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MCQ Question:

Vibrational frequency (f) of a diatomic molecule is given by

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where 𝑘 is the force constant and 𝜇 is the reduced mass. For a diatomic molecule (AX), the reduce mass is given by

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where 𝑚𝐴 and 𝑚𝑥 are mass of atom A and atom B respectively.

If vibrational frequencies (in wavenumber terms) of Cl2 and F2 in are 915 cm^-1 and 525 cm^-1 respectively, what is the ratio between the corresponding force constants of Cl2 and F2 (Cl2 : F2)?

(a) 5.7

(b) 6.0

(c) 6.3

(d) 6.7

(e) 7.1

The answer from MCQ is A) 5.7. I tried writing two equations for Cl2 and F2 and then dividing them but I got an answer of 1.74. How can I get 5.7?

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For diatomic molecules, $\mu = m/2$. This is $9.5$ for fluor and $17.75$ for chlorine. Now $\pu{k}$ is given by : $\pu{k = \mu* 4\pi^{2} f^2}$. If I use the index $1$ for chlorine and $2$ for fluorine, I can write : $$\pu{\frac{k_1}{k_2} = \frac{\mu_1 f_1^2}{\mu_2 f_2^2} = \frac{17.75}{9.5} * (\frac{915}{525})^2 = 5.67}$$

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  • $\begingroup$ How did you get the reduced mass as m/2. $\endgroup$
    – Jane902
    Commented Feb 13, 2023 at 9:40
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    $\begingroup$ Use the given formula for $\mu$. Replace $\pu{m_x}$ by $\pu{m_a}$. You will obtain :$\pu{\mu = \frac{m_a*m_a}{2m_a} = m_a/2}$ $\endgroup$
    – Maurice
    Commented Feb 13, 2023 at 16:08

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