Question

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

• A. \(\frac 14\)
• B. \(\frac 12\)
• C. \(\frac 32\)
• D. \(\frac 13\)

Answer 1

The Correct Option is B

The correct option is (B): \(\frac 12\).

Answer 2

The question asks for the nuclear spin quantum number of carbon-13, the isotope of carbon used in NMR spectroscopy. Let's check each proposed value against how nuclear spin actually arises.

1. \(\frac{1}{4}\): A spin quantum number of one quarter does not occur for any naturally occurring nucleus, since nuclear spin values are restricted to whole-number multiples of one half.
2. \(\frac{1}{2}\): Carbon-13 has an odd mass number, 6 protons plus 7 neutrons, which gives it a net unpaired nucleon and a spin quantum number of one half. This non-zero spin is exactly what makes carbon-13 NMR active and detectable, unlike the far more abundant but NMR-silent carbon-12, which has spin zero.
3. \(\frac{3}{2}\): A spin of three halves is seen in nuclei with more than one unpaired nucleon contributing angular momentum, such as sodium-23 or chlorine-35, not in carbon-13.
4. \(\frac{1}{3}\): Like one quarter, this is not a value nuclear spin can take, since spin quantum numbers are always whole-number multiples of one half.

Only one half matches the odd nucleon count of carbon-13 and its known behavior as an NMR active nucleus.

The correct answer is \(\frac{1}{2}\).