Question

A compound of vanadium has a magnetic moment of $1.73BM$ . The electronic configuration of vanadium ion in the compound is:
A: $\left[ {Ar} \right]3{d^2}$
B: $\left[ {Ar} \right]3{d^1}4{s^0}$
C: $\left[ {Ar} \right]3{d^0}$
D: $\left[ {Ar} \right]3{d^0}4{s^1}$

Answer 1

Hint: Some compounds have magnetic properties which arise due to spin and orbital angular momentum of the electrons contained in that compound. Unit used for defining magnetic moments is bohr magneton $\left( {BM} \right)$ .
Formula used: Magnetic moment$ = \sqrt {n\left( {n + 2} \right)} $
Where, $n$ is the number of unpaired electrons.

Complete step by step answer:Some chemical compounds show magnetic behavior due to spin and orbital angular momentum. An orbital can accommodate a maximum of two electrons that too with opposite spin. In this question we have given the magnetic moment of a compound of vanadium and we have to find the electronic configuration of vanadium. Magnetic moment is calculated by the formula:
Magnetic moment$ = \sqrt {n\left( {n + 2} \right)} $
Where, $n$ is the number of unpaired electrons.
The magnetic moment is $1.73BM$. Using this formula for a given compound we can calculate the number of unpaired electrons in vanadium. So,
$1.73 = \sqrt {n\left( {n + 2} \right)} $
Squaring both sides:
$3 = n\left( {n + 2} \right)$
${n^2} + 2n - 3 = 0$
This can be written as,
$\left( {n + 3} \right)\left( {n - 1} \right) = 0$
Solving this we get,
$n = 1$ (as number of electrons can never be negative in number)
So, the number of unpaired electrons in a vanadium is one. Atomic number of vanadium is $23$. Now, we will write actual configuration of vanadium, that is:
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^3}4{s^2}$
Configuration till $3{p^6}$ can be written as $\left[ {Ar} \right]$, as argon has stable electronic configuration. So the electronic configuration of vanadium is:
$\left[ {Ar} \right]3{d^3}4{s^2}$
We can see clearly that there are three unpaired electrons in vanadium but in the given compound of vanadium we have only one outermost shell electron. This means two electrons from the p orbital need to be removed. For this we have to remove two electrons from $s$ orbital as well because ${4^{th}}$ shell is the outermost shell, while making ionic compound electrons from this shell will be transferred first. Now, there is no electron in $s$ orbital. Also there is only one electron in $p$ orbital (calculated above for the given compound). This means now there is one electron in $p$ orbital and zero in $s$ orbital. So, electronic configuration of vanadium in this compound is:
$\left[ {Ar} \right]3{d^1}4{s^0}$

So, the correct answer is option B.

Note:
Remember while losing electrons, the electron which will be in the outermost shell will be lost first and then from the inner orbitals. While calculating magnetic moment only the electrons which are unpaired are used. Paired electrons are not used for calculation of magnetic moment.