The states of hybridisation of boron and oxygen atoms in boric acid (H3BO3) are respectively
(a) sp2

Q: Why doesn't the Boron atom have any lone pairs?

Boron is in Group 13 of the periodic table, meaning it only has 3 valence electrons. In boric acid, it uses all 3 of these electrons to form single bonds with the three oxygen atoms. Since all its valence electrons are involved in bonding, there are no electrons left to form a lone pair.

Q: I only see Oxygen making two bonds. Why is it sp3 hybridized?

Hybridization depends on the total number of electron domains (steric number), not just the number of bonds. While Oxygen does form 2 bonds, it also has 2 lone pairs of electrons. The sum of bonds (2) and lone pairs (2) is 4, which requires four hybrid orbitals, hence $\text{sp}^3$ hybridization.

Q: Does the hydrogen atom have a hybridization state here?

No, terminal hydrogen atoms do not undergo hybridization. Hydrogen only has a single 1s orbital, which it uses directly to overlap with the $\text{sp}^3$ hybrid orbital of oxygen to form the $\sigma$-bond.

Q: Because Boron is sp2 hybridized, is the entire boric acid molecule planar?

The core $\text{BO}_3$ unit is strictly trigonal planar due to the $\text{sp}^2$ hybridization of Boron. However, because the Oxygen atoms are $\text{sp}^3$ hybridized (bent geometry), the hydrogen atoms can rotate out of the plane. In its solid crystalline form, hydrogen bonding locks the molecules into a nearly planar sheet-like structure.

Q: Is boric acid a strong acid because it has three -OH groups?

Actually, no! Boric acid is a weak acid, and it doesn't even act as a typical Arrhenius acid (it doesn't release its own $H^+$ easily). Because the Boron atom is $\text{sp}^2$ hybridized, it has an empty p-orbital. It acts as a Lewis acid by accepting a lone pair from a water molecule, forming $\text{B(OH)}_4^-$ and releasing an $H^+$ from the water.

Question

The states of hybridisation of boron and oxygen atoms in boric acid ($	ext{H}_3	ext{BO}_3$) are respectively

Options:

(A) $	ext{sp}^2$ and $	ext{sp}^2$

(B) $	ext{sp}^2$ and $	ext{sp}^3$

(C) $	ext{sp}^3$ and $	ext{sp}^2$

(D) $	ext{sp}^3$ and $	ext{sp}^3$

Sigma Percentile · Accepted Answer

### Unveiling the Structure of Boric Acid

When we encounter a molecule like boric acid, $	ext{H}_3	ext{BO}_3$, the first step to understanding its chemical behavior is to visualize its molecular architecture. Imagine a central Boron ($	ext{B}$) atom acting as the core, symmetrically surrounded by three hydroxyl ($-	ext{OH}$) groups.

This structural foundation is absolutely critical because the geometry and reactivity of the molecule are dictated by the hybridization of its constituent atoms. To find the hybridization, we rely on a very simple yet powerful tool: the **Steric Number Rule**.

### The Steric Number Rule

The steric number ($Z$) of an atom in a molecule tells us how many electron domains (regions of electron density) are arranged around it. It is calculated as:

$$Z = 	ext{Number of } \sigma	ext{-bonds} + 	ext{Number of lone pairs}$$

Once we have the steric number, determining the hybridization is straightforward. A steric number of $2$ corresponds to $	ext{sp}$, $3$ corresponds to $	ext{sp}^2$, and $4$ corresponds to $	ext{sp}^3$.

### Analyzing the Boron Atom

Let's focus our attention on the central Boron atom. By looking at the Lewis structure, we can see that Boron forms exactly three single bonds (which are $\sigma$-bonds) with the three surrounding Oxygen atoms.

Does Boron have any lone pairs? Boron is in Group 13 and has exactly three valence electrons. Since it uses all three of these electrons to form the three $\sigma$-bonds, it has zero non-bonding electrons left.

Therefore, for Boron:
$$Z = 3 	ext{ (}\sigma	ext{-bonds)} + 0 	ext{ (lone pairs)} = 3$$

A steric number of $3$ means that the Boron atom is **$	ext{sp}^2$ hybridized**. This hybridization gives the $	ext{BO}_3$ core a flat, trigonal planar geometry.

### Analyzing the Oxygen Atom

Now, let's shift our focus to any one of the Oxygen atoms. This is where many students make a silly mistake by rushing.

Each Oxygen atom forms two $\sigma$-bonds: one with the central Boron atom and one with a Hydrogen atom. But we cannot stop there! Oxygen is in Group 16 and has six valence electrons. It uses two of these electrons for bonding, which leaves four electrons. These four electrons form two lone pairs.

Therefore, for Oxygen:
$$Z = 2 	ext{ (}\sigma	ext{-bonds)} + 2 	ext{ (lone pairs)} = 4$$

A steric number of $4$ means that the Oxygen atom is **$	ext{sp}^3$ hybridized**.

### The Final Verdict

By carefully applying the steric number rule and remembering to account for hidden lone pairs, we have successfully decoded the molecule. The Boron atom is $	ext{sp}^2$ hybridized, and the Oxygen atoms are $	ext{sp}^3$ hybridized. This beautiful combination of geometries is what gives boric acid its unique layered crystalline structure in the solid state!

Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: The states of hybridisation of boron and oxygen atoms in boric acid ( $H_{3} BO_{3}$ ) are respectively

Select Answer:

Visualized Solution

\text{Structure of Boric Acid}

\text{Steric Number Rule}

\text{Hybridisation of Boron}

\text{Hybridisation of Oxygen}

\text{Conclusion}

The Sigma Insight: Hybridisation and VSEPR Theory

Unveiling the Structure of Boric Acid

The Steric Number Rule

Analyzing the Boron Atom

Analyzing the Oxygen Atom

The Final Verdict

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Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The states of hybridisation of boron and oxygen atoms in boric acid (H3​BO3​) are respectively