Spin Quantum number of \( ^{13}C \) NMR is:
Show Hint
- \( \frac{1}{4} \)
- \( \frac{1}{2} \)
- \( \frac{3}{2} \)
- \( \frac{1}{3} \)
The Correct Option is B
Solution and Explanation
The spin quantum number (\( I \)) of a nucleus determines whether it is NMR active and affects its behavior in a magnetic field. The nucleus of Carbon-13 (\( ^{13}C \)) has a spin quantum number of \( \frac{1}{2} \).
- Carbon-12 (\( ^{12}C \)) has a spin quantum number of 0 and is NMR inactive. - Carbon-13 (\( ^{13}C \)) is NMR active because it has an odd number of neutrons, resulting in a nonzero spin quantum number \( I = \frac{1}{2} \).
- Nuclei with \( I = \frac{1}{2} \) (such as \( ^{1}H \) and \( ^{13}C \)) are the most commonly studied in NMR spectroscopy because they exhibit simple splitting patterns and good sensitivity.
Why Other Options Are Incorrect: - (A) \( \frac{1}{4} \), (C) \( \frac{3}{2} \), (D) \( \frac{1}{3} \): These values do not correspond to the spin quantum number of \( ^{13}C \). Most nuclei with nonzero spin have values of \( I = \frac{1}{2}, 1, \frac{3}{2}, 2, \) etc., but \( ^{13}C \) specifically has \( I = \frac{1}{2} \).
Thus, the correct answer is \( \frac{1}{2} \), confirming that \( ^{13}C \) is NMR active and useful in structural determination using \( ^{13}C \) NMR spectroscopy.
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