Question

A compound of vanadium has a magnetic moment of 1.73 BM work out electronic configuration of Vanadium ions in the compound is $[Ar]3d^x$. What is $x$?

Answer 1

Hint: Initially you must know what is the role of magnetic momentum in electronic configuration. Also, you must have configuration knowledge of elements and what is the formula of magnetic momentum.

Complete answer:
A moment of a magnet is typically a vector quantity, with a magnitude and a direction. An electron has an electron dipole moment, generated by the electron's intrinsic spin property, making it an electrical charge in motion. There are many alternative magnetic behaviours including paramagnetism, diamagnetism, and ferromagnetism.
Vanadium is one of the elements in the periodic table with the symbol V and number 23. It is a hard, silvery-grey, malleable transition metal. The elemental metal is never found in nature, but once isolated artificially the formation of an oxide layer (passivation) somewhat stabilizes the free metal against further oxidation. Vanadium has (Ar) $4s^2 3d^3$ configuration.
Magnetic moment,
Given,
$=1.73 \\
=3$
So, $n=1$
Unpaired electrons $= 1$
i.e., $4s^03d^1$configuration is formed. Therefore 3 electrons are removed
Now the ion is $V^{+3}$.
So, x is considered as 3 here.
So, the value of x is 3 i.e., $x=3$.

Note: The electronic configuration of a component may be a symbolic notation of the way during which the electrons of its atoms are distributed over different atomic orbitals. While writing electron configurations, a uniform notation is followed during which the energy state and therefore the sort of orbital are written first, followed by the number of electrons present in the orbital written in superscript. For example, the electronic configuration of carbon (atomic number: 6) is $1s^2 2s^2 2p^2$.