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)^@#

DISCLAIMER: LONG ANSWER!

Here's the initial setup:

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

#""^(79) Br" "" "" "" """^(35)Cl_((1))" "" "" """^(35)Cl_((2))" "" "" """^(35)Cl_((3))#
#"79.919 amu"" ""34.969 amu " ------>#

#x:" "0.0000" "" "1.6688" "" "-0.8344" "" "-0.8344#
#y:" "0.0000" "" "0.0000" "" "" "1.4450" "" "-1.4450#
#z:" "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 #(0,0,0)#.

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_(cm) = (sum_(i) m_i x_i)/(sum_i m_i) = (m_(Br)cdot0 + m_Ccdot0 + m_(Cl)(1.6688 + (-0.8344) + (-0.8344)))/(m_(Br) + m_C + 3m_(Cl))#

#= 0#

#y_(cm) = (sum_(i) m_i y_i)/(sum_i m_i) = (m_(Br)cdot0 + m_Ccdot0 + m_(Cl)(0 + (1.4450) + (-1.4450)))/(m_(Br) + m_C + 3m_(Cl))#

#= 0#

#z_(cm) = (sum_(i) m_i z_i)/(sum_i m_i) = (m_(Br)cdot1.9100 + m_Ccdot0 + m_(Cl)(3*-0.5900))/(m_(Br) + m_C + 3m_(Cl)) = 0.4611#

So, we have the center of mass at #(0,0,0.4611)#. To shift the center of mass to #(0,0,0)#, 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 #(0,0,0)#, the moments of inertia can be calculated from the inertia tensor:

#I = [(I_(x x),I_(xy),I_(xz)),(I_(xy),I_(yy),I_(yz)),(I_(xz),I_(yz),I_(zz))]#,

where #I_(pq)# 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_(yy)#, and #I_(zz)# 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_(xy) = I_(xz) = I_(yz) = 0#.

This means the tensor looks like this:

#I = [(I_(x x), 0, 0),(0, I_(yy), 0),(0, 0, I_(zz))]#

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

#I_(x x) = sum_i m_i(y_i^2 + z_i^2)#

#I_(yy) = sum_i m_i(x_i^2 + z_i^2)#

#I_(zz) = sum_i m_i(x_i^2 + y_i^2)#

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 = [(432.26, 0, 0),(0, 432.30, 0),(0, 0, 292.13)]#

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#
(#-># #0.76%# error)

#I_(zz) = "293.198 amu" cdot Å^2#
(#-># #0.37%# error)