List of H20 breakdown substances

[jake jot]

When salt is added to water, the water H20 molecule breaks apart. If you add apple juice to water, does it also happen? I'd like to see list of all substances that when added to water can break apart the H20 molecules and manner of breakup. Which of them is like salt added to water? Thank you.

 

[Borek]

When salt is added to water, the water H20 molecule breaks apart.

It doesn't. It can undergo some reactions in some cases for some salts, but in most cases these are the same reactions that occur in water even without any salt added, just the equilibrium is shifted.

I'd like to see list of all substances that when added to water can break apart the H20 molecules and manner of breakup.

It doesn't work this way. At best we can list some groups/classes of substances that can react with each other. Making a full list is impossible, as there is no such thing as a finite list of compounds.

 

[jake jot]

It doesn't. It can undergo some reactions in some cases for some salts, but in most cases these are the same reactions that occur in water even without any salt added, just the equilibrium is shifted.
It doesn't work this way. At best we can list some groups/classes of substances that can react with each other. Making a full list is impossible, as there is no such thing as a finite list of compounds.

Ok, let's not list billions or infinite number of compounds but just a dozen or so. What other substances are polar and can react with water molecules like the following descriptions?

"When table salt is placed in water, the slightly electropositive sodium portion is attracted to the slightly electronegative oxygen portion of water molecules. At the same time, the slightly electronegative chlorine portion of NaCl is attracted to the slightly electronegative oxygen portion of water.
In neither case is a true bond created, but the attractions set up a "tug-of-war" in which the ionic bonds of NaCl and the covalent bonds of H2O are both strained."

What other substances are polar like salt?

 

[HAYAO]

Ok, let's not list billions or infinite number of compounds but just a dozen or so. What other substances are polar and can react with water molecules like the following descriptions?

"When table salt is placed in water, the slightly electropositive sodium portion is attracted to the slightly electronegative oxygen portion of water molecules. At the same time, the slightly electronegative chlorine portion of NaCl is attracted to the slightly electronegative oxygen portion of water.
In neither case is a true bond created, but the attractions set up a "tug-of-war" in which the ionic bonds of NaCl and the covalent bonds of H2O are both strained."

What other substances are polar like salt?

Any polar compounds.

https://en.wikipedia.org/wiki/Chemical_polarity

 

[russ_watters]

When salt is added to water, the water H20 molecule breaks apart.

No, it's the salt molecules that break apart, not the water molecules.

 

[jake jot]

Any polar compounds.

https://en.wikipedia.org/wiki/Chemical_polarity

Ok. read them.

In the house, what are the common compounds that are polar besides water/salt and alcohol? Something safe to touch, as I want to put polar molecules in my fingers and explore them.

I also read all polar compounds are IR active. And IR spectrometer are cheap. What is the cheapest one out there that you can buy and experiment?

 

[sysprog]

Ok, let's not list billions or infinite number of compounds but just a dozen or so. What other substances are polar and can react with water molecules like the following descriptions?

"When table salt is placed in water, the slightly electropositive sodium portion is attracted to the slightly electronegative oxygen portion of water molecules. At the same time, the slightly electronegative chlorine portion of NaCl is attracted to the slightly electronegative oxygen portion of water.
In neither case is a true bond created, but the attractions set up a "tug-of-war" in which the ionic bonds of NaCl and the covalent bonds of H2O are both strained."

What other substances are polar like salt?

That description appears to be quoted from https://sciencing.com/happens-salt-added-water-5208174.html,
which says what you quoted:

When table salt is placed in water, the slightly electropositive sodium portion is attracted to the slightly electronegative oxygen portion of water molecules. At the same time, the slightly electronegative chlorine portion of NaCl is attracted to the slightly electronegative oxygen portion of water.

In neither case is a true bond created, but the attractions set up a "tug-of-war" in which the ionic bonds of NaCl and the covalent bonds of H2O are both strained.

but also immediately following that says:

The stronger covalent bonds of water (which is also held together generally by hydrogen bonds between water molecules) win out, and NaCl is pulled apart, with the Na+ and Cl_ ions setting loosely in place between the intact H2O molecules. NaCl is then dissolved.

That source cites this one as a reference $-$ https://www.usgs.gov/media/images/water-molecules-and-their-interaction-salt-molecules:

​

Water molecules and their interaction with salt molecules​

​

​

This diagram shows the positive and negative parts of a water molecule. It also depicts how a charge, such as on an ion (Na or Cl, for example) can interact with a water molecule.​

​

At the molecular level, salt dissolves in water due to electrical charges and due to the fact that both water and salt compounds are polar, with positive and negative charges on opposite sides in the molecule. The bonds in salt compounds are called ionic because they both have an electrical charge—the chloride ion is negatively charged and the sodium ion is positively charged. Likewise, a water molecule is ionic in nature, but the bond is called covalent, with two hydrogen atoms both situating themselves with their positive charge on one side of the oxygen atom, which has a negative charge. When salt is mixed with water, the salt dissolves because the covalent bonds of water are stronger than the ionic bonds in the salt molecules.​

​

The positively-charged side of the water molecules are attracted to the negatively-charged chloride ions and the negatively-charged side of the water molecules are attracted to the positively-charged sodium ions. Essentially, a tug-of-war ensues with the water molecules winning the match. Water molecules pull the sodium and chloride ions apart, breaking the ionic bond that held them together. After the salt compounds are pulled apart, the sodium and chloride atoms are surrounded by water molecules, as this diagram shows. Once this happens, the salt is dissolved, resulting in a homogeneous solution.​

​

Although both sources say that it is the sodium and chloride ions, and not the components of the water molecule, that are pulled apart (which is something that @russ_watters succinctly pointed out), there is an important difference in the description of the ionic interactions $-$ repeating from the sciencing.com source:

When table salt is placed in water, the slightly electropositive sodium portion is attracted to the slightly electronegative oxygen portion of water molecules. At the same time, the slightly electronegative chlorine portion of NaCl is attracted to the slightly electronegative oxygen portion of water.

