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The total spin and magnetic moment for the atom with atomic number 7 are:
A. \[ \pm 3,\sqrt 3 \,{\rm{BM}}\]
B. \[ \pm 1,\sqrt 8 \,{\rm{BM}}\]
C. \[ \pm \dfrac{3}{2},\sqrt {15} \,{\rm{BM}}\]
D. \[0,\sqrt 8 \,{\rm{BM}}\]

Answer
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Hint: The spin quantum number decides the spin of an electron in an atom. The magnetic moment is decided by the magnetic quantum number. Here, the magnetic moment and spin of the atom of the atomic number 7 are to be calculated using the formulas.

Formula used:The formula of spin is,
Total spin=\[ \pm \dfrac{1}{2} \times {\rm{Number}}\,{\rm{of}}\,{\rm{unpaired}}\,{\rm{electrons}}\]
The formula of magnetic moment is,
Magnetic moment=\[\sqrt {n\left( {n + 2} \right)} \]
Here, n represents the count of unpaired electrons.

Complete Step by Step Answer:
To calculate the spin and magnetic moment of the atom, we must know how many unpaired electrons are present in the atom. So, first, we have to write the electronic configuration of the atom which possesses the atomic number 7. The atomic number is 7 for the nitrogen atom.
The nitrogen atom has the following electronic configuration.
\[1{s^2}2{s^2}2{p^3}\]
So, we find that in the p orbital three electrons are present and are unpaired. So, now we can calculate spin.
Total number of spin=\[ \pm \dfrac{1}{2} \times 3 = \pm \dfrac{3}{2}\]
Now, we have to calculate the magnetic moment.
Magnetic moment=\[\sqrt {3\left( {3 + 2} \right)} = \sqrt {15} \,{\rm{BM}}\]
Therefore, option C is right.

Additional Information: The quantum numbers comprise four numbers namely principal, azimuthal, spin, and magnetic quantum number that decides an electron's position, energy, and orbital.

Note: The principal quantum number decides the principal shell of the electrons. The azimuthal quantum number gives the subshell of an electron. The magnetic quantum number gives the energy of the subshell and angular momentum and the spin quantum number gives the spin of an electron. And two electrons of an atom never possess the same quantum number set.