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Question

Arrange the following in increasing value of magnetic moments.
(i) [Fe(CN)6]4
(ii) Fe(CN)6]3
(iii) Cr(NH3)3+6
(iv) Ni(H2O)4]2+

A
(I)<(II)<(III)<(IV)
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B
(I)<(II)<(IV)<(III)
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C
(II)<(III)<(I)<(IV)
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D
(IV)<(III)<(II)<(I)
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Solution

The correct option is B (I)<(II)<(IV)<(III)
1) [Fe(CN)6]4
Here, Fe is present as Fe2+
Electronic configuration of Fe2+ is [Ar]3d64s0
Since CN is a strong field ligand so when it approaches Fe2+ it pair up the e s and result in the formation of low spin complex.
No. of unpaired e s =0
So, magnetic moment for n unpaired es
= n(n+2)BM
=0
2) [Fe(CN)6]3
Here, Fe is present as Fe3+
Electronic configuration of Fe3+ is [Ar]3d54s0
Since , CN is a strong field ligand so when it approaches Fe2+ it pair up the es so,
No. of unpaired e s =1
So, magnetic moment for n unpaired es
= 1×(1+2)
=3 BM.
3) [Cr(NH3)6]3+
Electronic configuration of Cr3+ is [Ar]3d34s0
Since , NH3 is a strong field ligand so it pairs up the electrons when it approaches when it approaches Cr3+
No. of unpaired e s =3
So, magnetic moment = 3(3+2)
=1×5
=15 BM
4)[Ni(H2O)4]2+
Electronic configuration of Ni2+ is [Ar]3d84s0
Since , H2O is a weak field ligand. it does not pair up the es and forms outer orbital complex.
No. of unpaired e s =2
So, magnetic moment =2(2+2)
=8 BM
But because of orbital effect it will be little lesser. So, correct order is iii)>iv)> ii)> i)

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