The second sentence is incorrect (due to a probably-inadvertent repetition of "electronegative oxygen portion" from the preceding sentence) $-$ it would not have been incorrect to say that the negatively charged chlorine portion of NaCl is attracted to the positively charged hydrogen portions of the water molecule, which is probably what the writer intended.

 

[jake jot]

That description appears to be quoted from https://sciencing.com/happens-salt-added-water-5208174.html,
which says what you quoted:

but also immediately following that says:That source cites this one as a reference $-$ https://www.usgs.gov/media/images/water-molecules-and-their-interaction-salt-molecules:

​

Water molecules and their interaction with salt molecules​

​

View attachment 272738​

​

This diagram shows the positive and negative parts of a water molecule. It also depicts how a charge, such as on an ion (Na or Cl, for example) can interact with a water molecule.​

​

At the molecular level, salt dissolves in water due to electrical charges and due to the fact that both water and salt compounds are polar, with positive and negative charges on opposite sides in the molecule. The bonds in salt compounds are called ionic because they both have an electrical charge—the chloride ion is negatively charged and the sodium ion is positively charged. Likewise, a water molecule is ionic in nature, but the bond is called covalent, with two hydrogen atoms both situating themselves with their positive charge on one side of the oxygen atom, which has a negative charge. When salt is mixed with water, the salt dissolves because the covalent bonds of water are stronger than the ionic bonds in the salt molecules.​

​

The positively-charged side of the water molecules are attracted to the negatively-charged chloride ions and the negatively-charged side of the water molecules are attracted to the positively-charged sodium ions. Essentially, a tug-of-war ensues with the water molecules winning the match. Water molecules pull the sodium and chloride ions apart, breaking the ionic bond that held them together. After the salt compounds are pulled apart, the sodium and chloride atoms are surrounded by water molecules, as this diagram shows. Once this happens, the salt is dissolved, resulting in a homogeneous solution.​

​

Although both sources say that it is the sodium and chloride ions, and not the components of the water molecule, that are pulled apart (which is something that @russ_watters succinctly pointed out), there is an important difference in the description of the ionic interactions $-$ repeating from the sciencing.com source:
The second sentence is incorrect (due to a probably-inadvertent repetition of "electronegative oxygen portion" from the preceding sentence) $-$ it would not have been incorrect to say that the negatively charged chlorine portion of NaCl is attracted to the positively charged hydrogen portions of the water molecule, which is probably what the writer intended.

Thanks for the enlightening explanations.

What common household substances can act like water and salt in the polar aspect?

 

[sysprog]

Thanks for the enlightening explanations.

What common household substances can act like water and salt in the polar aspect?

This list of polar-molecule substances (I underlined the ones that may be considered to be 'common household substances') is from https://sciencenotes.org/polar-and-nonpolar-molecules/:

• Water – H2O
• Ammonia – NH3
• Sulfur dioxide – SO2
• Hydrogen sulfide – H2S
• Carbon monoxide – CO
• Ozone – O3
• Hydrofluoric acid
• Ethanol – C2H6O (and other alcohols with an OH at one end)
• Sucrose – C12H22O11 (and other sugars with OH groups)

 

[jake jot]

That description appears to be quoted from https://sciencing.com/happens-salt-added-water-5208174.html,
which says what you quoted:

but also immediately following that says:That source cites this one as a reference $-$ https://www.usgs.gov/media/images/water-molecules-and-their-interaction-salt-molecules:

​

Water molecules and their interaction with salt molecules​

​

View attachment 272738​

​

This diagram shows the positive and negative parts of a water molecule. It also depicts how a charge, such as on an ion (Na or Cl, for example) can interact with a water molecule.​

​

At the molecular level, salt dissolves in water due to electrical charges and due to the fact that both water and salt compounds are polar, with positive and negative charges on opposite sides in the molecule. The bonds in salt compounds are called ionic because they both have an electrical charge—the chloride ion is negatively charged and the sodium ion is positively charged. Likewise, a water molecule is ionic in nature, but the bond is called covalent, with two hydrogen atoms both situating themselves with their positive charge on one side of the oxygen atom, which has a negative charge. When salt is mixed with water, the salt dissolves because the covalent bonds of water are stronger than the ionic bonds in the salt molecules.​

​

The positively-charged side of the water molecules are attracted to the negatively-charged chloride ions and the negatively-charged side of the water molecules are attracted to the positively-charged sodium ions. Essentially, a tug-of-war ensues with the water molecules winning the match. Water molecules pull the sodium and chloride ions apart, breaking the ionic bond that held them together. After the salt compounds are pulled apart, the sodium and chloride atoms are surrounded by water molecules, as this diagram shows. Once this happens, the salt is dissolved, resulting in a homogeneous solution.​

​

Although both sources say that it is the sodium and chloride ions, and not the components of the water molecule, that are pulled apart (which is something that @russ_watters succinctly pointed out), there is an important difference in the description of the ionic interactions $-$ repeating from the sciencing.com source:
The second sentence is incorrect (due to a probably-inadvertent repetition of "electronegative oxygen portion" from the preceding sentence) $-$ it would not have been incorrect to say that the negatively charged chlorine portion of NaCl is attracted to the positively charged hydrogen portions of the water molecule, which is probably what the writer intended.

