JEE Advance - Chemistry (1991 - No. 10)

Arrange the following :

Increasing strength of hydrogen bonding (X-H-X):

O, S, F, Cl, N
Cl < S < N < O < F
S < Cl < N < O < F
Cl < S < O < N < F
F < O < N < S < Cl
N < O < F < S < Cl

Explanation

The strength of hydrogen bonding is directly proportional to the electronegativity and inversely proportional to the size of the atom bonded to the hydrogen atom.

  • Electronegativity: A more electronegative atom attracts the electron density in the covalent bond more strongly, leaving the hydrogen atom with a partial positive charge (δ+). This partial positive charge allows for a stronger electrostatic attraction to the lone pair of electrons on another electronegative atom, forming a stronger hydrogen bond.

  • Size: A smaller atom has a more concentrated electron density, leading to a stronger electrostatic attraction and a stronger hydrogen bond.

Therefore, among the given atoms, the order of increasing hydrogen bond strength is:

Cl < S < N < O < F

With fluorine (F) forming the strongest hydrogen bonds due to its high electronegativity and small size.

Additionally, chlorine (Cl) is relatively large in size and less electronegative compared to the other atoms listed. This makes it less likely to participate in significant hydrogen bonding.

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