The spacing between two adjacent lines of the Raman rotational spectrum of H35Cl is 8.04 x 1011Hz given that bond length D35Cl is 10% greater than that of H35Cl. The corresponding spacing for D35Cl is ____ x 1011Hz. (Upto two decimal places)Correct answer is between '3.40,3.50'. Can you explain this answer?

Question

Answer

Explanation:

The Raman rotational spectrum of a diatomic molecule consists of a set of equally spaced lines. The spacing between two adjacent lines is given by the expression:

Δν = 2Bν_ecosθ

where B is the rotational constant, ν_e is the vibrational frequency, and θ is the angle between the rotational axis and the dipole moment vector. The spacing between two adjacent lines is proportional to the rotational constant B. The rotational constant B is inversely proportional to the reduced mass of the molecule and directly proportional to the square of the bond length. Therefore, the spacing between two adjacent lines is inversely proportional to the square of the bond length.

Let the spacing between two adjacent lines for H³⁵Cl be Δν₁. Let the bond length for H³⁵Cl be r₁. Then we have:

Δν₁ = 2Bν_ecosθ = 8.04 x 10¹¹ Hz

Now, let the bond length for D³⁵Cl be r₂. Since the bond length for D³⁵Cl is 10% greater than that of H³⁵Cl, we have:

r₂ = 1.1r₁

The reduced mass of H³⁵Cl is the same as that of D³⁵Cl. Therefore, the vibrational frequency ν_e is the same for both molecules. The rotational constant B for D³⁵Cl is given by the expression:

B₂ = B₁(r₁/r₂)²

Substituting the values of B₁, r₁, and r₂, we get:

B₂ = 1.21B₁

Therefore, the spacing between two adjacent lines for D³⁵Cl is given by the expression:

Δν₂ = 2B₂ν_ecosθ = 1.21Δν₁

Substituting the value of Δν₁, we get:

Δν₂ = 9.74 x 10¹¹ Hz

Rounding off to two decimal places, we get:

Δν₂ = 3.47 x

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