QJEEM20Ch5d70: JEE Main 2020 Chemistry - Class 12 JEEM20Ch5d70 Solution: The correct order of the spin-only magnetic moments of the…

(I) [Cr(H₂O)₆]Br₂

H₂O is weak field ligand so it can not pair up all the electrons.

Cr²⁺ : [Ar] 4s⁰3d⁴ ( $t_{2g}^3e{g^1}$ )

Unpaired e^– = 4

Magnetic moment = $\sqrt {24}$ = 4.89 BM

(II) Na₄[Fe(CN)₆]

CN^- is strong field ligand so it pair up all the electrons.

Fe²⁺ = [Ar] 4s⁰3d⁶ ( $t_{2g}^6e{g^0}$ )

Unpaired e^– = 0

Magnetic moment = 0 BM

(III) Na₃[Fe(C₂O₄)₃]

As $\Delta$ ₀ > P, so pairing of electrons happens.

Fe³⁺ = [Ar] 4s⁰3d⁵ ( $t_{2g}^5e{g^0}$ )

Unpaired e^– = 1

Magnetic moment = $\sqrt 3$ = 1.73 BM

(IV) (Et₄N)₂[CoCl₄]

Cl^-is weak field ligand so it can not pair up all the electrons.

Co²⁺ = [Ar] 4s⁰3d⁷ ( $e{g^4}t_{2g}^3$ )

Unpaired e^– = 3

Magnetic moment = $\sqrt {15}$ = 3.87 BM

Hence order of magnetic moment is
I > IV > III > II