Question

Magnetic moment of [Cu(NH3)4]2+ ion is
A . 1.414
B . 1.73
C . 2.23
D . 2.38

Answer 1

Hint: In this question we have to use the concept of the magnetic moment of a coordination compound to find out the magnetic moment of [Cu(NH3)4]2+ ion.
The magnetic field created by a magnet or other object, combined with its strength and direction, can be referred to as the magnetic moment. The magnetic moment is measured in terms of the Bohr magnetron, abbreviated as BM.


Complete answer:Coordination metals complexes are magnetic due to their unpaired electrons. The last electrons are in the d orbitals, thus this magnetism must be caused by the presence of unpaired d electrons. The quantity of unpaired electrons in a molecule controls its magnetic properties, and the electronic spin produces magnetism. The motion of magnetic or electric charges results in the creation of a magnetic field. These magnetic field lines of force possess several distinct characteristics. There are primarily three forms of magnetism displayed by the complexes: ferromagnetism, paramagnetism, and diamagnetism.

The magnetic moment of a complex having n-unpaired electrons is given by:
$\sqrt{n(n+2)}$, where n is the number of unpaired electrons.
With one unpaired electron, the hybridization of [Cu(NH3)4]2+ ion is sp2d. Therefore, the magnetic moment of [Cu(NH3)4]2+ ion is $\sqrt{1(1+2)}=1.73$BM.

The correct answer is B.

Note: The magnetic moment, μ is used to determine the degree of paramagnetism. The compound's paramagnetism increases with the magnitude of μ.
Spin and orbital angular momentum both contribute to a magnetic moment. The contribution from orbital angular momentum may be quenched in a non-spherical environment. The spin-only magnetic moment, however, endures in every scenario and is correlated with the overall quantity of unpaired electrons.