Question

Which among the following nuclei can exhibit NMR spectrum?

• A. ²⁸Si
• B. ¹⁶O
• C. ¹⁴C
• D. ¹⁴N

Answer 1

The Correct Option is D

The question is about identifying which nuclei can exhibit a Nuclear Magnetic Resonance (NMR) spectrum. NMR spectroscopy is a powerful analytical technique used to determine the structure of molecules. Whether a nucleus will exhibit an NMR spectrum depends on its spin quantum number. Here's the step-by-step explanation:

1. NMR spectroscopy involves the interaction of nuclear spins with an applied magnetic field. For a nucleus to be NMR active, it must have a non-zero nuclear spin quantum number (I).
2. The spin quantum number (I) is determined by the number of protons and neutrons in the nucleus:
   • If both the number of protons and the number of neutrons are even, I = 0 (NMR inactive).
   • If both are odd, or one is odd and the other is even, I is non-zero (NMR active).
3. Let's evaluate the options:
   • \(^{28}\text{Si}\): Silicon-28 has 14 protons and 14 neutrons, both even, so I = 0 (NMR inactive). 
   • \(^{16}\text{O}\): Oxygen-16 has 8 protons and 8 neutrons, both even, so I = 0 (NMR inactive).
   • \(^{14}\text{C}\): Carbon-14 has 6 protons and 8 neutrons, both even, so I = 0 (NMR inactive).
   • \(^{14}\text{N}\): Nitrogen-14 has 7 protons and 7 neutrons, both odd, so I = 1 (NMR active).
4. Conclusion: \(^{14}\text{N}\) has a non-zero spin quantum number and can exhibit an NMR spectrum.