Question

The 'spin only' magnetic moment of \[N{i^{2 + }}\] in aqueous solution would be : \[\left[ {Atomic{\text{ }}number{\text{ }}of{\text{ }}Ni{\text{ }} = {\text{ }}28} \right]\]
A.$\sqrt 6 BM$
B.$\sqrt {15} BM$
C.$\sqrt 2 BM$
D.$\sqrt 8 BM$

Answer 1

Hint:When we talk about magnetic moments we must know that it’s a determination of its tendency to get arranged through a magnetic field. Magnets are defined as poles which are North and South. The magnetic strength and orientation of a magnet or other object that produces a magnetic field. When we talk about objects which have magnetic moments are permanent magnets, elementary particles and loops of electric current.

Complete step-by-step answer:Magnetic moment is an entity which is a vector and is related to torque of an object to the magnetic field. The mathematical representation of the same is:
\[\tau {\text{ }} = {\text{ }}m\; \times {\text{ }}B\]
Here,
\[\tau \] is the torque acting on the dipole,
\[m\] is the magnetic moment,
\[B\] is the external magnetic field
When we talk about the current loop, the Magnetic moment is the product of the current flowing and the area, \[M{\text{ }} = {\text{ }}I{\text{ X}}\,A\]. A Magnetic Dipole comprises two unlike poles of equivalent strength and parted by a small distance. \[N{\text{ }}m{\text{ }}{T^{ - 1}}\] is the SI unit for magnetic moments. Magnetic Dipole Moment is described as the product of pole strength and the distance amidst the two poles. The distance between the two poles of a magnetic or a magnetic dipole is named as the magnet length and is given as the \[2\tau \].
Here we know that \[N{i^{2 + }}\] has \[3{d^8}\] as outer electronic configuration.
This means that it has two unpaired electrons.   The 'spin only' magnetic moment of \[N{i^{2 + }}\] in aqueous solution would be found by following equation: \[n\left( {n + 2} \right)\]
\[ = 2\left( {2 + 2} \right) = 8\;BM.\]​

Note:If an electric current I flows in a plane coil of area A then to find the torque which it will experience in a magnetic field is given by
\[\tau = IA \times B\]
Thus, the magnetic moment of the coil is given by
\[p = IA\]
Thus the unit \[{\text{A }}{{\text{m}}^2}\] is also a correct SI unit for magnetic moment, though, unless the concept of “current in a coil” needs to be emphasized in a particular context, it is perhaps better to stick to \[N{\text{ }}m{\text{ }}{T^{ - 1}}\]