
For the octahedral complexes of $F{{e}^{3+}}$ in $SC{{N}^{-}}$ (thiocyanato-S) and in $C{{N}^{-}}$ ligand environments, the difference between the spin- only magnetic moments in Bohr magnetrons (when approximated to the nearest integer) is:
[atomic number of Fe = 26]
Answer
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Hint: Electrons possess the property of spin under the influence of some magnetic field. The net magnetic moment of any substance is due to the presence of unpaired electrons. Strong and weak ligands affect the pairing of the electrons in a complex.
Formula used:
Magnetic moment, ${{\mu }_{eff}}=\sqrt{n(n+2)}$B.M., n is the number of unpaired electrons.
Complete answer:
Spin-only magnetic moments, tell us the net magnetic moment of unpaired electrons in any substance. We have given ferric, $F{{e}^{3+}}$ complexes, with ligands, $SC{{N}^{-}}$ and $C{{N}^{-}}$, we have to find the difference in the spin-only magnetic moments in Bohr’s magneton.
Ferric $F{{e}^{3+}}$has the electronic configuration of $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{5}}$ , so it has 5 electrons in its outer valence shell.
With the ligand, $SC{{N}^{-}}$ (thiocyanato-S), ferric will produce high spin complex, as this ligand is considered as a weak field ligand, so the pairing of electrons in $F{{e}^{3+}}$will not occur, and have 5 unpaired electrons, so n = 5, now the magnetic moment will be,
${{\mu }_{eff}}=\sqrt{5(5+2)}$
${{\mu }_{eff}}=\sqrt{35}$
${{\mu }_{eff}}=5.92\,B.M.$
With the ligand, $C{{N}^{-}}$, $F{{e}^{3+}}$ will form low spin complexes as this ligand is a strong field ligand and will result in pairing of electrons in $F{{e}^{3+}}$, so after pairing of 5 electrons, only 1 unpaired electron left, and n = 1. So magnetic moment will be,
${{\mu }_{eff}}=\sqrt{1(1+2)}$
${{\mu }_{eff}}=\sqrt{3}$
${{\mu }_{eff}}=1.73\,B.M.$
Now, the difference in the spin- only magnetic moment with both the ligands will be,
\[5.92\text{ }B.M.\text{ }-\text{ }1.73\text{ }B.M.\text{ }=\text{ }4.19\text{ }B.M.\simeq 4\,B.M.\]
Hence, the difference in spin- only magnetic moment in Bohr magneton with $SC{{N}^{-}}$ and $C{{N}^{-}}$ ligand attached with ferric atom, is 4 Bohr magneton.
Note:
Ligands that are able to produce large splitting are strong field ligands, and those which produce small splitting are weak field ligands. The field splitting is known as crystal field splitting. Some weak ligands are, ${{I}^{-}},B{{r}^{-}},C{{l}^{-}},O{{H}^{-}}$, etc. Some of the strong ligands are, $C{{N}^{-}},NC{{S}^{-}},CO,N{{H}_{3}}$, etc. They are strong due to the donor atom with a lone pair of electrons.
Formula used:
Magnetic moment, ${{\mu }_{eff}}=\sqrt{n(n+2)}$B.M., n is the number of unpaired electrons.
Complete answer:
Spin-only magnetic moments, tell us the net magnetic moment of unpaired electrons in any substance. We have given ferric, $F{{e}^{3+}}$ complexes, with ligands, $SC{{N}^{-}}$ and $C{{N}^{-}}$, we have to find the difference in the spin-only magnetic moments in Bohr’s magneton.
Ferric $F{{e}^{3+}}$has the electronic configuration of $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{5}}$ , so it has 5 electrons in its outer valence shell.
With the ligand, $SC{{N}^{-}}$ (thiocyanato-S), ferric will produce high spin complex, as this ligand is considered as a weak field ligand, so the pairing of electrons in $F{{e}^{3+}}$will not occur, and have 5 unpaired electrons, so n = 5, now the magnetic moment will be,
${{\mu }_{eff}}=\sqrt{5(5+2)}$
${{\mu }_{eff}}=\sqrt{35}$
${{\mu }_{eff}}=5.92\,B.M.$
With the ligand, $C{{N}^{-}}$, $F{{e}^{3+}}$ will form low spin complexes as this ligand is a strong field ligand and will result in pairing of electrons in $F{{e}^{3+}}$, so after pairing of 5 electrons, only 1 unpaired electron left, and n = 1. So magnetic moment will be,
${{\mu }_{eff}}=\sqrt{1(1+2)}$
${{\mu }_{eff}}=\sqrt{3}$
${{\mu }_{eff}}=1.73\,B.M.$
Now, the difference in the spin- only magnetic moment with both the ligands will be,
\[5.92\text{ }B.M.\text{ }-\text{ }1.73\text{ }B.M.\text{ }=\text{ }4.19\text{ }B.M.\simeq 4\,B.M.\]
Hence, the difference in spin- only magnetic moment in Bohr magneton with $SC{{N}^{-}}$ and $C{{N}^{-}}$ ligand attached with ferric atom, is 4 Bohr magneton.
Note:
Ligands that are able to produce large splitting are strong field ligands, and those which produce small splitting are weak field ligands. The field splitting is known as crystal field splitting. Some weak ligands are, ${{I}^{-}},B{{r}^{-}},C{{l}^{-}},O{{H}^{-}}$, etc. Some of the strong ligands are, $C{{N}^{-}},NC{{S}^{-}},CO,N{{H}_{3}}$, etc. They are strong due to the donor atom with a lone pair of electrons.
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