The rotational constant of HCN is 1.4782 cm^-1 , whereas on substitution of H by deuteriun atom , DCN its value is 1.2077 cm^-1 for C12 and N14 isotopes . Calculate the moment of inertia of the molecules and obtain rC-H and rC-N from it

To solve this problem, we'll start by calculating the moment of inertia (\(I\)) of the molecules using the rotational constant (\(B\)) for both HCN and DCN. Then, we'll use the moment of inertia to calculate the bond lengths \(r_{C-H}\) and \(r_{C-N}\).

### 1. Moment of Inertia Calculation
The rotational constant \(B\) is related to the moment of inertia \(I\) by the following equation:
\[
B = \frac{h}{8\pi^2 c I}
\]
where:
- \(B\) is the rotational constant in cm\(^{-1}\),
- \(h\) is Planck's constant (\(6.62607015 \times 10^{-34} \, \text{Js}\)),
- \(c\) is the speed of light (\(2.998 \times 10^{10} \, \text{cm/s}\)),
- \(I\) is the moment of inertia in \( \text{kg} \cdot \text{m}^2 \).

We can rearrange this equation to solve for \(I\):
\[
I = \frac{h}{8\pi^2 c B}
\]

#### For HCN:
Given \(B = 1.4782 \, \text{cm}^{-1}\), the moment of inertia \(I_{\text{HCN}}\) is:
\[
I_{\text{HCN}} = \frac{6.62607015 \times 10^{-34} \, \text{Js}}{8\pi^2 \times 2.998 \times 10^{10} \, \text{cm/s} \times 1.4782 \, \text{cm}^{-1}}
\]

#### For DCN:
Given \(B = 1.2077 \, \text{cm}^{-1}\), the moment of inertia \(I_{\text{DCN}}\) is:
\[
I_{\text{DCN}} = \frac{6.62607015 \times 10^{-34} \, \text{Js}}{8\pi^2 \times 2.998 \times 10^{10} \, \text{cm/s} \times 1.2077 \, \text{cm}^{-1}}
\]

Let's calculate these values.The moments of inertia for the molecules are as follows:

- **HCN**: \( I_{\text{HCN}} = 1.8937 \times 10^{-46} \, \text{kg} \cdot \text{m}^2 \)
- **DCN**: \( I_{\text{DCN}} = 2.3178 \times 10^{-46} \, \text{kg} \cdot \text{m}^2 \)

### 2. Bond Lengths Calculation

The moment of inertia \(I\) for a diatomic molecule can be expressed as:
\[
I = \mu r^2
\]
where:
- \( \mu \) is the reduced mass of the molecule,
- \( r \) is the bond length.

The reduced mass \( \mu \) is calculated using:
\[
\mu = \frac{m_1 m_2}{m_1 + m_2}
\]
where \(m_1\) and \(m_2\) are the masses of the two atoms.

Given that \(I_{\text{HCN}}\) corresponds primarily to the C-H bond and \(I_{\text{DCN}}\) corresponds primarily to the C-D bond, we can use these moments of inertia to calculate \(r_{C-H}\) and \(r_{C-N}\).

### Masses:
- \( m_{\text{C}} \) (Carbon): \( 12 \, \text{u} \)
- \( m_{\text{H}} \) (Hydrogen): \( 1 \, \text{u} \)
- \( m_{\text{D}} \) (Deuterium): \( 2 \, \text{u} \)
- \( m_{\text{N}} \) (Nitrogen): \( 14 \, \text{u} \)

1 atomic mass unit (u) = \( 1.660539 \times 10^{-27} \, \text{kg} \).

Let's calculate the bond lengths \(r_{C-H}\) and \(r_{C-N}\).The bond lengths calculated are:

- **C-H bond length (\(r_{C-H}\))**: \(3.515 \times 10^{-10} \, \text{m}\) or \(3.515 \, \text{Å}\)
- **C-D bond length (\(r_{C-D}\))**: \(2.853 \times 10^{-10} \, \text{m}\) or \(2.853 \, \text{Å}\)

These values represent the bond lengths in the HCN and DCN molecules, respectively.

### Summary
- The moment of inertia for HCN is \(1.8937 \times 10^{-46} \, \text{kg} \cdot \text{m}^2\).
- The moment of inertia for DCN is \(2.3178 \times 10^{-46} \, \text{kg} \cdot \text{m}^2\).
- The C-H bond length is \(3.515 \, \text{Å}\).
- The C-D bond length is \(2.853 \, \text{Å}\).

Would you like to delve into more details, or do you have any questions?

### 5 Related Questions:
1. How does the substitution of hydrogen with deuterium affect the bond length?
2. What role does the reduced mass play in determining the moment of inertia?
3. How do the bond lengths affect the vibrational frequencies of molecules?
4. Can this approach be used for other diatomic molecules?
5. How would isotopic substitution affect the rotational spectrum of a molecule?

### Tip:
The bond length and rotational constant are inversely related; a larger moment of inertia corresponds to a longer bond length and a smaller rotational constant.

Learn how to calculate the moment of inertia and bond lengths for HCN and DCN molecules using their respective rotational constants. Understand the impact of isotopic substitution on molecular properties. Detailed step-by-step solution provided.

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