Question

Which of the following has the magnetic moment of 1.75 BM?
$
  {\text{A}}{\text{. T}}{{\text{i}}^{3 + }} \\
  {\text{B}}{\text{. }}{{\text{V}}^{3 + }} \\
  {\text{C}}{\text{. C}}{{\text{r}}^{3 + }} \\
  {\text{D}}{\text{. F}}{{\text{e}}^{3 + }} \\
 $

Answer 1

Hint: First we need to know the valence electronic configuration of the given elements. Then using the number of unpaired electrons in the outermost shell in the formula for magnetic moment, we can find out the magnetic moment of the given elements.

Formula used:
The magnetic moment of an element can be calculated by using the following formula.
$\mu = \sqrt {n\left( {n + 2} \right)} {\mu _B}$

Complete step-by-step answer:
We are given certain elements for which we need to check which we need to measure the magnetic moment.
The magnetic moment for a given element can be found by using the following formula. It is given as
$\mu = \sqrt {n\left( {n + 2} \right)} {\mu _B}$
Here n signifies the total number of unpaired electrons in the outermost shell of an atom while ${\mu _B}$ is known as the Bohr magneton.
Now first of all we need to know the valence shell configuration of the given atoms. It is given as follows:
1. $T{i^{3 + }}{\text{ }}\left[ {Ar} \right]3{d^1}$
The total number of unpaired electrons is 1.
$\therefore n = 1$
Inserting this value in the formula for magnetic moment, we get
$\mu = \sqrt {1\left( {1 + 2} \right)} {\mu _B} = \sqrt 3 {\mu _B} = 1.732{\mu _B}$
2. ${V^{3 + }}{\text{ }}\left[ {Ar} \right]3{d^2}$
The total number of unpaired electrons is 2.
$\therefore n = 2$
Inserting this value in the formula for magnetic moment, we get
$\mu = \sqrt {2\left( {2 + 2} \right)} {\mu _B} = \sqrt 8 {\mu _B} = 2.828{\mu _B}$
3. $C{r^{3 + }}{\text{ }}\left[ {Ar} \right]3{d^3}$
The total number of unpaired electrons is 3.
$\therefore n = 3$
Inserting this value in the formula for magnetic moment, we get
$\mu = \sqrt {3\left( {3 + 2} \right)} {\mu _B} = \sqrt {15} {\mu _B} = 3.872{\mu _B}$
4. $F{e^{3 + }}{\text{ }}\left[ {Ar} \right]3{d^5}$
The total number of unpaired electrons is 5.
$\therefore n = 5$
Inserting this value in the formula for magnetic moment, we get
$\mu = \sqrt {5\left( {5 + 2} \right)} {\mu _B} = \sqrt {35} {\mu _B} = 5.916{\mu _B}$
So, based on the obtained value, the magnetic moment of $T{i^{3 + }}$ is closest to 1.75 BM. Hence, the correct answer is option A.

So, the correct answer is “Option A”.

Note: It should be noted that ${\mu _B}$ is pronounced as Bohr magneton and it is used as a unit for measuring the magnetic moment due to electrons in an atom. For a single unpaired electron, the formula for calculated the value of the Bohr magneton is given as follows:
${\mu _B} = \dfrac{{eh}}{{4\pi {m_e}}}$
Here e is the charge on the electron, h is known as the Planck’s constant while ${m_e}$ signifies the mass of an electron.