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# The magnetic moment of an octahedral homoleptic Mn(II) complex is 5.9 BM. The suitable ligand for this complex is:(a) $C{{N}^{-}}$ (b) $NC{{S}^{-}}$(c) CO(d) ethylenediamine

Last updated date: 20th Jun 2024
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By the octahedral complex we mean the complex which is made up at the center of the six spheres i.e. its coordination number is six. Magnetic moments tell us about the presence of unpaired electrons in the element. The magnetic moment of Mn is 5.9 BM(given) i.e. there are five unpaired electrons in the Mn. Atomic number of manganese is 25. The electronic configuration of Mn is: ${{[Ar]}^{18}}3{{d}^{5}}4{{s}^{2}}$. So, in the ground state electronic configuration of Mn there are five unpaired electrons and it will not undergo pairing unless and until any strong ligand is not attached to it because strong ligand results into the pairing of the electrons and thus, there will be no unpaired electrons and zero magnetic moment. Since, the magnetic moment is 5 given in the statement i.e. there are unpaired electrons in it , so the ligand attached to the Mn should be a weak ligand which doesn’t allow the pairing up of the electrons and hence, the electrons will be unpaired. Among all the ligands given above, only $NC{{S}^{-}}$is a weak ligand while the rest all others are strong ligands resulting in the pairing up of electrons and hence, there will be no unpaired electrons. Hence, the magnetic moment of an octahedral homoleptic Mn(II) complex is 5.9 BM and the suitable ligand for this complex is: $NC{{S}^{-}}$