bond angles, formal charges, hybridization, polarity and much more!
The hydroxide (OH–) ion consists of only an oxygen (O) atom and a hydrogen (H) atom single-covalently bonded to each other. There are 3 lone pairs of electrons on the central O-atom in the OH– Lewis structure.
You can draw the Lewis dot structure of OH– by following the simple steps given below:
Steps for drawing the Lewis dot structure of OH–
1. Count the total valence electrons present in OH–
OH– consists of two distinct elements, i.e., oxygen and hydrogen.
Oxygen (O) is located in Group VI A (or 16), containing 6 valence electrons in each atom.
Contrarily, hydrogen (H) lies at the top of the Periodic Table, having a single valence electron only.
Total number of valence electrons in hydrogen = 1
Total number of valence electrons in oxygen = 6
The OH– ion comprises 1 O-atom and 1 H-atom.
An important point to remember is that the OH– ion carries a negative (-1) charge, which means 1 extra valence electron is added in this Lewis structure.
∴ Therefore, the total valence electrons available for drawing the Lewis dot structure of OH– = 1(6) + 1(1) = 7 + 1 =8 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 or molecular ion.
Among the two types of atoms present in OH–, hydrogen (E.N = 2.20) is less electronegative than oxygen (E.N = 3.44).
However, the H-atom is an exception. It always occupies an outer position in any Lewis structure, as it can only form a single bond accommodating a total of 2 valence electrons max.
So, among the two atoms of OH–, the O-atom is considered a central atom while the H-atom is placed next to it, as shown below.
3. Connect the outer atom with the central atom
In this step, the terminal H-atom is joined to the central O-atom using a single straight line.
A straight line represents a single covalent bond, i.e., a bond pair containing 2 electrons.
2 valence electrons used out of the 8 initially available means we are now left with 6 valence electrons.
So let’s see in the next steps where we can place these remaining valence electrons.
4. Complete the duplet of the outer atom
An O-H bond represents 2 valence electrons surrounding the outer H-atom, which means it already has a complete duplet electronic configuration in the OH– Lewis structure.
So we do not need to make any changes w.r.t this O-atom in the above structure.
5. Complete the octet of the central atom
Total valence electrons used till step 4 = 1 single bond = 2 valence electrons.
Total valence electrons – electrons used till step 4 = 8 – 2 = 6 valence electrons.
Thus these 6 valence electrons are placed as 3 lone pairs on the central O-atom in the OH– Lewis structure, as shown below.
This automatically completes the octet of the central oxygen atom.
So, let’s move ahead and check the stability of the OH– 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 or molecular ion, 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 OH– bonded atoms.
Non-bonding electrons = no lone pair = 0 electrons
Formal charge = 1-0-2/2 = 1-0-1= 1-1 = 0
As per the above calculation, zero or no formal charge is present on the H-atom, while the O-atom carries a -1 formal charge in the OH– Lewis structure which is also the charge present on the hydroxide ion overall.
This ensures we have obtained the correct Lewis representation for the hydroxide ion.
The hydroxide ion Lewis structure is enclosed in square brackets, and a -1 charge is placed at the top right corner, as shown below.
Now, in the next section, we will discuss the electron and molecular geometry of OH–.
What are the electron and molecular geometry of OH-?
The molecular geometry or shape of the hydroxide (OH–) ion is linear. However, its electronic geometry w.r.t the O-atom considered the central atom is tetrahedral.
The presence of 3 lone pairs of electrons on oxygen leads to molecular distortion. The OH– ion thus occupies a molecular shape different from the ideal electron pair geometry.
Molecular geometry of OH–
The molecular geometry or shape of the hydroxide (OH–) ion is linear.
There are 3 lone pairs of electrons on the central O-atom that leads to strong lone pair-lone pair and lone pair-bond pair electronic repulsions.
To minimize the strong repulsive effect, the OH– ion occupies a linear shape, keeping the lone pairs as far apart from one another as possible, as shown below.
Electron geometry of OH–
According to the valence shell electron pair repulsion (VSEPR) theory of chemical bonding, the ideal electron geometry of a molecule or molecular ion containing a total of 4 electron density regions around the central atom is tetrahedral.
In OH–, the central O-atom is directly bonded to an H-atom only. There are 3 lone pairs of electrons on the central O-atom.
1 O-H bond pair + 3 lone pairs makes a total of 4 electron density regions surrounding the central oxygen atom in OH–. Hence its electron geometry is tetrahedral.
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 on the central atom.
It is used to predict the shape and geometry of a molecule or molecular ion using the VSEPR concept.
AXN notation for the hydroxide (OH–) ion
A in the AXN formula represents the central atom. In OH–, an oxygen (O) atom is present at the center, so A = O.
X denotes the atoms bonded to the central atom. In OH–, an H-atom is directly bonded to the central O-atom, so X= 1.
N stands for the lone pairs present on the central atom. As per the Lewis structure of OH–, the central O-atom has 3 lone pairs of electrons. Thus, N= 3 for OH–.
As a result, the AXN generic formula for OH– is AX1N3 or simply AXN3.
Now, you may have a look at the VSEPR chart below.
The VSEPR chart confirms that the molecular geometry or shape of a molecule or molecular ion with an AXN3 generic formula is linear, while its electron geometry is tetrahedral, as we already noted down for the hydroxide (OH–) ion.
Hybridization of OH–
The central O-atom is sp3 hybridized in OH–.
