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INTERMOLECULAR FORCES WORKSHEET, Study notes of Chemistry

Chemistry

London (dispersion) forces. 2) Predict which compound in each pair will have the higher melting point/boiling point. Explain your answer. (a) CS2 or CCl4.

Typology: Study notes

2022/2023

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CHEMISTRY 11 AP – INTERMOLECULAR FORCES WORKSHEET
1) Identify the main intermolecular force acting between the following molecules:
(a)
CH4
London (dispersion) forces
(g)
Dipole-Induced Dipole
(b)
Dipole-Dipole
(h)
Dipole-Dipole
(c)
AlCl3
Ionic Bonding
(i)
Dipole-Dipole
(d)
London (dispersion) forces
(j)
N2
London (dispersion) forces
(e)
Hydrogen bonding
(k)
Nitrogen tribromide
Dipole-Dipole
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polar
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Partial preview of the text

Download INTERMOLECULAR FORCES WORKSHEET and more Study notes Chemistry in PDF only on Docsity!

CHEMISTRY 11 AP – INTERMOLECULAR FORCES WORKSHEET

  1. Identify the main intermolecular force acting between the following molecules: (a) (^) CH 4 London (dispersion) forces (g) Dipole-Induced Dipole (b) Dipole-Dipole (h) Dipole-Dipole (c) (^) AlCl 3 Ionic Bonding (i) Dipole-Dipole (d) London (dispersion) forces (j) (^) N 2 London (dispersion) forces (e) Hydrogen bonding (k) (^) Nitrogen tribromide Dipole-Dipole B ) ) )
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(f) Ion-Dipole force (l) (^) SiF 4 London (dispersion) forces

  1. Predict which compound in each pair will have the higher melting point/boiling point. Explain your answer. (a) CS 2 or CCl 4 CCl 4 – Since both are non-polar, the compound that has the stronger London (dispersion) forces will have the higher MP/BP. London (dispersion) forces get stronger as the molecular mass increases. (b) HI or KI KI – KI has ionic bonds while HI possesses weaker dipole-dipole forces. (c) Cl 2 or F 2 Cl 2 – Since both are non-polar, the compound that has the stronger London (dispersion) forces will have the higher MP/BP. London (dispersion) forces get stronger as the molecular mass increases. valence E- 4+28=32-8--

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  • Cl : :^ F- F : , •^6 °^ a^ -^ °^0 non. polar non^ . polar

(h) BF 3 or P 4 P 4 – Since both are non-polar, the compound that has the stronger London (dispersion) forces will have the higher MP/BP. London (dispersion) forces get stronger as the molecular mass increases. (i) CaF 2 or HF CaF 2 – CaF 2 has ionic bonds while HF possesses weaker hydrogen bonding

  1. The industrial production of ammonia, NH 3 , from H 2 and N 2 is called the Haber process, named for Fritz Haber, the German chemist who developed it just before World War I. During the process, in a gaseous mixture of all three substances, NH 3 must be separated from H 2 and N 2. This is done by cooling the gaseous mixture so as to condense only the NH 3. This leaves the elemental nitrogen and hydrogen as gases to be recycled and produce more ammonia. Why does only the ammonia liquefy upon cooling, but not the H 2 or N 2? Since hydrogen bonds exist between ammonia molecules but only London (dispersion) forces exist between diatomic molecules of nitrogen and hydrogen, the ammonia has a higher boiling point and therefore liquefies (condenses) at a higher temperature than nitrogen and hydrogen that remain in the gaseous phase upon cooling.
  2. Rank the following compounds from weakest intermolecular forces to strongest. Justify your answers. H 2 S, I 2 , N 2 , H 2 O N 2 and I 2 are non-polar, so they only have London (dispersion) forces – I 2 has stronger forces since it is larger. H 2 S has dipole-dipole forces therefore it is stronger than I 2. H 2 O has hydrogen bonding therefore it is stronger than H 2 S. N 2 < I 2 < H 2 S < H 2 O Btzvalencee ' = (^) 3+21=24 -6=^

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  1. Rank the following from weakest intermolecular forces to strongest. Justify your answers. H 2 Se, H 2 S, H 2 Po, H 2 Te These compounds are all the same bent shape. Although H 2 S is slightly more polar than the others, it is not very polar so it has very weak dipole-dipole forces. Therefore, the differences in London (dispersion) forces are more important for these compounds. H 2 Po is the largest and therefore has the strongest London (dispersion) forces.
  2. Iodine is a non-polar diatomic molecule, yet its molecules have enough attraction for each other that the element exists as a solid at room temperature. Identify the attractive force and explain why it is strong enough to keep the molecules of I 2 attached to each other even at room temperature. I 2 is a relatively large and massive diatomic molecule and the strength of London dispersion forces increases as the size of the molecules involved increases. This is because large electron clouds are more loosely held than smaller clouds and therefore more easily deformed or polarized by a nearby dipole than compact tightly held clouds. In addition, large molecules with more surface area have electron clouds that are spread out and so are more easily distorted by neighbouring dipoles. As a result, the dispersion forces are strong enough to keep the molecules of I 2 attached to each other even at room temperature.
  3. Surface tension is the ability of a fluid to act as a thin elastic membrane at its surface. Explain why non-polar molecules usually have much lower surface tension than polar ones. Surface tension results when the molecules at a surface attract each other to be able to form a thin, elastic membrane. For example, H 2 O (due to their hydrogen bonding) has surface tension since the positive hydrogen pole of one water molecule is attracted to the negative oxygen pole of a neighbouring water molecule. Since non-polar molecules are not attracted to each other as much as in polar molecules, these molecules are much less likely to have high surface tension. all valence^

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