Question

Magnetic moment (spin only) of octahedral complex having $CFSE = - 0.8{\Delta _o}$ and surrounded by weak field ligands can be:
A.$\sqrt {15} BM$
B.$\sqrt 8 BM$
C.A and B
D.None of the above

Answer 1

Hint: We know that metal atoms in coordination compounds are from 3d transition series. So, we can calculate the magnetic moment of the compounds with the help of the formula,
$\mu = \sqrt {n\left( {n + 2} \right)} BM$
Here, $\mu $ is the magnetic moment
$n$ is the number of unpaired electrons

Complete step by step answer:
We can define complex structures as compounds that are made of a central metal ion bound by molecules or ions called ligands. These ligands are bonded to the central metal ion through coordinate covalent bonds. The central metal ion contains a positive charge. The ligands are rich with lone pairs of electrons. These electrons are given to the metal ion in order to decrease the positive charge on the metal ion. This kind of bonding is called a coordinate bond.
We can determine the magnetic properties of a compound from their electron configuration and the atomic size. Because electron spin generates the magnetic moments, the number of unpaired electrons in a specific compound represents the magnetic nature of the compound.
The octahedral complex that contains crystal field splitting energy has $ - 0.8{\Delta _o}$ enclosed by ligands which are weak and could contain one (or) more than one unpaired electron.
We can write the formula of spin magnetic moment as,
$\mu = \sqrt {n\left( {n + 2} \right)} BM$
Here, $\mu $ is the magnetic moment
$n$ is the number of unpaired electrons
If there two unpaired electrons, the formula turns
${\mu _{eff}} = \sqrt {n\left( {n + 2} \right)} BM$
On substituting the value of n=2 we get,
$ \Rightarrow {\mu _{eff}} = \sqrt {2\left( {2 + 2} \right)} BM$
On simplifying we get,
${\mu _{eff}} = \sqrt 8 BM$
The spin only magnetic moment for two unpaired electrons is $\sqrt 8 BM$.
If there three unpaired electrons, the formula turns
${\mu _{eff}} = \sqrt {n\left( {n + 2} \right)} BM$
Now we substitute the value of n as 3 we get,
$ \Rightarrow {\mu _{eff}} = \sqrt {3\left( {3 + 2} \right)} BM$
On simplifying we get,
${\mu _{eff}} = \sqrt {15} BM$
The spin only magnetic moment for three unpaired electrons is $\sqrt {15} BM$.
We can observe spin only magnetic moments are $\sqrt 8 BM$ and $\sqrt {15} BM$. Options (A) and (B) are correct.
So, the final correct option is Option (C).

Note:
We must remember that the spin magnetic moments generate a foundation for one of the most significant principles in chemistry, the Pauli Exclusion Principle. The theory plays additional roles than just the descriptions of doublets inside the electromagnetic spectrum. This extra quantum number, spin, developed the foundation for the modern standard model employed recently, that includes the support of Hund's rules.