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Example · Example 26

Q.The boiling points of the hydrogen halides are: HF=19.5∘C\text{HF} = 19.5^{\circ}\text{C}, HCl=−85∘C\text{HCl} = -85^{\circ}\text{C}, HBr=−66∘C\text{HBr} = -66^{\circ}\text{C}, HI=−35∘C\text{HI} = -35^{\circ}\text{C}. Explain both the anomalously high boiling point of HF\text{HF} and the rising trend from HCl\text{HCl} to HI\text{HI}.

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Among the hydrogen halides, only fluorine is both small enough and electronegative enough to support genuine hydrogen bonding with hydrogen; chlorine, bromine and iodine are all too large and insufficiently electronegative for the H–X bond to qualify. This is why HF\text{HF}'s boiling point (19.5∘C19.5^{\circ}\text{C}) is dramatically higher than a simple molar-mass trend would predict — its molecules are held together not just by dispersion and dipole-dipole forces but by strong, additional hydrogen bonds. For HCl\text{HCl}, HBr\text{HBr} and HI\text{HI}, hydrogen bonding is not possible, so their intermolecular attraction comes from dipole-dipole forces (all three are polar) plus, dominantly, London dispersion forces. Dispersion forces grow steadily stronger as the halogen atom gets larger and more electron-rich — chlorine to bromine to iodine — because a larger, more diffuse electron cloud is more easily polarised (more polarisable). This is why boiling point rises steadily from HCl\text{HCl} (−85∘C-85^{\circ}\text{C}) to HBr\text{HBr} (−66∘C-66^{\circ}\text{C}) to HI\text{HI} (−35∘C-35^{\circ}\text{C}), even thou …

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