Animated Solution for Chemistry - s and p-Block Elements: The correct statements(s) for orthoboric acid is/are-

Imagine you are in a chemistry lab, staring at a white, crystalline powder. This is orthoboric acid, $H_{3}B{O}_3$. It looks harmless, but its chemical behavior is full of surprises. Unlike the classic acids you first learn about in chemistry, boric acid does not play by the standard Arrhenius rules. Let's embark on a journey to decode its true nature by analyzing each statement in this classic JEE Advanced problem.

Is it an Arrhenius Acid?

Let's break down the first claim: Does orthoboric acid self-ionize in water to release protons? It is incredibly tempting to look at the formula $H_{3}B{O}_3$ and assume it simply sheds its hydrogen atoms like phosphoric acid ($H_{3}P{O}_4$) does.

However, we must look at the central boron atom. Boron is in Group 13, meaning it only has three valence electrons. In $H_{3}B{O}_3$, it forms three single bonds with oxygen, giving it a total of six electrons in its valence shell. It has an incomplete octet. Because it is electron-deficient, its primary chemical drive is not to throw away a proton, but rather to accept a pair of electrons to complete its octet.

When placed in water, it acts as a Lewis acid. It accepts a lone pair of electrons from the oxygen atom of a water molecule. This forms the tetrahedral tetrahydroxyborate ion, $[B(OH{)}_4{]}^{-}$, and in the process, a proton ($H^{+}$) from the water molecule is released into the solution.

Because it does not self-ionize but rather extracts a proton from water, statement (A) is incorrect.

The Chelate Effect

Now, let's introduce a fascinating twist: what happens when we add ethylene glycol to the aqueous solution? Ethylene glycol ($HO-C{H}_2-C{H}_2-OH$) is a 1,2-diol. Because its two hydroxyl groups are adjacent, it can adopt a cis-like conformation.

When ethylene glycol encounters the $[B(OH{)}_4{]}^{-}$ ion, a remarkable reaction occurs. The diol molecules attack the borate ion, kicking out water molecules and forming a highly stable, cyclic chelate complex.

Why does this matter for acidity? Enter Le Chatelier's Principle. Because this chelate complex is incredibly stable, it effectively acts as a sink, continuously removing $[B(OH{)}_4{]}^{-}$ ions from the equilibrium mixture. To compensate for this loss, the initial equilibrium ($H_{3}B{O}_3+H_{2}O\rightleftharpoons [B(OH{)}_4{]}^{-}+H^{+}$) is forced to shift aggressively to the right.

As the reaction shifts forward, more and more $H^{+}$ ions are generated. The acidity of the solution dramatically increases. This is a favorite concept of JEE examiners! Therefore, statement (B) is absolutely correct.

What about its physical structure? If you were to zoom in to the molecular level, would you see a sprawling 3D network like diamond or ice?

The individual $H_{3}B{O}_3$ molecules are trigonal planar. In the solid state, these flat molecules link together through an extensive network of hydrogen bonds. However, they do so by forming flat, hexagonal rings. This creates a two-dimensional sheet-like structure, very similar to graphite.

These 2D sheets stack on top of one another, held together only by weak van der Waals forces. This is exactly why solid boric acid feels soft and soapy to the touch! Because it forms 2D sheets and not a 3D network, statement (C) is false.

The Verdict

Finally, let's evaluate its electrical conductivity. Is it a weak electrolyte?

An electrolyte's strength is determined by how completely it dissociates into ions in solution. As we established earlier, boric acid is a weak Lewis acid. The equilibrium $H_{3}B{O}_3+H_{2}O\rightleftharpoons [B(OH{)}_4{]}^{-}+H^{+}$ lies heavily to the left. Only a very small fraction of the molecules react to form ions.

Because the concentration of mobile ions in the solution is very low, it conducts electricity poorly. It is, without a doubt, a weak electrolyte. Thus, statement (D) is correct.

By carefully analyzing the electron deficiency of boron, the thermodynamics of chelation, and the geometry of hydrogen bonding, we can confidently conclude that the correct statements are (B) and (D).