Question

Among the following, the compound that forms the strongest hydrogen bond is

• A. HF
• B. HCI
• C. HBr
• D. HI

Answer 1

The Correct Option is A

$\text{Explanation}$

Hydrogen bonding (H-bonding) is a special type of dipole-dipole attraction that occurs when a hydrogen atom is covalently bonded to a small, highly electronegative atom, typically Fluorine ($\text{F}$), Oxygen ($\text{O}$), or Nitrogen ($\text{N}$).

The strength of the hydrogen bond is primarily determined by two factors:

$\text{1. High Electronegativity of the Donor Atom}$

The H-bond forms due to the strong polarity of the $\text{X}-\text{H}$ bond. A greater difference in electronegativity ($\Delta\text{EN}$) between Hydrogen ($\text{H}$) and the electronegative atom ($\text{X}$) leads to a greater partial positive charge ($\delta^{+}$) on the hydrogen atom and a greater partial negative charge ($\delta^{-}$) on the $\text{X}$ atom.

This increased charge separation results in a stronger electrostatic attraction (the hydrogen bond) between the $\delta^{+}\text{H}$ and the lone pair of the neighboring molecule's $\delta^{-}\text{X}$ atom.

Comparing the halogens:

$$\text{Electronegativity Order: } \text{F} > \text{Cl} > \text{Br} > \text{I}$$

Since Fluorine ($\text{F}$) is the most electronegative element among the halogens (and the entire periodic table), the $\text{H}-\text{F}$ bond is the most polar ($\Delta\text{EN}$ is highest).

Consequently, the partial charges ($\delta^{+}\text{H}$ and $\delta^{-}\text{F}$) are maximized, leading to the strongest intermolecular hydrogen bond in liquid $\text{HF}$.

$\text{2. Small Size of the Donor Atom}$

Fluorine is the smallest atom among the halogens ($\text{F}, \text{Cl}, \text{Br}, \text{I}$).

A smaller size concentrates the negative charge ($\delta^{-}$) into a smaller volume, increasing the charge density and allowing the approaching $\delta^{+}\text{H}$ atom to get closer. This leads to a shorter and stronger hydrogen bond.

Due to the combined effect of the highest electronegativity and smallest atomic size, $\text{HF}$ forms the strongest hydrogen bonds among the hydrogen halides ($\text{HF}, \text{HCl}, \text{HBr}, \text{HI}$).

$$\text{Strength Order: } \mathbf{\text{H}-\text{F} \cdots \text{H}-\text{F}} > \text{H}-\text{O} \cdots \text{H}-\text{O} > \text{H}-\text{N} \cdots \text{H}-\text{N} > \text{H}-\text{Cl} \cdots \text{H}-\text{Cl } (\text{negligible})$$

Hydrogen bonding is not typically considered significant in $\text{HCl}$, $\text{HBr}$, and $\text{HI}$ because $\text{Cl}$, $\text{Br}$, and $\text{I}$ are not sufficiently electronegative, and their larger size reduces the charge density compared to $\text{F}$.