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No. Of unpaired electron in [FeF6]3- and magnetic moment of it also.

No. Of unpaired electron in [FeF6]3- and magnetic moment of it also.

Grade:12

2 Answers

Bhanu
40 Points
6 years ago
[FeF6]3-
oxidation state of Fe;
x+6(-1)=-3
x=+3
no. of un paired electrons=no ofun paired electrons in 3d orbital of Fe
now no. of electrons in Fe+3=23
=23-18=5
therefore there are 5 unpaired electrons in 3d orbital of Fe3+
therefore magnetic moment =square root of (n(n+2))
=square root of(5(5+2))
=5.91B.M
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the solution to your problem below.
 
[FeF6]3-
oxidation state of Fe;
x+6(–1) = – 3
x = +3
now no. of electrons in Fe+3 = 23
Fe+3 : 1s2 2s2 2p6 3s2 3p6 3d5
No of unpaired electrons in = 5 (in 3d orbital)
Therefore magnetic moment = √[n(n+2)]
=√[5(5+2)]
= 5.91 B.M.
 
Thanks and regards,
Kushagra

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