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Strength of a hydrogen bond depends mainly on two factors:
1. Electronegativity of the atoms directly attached to hydrogen (the donor X in X-H).
2. Electronegativity and small size of the lone-pair bearing atom that accepts the hydrogen (the acceptor Y in X-H···Y).
Greater electronegativity and smaller atomic radius strengthen the attractive interaction because the partial charges become larger and the dipole is shorter.
Among the common non-metals, the trend of electronegativity is
$$F \gt O \gt N \gt Cl$$
Apply this trend to the given pairs:
Case 1: $$O-H \cdots N$$ — donor O is very electronegative, but acceptor N is less electronegative than O and F, so the bond is moderate.
Case 2: $$F-H \cdots F$$ — both donor and acceptor are fluorine, the most electronegative and the smallest of all; hence the charge separation and electrostatic attraction are maximal.
Case 3: $$O-H \cdots O$$ — both atoms are oxygen. Although O is strongly electronegative, it is still weaker than F, so this bond is weaker than the F-H···F bond.
Case 4: $$O-H \cdots F$$ — donor O is less electronegative than F, but acceptor F is very electronegative. Even so, one side of the bond (the donor) involves oxygen, making it weaker than the fully fluorine-based $$F-H \cdots F$$ interaction.
Therefore, the hydrogen bond in which both the donor and the acceptor atoms are fluorine is the strongest.
Option B which is: $$F - H \cdots F$$
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