Question

The rotational spectrum of HF has lines
41.9 cm-apart. Calculate the bond
length of H--F bond in HF.
​

Answer 1

sorry I don't know

I can give another answer

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Answer 2

Answer: The bond length is $3.75\times 10^{-14}cm$.

Explanation: The term Bond length is the average distance between the nuclei of two atoms which took part in bonding and became a single molecule.

To calculate the bond length, we will use the formula-

                                       $\implies I=\mu\times r^{2}$

Step 1 out of 5:

in order to take out the value of I and μ we will use the formula-

                                       $\implies B= \frac{h}{8\pi^{2}Ic}cm^{-1}$

and,

                B= rotational constant

                h= Planck's constant= $6.626\times 10^{-27}gmcm^{2s^{-2}}$

                I= moment of inertia of the molecule along the axis of rotation

                c=speed of light= $3\times 10^{10}cms^{-1}$

Step 2 out of 5:

since,

                            $\implies2B=41.9cm^{-1} \\\implies B=\frac{41.9}{2}\\\implies B= 20.95cm^{-1}$

Step 3 out of 5:

and,

                           $\implies 20.95= \frac{h}{8\pi^{2}Ic}\\\implies I= \frac{6.6261\times10^{-27}}{8\times20.95\times (3.14)^{2}\times3\times10^{10}}\\\\\implies I= 1.3366\times10^{-33}gmcm^{2}s^{-2}$

also,

Step 4 out of 5:

                          $\implies \mu=\frac{m_{1}m_{2}}{m_{1}+m_{2}}$

where, $m_1=$ mass of Hydrogen and $m_{2}=$ mass of Fluorine

                          $\\\implies \mu= \frac{1\times 19}{ 1+19}\\\implies \mu= \frac{19}{20}\\\implies \mu=0.95gm$

Step 5 out of 5:

Now, putting values in the formula, we will get-

                          $\implies I= \mu\times r^{2}\\\implies 1.3366\times 10^{-33}=0.95 \times r^{2}\\\implies r= 3.75\times 10^{-14}cm$

Hence, bond length of H--F is $3.75\times 10^{-14}cm$