How do you interpret it in terms of quantum chemistry? What is the quantum chemistry or wave function of pure water? And what happens to the wave function when salt is added? What text strings to look for in the references, and do you have any?

 

[sysprog]

How do you interpret it in terms of quantum chemistry? What is the quantum chemistry or wave function of pure water? And what happens to the wave function when salt is added? What text strings to look for in the references, and do you have any?

I think that the applicability of quantum theory to ordinary chemical reactions is something to be aware of the existence of, but that for most practical purposes in the study of basic chemistry it can be safely disregarded.

I think that it should be examined in depth only after one has acquired a competent understanding of chemical interactions at a basic classical level, e.g. learning and working with and understanding oxidation and reduction, acids, bases, and salts, why carbon has the most reactions of any element, why flourine is the most reactive, why helium is chemically inert, etc., understanding Bohr and Mendeleev before digging into L. Pauling, etc..

Also, importantly, understanding quantum theory requires understanding of very advanced mathematics, while for the most part, at the undergraduate level, study of chemistry requires mathematiccs at about the level described in this booklet: https://www.birmingham.ac.uk/Docume...on-Internships/Maths-for-Chemists-Booklet.pdf

 

[jedishrfu]

Another good resource for basic chemistry is the Openstax online book:

https://openstax.org/details/books/chemistry-2e

 

[jake jot]

Another good resource for basic chemistry is the Openstax online book:

https://openstax.org/details/books/chemistry-2e

I plan to get this.

https://www.amazon.com/dp/B08JSGK97P/?tag=pfamazon01-20

Perhaps someone has tried it or a better package?

 

[jedishrfu]

What is your expectation from this chemistry set?

I saw there were two reviews and one said their 6 yr old daughter enjoyed the experiments.

If your expectation is to learn chemistry at the high school or college level, I think you will be sadly disappointed. However, for some light self learning it may be good.

A lot of chemistry involves measurement and calculation of moles, molarity, molality, solutions, balancing equations ... I don't think this set will even approach those things.

 

[phinds]

... IR spectrometer are cheap. What is the cheapest one out there that you can buy and experiment?

Do you not know how to use Google?

 

[jake jot]

Do you not know how to use Google?

Tried that. The articles were complex. So I was thinking those who have actually acquired one and tried it. Polar compound means the electrons can move like antenna so they absorb and emit IR. This is profound thing.

Also PF is a self contained team of multidisciplinary experts who can handle LHC or even build nukes or create new physics theories that work. If we were to colonize other planets. The PF team can recreate all parts of science, and hence Earth "designated survivor". Therefore asking personal tips of cheap IR spectrometers are justified. But then I think most have access to expensive ones only. I just want to try IR spectrometers on polar compound and see if they are very dense compared to nonpolar (which doesn't have IR) but then IR also work far from thermal equilibrium? I guess it is the IR frequencies?

 

[DrClaude]

Tried that. The articles were complex. So I was thinking those who have actually acquired one and tried it. Polar compound means the electrons can move like antenna so they absorb and emit IR. This is profound thing.

Also PF is a self contained team of multidisciplinary experts who can handle LHC or even build nukes or create new physics theories that work. If we were to colonize other planets. The PF team can recreate all parts of science, and hence Earth "designated survivor". Therefore asking personal tips of cheap IR spectrometers are justified. But then I think most have access to expensive ones only. I just want to try IR spectrometers on polar compound and see if they are very dense compared to nonpolar (which doesn't have IR) but then IR also work far from thermal equilibrium? I guess it is the IR frequencies?

There is a lot that is wrong in there. You should really pick up a chemistry textbook or dig into Openstax as @jedishrfu suggested.

All molecules absorb and emit IR radiation, whether polar or not. It is related to the vibration of molecules, i.e., the relative motion of the nuclei with respect to one another.

The only IR spectrometer I know of are laboratory equipment, and I don't think that you would be able to use one (or even prepare samples to be studied) without some guidance.

 

[snorkack]

There is a lot that is wrong in there. You should really pick up a chemistry textbook or dig into Openstax as @jedishrfu suggested.

All molecules absorb and emit IR radiation, whether polar or not. It is related to the vibration of molecules, i.e., the relative motion of the nuclei with respect to one another.

Not quite all. A molecule needs transition dipole moment around a bond, any bond.
Diatomic molecules with same atoms both side of bond cannot absorb or emit IR. Such as N₂, O₂ or I₂.
Diatomic molecules with two different atoms can have quite small dipole moment, but the dipole moment would have to be exactly zero to prevent IR. CO has only a small dipole moment, but does absorb IR.
Now, many molecules have several slightly polar bonds whose dipole moments cancel out by symmetry. In CO₂, the two C-O bonds are opposite in direction so dipole moments cancel and carbon dioxide is not polar. But the IR absorptions do not cancel, which is why CO₂ is a greenhouse gas. In CH₄, C-H bonds have little polarity - carbon and hydrogen are close but not exactly equal in electronegativity - and the 4 dipole moments cancel out anyway. But the IR do not cancel, and weak as the polarity is, it is enough to make methane a greenhouse gas.
To put alkane polarity into contect, while ethane also has no dipole moment due to symmetry, propane does not have such symmetry. Two hydrogens are on one side of central carbon, two methyl groups on the opposite side. Resulting in a nonzero dipole moment... of about 0,1 D. Compare dimethyl ether with the same bent geometry but two O-C bonds... dipole moment 1,3 D.

Oh, and NO, not all polar molecules are electrolytes. Dimethyl ether, acetone... fairly polar but not enough for electrolytic dissociation.

