JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Coordination Compounds

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

(A)ethylenediamine
(B) $CN^{-}$
(C) $NCS^{-}$ ✓ Correct
(D)CO

Answer: (C)

Solution

A magnetic moment of

5.9

BM points to five unpaired electrons, since

μ = n (n + 2)

with

n = 5

gives

5.92

-

the high-spin

d^{5}

arrangement of

Mn X^{2 +}

. High spin needs a weak-field ligand; of the choices,

CN X^{-}

and

CO

are strong-field and ethylenediamine is moderate, while

NCS X^{-}

is weak enough to leave all five electrons unpaired, forming

[Mn (NCS) X_{6}]^{4 -}

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option

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