JEE MAIN - Chemistry (2020 - 9th January Evening Slot - No. 20)
The correct order of the spin-only magnetic
moments of the following complexes is :
(I) [Cr(H2O)6]Br2
(II) Na4[Fe(CN)6]
(III) Na3[Fe(C2O4)3] ($$\Delta $$0 $$>$$ P)
(IV) (Et4N)2[CoCl4]
(I) [Cr(H2O)6]Br2
(II) Na4[Fe(CN)6]
(III) Na3[Fe(C2O4)3] ($$\Delta $$0 $$>$$ P)
(IV) (Et4N)2[CoCl4]
(III) > (I) > (II) > (IV)
(II) $$ \approx $$ (I) > (IV) > (III)
(III) > (I) > (IV) > (II)
(I) > (IV) > (III) > (II)
Explanation
(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
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
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