 

[symbolipoint]

jedishrfu #14 is an excellent response.

 

[sysprog]

I wrote an 'erratum' email to the writer of the sciencing.com article (omitting salutations in the quote):

Your article at https://sciencing.com/happens-salt-added-water-5208174.html article, contains the following:

When table salt is placed in water, the slightly electropositive sodium portion is attracted to the slightly electronegative oxygen portion of water molecules. At the same time, the slightly electronegative chlorine portion of NaCl is attracted to the slightly electronegative oxygen portion of water.

The second sentence is incorrect (due to a probably-inadvertent
repetition of "electronegative oxygen portion" from the preceding sentence) − it would not have been incorrect to say that the negatively charged chlorine portion of NaCl is attracted to the positively charged hydrogen portions of the water molecule, which is presumably what you intended.

He soon responded (again omitting salutations):

Hey, thanks for picking this up. You're correct in identifying the process that led to the mistake, but a mistake it is, and I will notify the people with access to the site to make the change. Unfortunately, Leaf Media discontinued its Sciencing arm in April, so it may be a while...

. . . just like a science guy to want to correct his mistake . . .

 

[DrClaude]

Not quite all. A molecule needs transition dipole moment around a bond, any bond.
Diatomic molecules with same atoms both side of bond cannot absorb or emit IR. Such as N₂, O₂ or I₂.
Diatomic molecules with two different atoms can have quite small dipole moment, but the dipole moment would have to be exactly zero to prevent IR. CO has only a small dipole moment, but does absorb IR.

Yes, homonuclear diatomic molecules are an exception, but I didn't want do get into those weeds.

By the way, what is important is not the dipole moment, but the variation of the dipole moment as the molecule vibrates.

 

[jake jot]

I think the simple rule is simply. IR spectroscopy spectrum (low infrared) come from interaction within a single molecule. While thermal IR (mid to far infrared) comes from interaction *between* molecules. Right? So in the case of thermal, all molecules produce IR unless at vacuum.

 

[DrClaude]

I think the simple rule is simply. IR spectroscopy spectrum (low infrared) come from interaction within a single molecule. While thermal IR (mid to far infrared) comes from interaction *between* molecules. Right?

No.

So in the case of thermal, all molecules produce IR unless at vacuum.

Do you mean "absolute zero" instead of "vacuum"?

View attachment 273060

This is related to my reply to @snorkack. The normal modes of CO₂ that are active in the infrared are the asymmetric stretch and the bending modes, which change the dipole moment of the molecule (in this case, induce a dipole since CO₂ is globally non-polar), while the symmetric stretch is not active in the infrared (dipole stays unchanged, = 0) and is only observable in Raman spectroscopy.

 

[jake jot]

No.Do you mean "absolute zero" instead of "vacuum"?This is related to my reply to @snorkack. The normal modes of CO₂ that are active in the infrared are the asymmetric stretch and the bending modes, which change the dipole moment of the molecule (in this case, induce a dipole since CO₂ is globally non-polar), while the symmetric stretch is not active in the infrared (dipole stays unchanged, = 0) and is only observable in Raman spectroscopy.

Where do the images from thermal imagers come from? Is it not from the vibrations of the molecules (in between the molecules) as bulk? And not from those IR active modes?

 

[DrClaude]

Where do the images from thermal imagers come from? Is it not from the vibrations of the molecules (in between the molecules) as bulk? And not from those IR active modes?

If you are looking at solid objects, then of course the IR radiation comes from bulk modes. Gases will emit thermal radiation only in narrow bands, while bulk matter will have a much broader spectrum. This only reflects the fact that bulk matter has many, tightly-packed vibrational modes, while in the gas phase vibrational modes are discrete. But the phenomenon is the same.

 

[snorkack]

van der Waals bonds are weak - and therefore should vibrate only in far IR. If they have transition moment. Note that if they don´ t because of symmetry, bulk matter will still have multiphonon modes where the interaction of several phonons breaks symmetry - whether the symmetric bond is a van der Waals or a covalent one.

 

[jake jot]

Guys, for home substances, what is the range of near IR or visible frequencies that is absorbance active? I am deciding what range of wavelength of the IR spectrometer to get (a one time purchase that lasts a lifetime, unlike a smartphone that you can change every year).

This is for Near Infrared Range.

Models	Number of Elements	Spectrometer Range (nm)	Grating(g/mm)	Range (nm)	Dispersion (nm/pixel)	Estimated Resolving Resolution
DWARF-Star-NIR	512	900-1700	250	800nm	1.562	3.1nm
DWARF-Star-NIRb	512	900-1600	300	650nm	1.269	2.5nm
DWARF-Star-NIR2	512	1250-1575	600	325nm	0.634	1.3nm
DWARF-Star-NIR2b	512	1150-1475	600	325nm	0.634	1.3nm
DWARF-Star-NIR	1024	1000-1700	600	700nm	0.683	1.4nm
DWARF-Star-NIR3-HR	512	1530-1605	1200	70nm	0.195	0.4nm
DWARF-Star-NIR3-HR	1024	1500-1640	1200	140nm	0.195	0.4nm
RED-Wave-NIRX	512	1500-2200	300	700nm	1.367	2.8nm
RED-Wave-NIRX	1024	1500-2200	600	700nm	0.683	1.4nm
RED-Wave-NIRX-SR	512	900-2300	300	1400nm	5.3	<13nm
NIRX-SR	1024	900-2300	600	1400nm	2.7	<7nm
Micro-NIR	1	1750-2150	NA	400nm	1	10nm

This is for visible and UV light spectrometer.

