1) since the moment of inertia for the end - over - end rotation of a polyatomic linear molecule is considerably greater than that of diatomic molecule, the B value will be much smaller and the spectral lines are more closely spaced. Thus the B values for a diatomic molecules are about 10 cm⁻¹ and while for triatomic molecules they can be 1 cm⁻¹ or less and for larger molecules smaller still.
2) The molecule must possess a dipole moment to be microwave active. Thus OCS would be microwave active while OCO will not be active. Isotopic substitution will not lead to a dipole moment since the bond length and atomic charges will not be changed. Thus ¹⁶OC¹⁸O is microwave inactive.
3) A non-cyclic polyatomic molecule with N natural atoms will have N-1 bond distances to be determined. Thus in triatomic OCS molecule there's CO distance r꜀ₒ and the CS distance r꜀ₛ. But there's only one moment of inertia to be determined from the spectrum. From one value, two unknown r꜀ₒ and r꜀ₛ cannot be determined. But this problem will be overcome by isotopic substitution.
Consider the rotation of OCS. Below the figure shows the molecule where r₀, r꜀ and rₛ represent the distance of the atoms from the centre of gravity C.
m₀r₀ + m꜀r꜀ = mₛrₛ ................................ (1)
The moment of inertia is,
I = m₀r₀² + m꜀r꜀² + mₛrₛ² ......................(2)
and
r₀= r꜀ₒ + r꜀ ; rₛ = r꜀ₛ - r꜀ ........................ (3)
Where r꜀ₒ and r꜀ₛ are the bond lengths. Substitute equation (3) in equation (1) and we get,
m₀ ( r꜀ₒ + r꜀) + m꜀r꜀ = m꜀ (r꜀ₛ - r꜀)
m₀ r꜀ₒ + mₒr꜀ + m꜀r꜀ = mₛr꜀ₛ - mₛr꜀
( m₀ + m꜀ + mₛ) r꜀ = mₛr꜀ₛ - mₒr꜀ₒ
Mr꜀ = mₛr꜀ₛ - mₒr꜀ₒ ................................ (4)
Where M is the total mass of th e molecule, substituting equation (3) in equation (2)
I = m₀ (r꜀ₒ + rc)² + m꜀r꜀² + mₛ(r ꜀ₛ - r꜀)²
= m₀r꜀ₒ² + m₀r꜀² + 2 m₀ r꜀ₒ r꜀ +m꜀r꜀² + mₛ r꜀ₛ² + mₛr꜀² −2mₛr꜀ₛ r꜀
= Mr꜀² + 2r꜀ ( m₀ r꜀ₒ - mₛ r꜀ₛ) + mₒ r² ꜀ₒ + mₛ r꜀ₛ² ............................(5)
From equation (4)
r꜀ = ( mₛ r꜀ₛ - m₀ r꜀ₒ) /M .................. (6)
Substitute the value of r꜀ in equation (5)
I = M (mₛm꜀ₛ − m₀m ꜀ₒ ) +
M²
2 (mₛr꜀ₛ − m₀r꜀ₒ)(m₀r꜀ₒ− mₛ r꜀ₛ) +
M
mₒr꜀ₒ² + mₛr꜀ₛ²
= m₀ r꜀ₒ² + mₛ r꜀ₛ² + [(m₀ r꜀ₒ)² − 2mₛ r꜀ₛ m₀ r꜀ₒ - 2( m꜀ ᵣ꜀ₛ)² − 2(m₀ r꜀ₒ )² +2m₀ r꜀ₒ mₛ r꜀ₛ ]/ M
I = m₀ r꜀ₒ² + mₛ r꜀ₛ - [ m₀ r꜀ₒ - mₛ r꜀ₛ ]/M
I = m₀ r꜀ₒ² + mₛ r ꜀ₛ² − [ m₀ r ꜀ₒ − mₛ r ꜀ₛ ]²/M
................(7)
Considering now the isotopic molecule
¹⁸OCS, we may write m₀ for m₀ in the above equation.
I' = m₀ r꜀ₒ² + mₛ r꜀ₛ² −[ m₀ r꜀ₒ − mₛ r꜀ₛ ]²/M' .................(8)
Using equation (7) and (8) we can solve for r꜀ₒ and r꜀ₛ provided we have obtained the value of l" from the microwave spectrum of the isotopic molecule.
Very nice
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