JEE Main 2020 · Chemistry · Coordination Compounds · medium

The correct increasing order of spin-only magnetic moment of [Fe(CN)6]^4-, [Mn(CN)6]^3-, [Cr(H2O)6]^3+ and [Fe(H2O)6]^2+ is (CN^- is a strong-field…

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The correct increasing order of spin-only magnetic moment of

\mathrm{[Fe(CN)_6]^{4-}}

\mathrm{[Mn(CN)_6]^{3-}}

\mathrm{[Cr(H_2O)_6]^{3+}}

and

\mathrm{[Fe(H_2O)_6]^{2+}}

is (CN

^-

is a strong-field ligand and H

_2

O is weak-field here)

A. $\mathrm{[Fe(CN)_6]^{4-}<[Mn(CN)_6]^{3-}<[Cr(H_2O)_6]^{3+}<[Fe(H_2O)_6]^{2+}}$
B. $\mathrm{[Fe(CN)_6]^{4-}<[Cr(H_2O)_6]^{3+}<[Mn(CN)_6]^{3-}<[Fe(H_2O)_6]^{2+}}$
C. $\mathrm{[Mn(CN)_6]^{3-}<[Fe(CN)_6]^{4-}<[Cr(H_2O)_6]^{3+}<[Fe(H_2O)_6]^{2+}}$
D. $\mathrm{[Fe(CN)_6]^{4-}<[Fe(H_2O)_6]^{2+}<[Mn(CN)_6]^{3-}<[Cr(H_2O)_6]^{3+}}$

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Answer: (A) $\mathrm{[Fe(CN)_6]^{4-}<[Mn(CN)_6]^{3-}<[Cr(H_2O)_6]^{3+}<[Fe(H_2O)_6]^{2+}}$

\mathrm{Fe^{2+}}

(

d^6

) with CN

^-

t_{2g}^6

, 0 unpaired electrons,

\mu=0

\mathrm{Mn^{3+}}

(

d^4

) strong field:

t_{2g}^4

, 2 unpaired,

\mu=\sqrt8=2.83

BM.

\mathrm{Cr^{3+}}

(

d^3

): 3 unpaired,

\sqrt{15}=3.87

BM.

\mathrm{Fe^{2+}}

(

d^6

) weak field:

t_{2g}^4e_g^2

, 4 unpaired,

\sqrt{24}=4.90

BM. Hence

0<2.83<3.87<4.90

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