-5
$\begingroup$

Suggest one reason why HCl has only one peak?

Does this have something to do with fact that peaks (in IR spectroscopy) represent areas of the spectrum where specific bond vibrations occur and therefore since HCl has a single bond it only has one peak to show only one bond vibration?

$\endgroup$
5
  • 2
    $\begingroup$ Sorry had the wrong spectra - Gas phase HCl $\endgroup$ – MaxW Jan 12 at 0:58
  • 2
    $\begingroup$ Please, don't use all caps in your title. It comes off as very rude, even though that may not have been your intention. Please also try to choose appropriate tags for your question. This is not about organic chemistry or reaction mechanisms. $\endgroup$ – orthocresol Jan 12 at 2:14
  • $\begingroup$ Do you recall that a linear diatomic molecule has only 3(2)-5=1, i.e., one degree of freedom. $\endgroup$ – M. Farooq Jan 12 at 3:22
  • $\begingroup$ @orthocresol it only comes off as rude to you please don't comment things that don't make sense and that don't help me answer my question. Also thank you to everyone who left a comment sorry I didn't reply sooner but I did end up figuring out the answer $\endgroup$ – VENDINGMACHINE Jan 15 at 1:45
  • 2
    $\begingroup$ @VENDINGMACHINE I'm afraid there is only your complaint that doesn't make sense. Before flagging a comment faithfully suggesting a change for better readability as rude, you could at least do a little research. Feel free to read about all caps usage on Wikipedia or check community guidelines on SO.Meta or CSE.Meta. $\endgroup$ – andselisk Jan 15 at 5:03
1
$\begingroup$

A molecule's vibrations are quantised making a ladder of vibrational energy levels. At very, very low resolution a single wide line (or peak) is seen from the $n=0 \to 1 $ transition. At high resolution rotational levels are also resolved as shown in the link in the comment by MaxW above. Additionally, v. weak overtones can be seen from $n\to 2$ and $n\to 3$ transitions (and so at higher energy) due to what is called anharmonicity in the bond's vibration. See a text such as 'Modern Spectroscopy' by Hollas (publ Wiley) for details but many Phys. Chem. textbooks will also explain this.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.