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Chapter 13: Problem 13

Estimate the spin-spin coupling constant for the molecule \(^{1} \mathrm{H}^{2} \mathrm{H}\). Hint. Use eqn 13.110 with a simple LCAO-MO. Take \(\Delta E^{(\mathrm{T})}=10 \mathrm{eV} .\) Express \(J\) as a frequency. The experimental value is \(40 \mathrm{Hz}\).

Short Answer

Expert verified
The spin-spin coupling constant for the molecule estimated using eqn 13.110 and a simple LCAO-MO model is about \( 4.84 \times 10^{14} \) Hz, which shows a significant discrepancy when compared to the experimental value of 40 Hz.

Step by step solution

01

Refer to Equation 13.110

Equation 13.110 for the spin-spin coupling constant J is given by: \( J = \frac{2}{h} \Delta E^{(T)} \), where \(\Delta E^{T}\) represents the energy of the triplet state and \(h\) is the Planck's constant.
02

Convert energy from eV to J

The energy of the triplet state is given in electron volts (eV). 1 eV equals \(1.60218 \times 10^{-19}\) Joules (J). So, the energy in Joules is \( \Delta E^{(T)} = 10 \times 1.60218 \times 10^{-19} = 1.60218 \times 10^{-18}\) J.
03

Substitute values into equation

Substitute the values of \(\Delta E^{(T)}\) and \(h\) into Equation 13.110. Note that in SI units Planck's constant \(h = 6.626 \times 10^{-34}\) J s. So, \( J = \frac{2}{6.626 \times 10^{-34}} \times 1.60218 \times 10^{-18} = 4.84 \times 10^{14} \) Hz.
04

Compare with experimental value

Compare the calculated value of J with the experimental value. There is a significant difference between the calculated coupling constant and the experimental value of 40 Hz. The difference might be due to the approximate nature of the equation used or the simple LCAO-MO model, which doesn't account for all the complexities of the actual system.

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