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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)
    CO
  • b)
    Ethylenediamine
  • c)
    NCS
  • d)
    CN
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
The magnetic moment of an octahedral homoleptic Mn(II) complex is 5.9 ...
Electronic configuration of Mn2+ is
Mn+2 : 3d5
It has 5 unpaired electrons which corresponds to magnetic moment of   This shows that the homoleptic complex of Mn2+ has only weak field ligands and that is NCS. The remaining three ligands are strong field ligands.
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Most Upvoted Answer
The magnetic moment of an octahedral homoleptic Mn(II) complex is 5.9 ...
Magnetic moment is 5.9 BM = √{n(n+2)}
n=5

EC of MN+2 = 3d5

i.e, 5 unpaired e

out of given ligands NCS- is a weak field ligand which cannot Pair up the e where rest are strong field ligand and they Pair up unpaired electrons and if it happens then BM will not be equal as given.
So, a/c question answer is C
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Community Answer
The magnetic moment of an octahedral homoleptic Mn(II) complex is 5.9 ...
Electronic configuration of Mn2+ is
Mn+2 : 3d5
It has 5 unpaired electrons which corresponds to magnetic moment of   This shows that the homoleptic complex of Mn2+ has only weak field ligands and that is NCS. The remaining three ligands are strong field ligands.
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