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Question

# Arrange the following in increasing value of magnetic moments.(i) [Fe(CN)6]4âˆ’(ii) Fe(CN)6]3âˆ’(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 =0So, magnetic moment for n unpaired e−s = √n(n+2)BM=02) [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 e−s so,No. of unpaired e− s =1So, magnetic moment for n unpaired e−s = √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 =3So, magnetic moment = √3(3+2)=√1×5=√15 BM4)[Ni(H2O)4]2+Electronic configuration of Ni2+ is [Ar]3d84s0 Since , H2O is a weak field ligand. it does not pair up the e−s and forms outer orbital complex. No. of unpaired e− s =2So, magnetic moment =√2(2+2)=√8 BMBut because of orbital effect it will be little lesser. So, correct order is iii)>iv)> ii)> i)

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