# Find moment of inertia around two axes of sulphur dioxide

Problem:

Find the moment of inertia for 2 axes of $\ce{SO2}$. The rotational constants are given in $\pu{MHz}$. The answer is required to be $10^{-46}\rm~ kg\cdot m^2$, 7 digit accuracy.

I used $$I = \frac{h}{8\times\pi^2\times B}$$ but the answer is marked wrong. I do not understand what I am doing incorrectly. I thought maybe that $\ce{SO2}$ is asymmetric so that formula will not work.

This is an online course that has minimal discussion assistance. I have scrounged on Google & physical chemistry textbook to see what the alternative could be. Did not see another formula. Where did I go wrong?

• Please take your time and format yuor post with the MathJax syntax so that we have a better idea of what you’re saying. – Jan Jun 30 '16 at 18:10
• First question for you is what is the geometry of sulfur dioxide? Next, how many unique principle axes does it have? – Zhe Jan 18 '17 at 17:45

Without explaining some more its hard to help, but the most likely error is in using the wrong units. The rotational constant is usually given in wavenumber units (cm$^{-1}$), e.g. 2.02736 cm$^{-1}$. In a linear molecule the equilibrium value is
$B_e = \frac{h}{(8\pi^2cI)}$
with $B_e$ in wavenumbers and c is the speed of light in cm/sec. In your case since the SO$_2$ is not linear (its planar bent at the S atom, C$_{2v}$ point group) there are three rotational constants usually called A$_0$, B$_0$, C$_0$.
(You can check the units by converting Joules in to base units, m, kg, s etc . The moment of inertia is in kg m$^2$ and $10^{-46}$ is of the correct order of magnitude. Seven decimal places seems too many as the rotational constants are not usually known to more than 5 places.)