Explain why hydrogen fluoride (HF) has a higher boiling temperature than hydrogen chloride (HCl) (19.4 vs. \(-85^{\circ} \mathrm{C}\) ), even though HF has a lower molecular weight.

Short Answer

Expert verified
Short Answer: Hydrogen fluoride (HF) has a higher boiling temperature than hydrogen chloride (HCl) because HF molecules experience stronger intermolecular forces due to hydrogen bonding. This stronger force requires more energy to overcome, leading to a higher boiling temperature for HF, despite its lower molecular weight compared to HCl.

Step by step solution

01

Identify the type of intermolecular forces in the compounds

HF and HCl are both polar covalent compounds, which means that they have a difference in electronegativity between the hydrogen and halogen atoms (fluorine in HF, chlorine in HCl). This electronegativity difference gives rise to a dipole-dipole interaction, which is an attractive force between the positive end of one polar molecule and the negative end of another polar molecule. Additionally, HF molecules exhibit hydrogen bonding, which is a special type of dipole-dipole interaction that occurs between hydrogen atoms covalently bonded to highly electronegative elements (such as fluorine, oxygen, or nitrogen) and the lone pairs of electrons on another highly electronegative atom. HCl, however, does not form hydrogen bonds because chlorine's electronegativity is not as high as fluorine's.
02

Compare the strength of the intermolecular forces in HF and HCl

Hydrogen bonding in HF is a significantly stronger intermolecular force compared to the dipole-dipole interaction present in HCl. This is because fluorine is the most electronegative element, and its large electronegativity difference with hydrogen results in a highly polarized bond and consequently a strong hydrogen bond. On the other hand, while HCl exhibits dipole-dipole interactions due to its polar nature, the electronegativity difference between hydrogen and chlorine is not large enough to result in hydrogen bonding. As a result, the intermolecular forces in HCl are weaker than those in HF.
03

Relate the strength of intermolecular forces to boiling temperature

The boiling temperature of a substance is related to the strength of its intermolecular forces. Greater intermolecular forces require more energy to break the interactions between molecules, which in turn leads to a higher boiling temperature. Since HF experiences stronger intermolecular forces due to hydrogen bonding, it requires more energy to break these forces than the weaker dipole-dipole interactions in HCl, which explains the higher boiling temperature of HF (19.4°C) compared to HCl (-85°C).
04

Conclusion

Hydrogen fluoride (HF) has a higher boiling temperature than hydrogen chloride (HCl) due to the presence of stronger intermolecular forces, specifically hydrogen bonding, in HF. This stronger intermolecular force requires more energy to overcome, resulting in a higher boiling temperature for HF, even though it has a lower molecular weight compared to HCl.

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Most popular questions from this chapter

(a) Briefly cite the main differences between ionic, covalent, and metallic bonding. (b) State the Pauli exclusion principle.

The net potential energy \(E_{N}\) between two adjacent ions is sometimes represented by the expression $$E_{N}=-\frac{C}{r}+D \exp \left(-\frac{r}{\rho}\right)$$ in which \(r\) is the interionic separation and \(C\) \(D,\) and \(\rho\) are constants whose values depend on the specific material. (a) Derive an expression for the bonding energy \(E_{0}\) in terms of the equilibrium interionic separation \(r_{0}\) and the constants \(D\) and \(\rho\) using the following procedure: 1\. Differentiate \(E_{N}\) with respect to \(r\) and set the resulting expression equal to zero 2\. Solve for \(C\) in terms of \(D, \rho,\) and \(r_{0}\) 3\. Determine the expression for \(E_{0}\) by substitution for \(C\) in Equation 2.12 (b) Derive another expression for \(E_{0}\) in terms of \(r_{0}, C,\) and \(\rho\) using a procedure analogous to the one outlined in part (a).

Allowed values for the quantum numbers of electrons are as follows: \\[ \begin{aligned} n &=1,2,3, \ldots \\ l &=0,1,2,3, \ldots, n-1 \\ m_{l} &=0,\pm 1,\pm 2,\pm 3, \ldots, \pm l \\ m_{s} &=\pm \frac{1}{2} \end{aligned} \\] The relationships between \(n\) and the shell designations are noted in Table \(2.1 .\) Relative to the subshells, \(l=0\) corresponds to an \(s\) subshell \(l=1\) corresponds to a \(p\) subshell \(l=2\) corresponds to a \(d\) subshell \(l=3\) corresponds to an \(f\) subshell For the \(K\) shell, the four quantum numbers for each of the two electrons in the 1 s state in the order of \(n l m_{l} m_{s},\) are \(100\left(\frac{1}{2}\right)\) and \(100\left(-\frac{1}{2}\right) .\) Write the four quantum numbers for all of the electrons in the \(L\) and \(M\) shells, and note which correspond to the \(s, p,\) and \(d\) subshells.

With regard to electron configuration, what do all the elements in Group IIA of the periodic table have in common?

The net potential energy between two adjacent ions, \(E_{N},\) may be represented by the sum of Equations 2.8 and \(2.9 ;\) that is, $$E_{N}=-\frac{A}{r}+\frac{B}{r^{n}}$$ Calculate the bonding energy \(E_{0}\) in terms of the parameters \(A, B,\) and \(n\) using the following procedure: 1\. Differentiate \(E_{N}\) with respect to \(r,\) and then set the resulting expression equal to zero, since the curve of \(E_{N}\) versus \(r\) is a minimum at \(E_{0}\) 2\. Solve for \(r\) in terms of \(A, B,\) and \(n,\) which yields \(r_{0},\) the equilibrium interionic spacing. 3\. Determine the expression for \(E_{0}\) by substitution of \(r_{0}\) into Equation 2.11

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