Question

The calculated spin-only magnetic moment (BM) of the anionic and cationic species of ${{[\text{Fe(}{{\text{H}}_{2}}\text{O}{{\text{)}}_{6}}]}_{2}}$ and $[\text{Fe(CN}{{\text{)}}_{6}}]$, respectively, are :
a- 4.9 and 0
b- 2.84 and 5.92
c- 0 and 4.9
d- 0 and 5.92

Answer 1

Hint: An attempt to this question can be made by determining the meaning of magnetic moment and then spin-only magnetic moment in particular. Spin-only magnetic moment (BM) for coordination compounds can be determined by following formula:
$\mu $ = \[\sqrt{n(n+2)}\]BM
Here $n$= number of unpaired electrons of central metal atoms.

Complete answer:
A spin magnetic moment is the magnetic moment caused by the spin of elementary particles (electrons, neutrino, etc).Elementary particles are conceived as point objects which have no axis to "spin" around.
-Diamagnetic materials are repelled by a magnetic field; an applied magnetic field creates an induced magnetic field in them in the opposite direction, causing a repulsive force.
-On the other hand, paramagnetic materials are attracted by a magnetic field.
Now, spin-only magnetic moment (BM) of ${{[\text{Fe(}{{\text{H}}_{2}}\text{O}{{\text{)}}_{6}}]}_{2}}$
- Configuration of $F{{e}^{2+}}$ = $t_{2g}^{4}$, $e_{g}^{2}$
- Number of unpaired electrons = 4
- spin-only magnetic moment (BM) : $\mu$ = $\sqrt{n(n+2)}$ BM
= $\sqrt{4(4+2)}$
= 4.92 BM

We will now calculate the spin-only magnetic moment (BM) of $[\text{Fe(CN}{{\text{)}}_{6}}]$
- Configuration of $F{{e}^{0}}$ = $t_{2g}^{6}$, $e_{g}^{0}$
- Number of unpaired electrons = 0
- Spin-only magnetic moment (BM): $\mu$ = $\sqrt{n(n+2)}$ BM
=$\sqrt{0(0+2)}$
=0 BM
So, the correct answer is “Option A”.

Note: Those coordination complexes which have 0 BM magnetic moment are diamagnetic in nature. Non zero value of magnetic moment makes the complex compounds paramagnetic.