# Given internal coordinates of a molecule (say, "CCl"_3"Br"), its relevant bond length(s), and bond angle(s), how do you calculate the moments of inertia in each direction? r_(C-Br) = "1.91 Å", r_(C-Cl) = "1.77 Å", /_ClCBr = 109.4bar(66)^@

Jun 29, 2017

Here's the initial setup:

To do this, suppose we already have the initial guess for the internal coordinates in $\text{Å}$ (which you would get by setting the central atom, $\text{C}$, at $\left(0 , 0 , 0\right)$ and using trigonometry to find each position relative to that):

${\text{^(79) Br" "" "" "" """^(35)Cl_((1))" "" "" """^(35)Cl_((2))" "" "" }}^{35} C {l}_{\left(3\right)}$
$\text{79.919 amu"" ""34.969 amu } - - - - - \to$

$x : \text{ "0.0000" "" "1.6688" "" "-0.8344" "" } - 0.8344$
$y : \text{ "0.0000" "" "0.0000" "" "" "1.4450" "" } - 1.4450$
$z : \text{ "1.9100" "-0.5900" "" "-0.5900" "" } - 0.5900$

The idea here is to first find the center of mass coordinates and shift the molecule so that the center of mass lies at $\left(0 , 0 , 0\right)$.

The center of mass coordinates are given by (where ${m}_{i}$ are the isotopic masses of each atom and ${q}_{i}$ are the $q$ coordinates of each atom):

${x}_{c m} = \frac{{\sum}_{i} {m}_{i} {x}_{i}}{{\sum}_{i} {m}_{i}} = \frac{{m}_{B r} \cdot 0 + {m}_{C} \cdot 0 + {m}_{C l} \left(1.6688 + \left(- 0.8344\right) + \left(- 0.8344\right)\right)}{{m}_{B r} + {m}_{C} + 3 {m}_{C l}}$

$= 0$

${y}_{c m} = \frac{{\sum}_{i} {m}_{i} {y}_{i}}{{\sum}_{i} {m}_{i}} = \frac{{m}_{B r} \cdot 0 + {m}_{C} \cdot 0 + {m}_{C l} \left(0 + \left(1.4450\right) + \left(- 1.4450\right)\right)}{{m}_{B r} + {m}_{C} + 3 {m}_{C l}}$

$= 0$

${z}_{c m} = \frac{{\sum}_{i} {m}_{i} {z}_{i}}{{\sum}_{i} {m}_{i}} = \frac{{m}_{B r} \cdot 1.9100 + {m}_{C} \cdot 0 + {m}_{C l} \left(3 \cdot - 0.5900\right)}{{m}_{B r} + {m}_{C} + 3 {m}_{C l}} = 0.4611$

So, we have the center of mass at $\left(0 , 0 , 0.4611\right)$. To shift the center of mass to $\left(0 , 0 , 0\right)$, we subtract $0.4611$ from each coordinate on the atoms to get:

Now that our molecule is set so that the center of mass is at $\left(0 , 0 , 0\right)$, the moments of inertia can be calculated from the inertia tensor:

$I = \left[\begin{matrix}{I}_{x x} & {I}_{x y} & {I}_{x z} \\ {I}_{x y} & {I}_{y y} & {I}_{y z} \\ {I}_{x z} & {I}_{y z} & {I}_{z z}\end{matrix}\right]$,

where ${I}_{p q}$ is the component of the inertia tensor that is a function of the $p$ and $q$ coordinates.

By diagonalizing this matrix, the components of the inertia ${I}_{x x}$, ${I}_{y y}$, and ${I}_{z z}$ can be obtained. Fortunately, we don't have to do this, because the molecule's center of mass is at the origin, i.e. ${I}_{x y} = {I}_{x z} = {I}_{y z} = 0$.

This means the tensor looks like this:

$I = \left[\begin{matrix}{I}_{x x} & 0 & 0 \\ 0 & {I}_{y y} & 0 \\ 0 & 0 & {I}_{z z}\end{matrix}\right]$

Instead, all we need to do is sum over all the atoms and evaluate:

${I}_{x x} = {\sum}_{i} {m}_{i} \left({y}_{i}^{2} + {z}_{i}^{2}\right)$

${I}_{y y} = {\sum}_{i} {m}_{i} \left({x}_{i}^{2} + {z}_{i}^{2}\right)$

${I}_{z z} = {\sum}_{i} {m}_{i} \left({x}_{i}^{2} + {y}_{i}^{2}\right)$

We obtain:

color(blue)(I_(x x)) = overbrace(79.919(0^2 +1.4489^2))^(Br) + overbrace(12(0^2+ 0.4583^2))^(C) + overbrace(34.969(0^2 + 1.0511^2) +2⋅34.969(1.445^2 +1.0511^2))^(Cl_((1)), Cl_((2)), Cl_((3)))

$=$ color(blue)("432.26 amu" cdot Å^2)

color(blue)(I_(yy)) = overbrace(79.919(0^2+1.4489^2))^(Br) + overbrace(12(0^2+0.4583^2))^(C) + overbrace(34.969(1.669^2+1.0511^2) +2⋅34.969(0.8344^2+1.0511^2))^(Cl_((1)), Cl_((2)), Cl_((3)))

$=$ color(blue)("432.30 amu" cdot Å^2)

color(blue)(I_(zz)) = overbrace(79.919(0^2+0^2))^(Br) + overbrace(12(0^2+0^2))^(C) + overbrace(34.969(1.669^2+0^2) + 2⋅34.969(0.8344^2+1.445^2))^(Cl_((1)), Cl_((2)), Cl_((3)))

$=$ color(blue)("292.13 amu" cdot Å^2)

For perspective, the evaluated tensor now looks like this:

$I = \left[\begin{matrix}432.26 & 0 & 0 \\ 0 & 432.30 & 0 \\ 0 & 0 & 292.13\end{matrix}\right]$

The three components of the moment of inertia are given on the diagonal ($x , y , z$). For reference, the calculated values from NIST were (using "CBS-Q", the complete basis set limit at the quadruple zeta level) were:

I_(x x) = I_(yy) = "435.603 amu" cdot Å^2
($\to$ 0.76% error)

I_(zz) = "293.198 amu" cdot Å^2
($\to$ 0.37% error)