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Crystal Field Split in Octahedral Complexes

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Question

Calculate the magnetic moment of a high-spin octahedral complex that has six electrons in $3 d -$ orbitals (in nearest integer).

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Solution

A high-spin octahedral complex that has six electrons in $3 d -$ orbitals contains 4 unpaired electrons.
Hence, the magnetic moment is:

$μ_{eff} = \sqrt{n (n + 2)} B . M . = \sqrt{4 (4 + 2)} B . M = \sqrt{24} B . M = 4.89 B . M ≃ 5 B . M$

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Q. For the octahedral complexes of

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Q. In which of the following configuration (s) the value of 'spin only' magnetic moment is 2.84 BM for octahedral complex:

Valence bond theory describes the bonding in complexes in terms of coordinate-covalent bonds resulting from the overlap of filled ligand orbitals with vacant metal hybrid orbitals. This theory explains the magnetic behavior and geometrical shape of coordination compounds. The magnetic moment of a complex compound can be determined experimentally and theoretically by using spin only formula.

Magnetic moment =

\sqrt{n (n + 2)}

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The value of spin only magnetic moment for an octahedral complex of the following configuration is 2.84 B.M. The correct statement is :

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Crystal Field Split in Octahedral Complexes

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