Question

What is the magnetic moment of \[{V^{ + 4}}\]?

Answer 1

Hint: Magnetic moments of the transition metal ion are of great help while determining the stereochemistry and oxidation of central metal ions in the coordination complex. Magnetic moment expresses the strength of the magnetic field and the orientation of a magnet or other objects producing magnetic fields.

Formula used:
$[mu = \sqrt {n(n + 2)} ]$
\[\mu \]= magnetic dipole moment of the metal ion
\[n = \] number of unpaired electrons in the metal ion

Complete step by step answer:
The transition metal, Vanadium denoted by the symbol V, has atomic number 23. It does not exist freely, but exists in the compound form in nature.
The electronic configuration of vanadium is given below.
\[\left[ {Ar} \right]3{d^3}4{s^2}\]or \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^3}4{s^2}\]

Step 1: To calculate the magnetic moment of \[{V^{ + 4}}\], we will use the formula given above. We first need to find out the number of unpaired electrons present in the \[{V^{ + 4}}\] ion.
Let us first write the electronic configuration of \[{V^{ + 4}}\]ion.
The \[{V^{ + 4}}\] ion has been formed by losing 4 electrons, so electronic configuration of \[{V^{ + 4}}\] will be:
\[\left[ {Ar} \right]3{d^1}\]or \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^1}\]
The atomic orbitals 1s, 2s, 2p, 3s and 2p of \[{V^{ + 4}}\] ion are completely filled, only 3d orbital of \[{V^{ + 4}}\]has one unpaired electron.
\[\therefore n = 1\] for \[{V^{ + 4}}\]
The \[{V^{ + 4}}\]ion is paramagnetic in nature as it contains 1 unpaired electron in its 3d orbital.

Step 2: Now we have got the value of \[n\] for \[{V^{ + 4}}\], we can calculate the value of magnetic moment for \[{V^{ + 4}}\] ion.
By substituting the required values in the formula for magnetic moment, we get
$[\mu = \sqrt {n(n + 2)} ]$
$[mu = \sqrt {1(1 + 2)} ]$
\[\mu = \sqrt 3 = 1.732BM\]
The BM represents Bohr magneton, the unit for denoting the magnetic moment of an electron caused by its spin. Every electron has its permanent magnetic moment that is equal to the value of Bohr's magneton that is \[9.274 \times {10^{ - 24}}J/T\]

So,the magnetic moment is 1.732BM.

Additional information:
A laboratory technique is also used for calculating the magnetic moment of the coordination complex and this method is termed as the Gouy method. In this method, first the weighing of samples of the complex in the presence of magnetic field is carried out. In the next step, the sample is again weighed in the absence of a magnetic field and then the difference between these two weights is calculated.

Note: The formula used for calculating spin only magnetic moment is written as:
$[\mu = \sqrt {n(n + 2)} ]$
This formula is based on the number of unpaired electrons present in the central metal ion in the complex.
It can also be expressed in different forms using the electron spin quantum number for each unpaired electron in the central metal ion of the complex.

If \[n = 1\] then S = $\dfrac{1}{2}
$[\mu = \sqrt {4S(S + 1)} ]$
Here, S denotes the electron spin quantum number of an unpaired electron.