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  • Give answer! The magnetic moment of an octahedral homoleptic Mn (II) complex is 5.9 BM. The suitable ligand for this complex is :

The magnetic moment of an octahedral homoleptic Mn (II) complex is 5.9 BM. The suitable ligand for this complex is :

  • Option 1)

    CO

  • Option 2)

    ethylenediamine 

  • Option 3)

    NCS-

  • Option 4)

    CN-

Answers (1)

best_answer

 

CFSE in an octahedral complex -

CFSE =\left [- \frac{2}{5}\left ( No. of\;electron\;in\;t_{2}g \right ) +\frac{3}{5}\left ( No. of\;\acute{e}s\;in\;eg \right ) \right ]\bigtriangleup \circ

 

-

 

Magnetic Moments (spin only) -

\sqrt{n\left ( n+2 \right )} where n= number of unpaired electron.

- wherein

Number of unpaired e^{-} s and corresponding magnetic moments

1\rightarrow 1.7\\2\rightarrow 2.8\\3\rightarrow 3.9\\4\rightarrow 4.9\\5\rightarrow 5.9

As we have learned in magnetic moment 

\mu=5.9 BM therefore no. of unpaired C^{-}=5cation  {Mn}''-3d^{5} configuration only possible for relatively weak ligand .

therefore NCS^{-}

 

  

 


Option 1)

CO

Option 2)

ethylenediamine 

Option 3)

NCS-

Option 4)

CN-

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