Question

Chose the right statement a. 1H and 13C nuclei are unique in their ability to exhibit the NMR phenomenon Nele with b. 1H and 13C nuclei are not unique in their ability to exhibit the NMMR phenomenon All c. 1H and 13C nuclei are not unique in their ability to exhibit the NMR phenomenon Ail Iuse d. None of them. an odd number of protons (3 2H, 14N 19, 31P) will not be applicable with an even number of protons are applicable with an odd number of protons (1N, 2N, 14N, 19F, 31P)

Answer 1

2.

For a nuclei to exhibit NMR phenomenon, or in other words, be NMR active, it must have non-zero nuclear spin.

The nuclear spin originates from the protons and neutrons in the nucleus which are again in turn made up of up and down quarks.

As a rule of thumb it is known that when the number of protons and neutrons are even, the net spin is zero, Hence, the nuclei is not NMR active. For example, is NMR inactive, as it has 6 protons and 6 neutrons giveing it a net nuclear spin of 0.

For a nuclei to exhibit NMR phenomenon, or in other words, be NMR active, it must have non - zero nuclear spin. The nuclear spin originates from the protons and neutrons in the nucleus which are again in turn made up of up and down quarks. As a rule of thumb it is known that when the number of protons and neutrons are even, the net spin is zero, Hence, the nuclei is not NMR active. For example, ^12 C is NMR inactive, as it has 6 protons and 6 neutrons giving it a net nuclear spin of 0. In any other case, where atleast one of proton or neutron is in odd number, then the spin of the nuclei is non - zero. Hence, these kind of nuclei become NMR active. For example, ^14 N has a 7 neutrons and 7 protons and a net nuclear spin 1. When the nuclear spin is I > 1/2, we encounter a problem of line broadening in spectra which is not desirable. Because of this these nuclei are not very useful in NMR. But nonetheless they are NMR active. A nuclei like ^15 N on the other hand is an excellent candidate for NMR, it has 8 neutrons and 7 protons with a net spin of +1/2. Now, coming to our options

In any other case, where atleast one of proton or neutron is in odd number, then the spin of the nuclei is non-zero. Hence, these kind of nuclei beceom NMR active.

For example, $^{14}N$ has a 7 neutrons and 7 protons and a net nuclear spin 1. When the nuclear spin is

$I>\frac{1}{2}$ , we encounter a problem of line broadening in spectra which is not desirable. Because of this these nuclei are not very useful in NMR. But nonetheless they are NMR active.

A nuclei like $^{15}N$ on the other hand is an excellent candidate for NMR, it has 8 neutrons and 7 protons with a net spin of +1/2.

Now, coming to our options

a.

are certainly not unique as there are plenty of other nuclei that are NMR active. Also nuclei with odd number of protons() are perfectly capable of being NMR active as we saw above.

Hence, the statement is incorrect.

b.

All nuclei with even number of protons does not exhibit NMR phenomenon. As we saw in case of , it has 6 protons(even) and 6 neutrons with a net spin of 0 which makes it NMR inactive.

For a nuclei to exhibit NMR phenomenon, or in other words, be NMR active, it must have non - zero nuclear spin. The nuclear spin originates from the protons and neutrons in the nucleus which are again in turn made up of up and down quarks. As a rule of thumb it is known that when the number of protons and neutrons are even, the net spin is zero, Hence, the nuclei is not NMR active. For example, ^12 C is NMR inactive, as it has 6 protons and 6 neutrons giving it a net nuclear spin of 0. In any other case, where atleast one of proton or neutron is in odd number, then the spin of the nuclei is non - zero. Hence, these kind of nuclei become NMR active. For example, ^14 N has a 7 neutrons and 7 protons and a net nuclear spin 1. When the nuclear spin is I > 1/2, we encounter a problem of line broadening in spectra which is not desirable. Because of this these nuclei are not very useful in NMR. But nonetheless they are NMR active. A nuclei like ^15 N on the other hand is an excellent candidate for NMR, it has 8 neutrons and 7 protons with a net spin of +1/2. Now, coming to our options

Hence, the statement is incorrect.

c. All nuclei with odd number of proton generally have a net nuclear spin greater than 0. Hence, they are NMR active. However some of them can give broad spectra like $^{14}N$ , because of the presence of quadrupole moment in them.

Inspite of that, the statement is correct.

Hence, the correct answer is option c.