Question

Chloro compound of vanadium has only spin magnetic moment of 1.73 BM. This vanadium chloride has the formula:
[Atomic number of V=23]
A.\[VC{l_2}\]
B.$VC{l_4}$
C.$VC{l_3}$
D.$VC{l_5}$

Answer 1

Hint:The spin magnetic moment is the magnetic moment generated by the spin of the electron. The elementary spin in an electron is ½ Fermion. The commonly used value to measure the Magnetic moment is Bohr Magneton (BM).

Complete step by step answer:
The natural unit of expressing the magnetic moment is the Bohr Magneton and it is the physical constant as well. It is determined by the formula:
${\mu _B} = \dfrac{{eh}}{{2me}}$
Where e = charge
h= Plank’s constant
m= rest mass
c= speed of light
Thus for the given chloro compound of Vanadium the spin magnetic moment is 1.73 BM and the atomic number of V is 23, thus the number of unpaired electrons need to be calculated in order to know its valency .Therefore:

Magnetic Moment=$\sqrt {n(n + 2)} $
Where n= number of unpaired electrons
$\sqrt {n(n + 2)} = 1.73$
Thus n=1
The electronic configuration of the V is $[Ar]4{s^2}3{d^3}$ since its atomic number is 23. Thus because of the one unpaired electron it can only be ${V^{4 + }}$ so the 4 Cl atoms can only combine with the V ion, Thus the formula of the compound will be $VC{l_4}$
Thus the option B is the correct answer.
A magnetic moment of the charged particle can be generated by two different ways, first by the moving electric current and by the inherent rotation or the spin of the magnetic moment.

Note:
The non- classical property of the elementary particles is the spin, it is just the real momentum of a material object. It is the constitution of the object about its rotation axis.