KCET · Chemistry · p Block Elements (Group 15, 16, 17 & 18)
The correct order of boiling point in the following compounds is
- A \(\mathrm{HF}>\mathrm{H}_{2} \mathrm{O}>\mathrm{NH}_{3}\)
- B \(\mathrm{H}_{2} \mathrm{O}>\mathrm{HF}^{2}>\mathrm{NH}_{3}\)
- C \(\mathrm{NH}_{3}>\mathrm{H}_{2} \mathrm{O}>\mathrm{HF}\)
- D \(\mathrm{NH}_{3}>\mathrm{HF}>\mathrm{H}_{2} \mathrm{O}\)
Answer & Solution
Correct Answer
(B) \(\mathrm{H}_{2} \mathrm{O}>\mathrm{HF}^{2}>\mathrm{NH}_{3}\)
Step-by-step Solution
Detailed explanation
The correct order of boiling point of given hydrides is as follows \(\mathrm{H}_{2} \mathrm{O}>\mathrm{HF}>\mathrm{NH}_{3}\)
Strength of H-bonding depends upon size and electronegativity of the atom. Smaller the size of the atom, greater is the electronegativity and stronger is the \(\mathrm{H}\)-bonding.
But each \(\mathrm{H}_{2} \mathrm{O}\) molecule has two hydrogen atom, whereas each HF molecule has only one hydrogen atom.
\(\therefore\) Hydrogen bonding is more strong in \(\mathrm{H}_{2} \mathrm{O}\) molecules. While \(\mathrm{NH}_{3}\) molecules due to less electronegativity of ' \(N\) ' atom than ' \(F^{\prime}\) atom is less polar and hence, have weak hydrogen bonding than HF molecules.
Strength of H-bonding depends upon size and electronegativity of the atom. Smaller the size of the atom, greater is the electronegativity and stronger is the \(\mathrm{H}\)-bonding.
But each \(\mathrm{H}_{2} \mathrm{O}\) molecule has two hydrogen atom, whereas each HF molecule has only one hydrogen atom.
\(\therefore\) Hydrogen bonding is more strong in \(\mathrm{H}_{2} \mathrm{O}\) molecules. While \(\mathrm{NH}_{3}\) molecules due to less electronegativity of ' \(N\) ' atom than ' \(F^{\prime}\) atom is less polar and hence, have weak hydrogen bonding than HF molecules.
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