Question

The transition from J = 0 to J'=1 for HCl takes place at v = 21.18 cm^-1.Calculate the bond length for 1H35Cl.

Answer 1

Answer on question #62345, Chemistry / Inorganic Chemistry

The transition from $J = 0$ to $J' = 1$ for HCl takes place at $21.18 \, \text{cm}^{-1}$. Calculate the bond length for 1H35Cl.

Solution:

The dependence of the rotational constant on bond length, $R_0$:

$$\bar{B}_0 = \frac{\hbar}{4\pi c l}$$

Where the moment of inertia $I = \mu R^2$ $o$ for a diatomic molecule.

$$\bar{B}_0 = \frac{\hbar}{4\pi\mu c R_0^2}$$

Where $\mu = m1m2/m1 + m2$

$$m_1(H) = 1.008 \, \text{amu}$$ $$m_2(Cl) = 35.453 \, \text{amu}$$ $$\mu = \frac{1.008 \, \text{amu} \times 35.453 \, \text{amu}}{1.008 \, \text{amu} + 35.453 \, \text{amu}} = 0.980 \, \text{amu}$$

Then

$$R_0 = \sqrt{\frac{\hbar}{4\pi\mu c \bar{B}_0}}$$

You need the units in kg, so you multiply by the constant, $1.66 \cdot 10^{-27} \, \text{kg/amu}$.

Finally

$$R_0 = \sqrt{ \frac{1.0546 \cdot 10^{-34} \, \text{Js}}{4\pi \times 0.980 \, \text{amu} \times 1.66 \cdot 10^{-27} \, \frac{\text{kg}}{\text{amu}} \times 3 \cdot 10^8 \, \frac{\text{m}}{\text{s}} \times 2118 \, \text{m}^{-1}} } = 9.01 \cdot 10^{-11} \, \text{m}$$ $$R_0 = 90.1 \, \text{pm}$$

Answer: 90.1 pm

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