Question

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

• A. $CO$
• B. ${NCS}^{-}$
• C. ethylenediamjne
• D. ${CN}^{-}$

Answer 1

The correct option is B ${NCS}^{-}$

We know that: Magnetic Moments (spin only) is given by: $m=\sqrt{n(n+2)}$ where n is number of unpaired $e^{-}s$

Now, corresponding magnetic moments for a number of unpaired electrons is as follows

$1)1\to 1.7$

$2)2\to 2.8$

$3)3\to 3.9$

$4)4\to 4.9$

$5)5\to 5.9$

So when Magnetic moment $=5.9BM$, means $5$ unpaired electrons and $Mn$ has $5$ d electrons which are unpaired so ligand must be weak field ligand. Now, Comparing with given options $CO,En,C{N}^{-}$ are strong field ligand.

So, option B is the correct answer