Question

Which of the following has the maximum theoretical magnetic moment?
(A) ${{V}^{3+}}$
(B) $C{{r}^{3+}}$
(D) $T{{i}^{3+}}$
(D) $C{{o}^{3+}}$

Answer 1

Hint: To solve this, firstly write the electronic configuration of the given ions. Find out the number of unpaired electrons in each of them. To find out the magnetic moment you can use the formula- $\mu =\sqrt{n\left( n+2 \right)}\text{ BM}$, where n is the number of unpaired electrons and BM is the unit of the magnetic moment.

Complete step by step solution:
We know that magnetic moment is the tendency of a certain pattern in which the atoms are arranged in an orbital.
To find out the maximum magnetic moments among the given ions, we will discuss each of them one by one.
Firstly we have ${{V}^{3+}}$.
Vanadium lies in the first transition series i.e. in the 3d series and we can write its electronic configuration as $\left[ Ar \right]3{{d}^{3}}4{{s}^{2}}$.
Now for the 3+ ion, it will become $\left[ Ar \right]3{{d}^{2}}$. We can see from the electronic configuration that there are two unpaired electrons.
We know that magnetic moment is given by, $\mu =\sqrt{n\left( n+2 \right)}\text{ BM}$.
Therefore, putting the value of n, which is the number of unpaired electrons, we will get-
$\mu =\sqrt{2\left( 2+2 \right)}\text{ BM=2}\text{.82 BM}$
Now, in the second option we have $C{{r}^{3+}}$.
Electronic configuration of chromium is- $\left[ Ar \right]3{{d}^{4}}4{{s}^{2}}$.
Electronic configuration of $C{{r}^{3+}}$ is -$\left[ Ar \right]3{{d}^{3}}$ . It has three unpaired electrons.
Therefore, magnetic moment = $\mu =\sqrt{3\left( 3+2 \right)}\text{ BM=3}\text{.87 BM}$
Then we have $T{{i}^{3+}}$.
Electronic configuration of titanium is -$\left[ Ar \right]3{{d}^{2}}4{{s}^{2}}$ .
Electronic configuration of $T{{i}^{3+}}$is -$\left[ Ar \right]3{{d}^{1}}$ . It has one unpaired electron.
Therefore, the magnetic moment is, $\mu =\sqrt{1\left( 1+2 \right)}\text{ BM=1}\text{.73 BM}$ .
And lastly, we have $C{{o}^{3+}}$.
Electronic configuration of cobalt is $\left[ Ar \right]3{{d}^{7}}4{{s}^{2}}$ .
Electronic configuration of $C{{o}^{3+}}$ is $\left[ Ar \right]3{{d}^{6}}$ . It has 4 unpaired electrons (as d-orbital have 5 subshells and the electrons fill up in the same spin first and then the opposite spin. One electron will be paired whereas the other 4 will be unpaired)
Therefore, magnetic moment is, $\mu =\sqrt{4\left( 4+2 \right)}\text{ BM=4}\text{.89 BM}$
We can see from the above calculation that the cobalt 3+ ion has the maximum theoretical magnetic moment among the given options.

Therefore, the correct answer is option [D] $C{{o}^{3+}}$.

Note: Here, the unit of the magnetic moment we have written BM which is Bohr magneton. It is the moment of the electrons. We have a similar unit for the moment of nuclei, proton and neutrons known as the neutron magneton. We can also solve the above question using the spin-only formula of the magnetic moment including spin number. The formula is $\mu =\sqrt{4S\left( S+1 \right)}\text{ BM}$ , where s is the spin of the electrons in the orbital.