3
$\begingroup$

What does it mean for our bond of interest if we have low bond length and high frequency? Trying to make this connection but am not understanding. If anyone could explain it intuitively, that would be much appreciated. Say we have the bond of interest of heptane which is C-C. What is the variable that is changing (bond order, atom mass, etc.)? Also if the IR frequency is 1400, what is the effect of variable? Also what does this say about the physical bonding in the molecular bonds of interest?

$\endgroup$
3
  • 6
    $\begingroup$ Are you referring to vibrational (IR) spectroscopy? Your question (and its title) are a bit unclear. Please revise and include an example of the phenomenon you are trying to understand. $\endgroup$ Commented Jul 22, 2013 at 2:44
  • 2
    $\begingroup$ Can you narrow down or more exactly specify the parameters of your question? I presume by the reference to frequency you're referring to the vibrational frequency of bonds in spectroscopy? $\endgroup$
    – Greg E.
    Commented Jul 22, 2013 at 2:44
  • $\begingroup$ Is this much more clear? $\endgroup$
    – user67527
    Commented Jul 22, 2013 at 2:55

2 Answers 2

7
$\begingroup$

In IR, the vibrational frequencies of a simple diatomic molecule can be described by the following equation:

$$ \bar\nu = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}} $$

Here, $k$ is the force constant, $\mu$ is the reduced mass of the molecule, and $c$ is the speed of light (which results in appropriate units for reporting the wavenumber, rather than ordinary units of frequency, e.g., Hz.). This equation is derived from treating the molecule as a simple harmonic oscillator, using Hooke's law, Newton's laws of motion, and the classic technique of reducing a two-body problem to a one-body problem.*

The calculation of the force constant is complex, but suffice to say that it can be viewed as being roughly proportional to the strength of the bond. Since the wavenumber is proportional to the square root of the force constant it can be said that, ceteris paribus, stronger bonds tend to yield higher wavenumbers. Since higher bond multiplicity and shorter bond length correlate positively to bond strength, those factors can also be correlated to higher wavenumbers. (There are also positive correlations between metrics such as, e.g., bond energies and wavenumber, etc.) These factors can be used to rationalize the trend that wavenumbers of triple bonds tend to be greater than those of double bonds, which in turn tend to be greater than those of single bonds. Similarly, the diminished wavenumber of conjugated bonds vs. analogous unconjugated bonds can be explained by the bonds of conjugated systems having lengths intermediate between single and double bonds. One source indicates the following approximate proportionality relationship between bond length, $r$, and the force constant, $k$:

$$ k \propto \frac{1}{r^6} $$

Clearly, even minute differences in bond length produce large variance in the force constant.

Conversely, wavenumber is inversely proportional to the square root of the reduced mass. Hence, as reduced mass grows, wavenumbers decrease. This largely explains, for example, why the highest wavenumbers are those of bonds between hydrogen and some other element, since hydrogen is the lightest element.

*I think the math involved is probably well beyond the necessary scope of this answer, so I won't elaborate on it here. I dealt with these topics in introductory physics and differential equations classes. McQuarrie's physical chemistry textbook specifically deals with the harmonic oscillator model vis-a-vis spectroscopy, and I can recommend it for anybody interested in further details.

$\endgroup$
7
$\begingroup$

To add on to Greg's well-written answer (since this is too long for a comment) we can write Hooke's Law in a slightly more practical form for purposes of predicting IR stretching frequencies:$$\bar\nu (cm^{-1}) = 4.12\sqrt{\frac{k}{\mu}}$$ with $k = 5, 10 \text{, or } 15$ $\times10^5$ dynes/cm for a single, double or triple bond. For example, a $\ce{C=C}$ bond: k = $10 \times 10^5$, $\mu=\frac{(12)(12)}{12+12}=6$ giving $\bar\nu=1682 cm^{-1}$ where the experimental value is $1650$. We can leave it to the OP to see that a $\ce{C-H}$ bond has a calculated frequency of $3032 cm^{-1}$ (experimental is $3000$) and a $\ce{C-D}$ bond has a calculated frequency of $2228 cm^{-1}$ (experimental $2206$).

$\endgroup$
3
  • $\begingroup$ +1, a very nice addition. I'm sure I've seen that formulation before, but can you explain briefly the significance of the 4.12 factor? $\endgroup$
    – Greg E.
    Commented Jul 23, 2013 at 12:21
  • 1
    $\begingroup$ @GregE. $\frac{\sqrt{N}}{2\pi c}$ with speed of light in cm/s and N = Avogadros number so that the values used in calculating the reduced mass can be in AMUs. $\endgroup$ Commented Jul 23, 2013 at 13:07
  • $\begingroup$ Thanks. I assumed 1/(2pi*c) would be factored in but it hadn't occurred to me to think of conversion to amus. $\endgroup$
    – Greg E.
    Commented Jul 23, 2013 at 13:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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