Comparing Boiling Pts of Bromine & Iodine Monochloride - Intermolecular Forces

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SUMMARY

The boiling points of bromine (Br2) and iodine monochloride (ICl) differ significantly due to their intermolecular forces. Bromine has a boiling point of 59 degrees Celsius, while iodine monochloride has a boiling point of 97 degrees Celsius. The analysis reveals that bromine is non-polar due to its identical electronegativities (Br-Br), whereas iodine monochloride is polar (I-Cl), resulting in stronger dipole-dipole interactions. Consequently, the higher boiling point of ICl is justified by its stronger intermolecular forces compared to bromine.

PREREQUISITES
  • Understanding of intermolecular forces, specifically London dispersion forces and dipole-dipole interactions.
  • Familiarity with VSEPR Theory and molecular shapes.
  • Knowledge of electronegativity and its role in determining molecular polarity.
  • Basic principles of quantum mechanics related to electron shielding and effective nuclear charge.
NEXT STEPS
  • Research the impact of molecular polarity on boiling points in various substances.
  • Study London dispersion forces in detail, including their dependence on molecular size and shape.
  • Explore VSEPR Theory applications in predicting molecular geometries and polarities.
  • Investigate the relationship between quantum numbers and electron behavior in atoms.
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Chemistry students, educators, and professionals interested in molecular interactions, boiling point analysis, and the effects of polarity on physical properties.

mujahid613
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I have been trying to get my head around this question, but do not understand it.

Consider the two isoelectronic substances, bromine (Br2) and Iodine monochloride (ICl). Based upon your knowledge of intermolecular forces, explain the difference in their boilings pts. (Bromine: 59 degrees C; iodine monochloride: 97 degrees C)

Well, I know that both bromine and iodine monochloride have 70 electrons per molecule, therefore, the strength of the London dispersion forces between the molecules of each should be the same. Now, what I'm not sure of is how to assess their polarities.

VSPER Theory doesn't really help me out because they are both of the same shape, so one would assume that they would both be either polar or non-polar.

However, if I assess their respective electronegativities...

BR-BR (must be the same; therefore has to be non-polar?)
I-CL (is different; therefore polar?)

Therefore, ICL should have stronger dipole-diple-force, and a higher boiling point.

Are these conclusions justified?

Thanks for your prompt response.
 
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Looks good to me! You may want to wait for a more qualified person to confirm, but that's probably how I would've gone about doing it.
 
Yeah, I think that's a good assessment. Although the london dispersion interactions relate to the polarity of the molecule, in that you'll need to consider the individual atoms and their quantum numbers. The farther away an electron is from the nucleus, the effective nuclear charge is relatively less, the electron is not being held as strongly. The orbit is farther away from the nucleus (quantum number) and there's relatively more electron shielding.
 

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