Miniature UV-VIS Spectrometer Models	Wavelength
Blue-Wave Miniature Series	200-1150nm
Concave Grating Series	190 – 900nm
Super Range Concave Grating Series	200 – 1100nm
High Resolution Concave Grating Series	200-750nm
SILVER-Nova High Performance Series	190 – 1110nm
High Resolution Series	200 – 1075nm
GREEN-Wave Low Cost Series	350-1150nm

I won't get UV because it's hazardous. I want to be more familiar with polar substances by studying their IR absorbance and spectrum properties. I couldn't find the answer at google what range is useful for home use.

 

[jake jot]

No.Do you mean "absolute zero" instead of "vacuum"?This is related to my reply to @snorkack. The normal modes of CO₂ that are active in the infrared are the asymmetric stretch and the bending modes, which change the dipole moment of the molecule (in this case, induce a dipole since CO₂ is globally non-polar), while the symmetric stretch is not active in the infrared (dipole stays unchanged, = 0) and is only observable in Raman spectroscopy.

Something I don't quite get even after googling. When molecules are asymmetric, the movement can cause dipole moments and infrared active, the explanation being it moves like antenna sending infrared photons. No problem about this.

But with regards to polarizability, which involves the formation of temporary dipoles due to momentary electron cloud distortions. How come polarizability is low in asymmetric molecules? Do they cancel? How? I heard about the rule of mutual exclusion. But I want visual explanation why polarizability is low in asymmetric molecules.

About number 6. So IR spectroscopy gives an indication of ionic character in the molecule. There is no exception to the rule where covalent character is indicated?

 

[DrClaude]

How come polarizability is low in asymmetric molecules?

Do you have a source for this?

About number 6. So IR spectroscopy gives an indication of ionic character in the molecule. There is no exception to the rule where covalent character is indicated?

The point is that the dipole is an indication of an uneven distribution of electrons. Hence a less covalent bond.

 

[jake jot]

Do you have a source for this?

Here is the exact passage:

Chapter5.pdf (uc.edu) (page 2)

"The mobility of electrons in a bond is called the Polarizability of the bond, i.e. a measure of how easy it is to move electrons and polarize a bond. For bonds with a strong dipole moment (which are IR active) the mobility or polarizability is
usually low.".

Do you think it's low or cancelled? In the following explanation, it's sort of cancelled.

5: Raman Spectroscopy - Chemistry LibreTexts

"The asymmetric stretch of carbon dioxide is IR active because there is a change in the net molecular dipole (Figure 5.2). In the asymmetric stretch, one bond is stretched and is now more polarizable while the other bond is compressed and is less polarizable. The change in polarizability of the longer bond is exactly offset by the change in the shorter bond such that the overall polarizability of the molecule does not change. Therefore, the asymmetric stretch is not Raman active.

I want to understand how exactly "The change in polarizability of the longer bond is exactly offset by the change in the shorter bond such that the overall polarizability of the molecule does not change.". Does it mean polarizability is also present but in the case of asymmetric molecules, it's cancelled? How is it canceled when the each side have separate electron cloud? How does the electron cloud in each side interact so that they are cancelled, and hence not Raman active?

 

[mjc123]

First, to be pedantic: "Sorta" is not a scientific concept. Second, you seem to be making a number of confusions (which are quite common with beginners): one, between a quantity (e.g. polarisability) and the change in that quantity. All bonds are polarisable to some extent; the point is that to be Raman active a vibration must result in a change of polarisability. It is this change, not polarisability itself, that is cancelled. Two, between molecules and bonds (or vibrational modes, which may involve more than one bond). A symmetrical molecule may have asymmetrical vibrations which have a change of dipole moment and are IR active (like the one illustrated).

A more precise answer to your question involves a more detailed consideration of symmetry, which I'm guessing you haven't studied yet. The polarisability has certain symmetry properties which mean that Raman scattering of a photon can induce a change in vibrational state only for modes of certain symmetry types. In particular, for a centrosymmetric molecule, only centrosymmetric modes can be Raman active.

A more qualitative argument may go like this: Bear in mind that the relative change in bond length during a vibration is much, much smaller than the above diagram suggests (for ease of illustration). We may reasonably consider that, for small changes in length, the change in polarisability with bond length will, to a first approximation, vary linearly with the deformation. If we write dP/dx as P', the change in polarisability in an asymmetric stretching mode like that shown, where one bond lengthens by dx and the other shortens by dx, will be P'dx - P'dx = 0. For a symmetric stretch, where both bonds lengthen, it will be P'dx + P'dx ≠ 0. (For the dipole moment, we must remember that the moments of the two bonds are pointing in opposite directions, so the change in dipole moment for the asymmetric stretch is μ'dx - (-μ')dx ≠ 0, while for the symmetric stretch it is μ'dx + (-μ')dx = 0, so the asymmetric stretch is IR active but the symmetric stretch is not.)

 

[jake jot]

First, to be pedantic: "Sorta" is not a scientific concept. Second, you seem to be making a number of confusions (which are quite common with beginners): one, between a quantity (e.g. polarisability) and the change in that quantity. All bonds are polarisable to some extent; the point is that to be Raman active a vibration must result in a change of polarisability. It is this change, not polarisability itself, that is cancelled. Two, between molecules and bonds (or vibrational modes, which may involve more than one bond). A symmetrical molecule may have asymmetrical vibrations which have a change of dipole moment and are IR active (like the one illustrated).

A more precise answer to your question involves a more detailed consideration of symmetry, which I'm guessing you haven't studied yet. The polarisability has certain symmetry properties which mean that Raman scattering of a photon can induce a change in vibrational state only for modes of certain symmetry types. In particular, for a centrosymmetric molecule, only centrosymmetric modes can be Raman active.

