H3BO3 Lewis structure

Q: How many valence electrons are in the H3BO3 molecule?

We can determine the number of valence electrons in H3BO3 by adding up the valence electrons of each atom in the molecule. Each hydrogen atom has one valence electron. Boron, which is in group 3 of the periodic table, has three valence electrons. And each oxygen atom has six valence electrons. Therefore, H3BO3 has a total of 1(3) + 3 + 6(3) = 24 valence electrons .

Q: What is the shape of H3BO3 according to its Lewis structure?

The H3BO3 molecule has a trigonal planar shape according to its Lewis structure. The boron atom in the center is surrounded by three oxygen atoms, each of which is bonded to the boron atom through a single covalent bond. The three hydrogen atoms are also bonded to the oxygen atoms through single covalent bonds. The trigonal planar shape is the result of the boron atom forming only three covalent bonds, rather than the four bonds required for a tetrahedral shape.

H3BO3 Lewis Structure — H₃BO₃ Lewis structure | Image: Learnool

H₃BO₃ (boric acid) has three hydrogen atoms, one boron atom, and three oxygen atoms.

In H₃BO₃ Lewis structure, there are three single bonds around the boron atom, with three oxygen atoms attached to it, and each oxygen atom is attached with one hydrogen atom. And on each oxygen atom, there are two lone pairs.

Alternative method: Lewis structure of H₃BO₃

Contents

Rough sketch

First, determine the total number of valence electrons

In the periodic table, hydrogen lies in group 1, boron lies in group 13, and oxygen lies in group 16.

Hence, hydrogen has one valence electron, boron has three valence electrons, and oxygen has six valence electrons.

Since H₃BO₃ has three hydrogen atoms, one boron atom, and three oxygen atoms, so…

Valence electrons of three hydrogen atoms = 1 × 3 = 3
Valence electrons of one boron atom = 3 × 1 = 3
Valence electrons of three oxygen atoms = 6 × 3 = 18

And the total valence electrons = 3 + 3 + 18 = 24

Learn how to find: Hydrogen valence electrons, Boron valence electrons, and Oxygen valence electrons

Second, find the total electron pairs

We have a total of 24 valence electrons. And when we divide this value by two, we get the value of total electron pairs.

Total electron pairs = total valence electrons ÷ 2

So the total electron pairs = 24 ÷ 2 = 12

Third, determine the central atom

Here hydrogen can not be 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.

Now we have to choose the central atom from boron and oxygen. Place the least electronegative atom at the center.

Since boron is less electronegative than oxygen, assume that the central atom is boron.

Therefore, place boron in the center and hydrogen and oxygen on either side.

And finally, draw the rough sketch

H3BO3 Lewis Structure (Step 1) — Rough sketch of H₃BO₃ Lewis structure | Image: Learnool

Lone pair

Here, we have a total of 12 electron pairs. And six bonds are already marked. So we have to only mark the remaining six electron pairs as lone pairs on the sketch.

Also remember that hydrogen is a period 1 element, so it can not keep more than 2 electrons in its last shell. And both (boron and oxygen) are the period 2 elements, so they can not keep more than 8 electrons in their last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are hydrogens and oxygens. But no need to mark on hydrogen, because each hydrogen has already two electrons.

So for each oxygen, there are two lone pairs, and for boron, there is zero lone pair because all six electron pairs are over.

Mark the lone pairs on the sketch as follows:

H3BO3 Lewis Structure (Step 2) — Lone pairs marked, and got the stable Lewis structure of H₃BO₃ | Image: Learnool

Formal charge

Use the following formula to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

For each hydrogen atom, formal charge = 1 – 0 – ½ (2) = 0

For boron atom, formal charge = 3 – 0 – ½ (6) = 0

For each oxygen atom, formal charge = 6 – 4 – ½ (4) = 0

Here, the atoms do not have charges, so no need to mark the charges.

Final structure

The final structure of H₃BO₃ comprises a central boron atom linked to three oxygen atoms through single covalent bonds, with each oxygen atom further bonded to a single hydrogen atom. In this arrangement, the boron atom maintains an incomplete octet with only six valence electrons, while each oxygen atom satisfies the octet rule by retaining two lone pairs alongside its two bonding pairs. Each hydrogen atom achieves stability with a duet of electrons. This specific configuration is the most stable because it results in formal charges of zero for all participating atoms, representing the most energetically favorable state for the molecule. Consequently, this electronic distribution serves as the definitive and most accurate Lewis representation of boric acid.

FAQs

How many valence electrons are in the H₃BO₃ molecule?

We can determine the number of valence electrons in H₃BO₃ by adding up the valence electrons of each atom in the molecule.

Each hydrogen atom has one valence electron. Boron, which is in group 3 of the periodic table, has three valence electrons. And each oxygen atom has six valence electrons.

Therefore, H₃BO₃ has a total of 1(3) + 3 + 6(3) = 24 valence electrons.

What is the shape of H₃BO₃ according to its Lewis structure?

The H₃BO₃ molecule has a trigonal planar shape according to its Lewis structure.

The boron atom in the center is surrounded by three oxygen atoms, each of which is bonded to the boron atom through a single covalent bond. The three hydrogen atoms are also bonded to the oxygen atoms through single covalent bonds.

The trigonal planar shape is the result of the boron atom forming only three covalent bonds, rather than the four bonds required for a tetrahedral shape.

Does H₃BO₃ Lewis structure follow the octet rule?

Boron is an exception to the octet rule and can have fewer than eight valence electrons in its outer shell.

In H₃BO₃ Lewis structure, the boron atom has only six valence electrons. This is because it forms only three bonds with the surrounding atoms, rather than the four bonds required for a full octet. Therefore, H₃BO₃ does not follow octet rule for all atoms.

However, despite the lack of a full octet, the boron atom can still be considered to have a complete outer shell if its formal charge is taken into account.

Next: CH₃I Lewis structure

External video

H3BO3 Lewis Structure: How to Draw the Lewis Structure for B(OH)3 – YouTube • Wayne Breslyn

External links

Deep

Learnool.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.