Question

Which of the following species has a magnetic moment value of $3.87BM$?
A.$F{e^{3 + }}$
B.$C{r^{3 + }}$
C.$C{o^{2 + }}$
D.$A{u^{3 + }}$

Answer 1

Hint: We have to know that the transition metals are d block elements which also contain lanthanides and actinides series as well. Transition metals generally show color due to the presence of unpaired electrons. These metals have partially filled or completely filled d orbitals which also lead to cation after losing electrons from the element.

Complete answer:
We will look at all the options to answer this question but first we need to know what magnetic moment is? Magnetic moment has contributions from spin and orbital angular momentum. A non spherical environment may lead to quenching of the contribution from orbital angular momentum. However, the spin-only magnetic moment survives in all cases and is related to the total number of unpaired electrons. The formula that is used to calculate magnetic moment is represented below
Magnetic moment\[ = \sqrt {n(n + 2)} BM\]
Where n= number of unpaired electrons.
Option A) this option is incorrect as \[F{e^{3 + }} = 3{d^3}\] it has $3$ unpaired electron so magnetic moment is not $3.87BM$.
Option B) this option is incorrect as \[C{r^{2 + }} = 3{d^4}\] it has $4$ unpaired electrons so the magnetic moment is not $3.87BM$.
Option C) this option is correct as \[C{o^{2 + }} = 3{d^5}\] it has $5$ unpaired electrons so magnetic moment can be calculated as given below:
\[ = \sqrt {5(5 + 2)} = \sqrt {5 \times 7} \]
$ = \sqrt {35} = 3.87BM$

Therefore, option (C) is the correct option.

Note:
We need to remember that if ions or molecules are considered instead of free atoms, it is expected that the orbital motions of the electrons are tied into the nuclear configuration of the molecule or the ion so tightly that they are unable to line up with the applied magnetic field. Hence the magnetic moment contribution due to the orbital motion is negligible.