A more qualitative argument may go like this: Bear in mind that the relative change in bond length during a vibration is much, much smaller than the above diagram suggests (for ease of illustration). We may reasonably consider that, for small changes in length, the change in polarisability with bond length will, to a first approximation, vary linearly with the deformation. If we write dP/dx as P', the change in polarisability in an asymmetric stretching mode like that shown, where one bond lengthens by dx and the other shortens by dx, will be P'dx - P'dx = 0. For a symmetric stretch, where both bonds lengthen, it will be P'dx + P'dx ≠ 0. (For the dipole moment, we must remember that the moments of the two bonds are pointing in opposite directions, so the change in dipole moment for the asymmetric stretch is μ'dx - (-μ')dx ≠ 0, while for the symmetric stretch it is μ'dx + (-μ')dx = 0, so the asymmetric stretch is IR active but the symmetric stretch is not.)

Thanks. Let's illustrate this.

We know that during the inelastic scattering process the molecules excite to higher vibrational energy levels and make transitions to the ground level by emitting photons with a frequency different from the incident photons.

What is faster, the transition from higher vibration energy level to ground level or the one asymmetric stretch from left to right?

If the asymmetric stretch is faster, then there is no time for inelastic scattering. But if inelastic scattering is faster, it should release the say strokes photon and this should get out of the molecules to reach the detectors. So how can this change of polarizability be canceled when the photons were already released? Unless the one instance of asymmetric stretch from say left to right is faster so no time for inelastic scattering to occur? (I know the stretching occurs many billions of time a second or so. I'm describing 1/1000000000000000 millisec or less.)

 

[sysprog]

(I know the stretching occurs many billions of time a second or so. I'm describing 1/1000000000000000 millisec or less.)

I'm not here denying your main point, but that number of miiliseconds is a few orders of magnitude too small $--$ I counted 15 instances of the zero numeral in your fractional number of milliseconds $-$ a billionth is 0.000000001 $-$ your number is a quadrillionth $-$ that's a millionth of a billionth $-$ of a millisecond.

 

[jake jot]

I'm not here denying your main point, but that number of miiliseconds is a few orders of magnitude too small $--$ I counted 15 instances of the zero numeral in your fractional number of milliseconds $-$ a billionth is 0.000000001 $-$ your number is a quadrillionth $-$ that's a millionth of a billionth $-$ of a millisecond.

I just typed 000 without counting them.. just to emphasize I knew the stretching didn't occur in seconds but in billions of billions of milliseconds. I don't know exactly how fast, so how fast does one stretching occurs compared to the transition from virtual energy state to ground state, any ideas?

 

[mjc123]

Your reference says "Raman scattering has a lifetime of 10^-14 s", which is comparable to the period of molecular vibrations - e.g. the antisymmetric stretch of CO₂ at 2350 cm^-1 has a period of 1.4 x 10^-14 s. But in any case I don't get your point. There is no change of polarisability during the vibration, so there can be no Raman transition. It is forbidden by symmetry.

 

[jake jot]

Your reference says "Raman scattering has a lifetime of 10^-14 s", which is comparable to the period of molecular vibrations - e.g. the antisymmetric stretch of CO₂ at 2350 cm^-1 has a period of 1.4 x 10^-14 s. But in any case I don't get your point. There is no change of polarisability during the vibration, so there can be no Raman transition. It is forbidden by symmetry.

Change in polarizability is the same as transition from virtual energy state to ground state, right?

So I want to understand in picture what exactly happens that prevents the change of polarizability or transition during the vibration. I know when we hit water with a Raman spectrometer, billions of atoms were inside the laser spot. I have numerous experiences with Raman, so I need to visualize what happens when it the laser hits the molecules that is asymmetric and no Raman inelastic scattering occurs. This means the laser is not inelastically scattered back. Any references of the selection rule with great illustrations to make it logical and intuitive?

 

[mjc123]

Change in polarizability is the same as transition from virtual energy state to ground state, right?

Wrong. It is a property of the molecule (more particularly, of a vibrational mode of the molecule), whether or not there are any photons around to get scattered. But that property is necessary for Raman scattering to occur. BTW, water has different symmetry from CO₂, and all its vibrational modes are Raman active.

 

[jake jot]

Wrong. It is a property of the molecule (more particularly, of a vibrational mode of the molecule), whether or not there are any photons around to get scattered. But that property is necessary for Raman scattering to occur. BTW, water has different symmetry from CO₂, and all its vibrational modes are Raman active.

Ok. I've been googling with these words "group theory molecule symmetry Raman" and reading articles about it. If you know of one with good illustrations and most intuitive even with interactive java or animation, just share it anytime. Thanks a lot.

 

[jake jot]

We know that in order for a molecule to be Raman active, there must be a change in the polarizability, meaning that there must be change in the size, shape or orientation of the electron cloud that surrounds the molecule. And this change said to occur in symmetric stretching, but not asymmetric stretching.

Without using any rules or tables, can anyone describe using plain words why in asymmetric stretching, there is no change in the size, shape or orientation of the electron cloud that surrounds the molecule? (maybe a PF challenge?)

 

[snorkack]

We know that in order for a molecule to be Raman active, there must be a change in the polarizability, meaning that there must be change in the size, shape or orientation of the electron cloud that surrounds the molecule. And this change said to occur in symmetric stretching, but not asymmetric stretching.

Without using any rules or tables, can anyone describe using plain words why in asymmetric stretching, there is no change in the size, shape or orientation of the electron cloud that surrounds the molecule? (maybe a PF challenge?)

