Spin Quantum number of $ ^{13}C $ NMR is:

Question

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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.

Answer

$ \frac{1}{4} $

Answer

$ \frac{1}{2} $

Answer

$ \frac{3}{2} $

Answer

$ \frac{1}{3} $

Spin Quantum number of \( ^{13}C \) NMR is:

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The Correct Option is B

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