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I'm wondering why a bond between an O and an H is going to absorb a higher frequency than a bond between a C and an H (any hybridization).
O is heavier than C in terms of atomic masses, and I'm assuming the O-H bond will be weaker because of the electronegativities, so why would it absorb a higher frequency?
Thank you.
First, let me just put down the approximate $\ce{X-H}$ stretching frequencies (where, $\ce{X = O, C}$) just as a reference.
$$\begin{array}{c} & \ce{X-H}\ \text{stretch}\ \mathrm{(cm^{-1})} \\ \hline \ce{X = O} & 3600 \\ \hline \ce{X = C} & 3000 \\ \hline \end{array}$$
Now, as a simplification I will first discuss a diatomic molecule, thought of as a two balls of some mass $\mathrm{m}$ linked together with a spring. Solving, Schrodinger's equation for this system (called a harmonic oscillator), yields the following relation:
$\mathrm{\bar{\nu}} = \frac{1}{2\pi c}\sqrt{\frac{k}{m}}$
For, dissimilar masses, (say, $\mathrm{m_1}$ and $\mathrm{m_2}$), we can switch to the centre of mass coordinates, and the $\mathrm{m}$ in the preceding equation is replaced with the reduced mass ($\mu$) defined as, $\frac{1}{\mu} \equiv \frac{1}{m_1}+\frac{1}{m_2} $
Consequently, $\ce{C-H,O-H}$ bonds in general have much higher stretching frequencies than do corresponding bonds to heavier atoms other than $\ce{H}$.The mass effect becomes evident when deuterated isotopes are examined. The stretching frequency of a free $\ce{O-H}$ bond is ca. $3600\ \mathrm{cm^{-1}}$, but the $\ce{O-D}$ equivalent is lowered to ca. $2600\ \mathrm{cm^{-1}}$. Since deuterium has a mass twice that of hydrogen, the mass term in the equation changes from 1 to 1/2, and the frequency is reduced by the square root of 2.
Naively, one would expect the frequency of the $\ce{C-H}$ bond to be higher, based on the preceding discussion, but there is another factor at play, the force constant $k$. This, is different for $\ce{O-H}$ and $\ce{C-H}$ bonds, and depends on the bond strength (stiffness of the spring)
Bond strengths can be estimated from bond disassociation energies (BDEs), which is defined as the standard enthalpy change when a bond is cleaved by homolysis, with reactants and products of the homolytic cleavage at $\mathrm{0\ K}$. This values are frequently tabulated, and can be easily found online (here for example; caveat the values here are at $\mathrm{298\ K}$)
$$\begin{array}{c} & \ce{X-H}\ \text{BDE}\ \mathrm{(kJ\ mol^{-1})} \\ \hline \ce{X = O} & 428.0\ (21) \\ \hline \ce{X = H} & 337.2\ (8) \\ \hline \end{array}$$
Anyway, this tells you that the $\ce{O-H}$ bond is stronger; Why? Well, simply look at the approximate atomic radii of Oxygen and Carbon. The Oxygen is about $\mathrm{48\ pm}$ while carbon is about $\mathrm{67\ pm}$. The longer atomic radii implies a longer $\ce{C-H}$ bond length, ergo a weaker bond.
Stronger bond, implies larger $k$, which in turn means greater frequency of vibration.
So, the $\ce{O-H}$ bond is NOT weaker. Remember, when we define bond disassociation energy, we are looking at homolytic cleavage. I really don't understand the point you try to make about electronegativities.
We know that C-H, O-H, and N-H bonds absorb at a higher frequency due to Hydrogen's lighter mass (using Hooke's Law, we know that a lighter mass means it has a higher vibration, and therefore a higher frequency). Now although Oxygen may be heavier than Carbon, remember that Oxygen-Hydrogen bonds are stronger than Carbon-Hydrogen bonds (as there is a higher electronegativity difference between the two, therefore it has a higher ionic character than C-H which instead has a higher covalent character). Again using Hooke's Law, a stronger bond means there is higher vibration and therefore a higher frequency (just as triple bonds have higher frequencies than double bonds). Think of the effect of this stronger bond as 'overpowering' the effect of the heavier Oxygen atom. Don't worry, OChem got me messed up too.