Simple reason: "size, shape or orientation" is not enough.
In asymmetric stretching, one bond stretches but the other bond shrinks, and does so by pretty much exactly the same amount as the first bond stretches. Because the polarizability is not a directed vector, it does not matter for polarizability that the two bonds are in opposite directions - the sum is still the same because changes of the halves are equal and opposite. Shape changes but "size" does not, and "size" staying constant as shape changes is what makes Raman scattering impossible.

 

[mjc123]

So I want to understand in picture what exactly happens that prevents the change of polarizability or transition during the vibration.

I guess we belong to different tribes; I am of those who are content to say "if that's what the maths says, that's how it is"; you want to be able to visualise what is happening physically. I suspect what you are asking is impossible, or at least very difficult (I would be happy to be proved wrong). I can visualise in a crude way what is going on with IR and dipole moments; I find it much harder to visualise what's going on with polarisability and Raman scattering. But the maths is the only accurate and scientific description we have; mental models and "visualisations" may help our understanding to some extent, but they are limited, and possibly misleading, especially if they are confined to analogies of classical effects we are familiar with in daily life, whereas there are some quantum effects that have no classical analogue. Ultimately, if you want a proper understanding, you have to grapple with the theory and the maths; purely qualitative descriptions, whether in words or pictures, can only take you so far.

 

[jake jot]

I guess we belong to different tribes; I am of those who are content to say "if that's what the maths says, that's how it is"; you want to be able to visualise what is happening physically. I suspect what you are asking is impossible, or at least very difficult (I would be happy to be proved wrong). I can visualise in a crude way what is going on with IR and dipole moments; I find it much harder to visualise what's going on with polarisability and Raman scattering. But the maths is the only accurate and scientific description we have; mental models and "visualisations" may help our understanding to some extent, but they are limited, and possibly misleading, especially if they are confined to analogies of classical effects we are familiar with in daily life, whereas there are some quantum effects that have no classical analogue. Ultimately, if you want a proper understanding, you have to grapple with the theory and the maths; purely qualitative descriptions, whether in words or pictures, can only take you so far.

I think what you are saying is we must use the lessons in quantum mechanics whose punchline is "In the absence of measurement to determine it's properties, there are no properties". For example, in Stern Gerlach experiment, there is no definite spin without measurement, and in double slit experiiment, there is no position before measurement. So instead of visualizing the asymmetrical molecules with atoms on either side. We must instead think of it like this?

Before I begin to think of solid objects as like that. Please others try to explain it in plain english, if you still can.

 

[mjc123]

I'm not saying "in the absence of measurement there are no properties". To use your example of spin, I'm not saying the spin has no value before it's measured (I leave that question tot he philosophers of QM), but that spin itself is a property that has no anlogue in classical mechanics (nothing that we can "visualise" from experience). To model the electron as a little particle orbiting the nucleus and spinning on its own axis may be helpful to an extent, but it is not "what is really happening", and will let us down at some point if we take it too literally. You ask for plain English, but can plain English describe what is outside our experience, other than by using such only-partly-adequate analogies?

 

[jake jot]

Simple reason: "size, shape or orientation" is not enough.
In asymmetric stretching, one bond stretches but the other bond shrinks, and does so by pretty much exactly the same amount as the first bond stretches. Because the polarizability is not a directed vector, it does not matter for polarizability that the two bonds are in opposite directions - the sum is still the same because changes of the halves are equal and opposite. Shape changes but "size" does not, and "size" staying constant as shape changes is what makes Raman scattering impossible.

Someone told me that in the asymmetric stretch the molecule retains the same shape because as one extends the other contracts and vice versa. But you said it was the size staying constant. Is it not the shape that stays constant?

Whatever. So we must not imagine the oxygen atoms as either on the left and right side of the carbon in the case of carbon dioxide, but smeared in the molecules like in 2 places at once, like in the quantum double slit?

Look. I own a Raman spectrometer and putting up a molecular scanning laboratory to monitor changes of molecular bonds to unknown field and hamiltonians. And everyday whenever the laser hits the sample, I need to visualize or have ideas how it can cause transitions only in symmetric molecules and not in asymmetric. Also need to look at this from the view of an electromagnetic wavelenght that is a thousand times as large

Also the laser beam is light which has electric and magnetic field, it has directed vector, so I'm trying to reconcile it with the polarization that has no directed vector as you described.

 

[jake jot]

First, to be pedantic: "Sorta" is not a scientific concept. Second, you seem to be making a number of confusions (which are quite common with beginners): one, between a quantity (e.g. polarisability) and the change in that quantity. All bonds are polarisable to some extent; the point is that to be Raman active a vibration must result in a change of polarisability. It is this change, not polarisability itself, that is cancelled. Two, between molecules and bonds (or vibrational modes, which may involve more than one bond). A symmetrical molecule may have asymmetrical vibrations which have a change of dipole moment and are IR active (like the one illustrated).

A more precise answer to your question involves a more detailed consideration of symmetry, which I'm guessing you haven't studied yet. The polarisability has certain symmetry properties which mean that Raman scattering of a photon can induce a change in vibrational state only for modes of certain symmetry types. In particular, for a centrosymmetric molecule, only centrosymmetric modes can be Raman active.

A more qualitative argument may go like this: Bear in mind that the relative change in bond length during a vibration is much, much smaller than the above diagram suggests (for ease of illustration). We may reasonably consider that, for small changes in length, the change in polarisability with bond length will, to a first approximation, vary linearly with the deformation. If we write dP/dx as P', the change in polarisability in an asymmetric stretching mode like that shown, where one bond lengthens by dx and the other shortens by dx, will be P'dx - P'dx = 0.

