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
Approach Solution - 1
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.
Approach Solution -2
Nuclear spin quantum number depends on the balance of protons and neutrons inside the nucleus, so this question can be worked out by counting nucleons in carbon-13 rather than simply recalling a fact:
- \( \frac{1}{4} \): Nuclear spin quantum numbers follow a set of allowed values built from whole or half-integer steps, such as 0, \( \frac{1}{2} \), 1, \( \frac{3}{2} \), and so on. A quarter-integer value like \( \frac{1}{4} \) does not appear in this allowed sequence for any known nucleus, so it cannot be correct.
- \( \frac{1}{2} \): Carbon-13 has 6 protons and 7 neutrons, giving it an odd mass number of 13. Nuclei with an odd mass number and an odd number of nucleons in one of the proton or neutron counts, like carbon-13's combination of even protons and odd neutrons, typically carry a nuclear spin of \( \frac{1}{2} \), the same simple spin state seen in the hydrogen nucleus.
- \( \frac{3}{2} \): A spin of \( \frac{3}{2} \) is seen in nuclei with a more complex arrangement of unpaired nucleons, such as sodium-23 or chlorine-35, not in carbon-13, which has a simpler nucleon pairing pattern.
- \( \frac{1}{3} \): Like \( \frac{1}{4} \), a value of \( \frac{1}{3} \) falls outside the set of physically allowed nuclear spin values, which always progress in steps of one-half, so it cannot describe any real nucleus.
Working from carbon-13's actual nucleon count, an even number of protons paired with an odd number of neutrons gives a simple half-integer spin, matching the value seen experimentally in NMR spectroscopy.
So the correct answer is \( \frac{1}{2} \).
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