The ground state electronic configuration of an oxygen atom is 1s2 2s2 2p4.
Upon excitation, the O-atom gains an electron, and its electronic configuration becomes 1s2 2s2 2p5.
As a result, during chemical bonding in OH–, the 2s atomic orbital of oxygen hybridizes with its three 2p orbitals to produce four sp3 hybrid orbitals.
Each sp3 hybrid orbital possesses a 33.3 % s-character and a 66.7 % p-character.
However, these sp3 hybrid orbitals of oxygen are not all equivalent, as one of these contains a single unpaired electron.
In contrast, paired electrons are present in the remaining three sp3 hybrid orbitals of oxygen, which are situated as 3 lone pairs on the central O-atom in OH–.
On the other hand, the sp3 hybrid orbital of oxygen-containing 1 electron forms the O-H sigma bond by overlapping with the half-filled s orbital of the adjacent H-atom, as shown below.
Another shortcut to finding the hybridization present in a molecule or molecular ion is using its steric number against the table below.
The steric number of the O-atom in OH– is 4, so it has sp3 hybridization.
Steric number
Hybridization
2
sp
3
sp2
4
sp3
5
sp3d
6
sp3d2
The bond angle of OH–
The concept of bond angle is irrelevant in OH– as it comprises only two atoms; however, a bond angle is formed by the intersection of two lines, i.e., three atoms A-B-C involved.
However, the O-H bond length is equal to 97 pm in the hydroxide ion.
Is OH- polar or nonpolar?
As per Pauling’s electronegativity scale, a polar covalent bond is formed between two dissimilar atoms with an electronegativity difference between 0.4 and 1.6 units.
In OH–, a high electronegativity difference of 1.24 units is present between the oxygen (E.N = 3.44) and hydrogen (E.N = 2.20) atoms in the O-H bond.
The more electronegative oxygen atom thus gains a partial negative (δ–) charge by strongly attracting the bonded electrons towards itself, while the adjacent hydrogen atom obtains a partial positive (δ+) charge.
The strong O-H dipole moment stays intact in the linear shape of the hydroxide ion as there are no other dipole moments to cancel it out.
The charged electron cloud stays non-uniformly distributed overall. Consequently, OH– is strongly polar in nature (net µ > 0).
The Lewis dot structure of the hydroxide (OH–) ion displays a total of 8 valence electrons, i.e., 8/2 = 4 electron pairs.
Out of the 4 electron pairs, there are 1 bond pair and 3 lone pairs.
The O-atom and the H-atom are directly bonded to each other.
All 3 lone pairs are present on the oxygen atom in the OH– Lewis structure.
Why is the shape of OH– different from its electron geometry?
The molecular shape of OH– is linear. However, its electronic geometry w.r.t the O-atom is tetrahedral.
The presence of 3 lone pairs of electrons on the O-atom leads to strong lone pair-lone pair and lone pair-bond pair electronic repulsions. OH– thus occupies a different shape as opposed to its electron geometry.
How is the shape of a hydronium ion different from that of the hydroxide (OH–) ion?
In the hydronium (H3O+) ion, three H-atoms are covalently bonded to the central O-atom. There is a lone pair of electrons on the central O-atom, which leads to strong lone pair-bond pair electronic repulsions. The molecular ion thus adopts a trigonal pyramidal shape to overcome this strong repulsive effect.
In contrast, the shape of the hydroxide (OH–) ion is linear. Only 1 H-atom is directly attached to the O-atom, and it has 3 lone pairs of electrons.
Which of the following molecule/molecular ion is non-polar?
A) H3O+
B) H2O
C) OH–
D) None of the above
Option D is the correct answer. All the above-mentioned molecule/molecular ions are polar.
In H3O+, the central O-atom is covalently bonded to three H-atoms, and there is 1 lone pair of electrons on the central O-atom. It is due to the asymmetric trigonal pyramidal shape of H3O+ that the O-H dipole moments stay uncancelled to yield an overall polar molecular ion (net µ > 0).
The shape of H2O is bent, angular or V-shaped. The central O-atom is directly bonded to two H-atoms. There are 2 lone pairs of electrons on the central oxygen atom, which leads to strong lone pair-lone pair and lone pair-bond pair electronic repulsions.
The O-H dipole moments do not get canceled. So H2O is a polar molecule possessing a resultant dipole moment value.
The shape of OH– is linear. There is no competition for the only dipole moment present, i.e., O-H; thus, OH– ion is also overall polar.
If the pH of a solution is given, how can we find its hydroxide (OH–) ion concentration?
The pOH value of a solution is related to its pH by the formula given below:
pH + pOH = 14
So if the pH of a solution is given, we can calculate its pOH by rearranging the above equation:
pOH = 14- pH
Once the pOH of a solution is determined, we can find the hydroxide ion concentration [OH–] as follows:
The total number of valence electrons available for drawing the hydroxide (OH–) ion Lewis structure is 8.
The molecular geometry or shape of OH– is linear.
The electron geometry of OH– w.r.t the central O-atom is tetrahedral.
The oxygen atom is sp3 hybridized in OH–.
The O-H bond length equals 97 pm in the hydroxide ion.
OH– is a polar cation as the O-H dipole moment stays uncancelled.
Zero or no formal charge is present on the H-atom, while the O-atom carries a -1 formal charge which is also the charge present on the OH– ion overall.
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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