The Lewis structure of HOF represents the molecular arrangement of hypofluorous acid, a molecule containing one hydrogen, one oxygen, and one fluorine atom. The structure depicts two single bonds with oxygen at the center and hydrogen and fluorine atoms positioned on the outside in a straight line. There are no lone pairs on the hydrogen atom, two lone pairs on the oxygen atom, and three lone pairs on the fluorine atom.
To accurately draw the Lewis structure of HOF, begin by sketching a rough outline of the molecule. Then, illustrate the chemical bonds between each atom and indicate lone pairs on the atoms. Check whether the octet rule is satisfied for the central atom as well as all other atoms. Finally, calculate formal charges for each atom to assess the stability of the structure and aim to achieve a stable Lewis structure with the least possible formal charges.
Steps
Draw skeleton
In this step, first calculate the total number of valence electrons. And then, decide the central atom.
- Let’s calculate the total number of valence electrons
We know that… hydrogen is a group 1 element, oxygen is a group 16 element, and fluorine is a group 17 element. Hence, hydrogen has one valence electron, oxygen has six valence electrons, and fluorine has seven valence electrons.
Now HOF has one hydrogen atom, one oxygen atom, and one fluorine atom.
So the total number of valence electrons = valence electrons of hydrogen atom + valence electrons of oxygen atom + valence electrons of fluorine atom
Therefore, the total number of valence electrons = 1 + 6 + 7 = 14
- Now decide the central atom
We can not assume hydrogen as the central atom, because the central atom is bonded with at least two other atoms. And hydrogen has only one electron in its last shell, so it can not make more than one bond.
Therefore, choose the central atom from oxygen and fluorine.
The atom with the least electronegative value is placed at the center. By looking at the periodic table, we get the electronegativity values for oxygen and fluorine as follows:
Electronegativity value of oxygen = 3.44
Electronegativity value of fluorine = 3.98
Obviously, oxygen is less electronegative than fluorine. Hence, assume that oxygen is the central atom.
So now, put oxygen in the center and hydrogen and fluorine on either side. And draw the rough skeleton structure for the Lewis structure of HOF something like this:
Also read: How to draw Lewis structure of XeCl4 (4 steps)
Show chemical bond
Place two electrons between the atoms to show a chemical bond. Since oxygen is surrounded by hydrogen and fluorine, use four electrons to show two chemical bonds as follows:
Also read: How to draw CH3+ Lewis structure (3 steps)
Mark lone pairs
As calculated earlier, we have a total of 14 valence electrons. And in the above structure, we have already used four valence electrons. Hence, ten valence electrons are remaining.
Two valence electrons represent one lone pair. So ten valence electrons = five lone pairs.
Note that hydrogen is a period 1 element, so it can not keep more than 2 electrons in its last shell. And both (oxygen and fluorine) are the period 2 elements, so they can not keep more than 8 electrons in their last shell.
Also, make sure that you start marking these lone pairs on outside atoms first. And then, on the central atom.
The outside atoms are hydrogen and fluorine. But hydrogen can not keep more than 2 electrons in its last shell. Hence, don’t mark the lone pairs on hydrogen.
So fluorine will get three lone pairs. And the central atom (oxygen) will get two lone pairs.
Now draw the Lewis structure of HOF something like this:
In the above structure, you can see that the octet is completed on the central atom (oxygen), and also on the outside atoms. Therefore, the octet rule is satisfied.
After completing the octet, one last thing we need to do is, calculate the formal charge and check the stability of the above structure.
Also read: How to draw Lewis structure of CH3NCO (5 steps)
Calculate formal charge and check stability
The following formula is used to calculate the formal charges on atoms:
Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons
Collect the data from the above structure and then, write it down below as follows:
- For hydrogen atom
Valence electrons = 1
Nonbonding electrons = 0
Bonding electrons = 2
Formal charge = 1 – 0 – ½ (2) = 0
- For oxygen atom
Valence electrons = 6
Nonbonding electrons = 4
Bonding electrons = 4
Formal charge = 6 – 4 – ½ (4) = 0
- For fluorine atom
Valence electrons = 7
Nonbonding electrons = 6
Bonding electrons = 2
Formal charge = 7 – 6 – ½ (2) = 0
Mention the formal charges of atoms on the structure. So the Lewis structure of HOF looks something like this:
In the above structure, you can see that the formal charges of all atoms are zero. Therefore, this is the stable Lewis structure of HOF.
And each horizontal line drawn in the above structure represents a pair of bonding valence electrons.
Related
- Lewis structure of XeCl4
- CH3+ Lewis structure
- Lewis structure of CH3NCO
- Lewis structure of CO2
- Lewis structure of H2O
External video
- How to Draw the Lewis Dot Structure for HOF (and Molecular Geometry) – YouTube • Wayne Breslyn
External links
- HOF Lewis Structure in 6 Steps (With Images) – Pediabay
- Write the correct Lewis structure for the compound. HOF – Homework.Study.com
- Lecture 12: Lewis Structures – Hypofluorous Acid, Hof – iLectureOnline
- Draw the Lewis Dot Structure for HOF – Chegg
- What is the Lewis Structure of HOF? – Answers
- What is the Lewis structure of HOF? – OneClass
- how many lone pair of electrons are assigned to the fluorine atom in the lewis dot structure of hypofluorous acid, hof? – Brainly
- Draw a Lewis dot structure for HOF and indicate the shape of the molecule – Quizlet
- Draw the Lewis Dot Structure for HOF – Numerade
- HOF (Hypofluorous Acid) Oxidation Number – ChemicalAid
- Hypofluorous acid – Wikipedia
- Oxidation state of fluorine in HOF – Chemistry Stack Exchange
Deep
Rootmemory.com was founded by Deep Rana, who is a mechanical engineer by profession and a blogger by passion. He has a good conceptual knowledge on different educational topics and he provides the same on this website. He loves to learn something new everyday and believes that the best utilization of free time is developing a new skill.