Solveeit Logo

Question

Question: For the octahedral complexes of \(F{{e}^{3+}}\) in \(SC{{N}^{-}}\) (thiocyanato-S) and in \(C{{N}^{-...

For the octahedral complexes of Fe3+F{{e}^{3+}} in SCNSC{{N}^{-}} (thiocyanato-S) and in CNC{{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]

Explanation

Solution

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, μeff=n(n+2){{\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, Fe3+F{{e}^{3+}} complexes, with ligands, SCNSC{{N}^{-}} and CNC{{N}^{-}}, we have to find the difference in the spin-only magnetic moments in Bohr’s magneton.
Ferric Fe3+F{{e}^{3+}}has the electronic configuration of 1s22s22p63s23p63d51{{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, SCNSC{{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 Fe3+F{{e}^{3+}}will not occur, and have 5 unpaired electrons, so n = 5, now the magnetic moment will be,
μeff=5(5+2){{\mu }_{eff}}=\sqrt{5(5+2)}
μeff=35{{\mu }_{eff}}=\sqrt{35}
μeff=5.92B.M.{{\mu }_{eff}}=5.92\,B.M.
With the ligand, CNC{{N}^{-}}, Fe3+F{{e}^{3+}} will form low spin complexes as this ligand is a strong field ligand and will result in pairing of electrons in Fe3+F{{e}^{3+}}, so after pairing of 5 electrons, only 1 unpaired electron left, and n = 1. So magnetic moment will be,
μeff=1(1+2){{\mu }_{eff}}=\sqrt{1(1+2)}
μeff=3{{\mu }_{eff}}=\sqrt{3}
μeff=1.73B.M.{{\mu }_{eff}}=1.73\,B.M.
Now, the difference in the spin- only magnetic moment with both the ligands will be,
5.92 B.M.  1.73 B.M. = 4.19 B.M.4B.M.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 SCNSC{{N}^{-}} and CNC{{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,Br,Cl,OH{{I}^{-}},B{{r}^{-}},C{{l}^{-}},O{{H}^{-}}, etc. Some of the strong ligands are, CN,NCS,CO,NH3C{{N}^{-}},NC{{S}^{-}},CO,N{{H}_{3}}, etc. They are strong due to the donor atom with a lone pair of electrons.