The boiling point of bromine, a halogen, is $\pu{58.8^\circ C}$, while the boiling point of krypton, the noble gas in the same period as bromine, is $\pu{-153.4 ^\circ C}$.

I thought that the larger the atomic radius of an element, the more loosely the electrons would be held by the nucleus. So it would be easier for the atom to develop an instantaneous dipole, strengthening the London forces between the atoms of the element, and increasing the boiling point of the substance.

Krypton has a larger atomic radius than bromine. Using my reasoning from above, I thought that the boiling point of krypton would be higher than that of bromine. However, bromine actually has a higher boiling point than krypton.

Why is this? And where is my reasoning incorrect?

Complete data for reference: notice that the boiling point of a halogen is always higher than that of the corresponding noble gas, and the difference increases down the group.

$$ \begin{array}{|c|c|c|}\hline \text{Period}&\text{Halogen boiling point} (\pu{^\circ C})&\text{Noble gas boiling point} (\pu{^\circ C})\\\hline 2&−188.11&−246.046 \\\hline 3&−34.04&−185.848 \\\hline 4&58.8&−153.415 \\\hline 5&184.3&−108.099 \\\hline \end{array} $$

  • 1
    $\begingroup$ Bromine is found in nature as Br2 while krypton is just Kr. $\endgroup$ – Frank Apr 26 '18 at 22:24
  • 2
    $\begingroup$ FYI: Your listed value $(\pu{-7.2 ^\circ C})$ for the boiling point of bromine $(\ce{Br2})$ is incorrect. It is actually even higher value: $\pu{58.8 ^\circ C}$ look here. $\endgroup$ – Mathew Mahindaratne Apr 26 '18 at 22:58
  • 1
    $\begingroup$ The both values you listed are not relevant boiling points. They are melting points. $\endgroup$ – Mathew Mahindaratne Apr 26 '18 at 23:10
  • $\begingroup$ Just a thought: if the interactions in $Br_2$ molecules is governed by dip-ind interactions, bromine can form bigger induced dipole moments -cause the charge will get polarized in the molecule- which will be > than the one between krypton atoms. $\endgroup$ – santimirandarp Apr 27 '18 at 1:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.