I'm still pondering what you said. Why did you use one single "dx" for either side where one bond lengthens and the other shortens? Wouldn't it be "dx_l" and "dx_s"?

For a symmetric stretch, where both bonds lengthen, it will be P'dx + P'dx ≠ 0. (For the dipole moment, we must remember that the moments of the two bonds are pointing in opposite directions, so the change in dipole moment for the asymmetric stretch is μ'dx - (-μ')dx ≠ 0, while for the symmetric stretch it is μ'dx + (-μ')dx = 0, so the asymmetric stretch is IR active but the symmetric stretch is not.)

 

[DrClaude]

Someone told me that in the asymmetric stretch the molecule retains the same shape because as one extends the other contracts and vice versa. But you said it was the size staying constant. Is it not the shape that stays constant?

Shape is too vague a term. What does it mean exactly?

Whatever. So we must not imagine the oxygen atoms as either on the left and right side of the carbon in the case of carbon dioxide, but smeared in the molecules like in 2 places at once, like in the quantum double slit?

No, the classical picture with a localized oxygen nucleus on both sides is a very good one here.

Look. I own a Raman spectrometer and putting up a molecular scanning laboratory to monitor changes of molecular bonds to unknown field and hamiltonians. And everyday whenever the laser hits the sample, I need to visualize or have ideas how it can cause transitions only in symmetric molecules and not in asymmetric.

There is no way to avoid the math here, meaning quantum mechanics and group theory for symmetry.

Also need to look at this from the view of an electromagnetic wavelenght that is a thousand times as large

This is mostly irrelevant here, except for the fact that it means that you can consider that the electromagnetic field at any given instant is uniform over the molecule (long-wavelength approximation).

Also the laser beam is light which has electric and magnetic field, it has directed vector, so I'm trying to reconcile it with the polarization that has no directed vector as you described.

Polarizability is a tensor, not a vector.

I'm still pondering what you said. Why did you use one single "dx" for either side where one bond lengthens and the other shortens? Wouldn't it be "dx_l" and "dx_s"?

There is a single x coordinate here. You could but labels on P if you want, but I think that what @mjc123 wrote was clear enough for a forum discussion.

 

[jake jot]

In our discussions. We jumped from change in polarizability to transitions from virtual energy state to ground state. We omitted the topic of induced dipole. IR activeness has to do with permanent dipole moment, and Raman has to do with induced dipole.

Can you give an example where induced dipole can cause changes in vibrational states? In plain QM, when you give electrons energy. It got excited to higher electronic energy level. When you cause an induced dipole, how come there is more tendency for the molecule to get to higher energy state? Can induced dipole itself bestow energy to the electrons, or cause distinctions in energy states enough to make transitions possible? But it's not intuitive how induced dipole can cause ease of electronic transition. Note in electronic transitions, it's not moving from negative to positive.

I'm trying to understand the connection between the two. I actually watch many videos at youtube and this is not so clear. An intuitive answer would make it become part of common sense understanding and memorizing formulas or equations.

 

[DrClaude]

Can you give an example where induced dipole can cause changes in vibrational states? In plain QM, when you give electrons energy. It got excited to higher electronic energy level. When you cause an induced dipole, how come there is more tendency for the molecule to get to higher energy state? Can induced dipole itself bestow energy to the electrons, or cause distinctions in energy states enough to make transitions possible? But it's not intuitive how induced dipole can cause ease of electronic transition. Note in electronic transitions, it's not moving from negative to positive.

Transitions between states in an atom or a molecule take place because of the coupling of the atom/molecule with the electromagnetic field. Using perturbation theory, one finds, to first order, that the coupling comes from the interaction of the EM field with the dipole moment of the atom/molecule. Polarizability comes in as a second-order term, where a dipole can be induced by the field, and this dipole then interacts again with the field, causing the transition.

In all cases, what the EM field can do is move charges around, hence its interaction with a permanent or induced dipole moment.

Note that the virtual transitions you mentioned above are simply a visualization of perturbation theory.

I'm trying to understand the connection between the two. I actually watch many videos at youtube and this is not so clear. An intuitive answer would make it become part of common sense understanding and memorizing formulas or equations.

Again, the answers can only be found by diving into the theory.

 

[jake jot]

Transitions between states in an atom or a molecule take place because of the coupling of the atom/molecule with the electromagnetic field. Using perturbation theory, one finds, to first order, that the coupling comes from the interaction of the EM field with the dipole moment of the atom/molecule. Polarizability comes in as a second-order term, where a dipole can be induced by the field, and this dipole then interacts again with the field, causing the transition.

It makes more sense now.

For permanent dipole moments that is IR active where IR spectroscopy is useful. Do all heat and energy increase the dipole moments so you will have immediate effects on absorbance? What kinds of molecules where even when you heat them or introduce energy, there will be no changes in the permanent dipole moments or asymmetric stretching?

When I got the Raman scanner, my priority was studying the bonding of water to see influence to external fields and energy.. Lately I realized the water electrons were so close to the nucleus so the change in polarizability is poorer, hence water is a bad target.

 

[mjc123]

As has been repeatedly stated, a permanent dipole moment is not necessary for IR activity. What is required is a dipole moment that changes during the vibration (even if the time-average is zero). Heating will not affect the dipole moment significantly, but it may affect the spectrum by populating higher rotational and vibrational states (that's one way of measuring the temperature of a gas).

Did you not do some homework before buying a Raman scanner, and realize that water is a bad Raman scatterer? That's one of the advantages of Raman - you can look at species in aqueous solution , which you can't with IR because of the massive absorbance of water.