Use the information and data given below to answer the questions (a) to (c):
• Stronger intermolecular forces result in higher boiling point.
• Strength of London forces increases with the number of electrons in the molecule.
• Boiling point of HF, HCl, HBr and HI are 293 K, 189 K, 206 K and 238 K respectively.
(a) Which type of intermolecular forces are present in the molecules HF, HCl, HBr and HI?
(b) Looking at the trend of boiling points of HCl, HBr and HI, explain out of dipole-dipole interaction and London interaction, which one is predominant here
(c) Why is boiling point of hydrogen fluoride highest while that of hydrogen chloride lowest?
Use the information and data given below to answer the questions (a) to (c):
• Stronger intermolecular forces result in higher boiling point.
• Strength of London forces increases with the number of electrons in the molecule.
• Boiling point of HF, HCl, HBr and HI are 293 K, 189 K, 206 K and 238 K respectively.
(a) Which type of intermolecular forces are present in the molecules HF, HCl, HBr and HI?
(b) Looking at the trend of boiling points of HCl, HBr and HI, explain out of dipole-dipole interaction and London interaction, which one is predominant here
(c) Why is boiling point of hydrogen fluoride highest while that of hydrogen chloride lowest?
This is a short answer type question as classified in NCERT Exemplar
(a) HCl, HBr and HI have dipole-dipole and london dispersion interaction whereas HF has hydrogen bonding in addition (due to the high electronegativity of the F atom).
(b) The electronegativity decreases from Cl, Br and I so the
Similar Questions for you
CH3COOH + NaOH → CH3COONa + H2O
ΔH = –50.6 kJ/mol
NaOH + SA [HCl] → NaCl + H2O
ΔH = –55.9 kJ/mol
the value of ΔH for ionisation of CH3COOH
⇒ ΔH = +55.9 – 50.6
5.3 kJ/mol
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Chemistry NCERT Exemplar Solutions Class 11th Chapter Five 2025
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