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  • The species that has spin only magnetic moment   (Td-Tetraherdral)   Option: 1 <img alt="\left [ NiCl_{4} \right ]^{

The species that has spin only magnetic moment 5.9\; BM  (Td-Tetraherdral)
 
Option: 1 \left [ NiCl_{4} \right ]^{2-}\left \left ( Td \right ) \right
Option: 2 \left [ MnBr_{4} \right ]^{2-}\left \left ( Td \right ) \right
Option: 3 Ni(CO)_{4}\left ( Td \right )
Option: 4 \left [ CN\left [ Ni(CN)_{4} \right ] ^{2-}\right ]
 

Answers (1)

best_answer

We have :

\mathrm{\sqrt{n(n+2)}\: =\: 5.9BM}

Thus, n = 5(unpaired electrons)

In [Mn(Br)_{4}]^{2-}, Mn^{2+} has 5 unpaired electrons due to weak (Br^{-}) ligand.

Therefore, Option(2) is correct.

Posted by

Kuldeep Maurya

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